Relativity 107e: General Relativity Basics - Stress-Energy-Momentum Tensor
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- Опубліковано 23 лип 2024
- Full relativity playlist: • Relativity by eigenchris
Powerpoint slide files: github.com/eigenchris/MathNot...
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0:00 Introduction
2:22 Number-flux 4-vector N
9:58 Conservation of Particle Number
11:11 Galilean Transformation for N
12:57 Lorentz Transformation for N
15:41 Energy-Momentum Tensor T
21:51 Interpreting Components of T
23:47 Conservation of Energy-Momentum
28:28 Dust and Perfect Fluid
32:21 Conclusion
A video after my own heart.
bruh
Please make a video on your research!!
"Genius Eigenchris is back with other important topic.
You are the best."
I neglected to mention that THIS is the best explanation video EVER! Thank you.
Error at 18:20, the top-right text should say "constant position x".
Error at 21:35, it would be T^ab = T(epsilon^a, epsilon^b), where the epsilons are covectors, not vectors. But the idea still works.
But if that's the case surely the statement that "T is a machine that takes two vector inputs" is also incorrect? A tensor with two upper indexes ought to accept two covectors as inputs so that the Einstein convention works?
@@whizzdome Yes, you're right. You'd need to use T with the lower indices if you wanted to apply vector inputs. Sorry for mixing that up.
Dear Chris, you are an awesome tutor. Such a talent! Thank you for your hard work.
Eigenchris is back from summer vacation! Awesome video, keep it up!
So perfectly clearly presented. It makes so clear how increase in velocity increases density
Excellent video. Superb explanation. Thank you "eigenchris". (I note that all your videos on general relativity are exceptionally good.)
A great job in making this presentation to teach the Relativity in an efficient way. Thank u very much for all the effort and keep growing.
32:40 that's some serious flex right there
Probably the best typo I've ever made tbh.
@@eigenchris lol
This is so well explained, and beautifully clear thanks to all the illustrations!
Thanks!
Amazing presentation, as always!
You really need to truly understand a subject to be abale to explain it so clearly.
man how can you be this good, im awestruck
Best relativity teacher!
Enjoy your videos! Thanks a lot!
understanding complex subjects of physics is helpful and necessary !! Your videos are nicely clear and well explained!!your efforts in preparing your videos are very grateful. Thanks very much. I subscribe, like and share. Good lucks.
I LOVE YOU SO MUCH EIGENCHRIS
This cleared up a whole lot of questions i didnt know i had from my GR class. Thank u!!
thanks teacher... I'm waiting you for along time.
Three things make the internet not a total waste of human life, Wikipedia, Craigslist, and this guy
Thank you, that was extremely useful.
Thank you very much, it is very clear !
My mind is open to new concepts. New to me, anyway. Free education.
why you did not make this wonderful and effective video two years back when I just started my grad schooling ??? I'll sue a file against you for this late making, jokes apart it's a wonderful video pal... I love it
@eigenchris
Ty for posting
Oh cool, that shows how to transform kinetic energy to potential enery pretty nice.
And if you imagine one of those spots suddenly hit a barrier you can see how they should compress over time and why would that create a wave.
Thank you so much 🙏
Good job.
You are back sir
we love you chris
Awesome !
Very nice video. I would like to make a question. How would the formulation change if we do not assume that the continuity equation is 0 on the left (as you said, because of particle conservation), but some sink or source of particles (for example, from creation of pairs matter-antimatter).
I haven't studied particle physics a ton, but I know that matter/anti-matter creation (aka "pair production") requires input particles like photons, and matter/anti-matter annihilation requires output particles (also often photons). So I assume the continuity equation still holds and energy-momentum is still conserved. It's just that the specific type of particles carrying the energy-momentum can change due to particle interactions.
I really like the summary diagram at 23:45. Thank you so much!
I more or less stole that from wikipedia article on the EM tensor, but it's a nice summary.
Could you explain to my why off diagonal terms such
ho ux uy gives shear stress. I understand why the Txy = px/(ct dxdz) means shear stress, still I cannot imagine dust particles would experience shear stress because they are moving on x and y directions (have ux and uy)
Vary the diameter of the sphere or the manifold density to model one of the physical dimensions
Quick question: Is there an expression for a Stress-Energy-Momentum tensor of a single pointlike particle? And if yes is the tensor symetrical? Because if you think of it as surfaces being crossed Txt and Ttx should not be the same.
