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The equivalence principle does not exist in nature because nature does not produce uniform gravitational fields. All gravitational fields necessarily produce tidal forces, thus breaking the equivalence. With the equivalence principle out of the way, there's no paradox.
solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.
This question requires a universe that does not have the same properties as our own. aka, sure in a hypothetical universe where gravity does not impart energy to objects, sure, but gravity DOES impart energy to objects... seriously man this is high school physics.... go read up on what happens to particles near black holes.
The paradox is completely gone, when you realize different observers see different electric and magnetic fields. The charged particle inside the lift doesn't radiate but gets accelerated from radiation from outside. That's consistent with the fact kinetic energy is different for different observers.
you're asking the wrong physicists if they all say "of course", this is a famous problem for QED in curved spacetime, with the vacuum transforming non trivially...leading to things like radiating horizons (famously: Hawking radiation, and less so: Unruh radiation).
@@fixgirodwhy do you make it sound so nefarious? I think he’s simplifying for the sake of the video narrative. To me, it’s kind of rude to assume he must not be talking to smart enough people, instead of assuming he is trying to make compelling content and incorporating the wrong opinions only for the sake of discussion
1. Standing on the Earth, what you see is in a non-inertial frame. 2. Maxwell’s equations require modification in a non-inertial frame. 3. Special relativity was devised to make Maxwell’s equations covariant. 4. Covariance implies an inertial frame. 5. Therefore in a non-inertial frame you cannot use Maxwell’s equations to predict an absence of a time-varying magnetic field. 6. Thus there’s no paradox. 7. What we need are measurements precise enough to detect radiation from particles that are stationary in a 1g non-inertial frame.
Frame of Earth is only slightly non-covariant, and Lienard formula (Larmor formula for relativity) says that it starts radiating only when Beta is close to 1. So there will never be an observable radiation for particles with insignificant speed. It will radiate only when particle falls into Black Hole. Non-inertiality of Earth does not matter, while this inertiality is with its Beta much much less then 1.
"3. Special relativity was devised to make Maxwell’s equations covariant." I'd say Maxwell's equations were already covariant. These equations were not however compatible with the Cartesian view of the world, so the latter had to change. Maxwell's equations stood unaltered. They're covariant now so they were covariant then. They were perhaps Einstein's main guide to special relativity.
Doesn't necessarily have to be a precise measurement to detect the effect if you can either or both increase the intensity of the radiation (multiple charged particles) or run it enough times for a chance detection or an average above background noise, if applicable. Afterall nature has provided the most precise photon measurement devices possible; electrons in their exclusion energy wells.
The question is simply to be answered: the Larmor formula relates to an accelerated electrically charged particle losing energy in an electromagnetic field which is the cause of the acceleration. So it does not relate to any such particle which is accelerated in a gravitational field or a curved space time. Therefore the Larmor formula does not tell us anything about radiation being generated in the described setting.
So there is no brehmsstralung if the acceleration is caused by the strong nuclear force or the weak force? It seems me that the QFT calculation of photon emission in a collision is very much the same independent of what force caused a 4.momentum exchange.
No. I'm a physics grad student, I know where to look for precise information. I watch Veritasium as a entertainment activity. Sometimes he talks about things that I don't know. Gravity is not my subject of research. Besides, he also have a PhD in physics.
@@karkaroff1617 He has a Ph.D. in physics education: _"Designing Effective Multimedia for Physics Education"._ So a Ph.D. in UA-cam physics presentation.
Hey, another great video! We once pondered over this exact question, and you know, there’s an EXTREMELY simple resolution - but you’re not going to like it. An illuminating paper we found on the topic was Rohrlich’s 1963 “The Principle of Equivalence”. The conclusion of the paper was that a co-accelerating observer does not observe any radiation - and that this was a direct consequence of the covariance of Maxwell’s equations. What does this tell us? Why - simply that acceleration is relative. An observer who constructs their coordinates via accelerating matter will not observe physical effects of that acceleration. Thus, much like it’s impossible to know one’s true velocity, it’s impossible to know one’s true acceleration. Hmmm, sounds an awful lot like being stuck in a sort of Matrix 🤔 Almost like there’s a theory about that… 🙃 At any rate, you’ve reminded us to do a video on this topic sometime soon. Great work again!
Co-accelerating observers will see each other steady (as we see while driving), but both will agree about feeling a gravitation-like force opposite to acceleration. Every observer irrespective of their state will agree on one of the two - acceleration or gravitation. But, even if we forget all that, how does relative acceleration prove your matrix theory?
Thank you for the comment. About the relativity of acceleration, there is still not consensus whether the General Relativity truly satisfy general principle of relativity maybe just qualitatively. What we know that it is not relative in the same sense as velocity because to transform among reference frames of different velocities you always use Lorentz transformations. So tranforming from stationary to moving you use LT and if you want to transform back then you also use LT but with opposite sign but you are still in the same group of transformations. If you want to transform from inertial frame to accelerated frame you use different transformation than when you transforming from accelerated frame to inertial frame. These transformations are different and therefore these frames are not equivalent quantitatively. (If they were the twin paradox would not be solvable) Anyway I can see a nice symmetry there particle on the surface + observer on the surface = no radiation particle falling + observer falling = no radiation particle on the surface + observer falling = radiation observer on the surface + particle falling = radiation!?!? would be nicely symmetric. I would love to see your video about this. Please make one :D
@@lukasrafajppsThanks for replying! We really appreciate that you’ve been choosing to tackle these interesting and more involved physics topics, rather than doing the usual circuit of broad-appeal subjects - we always learn something new watching your videos. You’re right to point out the transformations are different between two inertial frames versus an inertial and non-inertial frame - this is the distinction that separates the General Principle of Relativity (in which any frame of reference could be subsumed under a single set of transformation-invariant laws of physics) and the Principle of General Covariance (in which the inertial laws of physics are covariantly extended to non-inertial frames). So we wouldn’t disagree with you there - our point was slightly subtler. If you as observer B adopted as your inertial frame what was a non-inertial frame from the perspective of observer A, then to go between your frame and observer A’s frame, you would need covariant transformations (in each of you would describe the other’s laws as involving fictitious forces, etc.). However, if we consider a third observer C who shares A’s definition of the inertial and a fourth observer D who share’s B’s definition of the inertial frame, then to go from A to C or from B to D we need merely only use the Lorentz transformations, while going from C to D or A to D or vice versa we again need the covariant transformations. Thus, we see the nature of the transformations are untouched - it’s simply our starting point in choosing what constitutes the inertial frame that is subject to a certain arbitrariness. This is a point we’ve been trying (perhaps unsuccessfully) to communicate across various videos. Meanwhile, General Relativity asserts that all observers will unambiguously agree on what constitutes an inertial frame via use of an accelerometer. A rest frame on the surface of the earth is non-inertial, so GR definitively tells us a particle at rest on earth’s surface is “really” radiating. However, it appears Maxwell’s equations tells us we can’t actually measure this radiation if we are co-accelerating. This lands us, as you quite rightly assert at the end of the video, in a paradoxical situation, indicating something in our thinking needs revision. Hopefully we will get around to a video on this topic soon, and if we do so we will definitely cite and utilize the research you’ve done here!
@@aniksamiurrahman6365They would only “feel” a gravitational acceleration if they were using accelerometers that had been previously calibrated in some inertial frame - otherwise the accelerometers would register no change in dimension, and the observers would be unable to measure that force at all. We discuss this concept more thoroughly in our video “Why Relativity Doesn’t Add Up”. Interestingly, we actually first developed Matrix Theory to explain the conundrum of relative acceleration - how could acceleration be relative and yet still be responsible for causal propagation of information in our universe? The answer we finally hit upon was that it was our knowledge of acceleration which was relative to our coordinate constructions, and that we could still posit absolute acceleration, but we would have to concede it was ultimately unknowable. Later we realized we could say the same thing for absolute velocity (and absolute space), and then in turn realized that could be applied to resolve the puzzles of relativity.
Rhorlic's paper resolved this IMO: "The principle of equivalence". Annals of Physics. 22. This is described in the section _Resolution by Rohrlich_ in the Wikipedia article _Paradox of radiation of charged particles in a gravitational field._ When a charge radiates, the incoming Schott energy cancels the outgoing radiation giving a curved electric field, zero magnetic field and apparently no radiation which has tricked many experts. Here's a question: how would one levitate a charge so it's stationary on the surface of the Earth? When thinking about the issues you've raised in your video, I've found using a charged sphere as a model instead of a charged particle is very helpful in understanding what's going on physically.
You have to levitate it (counteract Earth's gravity) by either gravity or one of the other forces. If gravity, then the whole scenario is trivial. If electromagnetism, well that has to be accounted for somehow. I suspect the amount of EM force needed will change depending on the observer, and that's where the apparent discrepancy comes from. I also suspect that there's no way to keep the falling charge adjacent to the falling observer without some additional force.
many good comments so far. I just want to applaud your sketch of this as a kind of "paradox" along the lines of other relativity "paradoxes."
6 місяців тому+34
QED-wise, the radiation for a charged particle happens when it flips the spin and releases a photon. This can happen spontaneously (but also the recapture) OR due to an external field acting on the charge. In both cases the tree level approximation is that it flips the spin - tree level means, that is the most probable case. In gravity, there is no need to flip the spin (except diverging gravitational field may). So one problem is that the Maxwell equations and the Lorentz force does not take spin into account, the other is the lack of self-interaction (which is present even in the propagator of QED). In QED you have a vacuum full of virtual photons, representing the field of the particle, and when it accelerates the photon is not recaptured - however you need an other photon to accelerate the particle, whose remnant could also be the outgoing radiation. So in QED you have a convoluted answer, which does the radiated photon comes from, but you still need an external field that causes the acceleration.
Quite interesting. So can we calculate the reverse force? Wouldn't the same apply for acceleration caused by any external force other than electrostatic?
I loved the video. It's incredible to think that we understand so much complexities of how the universe behaves. From the orbitals of the quantum world to the hawking radiations of black holes and yet scientists disagree on what a plastic comb rubbed with my hair will do! Btw, I loved the presentation and the story telling. The way it starts with, 'of course free falling charges must radiate', and how it leads to 'falling charges having to fall slower' leading to the 'breaking the equivalence principe' leading to a need for a deeper investigation was brilliant (no pun intended). Also, congrats on getting Brilliant sponsorship :) Cheers.
@@lukasrafajpps solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.
I really like this question. I've always suggested that the equivalence principle should go both ways. Acceleration by electromagnetic fields should be indistinguishable from curved spacetime (gravity). Here's a thought experiment someone smarter than me could try playing with. Take yourself a universe. It has a charged black hole (let's say negatively charged). This black hole is in an especially clean corner of space, where it cannot easily find positive charge to balance out to a neutral charge. Now take an electron, and shoot it gently towards the black hole. Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's gravitational field (without needing to orbit) (at the point where the gravitational acceleration inward is equal to the electromagnetic acceleration outward). So with this setup, what does both the gravitational field and electromagnetic field look like?
Nice one. Not sure though, I'd expect the particle to radiate, since it is accelerated in the space-time picture. Just like a particle in an ion trap on earth, actually ... should be measureable this way ^^
The equivalence principle can't go both ways because the electromagnetic field differs between particles. The equivalence principle only holds because gravity behaves like a curvature, In the sense that It affects EVERYTHING the same.
"Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's..." Nonsense. You are claiming that the black hole can't 'see' the electron until it passes the event horizon at which point, the hole develops a sudden allergy.
@@nicolasreinaldet732 you dismiss this perspective too easily. it is possible to imagine that different particles experience different space-time curvature.
How do we know today for sure that linearly accelerated charges radiate, and jerk isn't needed fx.? I mean, what has happened since Pauli, Born and Feynman? At 8:50 : In this form, Maxwell's equations are only valid in flat spacetime, in an inertial frame, bc. those are not covariant derivatives. At 10:05 : This kind of Rindler horizon explanation only works in flat space time, but around the Earth, not neccessarily. In this case, the workings of such a horizon is not as trivial as in Minkowski, neither as in Schwarzschild, where every Schwarzschild observers Rindler horizon is just the event horizon.
@@moisessalazar4432 I don't know. Maxwell's equations in curved spacetime are nonlinear, so people can't/don't like working with them. Wikipédia says that Fritz Rohrlich examined the flat version of this problem in 1965 (a Rindler charged particle and an inertial observer). Based on how weak the curvature is around Earth, that approximation may be sufficient in our case.
15:33 The energy of the particle (or charged particle) is equal to the derivative of the action in time x(0)/c: and defined in world and proper time: E(0)=с∂S/∂x(0), E=∂S/∂t [momentum p(k)=∂S/∂x(k)]. Then E(0)=E√g(00)=const and E(0) is preserved, and E is not preserved. E=mc^2/√1-v^2/c^2, where in the static case v=dl/dt=-dl/dt√g(00). Thus, when a particle moves in a gravitational field, the energy E(0)=mc^2(√g(00)/√1-v^2/c^2 is preserved*. This formula remains valid in the case of a stationary field if, when determining the velocity v, one uses the proper time measured by the clock synchronized along the trajectory of the particle. P.S. "Due to the dependence of the speed of light and the speed of the clock on the gravitational potential, after the introduction of the gravitational potential, the laws of nature must be understood as the relationship between the gravitational potential and other physical quantities." (Pauli, RT). ---------- *) - So, it should always be remembered that according to GR, it is possible to make a general statement that: the free movement of the test body occurs along geodesic lines. And this is the most general formulation of Newton's first law. That is, in a Riemannian manifold, where formally there is no gravity and the masses move freely, free fall turns out to be not accelerated, but uniformly rectilinear motion. {Unless, of course, GR obeys a strong equivalence principle, that is, if in fact a locally Lorentzian system = a globally Lorentzian system.}
Note: Measuring a changing electric field does NOT necessarily mean magnetic field intensity is non-zero, only that there's a non-trivial curl of the magnetic field.
@@АндрейМирон-х2н No one is saying that there are two electromagnetic fields, but given the two Maxwell equations (one of which couples to the matter fields and the other which describes the free-electromagnetic field) we have 4 independent field components (the two excitation fields (D,H) and two that define a pair of field intensities (E,B) which can be related via the constitutive relations). What I am saying has nothing to do with this, rather, given some time-like curve what does the observer measure of the electromagnetic field components if the electric field intensity is a function of time (a non-trivial curl of the magnetic field intensity, and not necessarily a non-zero value of the field itself).