I think the tensor for a single pointlike particle is just the 4-momentum vector. The idea behind the energy-momentum tensor is that it measures 4-momentum densities. The idea of "density" doesn't make sense for a single point particle, so we just use the 4-momentum vector. This is similar to how the idea of mass density doesn't make sense for a single point particle... for a single particle we just use mass.
Can I ask a question? density that is in a Ttt is the rest density and it's related to the energy? For example,if we take a charge at rest,then find it's full energy as energy at rest+energy of it's electrical field,and find it's "relativistic" mass, because the energy will not only be mc²,then we should take the mass of a charge itself,or that "relativistic" mass for density?
hi thanks for your great job i watched all your videos please can you make a video which explains loop quantum gravity
1- why the loop?
2- what is a loop?
3- why is space-time discrete?
(especially) 4- how can a loop give us general relativity, that is to say the metric tensor / riemann tensor? (how to reconstruct classical general relativity from these loops?)
And thank you in advance
(I saw a lot of books but I did not understand this theory and these questions
I have some knowledge in general relativity, quantum mechanics, quantum field theory)
The stress energy tensor describes the flux of momentum which itself is the flux of mass/energy
Good.. helpful. Needed some solid examples w numbers.
Simple but a flow pattern of energy in components space in relativistic transformation appear as mass density.
Where momentum is assigned particle property in energy density state.
What this energy is like mass equivalent or density of space a state of matter .
Question sir. If energy and momentum arent conserved in large regions of spacetime, covariant_v(T^mu*v) is not zero, then what will happen to Contracted Bianchi Identity, covariant_v(R^mu*v - 0.5g^mu*v R) ?
I'm not familiar enough with what it would mean to integrate the energy-momentum tensor over a region of curved spacetime, so I can't answer.
genius !!!
So since the concepts of number density and other densities invoke the concept of "volumes", does that mean that one can formalize densities (and thus the energy-momentum tensor) using exterior calculus? I don't currently have a good idea of the exact way exterior calculus and tensor calculus are related to each other so I'd appreciate some pointers
I've been thinking about this question and I'm not sure how to answer it. A covector is used to measure a 1D density in a specific direction using a "stack" of hyperplanes. You measure a density by counting the number of hyperplanes pierced by a line segment. The wedge product of two covectors basically "overlays" the two covector stays on top of each other to give a kind of "grid" of parallelograms. This lets you measure a 2D density by counting the number of grid boxes covered by a 2D plane segment. You can extent this reasoning to any dimension that you like. But as for how this relates back to the EM tensor? A key property of the wedge product is that it's anti-symmetric, so a∧b = -b∧a.... which also means a∧a=0. This ensures everything created by the wedge product has an orientation. So if a∧b has a clockwise orientation, then b∧a will be the same geometry, but with a counter-clockwise orientation. Since the EM tensor is symmetric, you can't build it using the wedge product. I'm not quite sure what the "common sense" interpretation of this is, though. I'll have to think about it more.
@@eigenchris This is fascinating & links into musings I was having after watching the eigenbros' video on geometric algrebra earlier. I guess my "uncommon nonsense" interpretation would be something to do with time asymmetry idk? I'll have to spend several years learning physics then think about it more. XD
Yes you can! Like up the Current 3-form. Differential forms are just antisymmetric tensors
11:35-11:55, wouldn't the hashtags represent the number of particles going through different boxes? This would make the # numbers different depending on whether we are looking at an xyz box or an xyt, yzt, zxt box? I feel like this needs a different argument.
Hi Chris, I have a question.
is the Pt, Px, Py , Pz the total momentum of all particles inside the box through spacetime or the total momentum of a single particle inside the box through spacetime?
It's the total momentum contained in the box (so it includes multiple particles).
@@eigenchris Thank you!
Hi Chris, you mentioned T being symmetric. Why is that? I tried to look into it, but its a bit of a mess. really, somewhere it says its not symmetric, somewhere else that it follows from conservation of angular momentum, and somewhere else that it is just imposed by the EFE. Have you looked into it yet?