@@kylelochlann5053 But the author of the video clearly divides this into components. First, he needs to disassemble the physical meaning without confusing himself with a distorting perception of dividing formulas. That in different reference systems the electrical and magnetic calculations are different, but their resulting influence is identical. And after that it is more correct to express the meaning of the situations under consideration.
@@АндрейМирон-х2нThis is just semantics. Including the idea of their being a single electromagnetic field rather than two fields that induce eachother. What's crazy is you can't derive the magnetic field from the coloumb field, even given the fact the circular shape of the magnetic field around a moving electron is just the cross product of its velocity vector and the coloumb field. even given special relativity, physicists always presuppose the magnetic field already exists in order to separate them in lorentz force formula, and the em field tensor, etc it just makes the math 10x more elegant to represent it as two fields which separately induce one another. Theres a significantly more complicated double cross product representation where you treat the B field as only a lorentz boosted E field and B's dissapear but is it really worth it? Its like its only simple as two fields. Even the word electromagnetic implies this duality, if you were really capable of viewing it as a single 4d vector manifold thing youd just call it the electric field, because magnetic fields dont exist if you view its just "one" thing (the force transforms under lorentz boost).
Maxwell equations are not valid 'as is' to curved spacetime nor Larmur formula (nor QED). To tackle it, we need to use the curved spacetime version of the EM Lagrangian and derive the curved spacetime radiation condition. My (usually bad) intuition says the equivalence principle still holds.
It is not necessarily required to adhere to the equivalence of inertial frames either; observers in their own inertial systems can choose to always calibrate the speed of light as isotropic. This would mean that genuine Lorentz contraction and time dilation would occur relative to the 'stationary ether', and with proper synchronization of clocks, the space and time coordinates interpreted in different inertial frames could be connected by the Galilean transformation. It is just difficult to accept that in this case, the length of a stationary rod in its own inertial frame and the length of an identical rod moving relative to it in the moving inertial frame would not be the same.
One of those questions where you need to break out both the GR & EM tensors to answer correctly. The correct answer is "depends, relative to what observer?" As acceleration is observer dependent as is mode of radiated energy.
Right... you need a second charged particle for it to be interacting with, otherwise there is no force generated by the presence of the charge to act on it to cause it to be slowed down.
@@annoloki Not exactly. Photons can still be radiated & absorbed by non-charged observer. But you still need an observer & reference frame to describe the interaction.
10:55 I don't understand, shouldn't the Unruh effect be experienced by the person on the ground? The presence of the charged particle and the other, free falling observer shouldn't have any infuence on the observation of the grounded person, at least I don't think it should...
At last! I've been asking this question for decades. I appreciate your presentation, which brought out aspects of the problem I hadn't considered so clearly.
Good video but I think you should of included orbiting charge analysis too. For example, what's the difference between a charge moving in a circle in a magnetic field versus a gravitational field? (Like how you animated the electric field not being accelerated in the magnetic field, wouldn't that apply to an orbiting charge?)
1:40 ok I haven't watched the rest of the video nor read any comments but my gut feeling immediately is that gravity is a curvature in spacetime and does not accelerate the particle, the particle keeps moving in a straight line through the curved spacetime. So I would expect no radiation coming from it. Watching the rest of the video now...
I though about that problem when I wanted to apply the rain cycle we usually know on Earth, but for a hypothetical charged matter on the surface of a star. The basic hypothesis are a very strong gravity, a very hot heart and charged matter. The cycle of the charged matter would be: - from the heart, the very hot matter rises the gravity field (and an electric field may also) to gain altitude in the star. - at high altitude into the star, the charged slowed down by the gravity field, is at much lower temperature. - over that surface of charged electric matter at high altitude but at low temperature that creates near by an ocean of electric charged matter, there is a hole, and the charged electric matter falls into that hole to get back into the heart of the star. Of course, when the electric charged falls into the hole, it creates a huge beam of light. The is the image I've of a pulsar star. The beam we can see from a pulsar star is the hole where the charged electric matter falls into.
I'm a physicist and I think this is a great video in that it is seeking to move problem solving to a broader domain than the hyper conservative confines of academia and the scientific paper manufacturing industrial complex. Which is something that Hossenfelder aligns with. Now what I get out of this, and agree with, is that the electric field of a charge is an extrinsic property of that charge. This is an unavoidable conclusion from my own analysis that comes from a very different fundamental model that derives the formation of the electron and positron. In this model the electric field is a result of the capacitive electrical energy storage of the electrical permittivity of "empty" space. That capacitive energy stored in that "empty" space is due to the spin interaction of the electron ( a charge). So the fundamental intrinsic property of the electron is its spin and its magnetic dipole moment. The electric field, mass and gravitational field of the electron are all extrinsic properties of the electron. They are literally all "off particle" properties. ( by way of simple proof - the higgs field is an extrinsic field so the electron's mass and associated gravity field are all "off- particle " properties. The non- obvious realisation here is that the spin of the electron must extend out as far as the electric field it induces in the surrounding space - that generates that electric field. Quantum theory tells you that particle spin is non-physical. Not because it is proven not to be physical, but because no one can make sense of how it could be physical given the strangeness of spin half and g=2 for the magnetic moment. But that is an assumption and an assumption is a very very long way from a proof that these properties are not for example the result on an emergent physical property of a simple underlying but dynamic physical system. Not forgetting that we need to question the fundamentals of quantum theory - because it allows us to loose track of 96% of the universe. Anyway back to the problem. It is the extrinsic electric field of the charge that performs the function of ( electric and magnetic )energy storage of the electron. The electron at rest on the surface of the earth has an extrinsic electric field also at rest relative to the gravitational field it is in. The extrinsic electric field of the electron is not changing so there is no net energy exchange or emission from the electron. The proof of this is that all the electrons of the earth would be radiating and that would be detectable and wind down the existence of electrons pretty quickly due to persistent energy loss. This contradiction is pretty much the same as the strange non-radiative loss of an orbiting electron that is constantly changing direction ( accelerating). There is a very good simple reason for this behaviour. Which involves understanding the role of superfluids in this dynamic emergent behaviour. ...more on that very soon...
In my humble opinion, there is a quite simple solution for this problem without QED: The base of this thought experiment is the fact, gravity is really not a force, but is created by warping spacetime e.g. in the vicinity of heavy masses - or the warping time to be exact. Since force is F=ma, we can concentrate on acceleration. Since there are two absolutely (!) opposing acceleration vectors created by gravitation, they simply cancel in classical and relativistic electrodynamics. Hence free charges within gravitational fields do not radiate. Things however change completely within a rocket, since there or no longer opposing acceleration vectors but only one, that goes with the velocity vector, since a=dv/dt. Therefore, neither Einstein's strong equivalence principle is violated, nor Maxwell's laws, nor Larmor's formula. The only thing, that really happens, is things slow down within a gravitational field relative to a flat Minkovsky-spacetime. Every energy «lost» in a gravitational field (e.g. gravitational redshift) is gained, when a photon escapes the field. This saves even conservation of energy and momentum.
What do you mean by "two opposing acceleration vectors"; Are you referring to different reference frames or equal and opposite gravitational forces, or BOTH❔ 🤔
In relativity, force is not F=ma, it is F=d(mv)/dt and m is not constant. F would then have 2 components, one in the "a" direction and one in the "v" direction (or if you do the derivative from 1st principles & reinterpret as a difference equation you get the "v+dv" direction).
@@The_Green_Man_OAP This is correct, as far as relativistic velocities or strong gravitational forces are involved. In this case, masses must be multiplied by the Lorentz-Factor γ. Since this is not the case, we can use Newtonian dynamic. Furthermore, it is true, that on large scales - very week gravitational fields - strange things happen, that some people attempt to overcome with MOND (MOdified Newton Dynamics), while another hypothesis goes toward WIMPs. However, in case of elementary charges, we know, these are quantized, Therefore, they can't give up a tiny piece of their ground state energy. Furthermore, we would see a self-discharge of every capacitor on earth, since E=CU²/2=QU/2 because C=Q/U - but not in deep space or orbit. No such things have been observed.
A dumb question to which I don't know the answer: Does a mass in free fall increase its mass by gamma? I suspect the answer is no, but I'd happy to listen to arguments.
@@tonibat59 Yes, it does, since gamma is a function of velocity, while gained energy is a function of relativistic momentum - the product of velocity times rest mass times gamma: Einstein told it as follows: E²=(mc²)²+(pc)². This is his complete equation.
First things first: leaving out GR or QM, in flat space and using the Lorentz transformation to convert coordinates between inertial frames, what do Maxwell's equations say about about a uniformly accelerating charge and a non-accelerating observer vs. a non-accelerating charge and an accelerating observer?
Given the Lamour formula at 15:50, and the subsequent solutions, the amount of radiation emitted by charged particles in a gravity field would indeed be an undetectable fluctuation against the enormous electrostatic attraction any experimental setup would introduce. Such a small difference reminds me of gravity. Like if you turned the equation around, the gravitational constant would fall out.
The implication is, let me spell it out for you tinfoil hat guys, (IF true) bombarding a large mass of charged particles with a tiny amount of a specific type, wave shape, and polarization of non-ionizing radiation, should produce an acceleration not unlike gravity.
Did you mix up falling charge (constant speed) with accelerating charge (changing speed) at 3:38? To me it sounds almost like it should be similar to dropping a magnet down a copper tube; acceleration slowed in a similar way, also involving charges and photons.
@@lukasrafajpps I think what happened is I commented too early without watching the rest of the video. It might be confusing trying thought experiments with a single charge. Maybe a large group of charges would be easier to analyse - you could then confine them to a conductive object for experiments; an experiment along those lines is the one where length contraction of moving charges causes the same effect but by different mechanisms depending on your point of view. From my point of view is the question is: "does a static charge radiate (photons)" - The largest collection of static charge I can think of, that is easily testable, is a thunderstorm. I have no idea how much radiation there should be, but I can imagine it would have been detected by now if it were present.
@@lukasrafajpps I've been thinking about accelerating charges and wondered if surface charging of satellites causes them to slow down due to radiation, and whether that could be detected. Also ionised particles orbiting a black hole probably emit some radiation due to the huge acceleration they must experience, not just the friction with other particles.
The most Zen of knowledge: "A charged particle does not radiate across space, you radiate across the charged particle's space", I wanted to add (ᵃⁿᵈ ᶦᵗ ˢᵀᴱᴬᴸˢ ʸᴼᵁᴿ ᴱᴺᴱᴿᴳʸ ᵀᴼ ᴰᴼ ᴵᵀᵎ), But I don't wanna ruin it.
Questions: could it be that a charged particle is always radiating (and perhaps borrowing and returning energy from the vacuum), unless it is sitting on the surface of some body or other? And even if that were largely true, could it not be that few bodies are exactly at rest with respect to one another because of thermal motion? In free space, surely the particle's field is stretching out, and therefore at some point, some part of it may eventually encounter the weak gravitational field of a yet distant body, whereas the particle itself, further away, now moves relative to at least part of the field it produces. It could be I am havering, and/or maybe the effects are too minuscule to ever measure (or contemplate).
I'm trying to learn basic electricity/magnetism physics, I need not to enter this rabbit hole of madness. Still, interesting af, hopefully I will be able to understand the problem at some point. Good luck
The Lamour formula, named after physicist Louis de Launay Lamour, is a mathematical expression that relates the energy of a charged particle to its momentum and charge. The formula is given by: P = 2q²a²/3c³(4πεo) where: * P is the energy of the particle (W/m²) * q is the charge of the particle * a is the acceleration of the particle * c is the speed of light * εo is the vacuum permittivity constant This formula is a simplification of the more general Lorentz force equation, which describes the force experienced by a charged particle in a magnetic field. The Lamour formula is often used in situations where the magnetic field is strong and the particle's motion is relativistic, meaning that its speed is close to the speed of light. The Lamour formula is a useful tool for understanding the behavior of charged particles in high-energy environments, such as those found in particle accelerators and space plasmas. It is also a fundamental equation in the study of plasma physics and electromagnetic wave propagation.
This seems like a huge dialect W for me. Solipsism just cant describe Reality. Thank you so much for sharing this mystery with us. It was a very interesting video
1. Free-Falling charges don't radiate in their rest frame because they follow their geodetics. 2. Charges on the surface of the earth radiate because they are accelerated. We can't measure it in the rest frame of the charges because it's behind rindler horizon.
1:14 “…due to this acceleration it would radiate…”. no. a satellite 🛰 is in free fall. it’s not experiencing “acceleration”. [after listening to 12:41 … i’m now sooo confused 🫤]
That is also what I think but there are physicists that claim otherwise. On wiki you can find a resolution by Rohlich claiming it would radiate en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field
3:37 : must whether or not a particle radiates, be independent of frame of reference? Whether a detector detects the radiation has to be, I suppose, but isn’t Hawking radiation dependent on frame or reference?
@@JH-le4sd That's because it's hidden from plain sight. Recall that deriving all of the equations required using a point charge to determine the magnitude and direction of the magnetic and electric field strength.
@@GaussianEntity But a hypothetical magnitude (field point) and a physical detector aren't the same thing--e.g. in the examples, Maxwells equations should be valid for *all observers'* frames of reference, but in reality, the radiated photon will only be measured (or not) by the physical detector in its frame, but this still produces a physically observable fact which all will see (the count on the detector). Maxwell's equations have no built-in solution to this--no detector term, and that ambiguity of how you settle it is really what drives the overall problem.
Having no useful comment on this programme's content, I offer only my thanks for Lukas' stand-out presentation: it was my pleasure to have been confounded by his clearly-articulated conundrum, better than confused by so many presenters' verbal slurry, particularly when speaking critical points in sotto voce as though sharing a secret. On Internet. - Thanks, Lukas; I'm off to browse your channel. -g
I think this is a very interesting subject. I would be grateful if you could review the context of a transverse EM field (light wave) being emitted from an RF antenna. The emission arises from the oscillation of the electrons back and forth along the antenna- they are accelerating back and forth as the drive voltage is cycled. So far so good. The EM intensity is torus shaped being zero along the axis of the antenna - so there is a moving electric field back and forth along the axis but it is not a propagating light wave. I would argue the accelerating charge does not radiate energy along its direction of movement- in fact it cannot do so. Does this tally in any way with this discussion? The next question is whether we need the electron to reverse its path (at least once) to propagate the EM field. In other words simple harmonic motion. Apologies in advance if I’ve slipped a gear on this - just wanting to get to the very bottom of EM radiation at the fundamental level.