I thought the stress energy tensor was a pseudotensor, meaning it isn’t invariant in all coord systems. Is this correct?
I think the stress-energy tensor and stress-energy pseudotensor are different things. They each have their own article on wikipedia that you can read if you want to. This section might interest you: en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor#In_general_relativity_2
So energy is not conserved in general, but can I in general chose some coordinate system in which it is? If my coordinate system expands with space, the wavelengths of photons won't change, correct? In other words can I always chose some path for the parallel transport which doesn't change the vector?
I was mainly using this page as my source on conservation of energy: math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html . It states that a local (differential) law of conservation of energy works out, but a global (integral) law of conservation of energy will not, because of the parallel transport problem. It does mention towards the end that you can pick a specific coordinate system where you can get something *like* a global/integral energy conservation law, but since this depends on your choice of coordinates, I don't think you can consider it a "true"/"invariant" law. The article mentions the use of an energy-momentum pseudo-tensor (which is not a true invariant/tensor): en.wikipedia.org/wiki/Stress%E2%80%93energy%E2%80%93momentum_pseudotensor . To be honest, I don't understand all of this yet. It feels like it would require another entire video. In this video, I just wanted to include a warning about using conservation of energy in GR, since many viewer have probably already heard about redshift due to the expansion of the universe.
You might also be interested in this video by Sabine Hossenfelder. She talks about cosmological redshift at 6:40: ua-cam.com/video/ZYM6HMLgIKA/v-deo.html
Noether's theorem relates conservation laws to symmetries. Conservation of energy specifically is related to time symmetry. I think this means that in order to get energy conservation you would have to choose a reference frame with a time coordinate along which your spacetime is invariant.
At 16:00 the reason why a 4-vector is not enough to describe mass-density is that mass is not invariant. When you change coordinates, both the volume AND mass take up a lorentz factor. So two factors in total.
Contrast this with charge density where only the volume takes up a Lorentz factor (charge is invariant), so only one factor in total-->charge density can be treated as four vector.
Somehow keeping track of the "fluxes of fluxes", i.e. treating mass density as a tensor fixes things.
I would like to know how one could guess to treat density as a 4 by 4 tensor. Help? Referring to the dimensions of the Ricci tensor is not what im looking for. Thanks.
I've heard this explanation before, but in my videos I used (rest) mass as the mass, which is the same in all coordinate systems, just like charge, so I'm not sure how this reasoning applies to me. You could create a mass current density 4-vector "m*n*U" just as you could make an electric current density 4-vector "q*n*U".
I'm not sure what your question is in the last paragraph. In this video, mass density (energy density) is just the density of the time-component of 4-momentum. The EM tensor keeps track of all the densities of each component of 4-momentum. Is this what you mean?
But the EM tensor uses relativistic mass density, right? So it behaves differently from charge or number density.
If I was trying to formulate a relativistic theory, and needed to express mass density in some invariant form, my first try would of course be a density 4 vector... But then I would realize it is not invariant because of the relativistic mass. My Last questions was: how could I imagine that by keeping track also of the momentum fluxes by forming a 4 times 4 tensor would do the trick (be invariant)? I mean, is there some rule better than "if a quantity is not invariant, just make it tensor and keep adding indices untill it becomes invariant..."?
Anyway, this was the best introduction of the EM tensor I have seen. Thanks.
@@imaginingPhysics The dust energy momentum tensor can be defined as "m*n*(U⊗U)" where "m" is the rest mass, "n" is the rest number density, and U is the 4-velocity. All of these are either scalars or 4-vectors, so it's an invariant object that everyone agrees on. Invariance comes for free when you define something in terms of tensors. It just so happens that in any given basis, the tt component of this tensor is the energy density in that frame. I'm still not really sure why it's this tensor that defines spacetime curvature, though.
@eigenchris the dust tensor is correct. It is formed as the product of (four) momentum density and number density. The momentum part takes care of one lorentz factor (relativistic mass) and the density part takes care of the other (space conctraction). Its tt component is rho_0 gamma^2.
The General EM tensors uses relativistic mass/energy density which cannot be modeled as a single 4 vectors.
Now that I think of it, your presentation of the dust tensor makes it understandable why density might be handled as a tensor product of two 4 vectors each "keeping track of one Lorentz factor" - - > 4 by 4 tensor. Thanks again.
18:19 should that say "constant time" x?
Should be "constant position x". Whoops.
Maybe this is a dumb question but how do we know that numbers N^t, N^x, N^y, N^z can be put together to form components of a single 4-vector? I’m referring to 9:26. Generally I tend to think about vectors as arrows (direction + length) and I’m not sure how to apply this here.
The most straightforward reason is that (as I show later) it's just the 4-velocity vector times the rest-number density (scalar). So it should be guaranteed to transform like a 4-vector. I think part of the reason a vector feels weird for this concept is that we're sort of "cheating". Spatial density is an xyz quantity, but we're representing it using the vector orthogonal to xyz (time). x-current density is a tyz quantity, but we're representing it using the vector orthogonal to tyz (x-vector). This is sort of a similar to the "cheat" we use in mechanics for angular momentum, were angular momentum in the xy plane is represented using the orthogonal vector (z). It's more intuitive to represent angular momentum as a plane (e.g. xy, yz, zx) instead of the orthogonal vector (you can do this using the wedge product, if you want to read up on that). Similarly, it's probably more intuitive to represent the number densities as 3-volumes (e.g. xyz, tyz, txz, tyz). I'm not sure if you'd do this with the wedge product or tensor product though. I haven't looked into it.
Thanks Chris. Reading your comment I also realized that N^t, N^x, N^y, N^z is a collection of numbers, one for each plane/direction given by basis vectors (ie. How much flux we see in each basis direction?). And collection of numbers multiplying basis vectors just sound like components of a 4-vector :)
Форма объёма, квантов гравитационных волн, в основном должна периодически переходить, из вход - сфера симметрично, в выход - «яйцо курицы» не симметрично, только направленно, через период нарушая симметрию - может возникнуть гравитационное притяжение тел.
Но у Эйнштейна, почему-то симметрично изменяется, форма объёма гравитационной волны, и как при такой правильной симметрии, работает гравитационное притяжение тел???
also, is Pt the energy or gamma times rest mass times speed of light?
I mean, is momentum through time energy or energy/c.
Thanks for informative video!
Pt can be written as E/c or γmc (where E and m are the rest-energy and rest-mass). Both are equivalent, since E=mc^2.
@@eigenchris Thank you so much!!
that's why momentum density in the time component is called energy density by coincidence?
@@Mysoi123 Yes. It's normal to all Pt "Energy", similar to how we call ct "time". (In relativity we don't worry about factors of c too much.)
Great explained Video.... 👍👍..... Have you ever made a Video over covarient derivative of covarient vector component?Let me know if yes.. Thank you...
My "Tensor Calculus 20" video covers the covariant derivative of vectors, covectors, and all tensors. If you find it confusing, you might want to watch Tensor Calculus videos 17, 18, 19 first.
Awesome video. I'm jealous.
Thanks. This one was a headache to put together. So many spacetime diagrams and indices to keep track of.
Are you using minkowski space to describe energy momentum tensor?
In this video most of the examples are in flat space. But the idea can but used in curved space if you take infinitessimal volumes.
@@eigenchris Thank you for replying. But my specific question is, Shouldn't there be a negative sign in the spatial derivatives (24:47) when applying the divergence operator unlike normal cartesian coordinates?
@@AadityaVijayakumar Here we're doing a summation of derivatives (lower index) with the EM tensor (upper index), so no minkowski metric is needed, and so no negative sign comes up.
Sir, what would be the physical meaning of trace(T)=0
Thanks
I don't actually know. Sorry!
A trace is related to symmetries of a system; which is invariant under specific transformations.
According to page 40 of this PDF, trace(T) = 0 for massless particles (EM radiation) and non-zero for massive particles: www.lapasserelle.com/general_relativity/lesson_9.pdf
Sorry for another question on this video... I am confused by the notation you introduce at 11:20. Until that point all diagrams indicate that we have a stationary observer looking at a set of particles that are moving in her reference frame. All delta x,y,z,t , it seems to me, are considered from the reference frame of this stationary observer. But at 11:20 you define #/(delta-x delta-y delta-z) as density in the rest frame of the particles (and not the observer). Are delta x, y etc referring to small distances in the references frame of the particles as well?