This is a long comment, But gets in depth on the details so you can rest easier knowing it: I think it makes the most sense that a charge falling from a stationary observer on the ground would radiate just over time randomly. As in, Even if the Local reference frame didn't radiate given its gravity moves along with the electromagnetic field of the charge. Even then the electromagnetic field lines extend beyond the local scale of the particle to global scales. This entails it'll have a higher probability of creating radiation when the electromagnetic lines intersect the gravitational fields geodesic lines (aka where they move in different directions and aren't perfectly parallel even if on a extremely small scale). Similarly… The closer this intersection occurs, The higher chance it'll ACTUALLY release radiation. This would start to occur on molecular length scales just above the particles local scales, but would be so weak that it would be primarily virtual radiation and not legit radiation. This Virtual Radiation also is called “Virtual Photons”. Remember! Particles Technically are just localized Wavefunctions aka waves in disguise. This localization occurs because of the Quantum Decoherence as a result of the High Energies created by Strong Nuclear Forces Gluons. This high energy is akin to the energy created by acceleration inwards of the protons and neutrons (caused by strong nuclear force confinement battling against random wiggles of particles) in all directions and so the wave becomes compressed in all directions. Hence where the “wavefunction collapse” term mathematically comes in. You can even see this localization in the large hadron collider, Where Quarks that break away from the atom can be traced as if their particle-like due to the high velocities relative to the wavefunction. This pretty much compresses the quarks wavefunction like how you might compress a spring, reducing uncertainty in position. Now imagine this same compression but in all directions, THAT is what a atomic nucleus would be like! Yet even then. It can still interact non-locally at a distance just like any other wavefunction, just in this case via these electromagnetic field lines. What this would suggest, Is as the particles electromagnetic field lines intersect gravitational field lines at a distance the particle would release some virtual photons in bursts. With more virtual photons releasing the closer this intersection to the particles local reference frame is. It's just these virtual particles in the form of radiation would be so weak there isn't enough stuff to quantize it, and likely is quickly absorbed by the surrounding virtual particles on miniscule scales. In this sense it's kinda like the Weak Nuclear force. Just instead of rolling a dice to split a piece of the atom off over time, We are rolling the dice to release a bit of actual radiation over time. Such timescales are gigantic though, hence why evaluating a larger area around the particle is the better bet although more challenging. _________ Summary: It's almost guaranteed that particles falling relative to you on the ground releases virtual photon radiation. It's just that it doesn't become actual radiation unless you get lucky and one of the electromagnetic field lines intersects the gravitational field line at a close distance to the particle (instead of going around as it usually does). This means the radiation likely will occur in a very weak form (aka at a lower energy scale) further away from the particle. It's just correlating such low energy radiation further away, to the particle itself is extremely difficult. Soooo, We are running into the same situation we ran into with the Weak Nuclear force. Just now with virtual radiation becoming actual radiation. How do we induce this actual radiation though? Well… Perhaps accelerating the particle through a high density substance will work (to force more field intersections and closer to the particle). But then we need to find a dense substance which doesn't absorb the particle and radiation. And we'll also need to be capable of capturing what is going on at high FPS from multiple angles to accurately correlate the radiation to the particle. Needless to say this is one of those “Easier Said than done” type of things. No wonder a successful and repeatable experiment has yet to be done ooof.
Physics is Phun I've had a bad feeling about the EP for a long while, and this (new to me - I'm surprised) perspective on it has caused me a good bit of excitement Thank you
Page 0:30 A point charge in linear motion within vacuum-Aether space produces electromagnetic radiation? False. It only induce static B field that tracks the point charge in motion.
The charge does not radiate in its frame of reference, but does radiate in a 'rest' frame. Consider the falling charge as a current, which will generate a B field. As the charge accelerates, the current increases, which increases the B field. Couple that with the charge's natural E field (which is moving), and there's radiation. However, in reality, the effects of gravity are tiny relative to electromagnetic effects for charged particles. For macroscopic objects with appropriate charges, these forces can be comparable.
Interesting question. Does whether a charge radiates or not have to be frame independant? I don't think so. Neither the Electric nor Magnetic fields are tensors, meaning they can be 0 in one frame but not in another. In one frame you can see an electric current and therefore a magnetic field, but not in another.
15:28 so are you suggesting that the EM field is direction in the case of a falling charge? The overall flux is 0, but pick a directionally there are different flux?
em field is caused by moving charges, motion is relative to the observer therefore one observer could measure this field to be electric but the other one could measure the very same foeld to be magnetic
Request for another video! People often say you can't travel faster than light, but it seems to me that relative to the speed of light, you can't travel at all. No matter how fast you go, any light you emit always travels away at the same relative speed. Since in common parlance we have no trouble moving, this demonstrates that there is no fundamental limit to the rate at which we can get closer to a destination because our speed relative to light is always zero. It is merely that one can never *observe* something moving faster than light, including the launch pad you left behind you, or your destination. Instead what happens the faster you try to go is that distances start to contract. It is perfectly plausible to reach a four light-year destination in less than four years. It would be great to get a proper explanation of that. And, what does length contraction mean for the size of the observable universe if you can bring the unobervable parts closer by moving fast towards them?
6:28 - the field, which is made up of photons, doesn't generally fall through the ground, unless it is transparent to some frequencies (mostly it's opaque, if not totally, but guess you can try experimenting on a glass bridge and see if there's any difference).
As a charged particle falls, a man on the ground will detect a magnetic field created by the falling particle. A man falling with the charged particle will detect no magnetic field.
Not so weird if one really understands general relativity. Especially as well as Rohrlich who solved this paradox in 1965. The "sitting" particle does radiate but only if observed from a free falling frame of reference. This is described in Section 8.3 of his handbook.
@@lukasrafajpps His point is that the observer on the surface is not inertial one so Maxwell's equations do not work correctly in this reference frame. The one we discovered them in!
Hey buddy, I'm going to ask you an off-topic question. Where or with which software do you make animations like the one at 6:25? I want to express my ideas with such animations, just like you do. I would be very happy if you tell me how you do it.
Hi, I use adobe animate but there is some learning curve to it but if I had stronger computer I would try after effects since it supports 3D animations :D
your videos are so amazing. if only because of how humble you come across. your pace is just so perfectly slow. it gives me just enough time to digest each word. your animations are also slow and crisp. an absolute joy. 🥹 ❤❤❤
Is it possible that the accelerating charge appears to emit photons in the direction of its acceleration when viewed from the ground, but that the charge sees the ground emitting photons at it (or vice versa)? Or would that just make the paradox worse?
Can you uniformly accelerate charge, without acting with force on it? And there are 4 basic forces, electron will first interact with EM, and only EM force. Thus i would say entire discussion is about hypotetic object as accelerated electron without external field (force). Such object cannot exist under physical laws. Acceleration must comes from fundamental force, which has it's relative field, which means there must be external field to accelerate electron, thus it's moving in it's and something elses field.
Thank you for this great video. As the electron emits very low energy when falling towards earth the wavelength must be long. Synchrotron radiation has wavelengths in the nano meter range, close to the size of the electron, and it's real. From a radio-engineering standpoint that seems logical as size of an antenna must not be to far off from the wavelength, at least not many orders of magnitude. If the antenna is way to small compared to the wavelength, it is impossible to get the energy out of the antenna because there is no way to match the transmitter impedance with the impedance of vacuum. All the energy you push into the antenna bounces back to the transmitter. It becomes very unlikely to get a single photon out.
I like the idea of the electron being a single atomic point (2D space; mean all and null orientation at all times) that produces an “antimatter” field or electric field based off quantum spin, and when with a nucleus of protons and neutrons, it creates the electron field (now 3D space, all orientations and positions around the nucleus within the field limit and required energy levels) and it produces the electron field; electron=field which is why we cannot observe an electron around an atom cause the field is the electron(s). When there are many electrons or atoms collected together it creates a blanket field where the N and S poles of the quantum spin self align thus producing a 3D object. Gravity in my eyes is antimatter as it can produce force without physical interaction (mass on mass) and gravitons and positrons and gamma rays etc. may interact with the electron/field. Positron interaction with electrons produce gamma rays (cancer scanning) it goes from 2D to 3D, a sin wave is 2D and will represent positrons, and the rotational vector (the electron) will be what allows for spin(min required points for rotation=2). So based on orientation of the electromagnetic field gravity could decide the quantum spin (so could all other antimatter’s), knowing this we could induce different styles of energy, frequency, antimatter’s, and matters to detect radiation and energy levels of the electron in any reference point. Possibly able to move the field around the electron (push it left or right but can never pass outside of the electron or it would not exist and the electron would have to radiate another field around itself, gravity could be stripping away a layer of field and its perceived and radiation. with the right kind of induced process I’m sure something could be observed… OBSERVATION! Particles act differently when being observed as they have to reproduce the action for every observation (1 for its perspective and X for X many observations.) making the action of a particle 3D and materialistically observable. There’s a cool photon experiment done where they found that using a camera and itd act differently. Just my theory though lol
Simple: what do meteoroids do when they fall through the atmosphere? They explode. Not because of heat but because of charge equilibration. Overcharging and explosions happen before heat should become a problem. They (try to) radiate.
Note 2: Einstein Equivalence is a statement that the gravitational field is a metric field. The back reaction of the electromagnetic field would have nothing to do with the EEP.
@@astro_male Author should link the references properly.... He already has them, no need to waste time searching. Or maybe he misinterpreted them, and he knows it subconsciously, and now he hides the references, not being aware of it? Just do the references...
@@Kraflyn , okay, I agree. We need to assume the possibility that some of the viewers were banned from Google. You can copy the title from my comment and paste it into Google: 1) Paul Bracken "Falling charge in a gravitational field and radiation reaction" (2024) 2) Camila de Almeida, Alberto Saa "The radiation of a uniformly accelerated charge is beyond the horizon: a simple derivation" (2006) Open access papers (in Nature or in arXiv). In general, it was enough to google the first words of the title.
I think the solution is in that "radiation" as we have thought of it is only a coupling effect, not an objective reality. It is not the case that charged particles do not broadcast anything when not accelerated, but it is the case that when they are accelerated, they produce a coupling effect we call EM radiation. Energy is not a real quantity of something that gets "transferred" from one charged particle to another by this "Radiation", instead it is just a convenience calculation of totals. what's really going on is every charged particle is always "Radiating" the capability to change momentum per unit charge in all directions at speed c. that's happening regardless of whether the particle is accelerating or not. To test whether there is "radiation", we hold a second charge at a fixed distance away from a charged particle, and therefore there must be many many many other charged particles holding the second charge in place, held in position by EM oscillatory systems we call "atoms". Specific derivatives in the charge field due to the acceleration of the first charge particle will influence the second within the context of the oscillatory system's they're trapped in. if the derivatives don't happen in a way that resonates with the trap, then the second charged particle gets a net 0 acceleration because the change cancels out with it's oscillatory motion half the time, on average. only if the derivative of the first particle has a rate that resonates with the second will it produce enough change in the second to register a change in our instruments.
I am glad you touched upon this question. I have the feeling that when analysing this process we cannot view the electron as a classical object but as a quantum system which radiates only upon change of quantum state. Some people mention QED in the comments but from the little I have read it seemed that there is still debate when using the theory. There is though another way. What if you used the electron's zitterbewegung? I have worked a bit on this and it appears that the electron will radiate only if there is a change in the zitter frequency of the electron as viewed from the lab frame. Does this statement apply to all the cases you have shown? I will think about it..
This problem is older than general relativity. When Einstein was devising the equivalence principle before the announcement of general relativity in 1909, this problem was pointed out and refuted by Max Born. Nevertheless, Einstein had to stick to the equivalence principle to create general relativity. My answer to this problem is “General relativity itself is wrong.” I succeeded in producing the same result as the three early evidences of general relativity, using special relativity and Maxwell gravity, and in particular, I could remove any doubt by showing it through simulation rather than mathematical tricks. In that method, the equivalence principle had to be denied. If you want to know more about it, look for the book "Relativistic Universe and Forces" or related posts or site "金睛塔". I will not post a link because UA-cam prohibits links.
This was really awesome thanks. But it falls in the category of one of those unphysical classical EM puzzles. (From the "there is no spoon" department:) There is no electromagnetic or Faraday field, there are only electron-photon vertex interactions, so that'd be the proper setting for a definitive answer (if that's what you want). For the classical EM puzzle (as a nice puzzle in its own right) don't you want to conduct a fully covariant analysis, hence use the Faraday field, not separate (E, B, ρ, J). Justathought. Unruh effect seems to be the crux, no?
Correct me if I'm wrong (which it looks like I might be) but... isn't the particle in free-fall experiencing zero acceleration since it is simply following the curve of spacetime? And isn't the observer on the ground the one being accelerated by the surface of the earth by 9.2m/s^2 away from Earth's barycenter? So, wouldn't the person on the ground be the one radiating from the perspective of the charged particle? Doesn't the same thing apply to a charge sitting "still" on the surface of the Earth, as in, the particle is only accelerating if something is preventing it from following its free-fall trajectory?
I really find interesting what you said, but i can not deal with the sentence “ the electric field is separered from the change”. Also the fact that electric field is attracted from the gravitational pole. If you check in “Purcell or Griffiths” (2 well know books on elecromagneticity) nothing about what i mentioned is written
The "field" is only a mathematical description of the pattern of the effect that you would get in different places, for example, if there was an electron there... it makes no sense to talk about the field as if it is a thing in its own right, it isn't, it's a way of modelling potential outcomes of interactions that could occur, but it is only a model... for example, you don't need to model balanced forces that cancel out, like the charges in a neutron, but that doesn't mean that the universe doesn't "do" them.
Gravity is a byproduct of the electromagnetic force. The greater the mass on a planet the larger the gravity field. And mass is dependent on the accumulation of charge. So the total amount of charge determines the gravity of a given object. The standard model architects lifted their noses at observational data and chose to use conceptually based mathematical models. And today there are millions of intelligent people fighting to find out who will have the next impossible answer. There will be no explanation in any text book after the 1920’s. We’ve been duped. There is no room in today’s science for antigravity, transmutation, cold fusion. All key words we’re taught to dismiss. They are all required in an electric universe. The Theory of Everything is just another way to describe Natural Philosophy.
3:59 Maybe you’re going to hit upon this, but at relativistic speeds radiation may appear or disappear. Like how the Hawking radiation of a black hole disappears if you’re in free-fall around it (but just in your reference frame, because everyone else will still see it), or how sometimes relativistic observers can’t even agree on what particles exist.
The latter is very easily resolved: there is no such thing as a particle in quantum mechanics. There are only quanta of energy, but they are NOT atomistic constituents of the field. To be more precise, in quantum field theory the only states one can agree on are asymptotic states at infinity. This gets slightly more complicated in GR, but it's not fundamentally different. There is simple one additional reason why the only "real" states are asymptotic: they only exist in flat spacetime.
I'm not so sure on the idea that a field is falling gravitationally, and I have a feeling that the resolution will come from a better understanding of fields under GR. The EM math you're using throughout the video is not Lorentz invariant, which is well known to cause apparent paradoxes.
12:56 wait how can a charge sitting on earth radiate for a freely falling observer. It's losing energy? But not objectively for all observers? And with no limit?
5:20 - hence the resolution that "gravity is not a force" - gravity is space itself moving - see veritasium, pbs space-time, or others who talk about this
So is the photon observed real? If gravity is just space, this observed photon should be an illusion. Yet I'm positive that the radiation measured via experiments show that it is real.
I do have a few comments: - The tidal effects shouldn't intervene in the problem. You can imagine a similar problem without these, theoretically with a constant gravitational field, or practically with a supermassive black hole the size of the solar system, with an electron falling a few mm. You would have the same questions, and most likely very similar answers. - The mathematical tools we use should not be treated so much as "reality". We know that E or B depends on the frame of reference, so their "reality" looks dubious. At the end of the day, all the mathematical symbols we use are interpretations. The equivalence principle on the other hand is a law which doesn't depend too much on the labels and symbols used for electromagnetism. To me, that makes it more fundamental. - Radiation, as pointed out later in the video, is not a fully defined concept. - Even in the current framework, the reality of force-bearing "particles" is unclear. Photons in QED are, when "force" is looked at, virtual particles. - QED is a quantum theory. Relativity is not. Thus your emissions are quantized, but your acceleration is not, in the current theories.
I got a question. Let's imagine a thought experiment. Let's say we put a ccd beside a charged something and accelerate the whole setup to, say 60g using a solid rocket motor. Let's say we got another ccd, which is not moving with respect to the lab. What will our two CCDs observe? We can do the opposite too. Accelerate (at high g) the ccd towards a charged thing and see if the charge srarts to glow. Earth's gravitation might not be intense enough to cause detectable radiation, but we can perform equivalent experiments wiyh accelerating frame, accelerate them as much as we wish and see if we see anything.
I'm not an expert at all but I'm questioning if maybe we should consider the time dilation in a gravitational field. A particle sitting on the surface is accelerated up and it's field is falling down, but the part of the field that is under the particle is on a slower time frame and the part of the field that is over the particle is in a faster time frame. So the down field is receding slower from the particle than the upper field restoring the field shape to a perfect sphere, so all actions combined, you have no energy liberation. I can imagine that for a falling particle it should be something similar, but the other way around. If you consider an observer that is falling in the same reference frame as the particle, for that observer the time dilation/contraction should compensate for the radiation too (but in that case you should consider the differences in the time frame not only for the particle but also for the observer if it is in a gravitational field), so both particles (ground and free falling) don't radiate. For a particle and an observer that are in space (not subject to a gravitational field) there is no time frame dilation/contraction, so there is no need to correct the shape of the electric field. Where the energy comes from the opposite observer (the one that is on the ground looking at the falling particle or the one that is falling looking at the particle on the ground) ? The particles are sitting "still" with a perfect spherical field around, the one that is moving through the field is the observer, so the energy of the photons simply comes from the kinetic energy of the observer itself, not from the particle.
There are multiple things I need to correct: when ever the charge deviates from its original trajectory, there will be a change in the vector potential, which induces a change in the field. This change in the field when decomposed into electric and magnetic field, and have multi pole decomposition done onto it, we see that part of the change remains even at long distances. This is what Lamour radiation is, not the field lagging behind the particle, but the change in the particle itself propagating through the field. Now if we look at this in a semi classical manner, and interpret the particle as a spread out wave of field excitations, we see the geodesic deviation of a gravitational source does cause tidal effects that compresses and elongate the wave in various directions, partially accelerating the charge. This effect should still happen on the ground if we purely look in this perspective. This is obviously wrong, hence the need for a more quantum theory. Since the Lamour formula comes from any change in 4 velocity. This generalizes to any deviation away from a geodesic, even uniform acceleration should cause radiation in an external frame. Instantaneous comoving frames sees nothing. This should be consistent with universal expansion, and what you found.
Radiation and E&M fields are not absolute, they depend on reference frames. You notice Lorentz force law has velocity vector component. So in one ref frame you may have magnetic field whereas in the other you don't for the same moving charge. Furthermore that formula for radiation is incomplete as it's only the scalar component. The acceleration is also a vector, so radiation will also depend on if you're in an accelerating ref frame. Gravity's effect is already an accelerating ref frame. So a "falling charge" may experience zero net acceleration in that ref frame. If so it will not radiate when observed from such ref frame. As for the equivalence, if I remember correctly it's un-proven in the same way as inertial mass and and gravitational mass is considered equal.
Hi, yes E&M fields depend on the observer but you can't transform away an electromagnetic wave using lorentz transformations. So if there is an EM wave in one frame there is also in other.
@@lukasrafajpps The EM wave would be different frequency depending on the ref frame, but it would be non-vanishing in all ref frames. So that would mean the the falling charge is either radiating in all ref frames or it isn't. The question is, does a perfectly uniform gravitational potential (doesn't exist) cause a charge falling in it to radiate due to acceleration? I would say no because the acceleration is effectively zero due to the equivalence principle. The charge does not experience acceleration it is moving along a geodesic. A charge standing on the "ground" would observe the "falling" charge as accelerating. I'd argue that the charge on the ground is radiating not the falling charge.
@@narfwhals7843 I know I wasn't talking about such transformations. As I understood the author was talking about the mixing of the electric and magnetic field of moving charges since he was talking about the velocity component of the Lorentz force.
@@markattila9835 That is something I would be fine with but there are good arguments that a charge in freefall would radiate. Although Lorentz transformations can't create or destroy EM wave, the transformations between inertial and accelerating observers are not LT.
I also think a lot about this problem, although it lies outside my area of research. I think I'll spend a couple more years understanding and expanding my knowledge :) I suggest keeping in mind one more non-obvious thing related to relativism. The distances between energy levels of atoms vary depending on the height. This does not answer the questions, but it is also worth keeping in mind when analyzing.
Gravitation acts on the photon of starlight as starlight passes through a gravitational field I think the photon accelerates as it passes through and is deflected by the gravitational field Gravitation pushes; it does not pull Gravitation is a manifestation of space time curvature which distorts (bends, warps) the fabric of space and time also referred to as space-time Charged particles are deflected as they pass through a gravitational fired Gravitational fields are comprised of gravitational field waves As gravitational fields interact they produce gravitational field waves Charged particles are also deflected by magnetic field; they are also influenced by an electric field Negative electric charged particles such as electron are attracted to positive electric charge particles such as proton The proton positive charge is a net charge produced by interacting quarks (2 up and 1 down) comprising the proton An electron is attracted and becomes captured by proton via coulomb charge (also referred to as coulomb electric charge)
Vivian Robertson seems to address this directly in his rotating photon particle model. I think he suggests that permittivity of free space is EF free yes by charge density, somehow, this derives relationship can be used to derive gravity, and within that explanation, there was a throwaway comment that charges can be effected by gravitation, and radiate EM fields, but not lose any energy. If his model has any validity, it’s an intriguing and straightforward solution.
This is a very subtle and complicated problem. I looked at it in the context of an electromagnetism course I teach. I hope one day I understand it enough to be able to teach it as an example.
If a charge moving in a curved space-time radiates energy, and assuming radiates photons (aka energy and assuming it has mass) 1st…would it not require to meet certain frequencies as in hf so it would only occur if allowed, and 2nd, if it did radiate, would it not evaporate eventually due to E/c^2= m? Also, why would it slow falling from the tower? Regardless of mass, even if lost due to radiation, all mass accelerates the same in the same curved space-time?
Questions that arise: What does local mean in a practical sense? What physical distance is local? Why is an electric field, or any field of force affected by gravity? Could we accelerate the particle with laser pushing and differentiate between emissions from acceleration and emissions absorption? I would love to hear some answers if you could provide them.
Maybe it has to do with the curvature of spacetime? If the observer and the charge particle were to be in a same reference frame, they don’t experience any differences between their spacetime so they don’t feel any radiation or any other form of resistance. But if this were not to be the case, then the charged particle and the observer A don’t agree in the length of the spacetime and the differences within these lengths may lead to some form of resistance: in the case of the observer A, it would be radiation. If the observer B with the same reference frame as the charged particle sees the observer A, then observer B would sense gravitational waves, which would lead to similar effects. The energy, which causes all of this, is from the decrease in the potential energy. Does the acceleration change? No, because it’s only spacetime contraction and expansion. One can measure something similar. Of course, if you push it a bit hard, I’m willing to yield.
Acceleration does not cause time dilation (other than the resulting change in velocity which makes the dilation have a delta) the time dilation in gravity is constant. The time dilation of gravity is equivalent to moving at the escape.velocity
I guess larmor precession works in bulk where there is an energy source and a circuit. You can then track where the work is done. Scenarios for one individual particle doesn't complete a circuit, ie we don't have before and after measured states. suspect, the gedanken scenarios out of GR are masking assumptions.
I've a side quest-ion: looking at the Lorentz force, a charged particle in a magnetic field drifts thus gaining an acceleration (linear momentum becomes angular). It is correct that velocity isn't changing and the particle doesn't radiate even if it's accelerating?
well.. anyway, the free falling particle does not radiates because it is not accelerating. The same happens if the particle stays still and the observer is in free fall. To me the question can be: "Can I see a radiation from a particle (even if it's not radiating) just because I'm accelerating towards its?"
Are there quantum effects that stop a charged particle from radiating photons with power levels as low as one accelerating at 9.8m/s2, or the even slower tidal accelerations talked about in the video?
2:36 this is the point where i completely lost it lol Why does a particle emiting energy shoud deacelerate? Is it bc some energy is being radiated in the opposite direction of the aceleration? I have a strong feeling that im wrong lol
3:34 that's a false premise: "a particle either radiates or doesn't radiate". The number of particles (e.g. photons) is not conserved in accelerating vs. not accelerating frames of reference. Moreover, in General Theory of Relativity, the energy is not conserved.
If we move photons to field phenomena then the solution becomes obvious: Radiation of a charged particle is a relative phenomena. If it accelerates relative to an observer, it radiates. In particular, the comoving observer doesn't detect the photons as they don't exist to them. Their effect can be observed on the accelerating object. The change in kinetic energy must compensate this absorption to conserve the total energy of the system. Extrapolating, this would mean that the effective gravitational potential would be slightly less if the massive object had charge. Most likely, the difference would be insignificant. Besides the effect would be complicated by the emitted fields and particles carrying the charge.
You've forget that non-charged matter consist of charged particles. First time when you compare charge particle falling from tower. In this case both charged particle and neutral atom will radiate or not in the same way. The difference could be only in case of truly neutral particles like fotons, neutrinos or graviton. Second time in the last experiment: 10^50 atoms times 10^-52W = 0.01W . But you shuld multiply this by average number of electrons. So free falling Moon to the Earth should radiate weakly. This is still very subtle effect but at least there is some hope that in future it could be detectable.
Equivalency principle isn't exactly what it says. If you're in a gravity well the force of acceleration at the bottom of the box will be greater than at the top. If you're just accelerating it will be the same. Therefore you COULD do a test and realize if you're accelerating or in a gravity well.
If every point continuously along the height of the elevator was each experiencing proper acceleration, rather than being pushed from the bottom and the rigid material compressing, or pulled from the top, and the rigid material of the elevator streached... _then_ it would be indistinguishable from gravity. (Gravity where the gradient is uniformly parallel (towards a line or extremely distant point, not radial towards a single nearby point.) Although, there is also a way to measure the mass of the elevator vs. the mass of the Earth from inside the elevator, by jumping up and measuring how much the elevator's proper acceleration decreased. The ratio of your mass to the masses of the elevator and the Earth will give you the answer. But... this principle is about coordinate systems and not physical elevators.
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Equivalance principle is valid only in general relativity. Special relativity is incompatible with equivalence principle.
The equivalence principle does not exist in nature because nature does not produce uniform gravitational fields. All gravitational fields necessarily produce tidal forces, thus breaking the equivalence. With the equivalence principle out of the way, there's no paradox.
solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.
This question requires a universe that does not have the same properties as our own. aka, sure in a hypothetical universe where gravity does not impart energy to objects, sure, but gravity DOES impart energy to objects... seriously man this is high school physics.... go read up on what happens to particles near black holes.
The paradox is completely gone, when you realize different observers see different electric and magnetic fields.
The charged particle inside the lift doesn't radiate but gets accelerated from radiation from outside. That's consistent with the fact kinetic energy is different for different observers.
you're asking the wrong physicists if they all say "of course", this is a famous problem for QED in curved spacetime, with the vacuum transforming non trivially...leading to things like radiating horizons (famously: Hawking radiation, and less so: Unruh radiation).
exactly ....
Or capitalizing on a misrepresentation on purpose. He knows he doesn't have access to competent people
@@fixgirod Exactly! Clearly, he thinks his audience is as dumb as _he_ is.
@@fixgirodwhy do you make it sound so nefarious? I think he’s simplifying for the sake of the video narrative.
To me, it’s kind of rude to assume he must not be talking to smart enough people, instead of assuming he is trying to make compelling content and incorporating the wrong opinions only for the sake of discussion
@@chrisvinciguerra4128 how is that LESS nefarious? lmao
1. Standing on the Earth, what you see is in a non-inertial frame.
2. Maxwell’s equations require modification in a non-inertial frame.
3. Special relativity was devised to make Maxwell’s equations covariant.
4. Covariance implies an inertial frame.
5. Therefore in a non-inertial frame you cannot use Maxwell’s equations to predict an absence of a time-varying magnetic field.
6. Thus there’s no paradox.
7. What we need are measurements precise enough to detect radiation from particles that are stationary in a 1g non-inertial frame.
Just came to write this but you said it better than I would have.
Frame of Earth is only slightly non-covariant, and Lienard formula (Larmor formula for relativity) says that it starts radiating only when Beta is close to 1. So there will never be an observable radiation for particles with insignificant speed. It will radiate only when particle falls into Black Hole. Non-inertiality of Earth does not matter, while this inertiality is with its Beta much much less then 1.