I am also confused why delta-x/delta-t gives the x component of velocity of particles - after all both delta-x and delta-t are well defined even if there are no particles at all (as in: they refer to a small distance along each dimension from the point of view of the stationary observer). I feel like I am missing something simple here or maybe I am confusing things. Can you comment?
For the first question: all components of the vector are measured in the frame of the observer, not the frame of the particles. But in Galilean relativity, the spatial density #/(delta-x delta-y delta-z) is the same in all reference frames. So we can just set it to the density in the frame of the particles.
@@eigenchris thank you Chris - this is helpful. Do you have any thoughts on my second question? How come we interpret the delta-x/delta-t as the x-component of the velocity of the particles? I feel like delta-x and delta-t has nothing to do with the particles. We introduced them as small quantities along x,t dimensions for the stationary observer. When I see them in a formula describing the movement of particles my brain breaks :) Shouldn’t we have some other delta-tilda-x and delta-tilda-t to calculate the velocity of the particles? This is the only part that is still confusing for me in the whole video.
@@przadka Yeah, looking at that slide again, I may have just slipped in delta-x/delta-t and re-wrote it as the x-velocity without thinking too hard, but the end formula does make sense. If you imagine a 2D spacetime diagram for time and x, if we have a spatial density of 4 particle per unit x, and they are travelling at a velocity of 1/2 (measured in coordinate boxes), geometrically, it makes sense that 4*(1/2) = 2 particles will cross a given x-location per unit time.
@@eigenchris I think the problem is that until that point you were considering delta-x, delta-t as SQUARE and delta-x,delta-y, delta-t as CUBE. You refer to them like this multiple times. But at 11:20 you change your perspective: delta-t,delta-y,delta-z is no longer a cube - for this volume, the size along the t dimension is given by velocity of the particles, given by Ux. We are allowed to do that, but I was confused because it was never explicitly mentioned that at some point delta-t changes its meaning.
If it is possible to provide the PPT of error correcting part 1,2,3 for notes purpose
I admit to being thick as pig manure but what is your definition of space ?
Does space only exist if combined with time to provide space time which can then be manipulated ad infintum : even to the point of curvature !.
I'm not really sure how to answer this. Mathematically spacetime is a set of points where each point has 4 labels (t,x,y,z). In Galilean relativity you can "slice up" spacetime 3D "sheets" of space, but there's no way to do this in special relativity that everyone will agree on.
@@eigenchris Though mathematical modelling may allow spacetime to curve (influenced by that miniscule force, gravity) thickos like me cannot believe that real space can change shape.
Linguistically it certainly cannot !
Many physicists also claim that sending one photon (a massless particle) repeatedly at slits of the right dimension will eventually produce wave patterns on a recorder.
How a massless particle can produce the photo electric effect while at other times produce wave patterns is too much for my tiny mind.
The formula Energy = Hubble * frequency eliminates the wave particle duality does it not ?
I dont believe experts know as much about electro magnetism as they believe though a concensus of opinion does exist,
Why does EM radiation travel across free space ?
@23:47 Why do we need to set this row (Tti) to zero divergence? For a second I got all excited that this was the moment we were going to include the expansion of space, such that any xyz box set in an expanding universe will have a positive divergence.
Later in the next video on the E.F.E. there's a big move to include the 2nd Bianchi as a way to balance out the 0 divergence set here. If we let divergence happen here, can we ignore the 2nd Bianchi later?
I'm not a math guy (clearly) but I love these videos. Cheers.
Zero divergence of the first row basically means that energy is not created or destroyed. Any energy that comes into the box must increase the energy density, and any energy that leaves the box must decrease the energy. It's impossible to increase energy density for free unless it travels into the box from outside. So if we allowed for non-zero divergence, it would allow for energy to spontaneously be created or destroyed at a point for no apparent reason.
The expansion of the universe will come later using something called the "FLRW metric", which gives us the Friedmann Equations when we plug it into the EFE. The "a-dot" in the equations is the rate of universe expansion, and "a-double-dot" is the acceleration rate of expansion: en.wikipedia.org/wiki/Friedmann_equations#Equations
@@eigenchris Thanks for the reply. I understand what the conservation of energy law is and how it works, but I guess I'm not entirely sure why we need to apply it here.
For instance, couldn't we describe a stress energy tensor for an exploding star, in which the xyz box (or we could use spherical coords) is the size of the star. Thus energy would be leaving the box. Or, in the case of a star collapsing, if we choose the xyz box to be somewhere in the interior, energy would be entering.
Maybe more broadly, if we allow for non-zero divergence, do we get some wild spacetime geometry on the left side?
You say, "So if we allowed for non-zero divergence, it would allow for energy to spontaneously be created or destroyed at a point for no apparent reason." The reason is that energy is coming in to (or leaving) our chosen xyz coords, it is not spontaneous.
I guess my question is, why does the stress energy momentum tensor need to describe a closed system? What about it requires that the system be closed and thus conserve energy?
@Dan Sternof Beyer It's acceptable for energy to leave an xyz box (dT^tx/dx + dT^ty/dy + dT^tz/dz > 0) but if that happens, then the energy density of that box must decrease by the same amount (dT^tt/dct < 0). This is equivalent to setting the 4D divergence of the first row to zero. Does that add clarity or am I just repeating what you already know?
@@eigenchrisOkay, I've done a very minimal amount of looking into this and have gleamed a 'bad' intuition as to what is going on and why divergence is zero.
As I understand it (and good lord please jump in if this is crazy talk), the divergence is zero because of 'locality'. Such that, we are trying to figure out the mass/energy at a point on a pseudo Riemannian manifold, and as such the spacetime we are investigating reduces to flat Minkowski spacetime. This reduction to flat space can only occur with a tight lens, ie a small lab, ie local measurement.
When we are looking at the stress energy tensor of Dust, or a Perfect fluid, we also have to zoom way in, to get to locality. Similar to the manifold idea (where the surface of a sphere looks flat zoomed in, and can be investigated as a manifold, ie flat) we have to zoom into the flux of the stress energy tensor.
When we zoom way into the flux, whether the flux was positive or negative divergence globally, we will get, at a very small level, 'straight lines' of flux, or zero divergence.
So, as I understand it, zero divergence is a result of local measurement, specifically very small local measurement.
(okay, did I go off the rails with that interpretation?)
If this is close, then it begs the question : "What is the scale of locality in the stress energy tensor?"
Which is basically a question that try to combine the ideals of "point like" measurement, with the reality that nothing is "point like".
Thoughts?
I think you're touching on two separate ideas here. The first idea is that when you zoom in on a continuous field (scalar field, vector field, covector field, tensor field, etc.) it looks "constant" in a very small region if the region you pick is small enough. I don't think you can use this way of thinking 100% of the time though. For example, the electric field near an electron's center goes to infinity, so the field is not continuous here, and you can't "zoom in" to get a field that looks constant at this point. The second (separate) idea is the divergence of a field being zero. If you take a vector field, the divergence is a number is defined at every single point in the field. Some points might have zero divergence (e.g. all the vectors around that point are parallel), other points have non-zero divergence (e.g. all the vectors "point outward" from that point). In the case of the statement "the divergence of the energy-momentum tensor is zero", this is a law of physics that places constraints on the types of EM tensors that we allow if physics. In our universe, if the energy density of a 3D spatial box decreases, it must be because a particle left the box through one of the walls (3D divergence is positive). This rule forces the rate of change of the time derivative energy density + the spatial divergence of energy density to be zero. This rule is called the "continuity equation", or equivalently, the divergence of the first row of the EM tensor being zero. If we lived in some alternate universe where particles could vanish for no reason, the divergence of the EM tensor could be non-zero. If we look at our 3D spatial box again, and see the energy density decrease, it might not be the case that a particle left through one of the box walls... the particle may just have disappeared out of existence. So it's not the case that time derivative energy density + the spatial divergence of energy density is zero. The continuity equation can be violated, and the divergence of the first row of the EM tensor could be non-zero.