"3. Special relativity was devised to make Maxwell’s equations covariant." I'd say Maxwell's equations were already covariant. These equations were not however compatible with the Cartesian view of the world, so the latter had to change. Maxwell's equations stood unaltered. They're covariant now so they were covariant then.
They were perhaps Einstein's main guide to special relativity.
EXACTLY, CORRECT COMMENT
Doesn't necessarily have to be a precise measurement to detect the effect if you can either or both increase the intensity of the radiation (multiple charged particles) or run it enough times for a chance detection or an average above background noise, if applicable. Afterall nature has provided the most precise photon measurement devices possible; electrons in their exclusion energy wells.
My brain is now a paradox, I both feel smarter and dumber listening to this xD
You're already starting to see it work! 😎
The more you know, the more you realize you you don't know.
@@livingcorpse5664stupidoo
The question is simply to be answered: the Larmor formula relates to an accelerated electrically charged particle losing energy in an electromagnetic field which is the cause of the acceleration. So it does not relate to any such particle which is accelerated in a gravitational field or a curved space time. Therefore the Larmor formula does not tell us anything about radiation being generated in the described setting.
But they do get accelerated via Newton force for gravity, right?
So there is no brehmsstralung if the acceleration is caused by the strong nuclear force or the weak force? It seems me that the QFT calculation of photon emission in a collision is very much the same independent of what force caused a 4.momentum exchange.
@@perarve2463those are forces but gravity is not.
I've been thinking about this problem ever since i saw Veritasium talking about it, but your video goes in much more detail. Good job.
Veratasium is a click teaser.
Block his channel.
No.
I'm a physics grad student, I know where to look for precise information. I watch Veritasium as a entertainment activity. Sometimes he talks about things that I don't know. Gravity is not my subject of research.
Besides, he also have a PhD in physics.
@@karkaroff1617 He has a Ph.D. in physics education: _"Designing Effective Multimedia for Physics Education"._ So a Ph.D. in UA-cam physics presentation.
Sure, and?
Are you aware that he doesn't do his videos alone, right? There are physicists and other professionals as consultants.
@@karkaroff1617 let him have it karkaroff , its the only thing he has about him .
Hey, another great video! We once pondered over this exact question, and you know, there’s an EXTREMELY simple resolution - but you’re not going to like it.
An illuminating paper we found on the topic was Rohrlich’s 1963 “The Principle of Equivalence”. The conclusion of the paper was that a co-accelerating observer does not observe any radiation - and that this was a direct consequence of the covariance of Maxwell’s equations.
What does this tell us? Why - simply that acceleration is relative. An observer who constructs their coordinates via accelerating matter will not observe physical effects of that acceleration. Thus, much like it’s impossible to know one’s true velocity, it’s impossible to know one’s true acceleration.
Hmmm, sounds an awful lot like being stuck in a sort of Matrix 🤔 Almost like there’s a theory about that…
🙃 At any rate, you’ve reminded us to do a video on this topic sometime soon. Great work again!
Co-accelerating observers will see each other steady (as we see while driving), but both will agree about feeling a gravitation-like force opposite to acceleration. Every observer irrespective of their state will agree on one of the two - acceleration or gravitation.
But, even if we forget all that, how does relative acceleration prove your matrix theory?
Acceleration just means a deviation from the geodesic it doesn't mean anything else.
Thank you for the comment. About the relativity of acceleration, there is still not consensus whether the General Relativity truly satisfy general principle of relativity maybe just qualitatively.
What we know that it is not relative in the same sense as velocity because to transform among reference frames of different velocities you always use Lorentz transformations.
So tranforming from stationary to moving you use LT and if you want to transform back then you also use LT but with opposite sign but you are still in the same group of transformations.
If you want to transform from inertial frame to accelerated frame you use different transformation than when you transforming from accelerated frame to inertial frame.
These transformations are different and therefore these frames are not equivalent quantitatively. (If they were the twin paradox would not be solvable)
Anyway I can see a nice symmetry there
particle on the surface + observer on the surface = no radiation
particle falling + observer falling = no radiation
particle on the surface + observer falling = radiation
observer on the surface + particle falling = radiation!?!? would be nicely symmetric.
I would love to see your video about this. Please make one :D
@@lukasrafajppsThanks for replying! We really appreciate that you’ve been choosing to tackle these interesting and more involved physics topics, rather than doing the usual circuit of broad-appeal subjects - we always learn something new watching your videos.
You’re right to point out the transformations are different between two inertial frames versus an inertial and non-inertial frame - this is the distinction that separates the General Principle of Relativity (in which any frame of reference could be subsumed under a single set of transformation-invariant laws of physics) and the Principle of General Covariance (in which the inertial laws of physics are covariantly extended to non-inertial frames).
So we wouldn’t disagree with you there - our point was slightly subtler. If you as observer B adopted as your inertial frame what was a non-inertial frame from the perspective of observer A, then to go between your frame and observer A’s frame, you would need covariant transformations (in each of you would describe the other’s laws as involving fictitious forces, etc.). However, if we consider a third observer C who shares A’s definition of the inertial and a fourth observer D who share’s B’s definition of the inertial frame, then to go from A to C or from B to D we need merely only use the Lorentz transformations, while going from C to D or A to D or vice versa we again need the covariant transformations.
Thus, we see the nature of the transformations are untouched - it’s simply our starting point in choosing what constitutes the inertial frame that is subject to a certain arbitrariness. This is a point we’ve been trying (perhaps unsuccessfully) to communicate across various videos.
Meanwhile, General Relativity asserts that all observers will unambiguously agree on what constitutes an inertial frame via use of an accelerometer. A rest frame on the surface of the earth is non-inertial, so GR definitively tells us a particle at rest on earth’s surface is “really” radiating. However, it appears Maxwell’s equations tells us we can’t actually measure this radiation if we are co-accelerating. This lands us, as you quite rightly assert at the end of the video, in a paradoxical situation, indicating something in our thinking needs revision.
Hopefully we will get around to a video on this topic soon, and if we do so we will definitely cite and utilize the research you’ve done here!
@@aniksamiurrahman6365They would only “feel” a gravitational acceleration if they were using accelerometers that had been previously calibrated in some inertial frame - otherwise the accelerometers would register no change in dimension, and the observers would be unable to measure that force at all. We discuss this concept more thoroughly in our video “Why Relativity Doesn’t Add Up”.
Interestingly, we actually first developed Matrix Theory to explain the conundrum of relative acceleration - how could acceleration be relative and yet still be responsible for causal propagation of information in our universe? The answer we finally hit upon was that it was our knowledge of acceleration which was relative to our coordinate constructions, and that we could still posit absolute acceleration, but we would have to concede it was ultimately unknowable. Later we realized we could say the same thing for absolute velocity (and absolute space), and then in turn realized that could be applied to resolve the puzzles of relativity.
Rhorlic's paper resolved this IMO: "The principle of equivalence". Annals of Physics. 22. This is described in the section _Resolution by Rohrlich_ in the Wikipedia article _Paradox of radiation of charged particles in a gravitational field._ When a charge radiates, the incoming Schott energy cancels the outgoing radiation giving a curved electric field, zero magnetic field and apparently no radiation which has tricked many experts.
Here's a question: how would one levitate a charge so it's stationary on the surface of the Earth? When thinking about the issues you've raised in your video, I've found using a charged sphere as a model instead of a charged particle is very helpful in understanding what's going on physically.
You have to levitate it (counteract Earth's gravity) by either gravity or one of the other forces. If gravity, then the whole scenario is trivial. If electromagnetism, well that has to be accounted for somehow. I suspect the amount of EM force needed will change depending on the observer, and that's where the apparent discrepancy comes from. I also suspect that there's no way to keep the falling charge adjacent to the falling observer without some additional force.
Do you mean like the Millikin Oil Drop Experiment?
yikes another person who apparently uses ChatGPT
many good comments so far. I just want to applaud your sketch of this as a kind of "paradox" along the lines of other relativity "paradoxes."
QED-wise, the radiation for a charged particle happens when it flips the spin and releases a photon. This can happen spontaneously (but also the recapture) OR due to an external field acting on the charge. In both cases the tree level approximation is that it flips the spin - tree level means, that is the most probable case. In gravity, there is no need to flip the spin (except diverging gravitational field may). So one problem is that the Maxwell equations and the Lorentz force does not take spin into account, the other is the lack of self-interaction (which is present even in the propagator of QED). In QED you have a vacuum full of virtual photons, representing the field of the particle, and when it accelerates the photon is not recaptured - however you need an other photon to accelerate the particle, whose remnant could also be the outgoing radiation. So in QED you have a convoluted answer, which does the radiated photon comes from, but you still need an external field that causes the acceleration.
where is you username?
@@weeb3277 Honestly, I don't know
when it .... what?
@@weeb3277 the username is there, but it's behind event horizon of your reference frame 😜
Quite interesting. So can we calculate the reverse force? Wouldn't the same apply for acceleration caused by any external force other than electrostatic?
I loved the video. It's incredible to think that we understand so much complexities of how the universe behaves. From the orbitals of the quantum world to the hawking radiations of black holes and yet scientists disagree on what a plastic comb rubbed with my hair will do!
Btw, I loved the presentation and the story telling. The way it starts with, 'of course free falling charges must radiate', and how it leads to 'falling charges having to fall slower' leading to the 'breaking the equivalence principe' leading to a need for a deeper investigation was brilliant (no pun intended). Also, congrats on getting Brilliant sponsorship :) Cheers.
Thank you very much :) I am very glad you liked the video :) :)
I love both you guys.... I guess it would be incredible awesome and extremely funny when you get collaborate.... really looking forward to it! Cheers!
@@lukasrafajpps solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.
I really like this question. I've always suggested that the equivalence principle should go both ways. Acceleration by electromagnetic fields should be indistinguishable from curved spacetime (gravity).
Here's a thought experiment someone smarter than me could try playing with. Take yourself a universe. It has a charged black hole (let's say negatively charged). This black hole is in an especially clean corner of space, where it cannot easily find positive charge to balance out to a neutral charge. Now take an electron, and shoot it gently towards the black hole. Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's gravitational field (without needing to orbit) (at the point where the gravitational acceleration inward is equal to the electromagnetic acceleration outward). So with this setup, what does both the gravitational field and electromagnetic field look like?
Nice one.
Not sure though, I'd expect the particle to radiate, since it is accelerated in the space-time picture. Just like a particle in an ion trap on earth, actually ... should be measureable this way ^^
The equivalence principle can't go both ways because the electromagnetic field differs between particles.
The equivalence principle only holds because gravity behaves like a curvature, In the sense that It affects EVERYTHING the same.
"Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's..."
Nonsense. You are claiming that the black hole can't 'see' the electron until it passes the event horizon at which point, the hole develops a sudden allergy.
@@nicolasreinaldet732 you dismiss this perspective too easily. it is possible to imagine that different particles experience different space-time curvature.
@@michaelcarey26no it isn't without creating a completely new theory.
How do we know today for sure that linearly accelerated charges radiate, and jerk isn't needed fx.? I mean, what has happened since Pauli, Born and Feynman?
At 8:50 : In this form, Maxwell's equations are only valid in flat spacetime, in an inertial frame, bc. those are not covariant derivatives.
At 10:05 : This kind of Rindler horizon explanation only works in flat space time, but around the Earth, not neccessarily. In this case, the workings of such a horizon is not as trivial as in Minkowski, neither as in Schwarzschild, where every Schwarzschild observers Rindler horizon is just the event horizon.
Q: If you transform the electric and magnetic field to curved spacetime it would solve the problem?
@@moisessalazar4432 I don't know. Maxwell's equations in curved spacetime are nonlinear, so people can't/don't like working with them. Wikipédia says that Fritz Rohrlich examined the flat version of this problem in 1965 (a Rindler charged particle and an inertial observer). Based on how weak the curvature is around Earth, that approximation may be sufficient in our case.
15:33 The energy of the particle (or charged particle) is equal to the derivative of the action in time x(0)/c: and defined in world and proper time: E(0)=с∂S/∂x(0), E=∂S/∂t [momentum p(k)=∂S/∂x(k)].
Then E(0)=E√g(00)=const and E(0) is preserved, and E is not preserved. E=mc^2/√1-v^2/c^2, where in the static case v=dl/dt=-dl/dt√g(00).
Thus, when a particle moves in a gravitational field, the energy E(0)=mc^2(√g(00)/√1-v^2/c^2 is preserved*.
This formula remains valid in the case of a stationary field if, when determining the velocity v, one uses the proper time measured by the clock synchronized along the trajectory of the particle.
P.S.
"Due to the dependence of the speed of light and the speed of the clock on the gravitational potential, after the introduction of the gravitational potential, the laws of nature must be understood as the relationship between the gravitational potential and other physical quantities." (Pauli, RT).
----------
*) - So, it should always be remembered that according to GR, it is possible to make a general statement that:
the free movement of the test body occurs along geodesic lines. And this is the most general formulation of Newton's first law. That is, in a Riemannian manifold, where formally there is no gravity and the masses move freely,
free fall turns out to be not accelerated, but uniformly rectilinear motion.
{Unless, of course, GR obeys a strong equivalence principle, that is, if in fact a locally Lorentzian system = a globally Lorentzian system.}
Note: Measuring a changing electric field does NOT necessarily mean magnetic field intensity is non-zero, only that there's a non-trivial curl of the magnetic field.
It’s enough to just stop saying that there are two forces and two fields. There is only one electromagnetic field.
@@АндрейМирон-х2н No one is saying that there are two electromagnetic fields, but given the two Maxwell equations (one of which couples to the matter fields and the other which describes the free-electromagnetic field) we have 4 independent field components (the two excitation fields (D,H) and two that define a pair of field intensities (E,B) which can be related via the constitutive relations).
What I am saying has nothing to do with this, rather, given some time-like curve what does the observer measure of the electromagnetic field components if the electric field intensity is a function of time (a non-trivial curl of the magnetic field intensity, and not necessarily a non-zero value of the field itself).
@@kylelochlann5053 But the author of the video clearly divides this into components. First, he needs to disassemble the physical meaning without confusing himself with a distorting perception of dividing formulas. That in different reference systems the electrical and magnetic calculations are different, but their resulting influence is identical.
And after that it is more correct to express the meaning of the situations under consideration.
@@АндрейМирон-х2нThis is just semantics. Including the idea of their being a single electromagnetic field rather than two fields that induce eachother. What's crazy is you can't derive the magnetic field from the coloumb field, even given the fact the circular shape of the magnetic field around a moving electron is just the cross product of its velocity vector and the coloumb field. even given special relativity, physicists always presuppose the magnetic field already exists in order to separate them in lorentz force formula, and the em field tensor, etc it just makes the math 10x more elegant to represent it as two fields which separately induce one another. Theres a significantly more complicated double cross product representation where you treat the B field as only a lorentz boosted E field and B's dissapear but is it really worth it? Its like its only simple as two fields. Even the word electromagnetic implies this duality, if you were really capable of viewing it as a single 4d vector manifold thing youd just call it the electric field, because magnetic fields dont exist if you view its just "one" thing (the force transforms under lorentz boost).
UA-camrs famous words "there are some experts in the comments" :D
Great topic and video btw.
Maxwell equations are not valid 'as is' to curved spacetime nor Larmur formula (nor QED).
To tackle it, we need to use the curved spacetime version of the EM Lagrangian and derive the curved spacetime radiation condition.
My (usually bad) intuition says the equivalence principle still holds.
The equivalence principle has been proven over and over since GR requires it and we know GR is correct.
Nice try ChatGPT you failed again.
@@Saki630 not chat GPT but I'll take it as a compliment 😊
It is not necessarily required to adhere to the equivalence of inertial frames either; observers in their own inertial systems can choose to always calibrate the speed of light as isotropic. This would mean that genuine Lorentz contraction and time dilation would occur relative to the 'stationary ether', and with proper synchronization of clocks, the space and time coordinates interpreted in different inertial frames could be connected by the Galilean transformation. It is just difficult to accept that in this case, the length of a stationary rod in its own inertial frame and the length of an identical rod moving relative to it in the moving inertial frame would not be the same.
@@szilardecsenyi516but an observer sitting on earth's surface is not in an inertial frame. It is in an accelerated frame.
One of those questions where you need to break out both the GR & EM tensors to answer correctly.
The correct answer is "depends, relative to what observer?"
As acceleration is observer dependent as is mode of radiated energy.
Right... you need a second charged particle for it to be interacting with, otherwise there is no force generated by the presence of the charge to act on it to cause it to be slowed down.
@@annoloki Not exactly. Photons can still be radiated & absorbed by non-charged observer. But you still need an observer & reference frame to describe the interaction.
10:55 I don't understand, shouldn't the Unruh effect be experienced by the person on the ground? The presence of the charged particle and the other, free falling observer shouldn't have any infuence on the observation of the grounded person, at least I don't think it should...
At last! I've been asking this question for decades. I appreciate your presentation, which brought out aspects of the problem I hadn't considered so clearly.
Good video but I think you should of included orbiting charge analysis too. For example, what's the difference between a charge moving in a circle in a magnetic field versus a gravitational field? (Like how you animated the electric field not being accelerated in the magnetic field, wouldn't that apply to an orbiting charge?)
1:40 ok I haven't watched the rest of the video nor read any comments but my gut feeling immediately is that gravity is a curvature in spacetime and does not accelerate the particle, the particle keeps moving in a straight line through the curved spacetime. So I would expect no radiation coming from it. Watching the rest of the video now...
I though about that problem when I wanted to apply the rain cycle we usually know on Earth, but for a hypothetical charged matter on the surface of a star. The basic hypothesis are a very strong gravity, a very hot heart and charged matter. The cycle of the charged matter would be:
- from the heart, the very hot matter rises the gravity field (and an electric field may also) to gain altitude in the star.
- at high altitude into the star, the charged slowed down by the gravity field, is at much lower temperature.
- over that surface of charged electric matter at high altitude but at low temperature that creates near by an ocean of electric charged matter, there is a hole, and the charged electric matter falls into that hole to get back into the heart of the star.
Of course, when the electric charged falls into the hole, it creates a huge beam of light.
The is the image I've of a pulsar star.
The beam we can see from a pulsar star is the hole where the charged electric matter falls into.
10:10 Brazil mentioned🍺
Unruh was mentioned too!!! Good grief.
?
Brasileiro é foda
I'm a physicist and I think this is a great video in that it is seeking to move problem solving to a broader domain than the hyper conservative confines of academia and the scientific paper manufacturing industrial complex.
Which is something that Hossenfelder aligns with.
Now what I get out of this, and agree with, is that the electric field of a charge is an extrinsic property of that charge.
This is an unavoidable conclusion from my own analysis that comes from a very different fundamental model that derives the formation of the electron and positron.
In this model the electric field is a result of the capacitive electrical energy storage of the electrical permittivity of "empty" space. That capacitive energy stored in that "empty" space is due to the spin interaction of the electron ( a charge).
So the fundamental intrinsic property of the electron is its spin and its magnetic dipole moment.
The electric field, mass and gravitational field of the electron are all extrinsic properties of the electron. They are literally all "off particle" properties. ( by way of simple proof - the higgs field is an extrinsic field so the electron's mass and associated gravity field are all "off- particle " properties.
The non- obvious realisation here is that the spin of the electron must extend out as far as the electric field it induces in the surrounding space - that generates that electric field.
Quantum theory tells you that particle spin is non-physical. Not because it is proven not to be physical, but because no one can make sense of how it could be physical given the strangeness of spin half and g=2 for the magnetic moment.
But that is an assumption and an assumption is a very very long way from a proof that these properties are not for example the result on an emergent physical property of a simple underlying but dynamic physical system.
Not forgetting that we need to question the fundamentals of quantum theory - because it allows us to loose track of 96% of the universe.
Anyway back to the problem.
It is the extrinsic electric field of the charge that performs the function of ( electric and magnetic )energy storage of the electron.
The electron at rest on the surface of the earth has an extrinsic electric field also at rest relative to the gravitational field it is in.
The extrinsic electric field of the electron is not changing so there is no net energy exchange or emission from the electron.
The proof of this is that all the electrons of the earth would be radiating and that would be detectable and wind down the existence of electrons pretty quickly due to persistent energy loss.
This contradiction is pretty much the same as the strange non-radiative loss of an orbiting electron that is constantly changing direction ( accelerating).
There is a very good simple reason for this behaviour. Which involves understanding the role of superfluids in this dynamic emergent behaviour.
...more on that very soon...
In my humble opinion, there is a quite simple solution for this problem without QED: The base of this thought experiment is the fact, gravity is really not a force, but is created by warping spacetime e.g. in the vicinity of heavy masses - or the warping time to be exact. Since force is F=ma, we can concentrate on acceleration. Since there are two absolutely (!) opposing acceleration vectors created by gravitation, they simply cancel in classical and relativistic electrodynamics. Hence free charges within gravitational fields do not radiate. Things however change completely within a rocket, since there or no longer opposing acceleration vectors but only one, that goes with the velocity vector, since a=dv/dt. Therefore, neither Einstein's strong equivalence principle is violated, nor Maxwell's laws, nor Larmor's formula. The only thing, that really happens, is things slow down within a gravitational field relative to a flat Minkovsky-spacetime. Every energy «lost» in a gravitational field (e.g. gravitational redshift) is gained, when a photon escapes the field. This saves even conservation of energy and momentum.
What do you mean by "two opposing acceleration vectors"; Are you referring to different reference frames or equal and opposite gravitational forces, or BOTH❔ 🤔
In relativity, force is not F=ma, it is
F=d(mv)/dt and m is not constant.
F would then have 2 components, one in the "a" direction and one in the "v" direction (or if you do the derivative from 1st principles & reinterpret as a difference equation you get the "v+dv" direction).
@@The_Green_Man_OAP This is correct, as far as relativistic velocities or strong gravitational forces are involved. In this case, masses must be multiplied by the Lorentz-Factor γ. Since this is not the case, we can use Newtonian dynamic. Furthermore, it is true, that on large scales - very week gravitational fields - strange things happen, that some people attempt to overcome with MOND (MOdified Newton Dynamics), while another hypothesis goes toward WIMPs. However, in case of elementary charges, we know, these are quantized, Therefore, they can't give up a tiny piece of their ground state energy. Furthermore, we would see a self-discharge of every capacitor on earth, since E=CU²/2=QU/2 because C=Q/U - but not in deep space or orbit. No such things have been observed.
A dumb question to which I don't know the answer: Does a mass in free fall increase its mass by gamma?
I suspect the answer is no, but I'd happy to listen to arguments.
@@tonibat59 Yes, it does, since gamma is a function of velocity, while gained energy is a function of relativistic momentum - the product of velocity times rest mass times gamma: Einstein told it as follows: E²=(mc²)²+(pc)². This is his complete equation.
First things first: leaving out GR or QM, in flat space and using the Lorentz transformation to convert coordinates between inertial frames, what do Maxwell's equations say about about a uniformly accelerating charge and a non-accelerating observer vs. a non-accelerating charge and an accelerating observer?
Given the Lamour formula at 15:50, and the subsequent solutions, the amount of radiation emitted by charged particles in a gravity field would indeed be an undetectable fluctuation against the enormous electrostatic attraction any experimental setup would introduce. Such a small difference reminds me of gravity. Like if you turned the equation around, the gravitational constant would fall out.
The implication is, let me spell it out for you tinfoil hat guys, (IF true) bombarding a large mass of charged particles with a tiny amount of a specific type, wave shape, and polarization of non-ionizing radiation, should produce an acceleration not unlike gravity.
Did you mix up falling charge (constant speed) with accelerating charge (changing speed) at 3:38?
To me it sounds almost like it should be similar to dropping a magnet down a copper tube; acceleration slowed in a similar way, also involving charges and photons.
isn't a falling charge also accelerating?
@@lukasrafajpps I think what happened is I commented too early without watching the rest of the video.
It might be confusing trying thought experiments with a single charge. Maybe a large group of charges would be easier to analyse - you could then confine them to a conductive object for experiments; an experiment along those lines is the one where length contraction of moving charges causes the same effect but by different mechanisms depending on your point of view. From my point of view is the question is: "does a static charge radiate (photons)" - The largest collection of static charge I can think of, that is easily testable, is a thunderstorm. I have no idea how much radiation there should be, but I can imagine it would have been detected by now if it were present.
@@lukasrafajpps I've been thinking about accelerating charges and wondered if surface charging of satellites causes them to slow down due to radiation, and whether that could be detected. Also ionised particles orbiting a black hole probably emit some radiation due to the huge acceleration they must experience, not just the friction with other particles.
The most Zen of knowledge: "A charged particle does not radiate across space, you radiate across the charged particle's space", I wanted to add (ᵃⁿᵈ ᶦᵗ ˢᵀᴱᴬᴸˢ ʸᴼᵁᴿ ᴱᴺᴱᴿᴳʸ ᵀᴼ ᴰᴼ ᴵᵀᵎ), But I don't wanna ruin it.
Questions: could it be that a charged particle is always radiating (and perhaps borrowing and returning energy from the vacuum), unless it is sitting on the surface of some body or other? And even if that were largely true, could it not be that few bodies are exactly at rest with respect to one another because of thermal motion?
In free space, surely the particle's field is stretching out, and therefore at some point, some part of it may eventually encounter the weak gravitational field of a yet distant body, whereas the particle itself, further away, now moves relative to at least part of the field it produces.
It could be I am havering, and/or maybe the effects are too minuscule to ever measure (or contemplate).
I'm trying to learn basic electricity/magnetism physics, I need not to enter this rabbit hole of madness. Still, interesting af, hopefully I will be able to understand the problem at some point. Good luck
The Lamour formula, named after physicist Louis de Launay Lamour, is a mathematical expression that relates the energy of a charged particle to its momentum and charge. The formula is given by:
P = 2q²a²/3c³(4πεo)
where:
* P is the energy of the particle (W/m²)
* q is the charge of the particle
* a is the acceleration of the particle
* c is the speed of light
* εo is the vacuum permittivity constant
This formula is a simplification of the more general Lorentz force equation, which describes the force experienced by a charged particle in a magnetic field. The Lamour formula is often used in situations where the magnetic field is strong and the particle's motion is relativistic, meaning that its speed is close to the speed of light.
The Lamour formula is a useful tool for understanding the behavior of charged particles in high-energy environments, such as those found in particle accelerators and space plasmas. It is also a fundamental equation in the study of plasma physics and electromagnetic wave propagation.
This seems like a huge dialect W for me. Solipsism just cant describe Reality.
Thank you so much for sharing this mystery with us. It was a very interesting video
1. Free-Falling charges don't radiate in their rest frame because they follow their geodetics.
2. Charges on the surface of the earth radiate because they are accelerated. We can't measure it in the rest frame of the charges because it's behind rindler horizon.
Really enjoyed this video and greatly appreciated the math done at the end for some perspective
1:14 “…due to this acceleration it would radiate…”. no. a satellite 🛰 is in free fall. it’s not experiencing “acceleration”. [after listening to 12:41 … i’m now sooo confused 🫤]
That is also what I think but there are physicists that claim otherwise. On wiki you can find a resolution by Rohlich claiming it would radiate en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field
Good to see I'm not the only one confused 😮
3:37 : must whether or not a particle radiates, be independent of frame of reference? Whether a detector detects the radiation has to be, I suppose, but isn’t Hawking radiation dependent on frame or reference?
Another problem is there is no "detector" term in any of Maxwell's equations.
@@JH-le4sd That's because it's hidden from plain sight. Recall that deriving all of the equations required using a point charge to determine the magnitude and direction of the magnetic and electric field strength.
@@GaussianEntity But a hypothetical magnitude (field point) and a physical detector aren't the same thing--e.g. in the examples, Maxwells equations should be valid for *all observers'* frames of reference, but in reality, the radiated photon will only be measured (or not) by the physical detector in its frame, but this still produces a physically observable fact which all will see (the count on the detector). Maxwell's equations have no built-in solution to this--no detector term, and that ambiguity of how you settle it is really what drives the overall problem.
Having no useful comment on this programme's content, I offer only my thanks for Lukas' stand-out presentation: it was my pleasure to have been confounded by his clearly-articulated conundrum, better than confused by so many presenters' verbal slurry, particularly when speaking critical points in sotto voce as though sharing a secret. On Internet. - Thanks, Lukas; I'm off to browse your channel. -g
Thank you for kind words :)
I think this is a very interesting subject. I would be grateful if you could review the context of a transverse EM field (light wave) being emitted from an RF antenna. The emission arises from the oscillation of the electrons back and forth along the antenna- they are accelerating back and forth as the drive voltage is cycled. So far so good. The EM intensity is torus shaped being zero along the axis of the antenna - so there is a moving electric field back and forth along the axis but it is not a propagating light wave. I would argue the accelerating charge does not radiate energy along its direction of movement- in fact it cannot do so. Does this tally in any way with this discussion? The next question is whether we need the electron to reverse its path (at least once) to propagate the EM field. In other words simple harmonic motion. Apologies in advance if I’ve slipped a gear on this - just wanting to get to the very bottom of EM radiation at the fundamental level.
This is a long comment, But gets in depth on the details so you can rest easier knowing it:
I think it makes the most sense that a charge falling from a stationary observer on the ground would radiate just over time randomly.
As in, Even if the Local reference frame didn't radiate given its gravity moves along with the electromagnetic field of the charge. Even then the electromagnetic field lines extend beyond the local scale of the particle to global scales.
This entails it'll have a higher probability of creating radiation when the electromagnetic lines intersect the gravitational fields geodesic lines (aka where they move in different directions and aren't perfectly parallel even if on a extremely small scale).
Similarly… The closer this intersection occurs, The higher chance it'll ACTUALLY release radiation.
This would start to occur on molecular length scales just above the particles local scales, but would be so weak that it would be primarily virtual radiation and not legit radiation. This Virtual Radiation also is called “Virtual Photons”.
Remember! Particles Technically are just localized Wavefunctions aka waves in disguise.
This localization occurs because of the Quantum Decoherence as a result of the High Energies created by Strong Nuclear Forces Gluons.
This high energy is akin to the energy created by acceleration inwards of the protons and neutrons (caused by strong nuclear force confinement battling against random wiggles of particles) in all directions and so the wave becomes compressed in all directions. Hence where the “wavefunction collapse” term mathematically comes in.
You can even see this localization in the large hadron collider, Where Quarks that break away from the atom can be traced as if their particle-like due to the high velocities relative to the wavefunction. This pretty much compresses the quarks wavefunction like how you might compress a spring, reducing uncertainty in position.
Now imagine this same compression but in all directions, THAT is what a atomic nucleus would be like!
Yet even then. It can still interact non-locally at a distance just like any other wavefunction, just in this case via these electromagnetic field lines.
What this would suggest, Is as the particles electromagnetic field lines intersect gravitational field lines at a distance the particle would release some virtual photons in bursts. With more virtual photons releasing the closer this intersection to the particles local reference frame is.
It's just these virtual particles in the form of radiation would be so weak there isn't enough stuff to quantize it, and likely is quickly absorbed by the surrounding virtual particles on miniscule scales. In this sense it's kinda like the Weak Nuclear force. Just instead of rolling a dice to split a piece of the atom off over time, We are rolling the dice to release a bit of actual radiation over time.
Such timescales are gigantic though, hence why evaluating a larger area around the particle is the better bet although more challenging.
_________
Summary:
It's almost guaranteed that particles falling relative to you on the ground releases virtual photon radiation.
It's just that it doesn't become actual radiation unless you get lucky and one of the electromagnetic field lines intersects the gravitational field line at a close distance to the particle (instead of going around as it usually does).
This means the radiation likely will occur in a very weak form (aka at a lower energy scale) further away from the particle.
It's just correlating such low energy radiation further away, to the particle itself is extremely difficult.
Soooo, We are running into the same situation we ran into with the Weak Nuclear force. Just now with virtual radiation becoming actual radiation.
How do we induce this actual radiation though?
Well… Perhaps accelerating the particle through a high density substance will work (to force more field intersections and closer to the particle). But then we need to find a dense substance which doesn't absorb the particle and radiation.
And we'll also need to be capable of capturing what is going on at high FPS from multiple angles to accurately correlate the radiation to the particle.
Needless to say this is one of those “Easier Said than done” type of things.
No wonder a successful and repeatable experiment has yet to be done ooof.
Physics is Phun
I've had a bad feeling about the EP for a long while, and this (new to me - I'm surprised) perspective on it has caused me a good bit of excitement
Thank you
Page 0:30 A point charge in linear motion within vacuum-Aether space produces electromagnetic radiation? False.
It only induce static B field that tracks the point charge in motion.
The charge does not radiate in its frame of reference, but does radiate in a 'rest' frame. Consider the falling charge as a current, which will generate a B field. As the charge accelerates, the current increases, which increases the B field. Couple that with the charge's natural E field (which is moving), and there's radiation. However, in reality, the effects of gravity are tiny relative to electromagnetic effects for charged particles. For macroscopic objects with appropriate charges, these forces can be comparable.
Interesting question. Does whether a charge radiates or not have to be frame independant? I don't think so. Neither the Electric nor Magnetic fields are tensors, meaning they can be 0 in one frame but not in another. In one frame you can see an electric current and therefore a magnetic field, but not in another.
Can you make a video about Amanda Gefter's book
"Trespassing on Einstein's Lawn"? Can't stop thinking about it.
15:28 so are you suggesting that the EM field is direction in the case of a falling charge? The overall flux is 0, but pick a directionally there are different flux?
em field is caused by moving charges, motion is relative to the observer therefore one observer could measure this field to be electric but the other one could measure the very same foeld to be magnetic
Request for another video! People often say you can't travel faster than light, but it seems to me that relative to the speed of light, you can't travel at all. No matter how fast you go, any light you emit always travels away at the same relative speed. Since in common parlance we have no trouble moving, this demonstrates that there is no fundamental limit to the rate at which we can get closer to a destination because our speed relative to light is always zero. It is merely that one can never *observe* something moving faster than light, including the launch pad you left behind you, or your destination. Instead what happens the faster you try to go is that distances start to contract. It is perfectly plausible to reach a four light-year destination in less than four years. It would be great to get a proper explanation of that. And, what does length contraction mean for the size of the observable universe if you can bring the unobervable parts closer by moving fast towards them?
6:28 - the field, which is made up of photons, doesn't generally fall through the ground, unless it is transparent to some frequencies (mostly it's opaque, if not totally, but guess you can try experimenting on a glass bridge and see if there's any difference).
As a charged particle falls, a man on the ground will detect a magnetic field created by the falling particle. A man falling with the charged particle will detect no magnetic field.
My first answer was "Of course not". The surrounding of a free falling particle is actually an inertial frame of reference in General Relativity.
Mine too. But then the conclusion that a sitting charged particle radiates is super weird. 🤷♂️
Not so weird if one really understands general relativity. Especially as well as Rohrlich who solved this paradox in 1965. The "sitting" particle does radiate but only if observed from a free falling frame of reference. This is described in Section 8.3 of his handbook.
but what is weird that as far as I know Rohrlich also claims that observer on the surface would see radiation from freelly falling charge.
@@lukasrafajpps His point is that the observer on the surface is not inertial one so Maxwell's equations do not work correctly in this reference frame. The one we discovered them in!
@@arctic_haze I guess, I'm just an amateur. 🙂
Hey buddy, I'm going to ask you an off-topic question. Where or with which software do you make animations like the one at 6:25? I want to express my ideas with such animations, just like you do. I would be very happy if you tell me how you do it.
Hi, I use adobe animate but there is some learning curve to it but if I had stronger computer I would try after effects since it supports 3D animations :D
16:14 Wouldn't it be just sqrt( 5*10^52)= 2.24 *10^26 charges, because the charge in the formular is squared?
your videos are so amazing. if only because of how humble you come across. your pace is just so perfectly slow. it gives me just enough time to digest each word. your animations are also slow and crisp. an absolute joy. 🥹 ❤❤❤
Is it possible that the accelerating charge appears to emit photons in the direction of its acceleration when viewed from the ground, but that the charge sees the ground emitting photons at it (or vice versa)? Or would that just make the paradox worse?
Can you uniformly accelerate charge, without acting with force on it? And there are 4 basic forces, electron will first interact with EM, and only EM force. Thus i would say entire discussion is about hypotetic object as accelerated electron without external field (force). Such object cannot exist under physical laws. Acceleration must comes from fundamental force, which has it's relative field, which means there must be external field to accelerate electron, thus it's moving in it's and something elses field.
I have been asking this question myself for very long time. Thanks for making this video
Can't believe i am watching it again 😂
Thank you for this great video. As the electron emits very low energy when falling towards earth the wavelength must be long. Synchrotron radiation has wavelengths in the nano meter range, close to the size of the electron, and it's real. From a radio-engineering standpoint that seems logical as size of an antenna must not be to far off from the wavelength, at least not many orders of magnitude. If the antenna is way to small compared to the wavelength, it is impossible to get the energy out of the antenna because there is no way to match the transmitter impedance with the impedance of vacuum. All the energy you push into the antenna bounces back to the transmitter. It becomes very unlikely to get a single photon out.
I like the idea of the electron being a single atomic point (2D space; mean all and null orientation at all times) that produces an “antimatter” field or electric field based off quantum spin, and when with a nucleus of protons and neutrons, it creates the electron field (now 3D space, all orientations and positions around the nucleus within the field limit and required energy levels) and it produces the electron field; electron=field which is why we cannot observe an electron around an atom cause the field is the electron(s). When there are many electrons or atoms collected together it creates a blanket field where the N and S poles of the quantum spin self align thus producing a 3D object. Gravity in my eyes is antimatter as it can produce force without physical interaction (mass on mass) and gravitons and positrons and gamma rays etc. may interact with the electron/field. Positron interaction with electrons produce gamma rays (cancer scanning) it goes from 2D to 3D, a sin wave is 2D and will represent positrons, and the rotational vector (the electron) will be what allows for spin(min required points for rotation=2). So based on orientation of the electromagnetic field gravity could decide the quantum spin (so could all other antimatter’s), knowing this we could induce different styles of energy, frequency, antimatter’s, and matters to detect radiation and energy levels of the electron in any reference point. Possibly able to move the field around the electron (push it left or right but can never pass outside of the electron or it would not exist and the electron would have to radiate another field around itself, gravity could be stripping away a layer of field and its perceived and radiation. with the right kind of induced process I’m sure something could be observed…
OBSERVATION! Particles act differently when being observed as they have to reproduce the action for every observation (1 for its perspective and X for X many observations.) making the action of a particle 3D and materialistically observable. There’s a cool photon experiment done where they found that using a camera and itd act differently. Just my theory though lol
Fantastic video! Many thanks!
Fantastic analysis!
Simple: what do meteoroids do when they fall through the atmosphere? They explode. Not because of heat but because of charge equilibration. Overcharging and explosions happen before heat should become a problem. They (try to) radiate.
Note 2: Einstein Equivalence is a statement that the gravitational field is a metric field. The back reaction of the electromagnetic field would have nothing to do with the EEP.
Nice Video !
This paradox has confused me for a long time ⚡
This thing EM wave never gets old. 300 years of discussion and still no end in sight.
links to papers pls
Agree
just google it, you have the author's name and the title of the paper. I found all the papers, of course.
@@astro_male Author should link the references properly.... He already has them, no need to waste time searching. Or maybe he misinterpreted them, and he knows it subconsciously, and now he hides the references, not being aware of it? Just do the references...
@@Kraflyn , okay, I agree. We need to assume the possibility that some of the viewers were banned from Google.
You can copy the title from my comment and paste it into Google:
1) Paul Bracken "Falling charge in a gravitational field and radiation reaction" (2024)
2) Camila de Almeida, Alberto Saa "The radiation of a uniformly accelerated charge is beyond the horizon: a simple derivation" (2006)
Open access papers (in Nature or in arXiv).
In general, it was enough to google the first words of the title.
@@astro_male There is no excuse for skipping bibliography. I don't understand why would anyone try justifying it...
I think the solution is in that "radiation" as we have thought of it is only a coupling effect, not an objective reality. It is not the case that charged particles do not broadcast anything when not accelerated, but it is the case that when they are accelerated, they produce a coupling effect we call EM radiation. Energy is not a real quantity of something that gets "transferred" from one charged particle to another by this "Radiation", instead it is just a convenience calculation of totals. what's really going on is every charged particle is always "Radiating" the capability to change momentum per unit charge in all directions at speed c. that's happening regardless of whether the particle is accelerating or not. To test whether there is "radiation", we hold a second charge at a fixed distance away from a charged particle, and therefore there must be many many many other charged particles holding the second charge in place, held in position by EM oscillatory systems we call "atoms". Specific derivatives in the charge field due to the acceleration of the first charge particle will influence the second within the context of the oscillatory system's they're trapped in. if the derivatives don't happen in a way that resonates with the trap, then the second charged particle gets a net 0 acceleration because the change cancels out with it's oscillatory motion half the time, on average. only if the derivative of the first particle has a rate that resonates with the second will it produce enough change in the second to register a change in our instruments.
This is a good perspective.
The answer, my friend, is blowin' in the aether wind
The answer is blowin' in the wind
I am glad you touched upon this question. I have the feeling that when analysing this process we cannot view the electron as a classical object but as a quantum system which radiates only upon change of quantum state. Some people mention QED in the comments but from the little I have read it seemed that there is still debate when using the theory. There is though another way. What if you used the electron's zitterbewegung? I have worked a bit on this and it appears that the electron will radiate only if there is a change in the zitter frequency of the electron as viewed from the lab frame. Does this statement apply to all the cases you have shown? I will think about it..
This problem is older than general relativity. When Einstein was devising the equivalence principle before the announcement of general relativity in 1909, this problem was pointed out and refuted by Max Born. Nevertheless, Einstein had to stick to the equivalence principle to create general relativity. My answer to this problem is “General relativity itself is wrong.” I succeeded in producing the same result as the three early evidences of general relativity, using special relativity and Maxwell gravity, and in particular, I could remove any doubt by showing it through simulation rather than mathematical tricks. In that method, the equivalence principle had to be denied. If you want to know more about it, look for the book "Relativistic Universe and Forces" or related posts or site "金睛塔". I will not post a link because UA-cam prohibits links.
This was really awesome thanks. But it falls in the category of one of those unphysical classical EM puzzles. (From the "there is no spoon" department:) There is no electromagnetic or Faraday field, there are only electron-photon vertex interactions, so that'd be the proper setting for a definitive answer (if that's what you want). For the classical EM puzzle (as a nice puzzle in its own right) don't you want to conduct a fully covariant analysis, hence use the Faraday field, not separate (E, B, ρ, J). Justathought. Unruh effect seems to be the crux, no?
Correct me if I'm wrong (which it looks like I might be) but... isn't the particle in free-fall experiencing zero acceleration since it is simply following the curve of spacetime? And isn't the observer on the ground the one being accelerated by the surface of the earth by 9.2m/s^2 away from Earth's barycenter? So, wouldn't the person on the ground be the one radiating from the perspective of the charged particle? Doesn't the same thing apply to a charge sitting "still" on the surface of the Earth, as in, the particle is only accelerating if something is preventing it from following its free-fall trajectory?
I really find interesting what you said, but i can not deal with the sentence “ the electric field is separered from the change”. Also the fact that electric field is attracted from the gravitational pole. If you check in “Purcell or Griffiths” (2 well know books on elecromagneticity) nothing about what i mentioned is written
The "field" is only a mathematical description of the pattern of the effect that you would get in different places, for example, if there was an electron there... it makes no sense to talk about the field as if it is a thing in its own right, it isn't, it's a way of modelling potential outcomes of interactions that could occur, but it is only a model... for example, you don't need to model balanced forces that cancel out, like the charges in a neutron, but that doesn't mean that the universe doesn't "do" them.
Gravity is a byproduct of the electromagnetic force.
The greater the mass on a planet the larger the gravity field.
And mass is dependent on the accumulation of charge.
So the total amount of charge determines the gravity of a given object.
The standard model architects lifted their noses at observational data and chose to use conceptually based mathematical models.
And today there are millions of intelligent people fighting to find out who will have the next impossible answer.
There will be no explanation in any text book after the 1920’s.
We’ve been duped.
There is no room in today’s science for antigravity, transmutation, cold fusion.
All key words we’re taught to dismiss.
They are all required in an electric universe.
The Theory of Everything is just another way to describe Natural Philosophy.
3:59 Maybe you’re going to hit upon this, but at relativistic speeds radiation may appear or disappear. Like how the Hawking radiation of a black hole disappears if you’re in free-fall around it (but just in your reference frame, because everyone else will still see it), or how sometimes relativistic observers can’t even agree on what particles exist.
The latter is very easily resolved: there is no such thing as a particle in quantum mechanics. There are only quanta of energy, but they are NOT atomistic constituents of the field. To be more precise, in quantum field theory the only states one can agree on are asymptotic states at infinity. This gets slightly more complicated in GR, but it's not fundamentally different. There is simple one additional reason why the only "real" states are asymptotic: they only exist in flat spacetime.
I'm not so sure on the idea that a field is falling gravitationally, and I have a feeling that the resolution will come from a better understanding of fields under GR. The EM math you're using throughout the video is not Lorentz invariant, which is well known to cause apparent paradoxes.
12:56 wait how can a charge sitting on earth radiate for a freely falling observer. It's losing energy? But not objectively for all observers? And with no limit?
5:20 - hence the resolution that "gravity is not a force" - gravity is space itself moving - see veritasium, pbs space-time, or others who talk about this
So is the photon observed real? If gravity is just space, this observed photon should be an illusion. Yet I'm positive that the radiation measured via experiments show that it is real.
I do have a few comments:
- The tidal effects shouldn't intervene in the problem. You can imagine a similar problem without these, theoretically with a constant gravitational field, or practically with a supermassive black hole the size of the solar system, with an electron falling a few mm. You would have the same questions, and most likely very similar answers.
- The mathematical tools we use should not be treated so much as "reality". We know that E or B depends on the frame of reference, so their "reality" looks dubious. At the end of the day, all the mathematical symbols we use are interpretations. The equivalence principle on the other hand is a law which doesn't depend too much on the labels and symbols used for electromagnetism. To me, that makes it more fundamental.
- Radiation, as pointed out later in the video, is not a fully defined concept.
- Even in the current framework, the reality of force-bearing "particles" is unclear. Photons in QED are, when "force" is looked at, virtual particles.
- QED is a quantum theory. Relativity is not. Thus your emissions are quantized, but your acceleration is not, in the current theories.
I got a question. Let's imagine a thought experiment. Let's say we put a ccd beside a charged something and accelerate the whole setup to, say 60g using a solid rocket motor. Let's say we got another ccd, which is not moving with respect to the lab. What will our two CCDs observe?
We can do the opposite too. Accelerate (at high g) the ccd towards a charged thing and see if the charge srarts to glow. Earth's gravitation might not be intense enough to cause detectable radiation, but we can perform equivalent experiments wiyh accelerating frame, accelerate them as much as we wish and see if we see anything.
I'm not an expert at all but I'm questioning if maybe we should consider the time dilation in a gravitational field.
A particle sitting on the surface is accelerated up and it's field is falling down, but the part of the field that is under the particle is on a slower time frame and the part of the field that is over the particle is in a faster time frame.
So the down field is receding slower from the particle than the upper field restoring the field shape to a perfect sphere, so all actions combined, you have no energy liberation.
I can imagine that for a falling particle it should be something similar, but the other way around.
If you consider an observer that is falling in the same reference frame as the particle, for that observer the time dilation/contraction should compensate for the radiation too (but in that case you should consider the differences in the time frame not only for the particle but also for the observer if it is in a gravitational field), so both particles (ground and free falling) don't radiate.
For a particle and an observer that are in space (not subject to a gravitational field) there is no time frame dilation/contraction, so there is no need to correct the shape of the electric field.
Where the energy comes from the opposite observer (the one that is on the ground looking at the falling particle or the one that is falling looking at the particle on the ground) ?
The particles are sitting "still" with a perfect spherical field around, the one that is moving through the field is the observer, so the energy of the photons simply comes from the kinetic energy of the observer itself, not from the particle.
The electromagnetic four-potential is invariant.
@@juliavixen176 It is _Lorentz_ invariant. But the change to an accelerated frame is not a Lorentz boost.
There are multiple things I need to correct: when ever the charge deviates from its original trajectory, there will be a change in the vector potential, which induces a change in the field. This change in the field when decomposed into electric and magnetic field, and have multi pole decomposition done onto it, we see that part of the change remains even at long distances. This is what Lamour radiation is, not the field lagging behind the particle, but the change in the particle itself propagating through the field.
Now if we look at this in a semi classical manner, and interpret the particle as a spread out wave of field excitations, we see the geodesic deviation of a gravitational source does cause tidal effects that compresses and elongate the wave in various directions, partially accelerating the charge. This effect should still happen on the ground if we purely look in this perspective.
This is obviously wrong, hence the need for a more quantum theory.
Since the Lamour formula comes from any change in 4 velocity. This generalizes to any deviation away from a geodesic, even uniform acceleration should cause radiation in an external frame. Instantaneous comoving frames sees nothing. This should be consistent with universal expansion, and what you found.
Greatest stuff inside this video❤
Radiation and E&M fields are not absolute, they depend on reference frames. You notice Lorentz force law has velocity vector component. So in one ref frame you may have magnetic field whereas in the other you don't for the same moving charge. Furthermore that formula for radiation is incomplete as it's only the scalar component. The acceleration is also a vector, so radiation will also depend on if you're in an accelerating ref frame. Gravity's effect is already an accelerating ref frame. So a "falling charge" may experience zero net acceleration in that ref frame. If so it will not radiate when observed from such ref frame. As for the equivalence, if I remember correctly it's un-proven in the same way as inertial mass and and gravitational mass is considered equal.
Hi, yes E&M fields depend on the observer but you can't transform away an electromagnetic wave using lorentz transformations. So if there is an EM wave in one frame there is also in other.
@@lukasrafajpps The EM wave would be different frequency depending on the ref frame, but it would be non-vanishing in all ref frames. So that would mean the the falling charge is either radiating in all ref frames or it isn't. The question is, does a perfectly uniform gravitational potential (doesn't exist) cause a charge falling in it to radiate due to acceleration? I would say no because the acceleration is effectively zero due to the equivalence principle. The charge does not experience acceleration it is moving along a geodesic. A charge standing on the "ground" would observe the "falling" charge as accelerating. I'd argue that the charge on the ground is radiating not the falling charge.
@lukasrafajpps the transformation to an accelerated frame is not a Lorentz transformation.
@@narfwhals7843 I know I wasn't talking about such transformations. As I understood the author was talking about the mixing of the electric and magnetic field of moving charges since he was talking about the velocity component of the Lorentz force.
@@markattila9835 That is something I would be fine with but there are good arguments that a charge in freefall would radiate. Although Lorentz transformations can't create or destroy EM wave, the transformations between inertial and accelerating observers are not LT.
I also think a lot about this problem, although it lies outside my area of research. I think I'll spend a couple more years understanding and expanding my knowledge :)
I suggest keeping in mind one more non-obvious thing related to relativism. The distances between energy levels of atoms vary depending on the height. This does not answer the questions, but it is also worth keeping in mind when analyzing.
Gravitation acts on the photon of starlight as starlight passes through a gravitational field
I think the photon accelerates as it passes through and is deflected by the gravitational field
Gravitation pushes; it does not pull
Gravitation is a manifestation of space time curvature which distorts (bends, warps) the fabric of space and time also referred to as space-time
Charged particles are deflected as they pass through a gravitational fired
Gravitational fields are comprised of gravitational field waves
As gravitational fields interact they produce gravitational field waves
Charged particles are also deflected by magnetic field; they are also influenced by an electric field
Negative electric charged particles such as electron are attracted to positive electric charge particles such as proton
The proton positive charge is a net charge produced by interacting quarks (2 up and 1 down) comprising the proton
An electron is attracted and becomes captured by proton via coulomb charge (also referred to as coulomb electric charge)
Vivian Robertson seems to address this directly in his rotating photon particle model. I think he suggests that permittivity of free space is EF free yes by charge density, somehow, this derives relationship can be used to derive gravity, and within that explanation, there was a throwaway comment that charges can be effected by gravitation, and radiate EM fields, but not lose any energy. If his model has any validity, it’s an intriguing and straightforward solution.
This is a very subtle and complicated problem. I looked at it in the context of an electromagnetism course I teach. I hope one day I understand it enough to be able to teach it as an example.
If a charge moving in a curved space-time radiates energy, and assuming radiates photons (aka energy and assuming it has mass) 1st…would it not require to meet certain frequencies as in hf so it would only occur if allowed, and 2nd, if it did radiate, would it not evaporate eventually due to E/c^2= m? Also, why would it slow falling from the tower? Regardless of mass, even if lost due to radiation, all mass accelerates the same in the same curved space-time?
this is some serious physics !!
Bravo !!!!
Questions that arise:
What does local mean in a practical sense? What physical distance is local?
Why is an electric field, or any field of force affected by gravity?
Could we accelerate the particle with laser pushing and differentiate between emissions from acceleration and emissions absorption?
I would love to hear some answers if you could provide them.
Maybe it has to do with the curvature of spacetime? If the observer and the charge particle were to be in a same reference frame, they don’t experience any differences between their spacetime so they don’t feel any radiation or any other form of resistance. But if this were not to be the case, then the charged particle and the observer A don’t agree in the length of the spacetime and the differences within these lengths may lead to some form of resistance: in the case of the observer A, it would be radiation. If the observer B with the same reference frame as the charged particle sees the observer A, then observer B would sense gravitational waves, which would lead to similar effects. The energy, which causes all of this, is from the decrease in the potential energy. Does the acceleration change? No, because it’s only spacetime contraction and expansion. One can measure something similar. Of course, if you push it a bit hard, I’m willing to yield.
Acceleration does not cause time dilation (other than the resulting change in velocity which makes the dilation have a delta) the time dilation in gravity is constant. The time dilation of gravity is equivalent to moving at the escape.velocity
Just saying equivalence principal is good for thought experiments,but there are flaws that can be found in reality
I guess larmor precession works in bulk where there is an energy source and a circuit. You can then track where the work is done. Scenarios for one individual particle doesn't complete a circuit, ie we don't have before and after measured states. suspect, the gedanken scenarios out of GR are masking assumptions.
I've a side quest-ion: looking at the Lorentz force, a charged particle in a magnetic field drifts thus gaining an acceleration (linear momentum becomes angular). It is correct that velocity isn't changing and the particle doesn't radiate even if it's accelerating?
well.. anyway, the free falling particle does not radiates because it is not accelerating. The same happens if the particle stays still and the observer is in free fall.
To me the question can be: "Can I see a radiation from a particle (even if it's not radiating) just because I'm accelerating towards its?"
Are there quantum effects that stop a charged particle from radiating photons with power levels as low as one accelerating at 9.8m/s2, or the even slower tidal accelerations talked about in the video?
2:36 this is the point where i completely lost it lol
Why does a particle emiting energy shoud deacelerate? Is it bc some energy is being radiated in the opposite direction of the aceleration?
I have a strong feeling that im wrong lol
3:34 that's a false premise: "a particle either radiates or doesn't radiate". The number of particles (e.g. photons) is not conserved in accelerating vs. not accelerating frames of reference. Moreover, in General Theory of Relativity, the energy is not conserved.
If we move photons to field phenomena then the solution becomes obvious: Radiation of a charged particle is a relative phenomena. If it accelerates relative to an observer, it radiates.
In particular, the comoving observer doesn't detect the photons as they don't exist to them. Their effect can be observed on the accelerating object. The change in kinetic energy must compensate this absorption to conserve the total energy of the system.
Extrapolating, this would mean that the effective gravitational potential would be slightly less if the massive object had charge. Most likely, the difference would be insignificant. Besides the effect would be complicated by the emitted fields and particles carrying the charge.
You've forget that non-charged matter consist of charged particles. First time when you compare charge particle falling from tower. In this case both charged particle and neutral atom will radiate or not in the same way. The difference could be only in case of truly neutral particles like fotons, neutrinos or graviton. Second time in the last experiment: 10^50 atoms times 10^-52W = 0.01W . But you shuld multiply this by average number of electrons. So free falling Moon to the Earth should radiate weakly. This is still very subtle effect but at least there is some hope that in future it could be detectable.
Equivalency principle isn't exactly what it says. If you're in a gravity well the force of acceleration at the bottom of the box will be greater than at the top. If you're just accelerating it will be the same. Therefore you COULD do a test and realize if you're accelerating or in a gravity well.
That is why we often say the word "locally" meaning the box is small enough so that tidal forces are negligible
If every point continuously along the height of the elevator was each experiencing proper acceleration, rather than being pushed from the bottom and the rigid material compressing, or pulled from the top, and the rigid material of the elevator streached... _then_ it would be indistinguishable from gravity. (Gravity where the gradient is uniformly parallel (towards a line or extremely distant point, not radial towards a single nearby point.)
Although, there is also a way to measure the mass of the elevator vs. the mass of the Earth from inside the elevator, by jumping up and measuring how much the elevator's proper acceleration decreased. The ratio of your mass to the masses of the elevator and the Earth will give you the answer.
But... this principle is about coordinate systems and not physical elevators.