Space is thixotropic. With energy the density of space increases but it also becomes less viscous. At c space is maximal density but the viscosity is zero
This is why gravity
👍👍
Sir Can u plzz Provide the PPT of these video including Part 1,2,3 of Error Correcting it will be really helpful for us we can make notes
I don’t think it’s correct to say the energy-momentum tensor is “invariant” since this term usually refers to scalars. A tensor is a collection of numbers/components which can be expressed covariantly, contravariantly, or a mixture thereof. It has a physical significance that is independent of frames of reference but this is not the same meaning as invariant, and that loss of distinction will inevitably confuse those who are still trying to grasp the difference between invariant, covariant and contravariant.
The "invariance" part is what I was trying to explain at 30:45. The "T" itself (the multilinear map) is what I would call "invariant". The components shown at the bottom of the screen transform "cvoariantly".
Isn't this only the case in inertial frames?
Which part of the video are you asking about?
Wikipedia has T01, T02, T03 labeled as momentum density. And T10, T11, T12 as energy flux. Is this because they are assuming (-+++) and you are using (+---)?
The tensor is symmetric, so both labelings are equivalent (it has nothing to do with the metric). The image history in wikipedia says that before March 2021, it was labelled using my convention. Not sure why it was changed.
@@eigenchris thank you sir. Much appreciated.
at 14 minute why nx is negative?
It's because the current appears to be travelling in the -x~ direction. (This is what the car moving in the +x direction sees if things it's counting the are stationary.)
@@eigenchris thank u. So its like the counting at numerator has this sign conventions incorporated. For the same reason its correct to say that pressure terms (later in the video) are positive when momentum exit the surface?
If t x y and z are all contravarient, then the number flux would be 3x covariant
Why can we get away with using a 4-vector in electrodynamics but need a rank 2 tensor for gravity?
At first glance it would seem we can formulate a very similar, fully relativistic theory by using an energy flux 4-vector to construct "Gravitomagnetism".
At some point you mention that mass density is not Lorentz invariant, but neither is charge density.
My guess is that it is the equivalence principle. For gravity we require the field to vanish locally, and I guess this can't be done in 4-vector form, but I'm not sure.
I'm not sure I ever knew the answer to this question. GR does contain "garitomagnetism"-like forces (such as frame dragging with Kerr black holes). But I don't know why you need a rank-2 tensor to make it work off the top of my head. It might related to the fact that the generalization of the gravitational potential's Laplacian in the Poisson Equation is a rank-2 tensor, and so the mass density needs a rank-2 generalization as well.
I'M BUILDING A WARP DRIVE, AND NO ONE CAN STOP ME MWUHAHAHAHAHHA
slay
May I know you name as well as age?
My name is Chris and I'm 29.
Can i ask u what you do for a living apart from youtube
@@bharatvardhamane2406 I work as a software developer.
doesn't Bianchi identity conserve energy? 0o
Locally, yes. But the problem is that "energy" is a local concept--it's the time component of the 4-momentum vector, and different observers will disagree on its value. In an expanding universe, every point in the universe will have its own arbitrary local frame, "adding up" all the energies in each local frame isn't something that makes sense. There's a special frame I talk about in video 110e where we can try to compute the energy for the whole universe, but it's not a conserved quantity. This is related to Noether's Theorem, which says that conserved quantities are related to symmetries. In an expanding universe there's no time-translation symmetry, and so there's no conservation of energy.
@@eigenchris Wow! This is super disturbing! :D Thanx!
@@eigenchris OK, so here's the deal. The point of discontent is the statement that "Energy is not conserved", or that "the photon energy is lost". These statements are wrong. Here is why. Consider the first law of therma: dQ=dE+pdV, with the standard notation, Q is the energy going through the boundary of volume V, and E is the internal gas energy. Now, if the system is conserved, then dQ=0, no energy enters nor leaves the system, there is no boundary transfer. So in the conserved system, we are left with dE+pdV=0, or equivalently with dE=-pdV. So, since pressure p is positive by definition, when volume V increases, the energy E decreases. So the internal energy E isn't conserved. So it is true that the internal energy E isn't conserved. Photons do "lose" energy. But... this energy isn't lost. After all, the starting assumption is -- the system is conserved. So, the system energy cannot be lost. But, the system consists of a bunch of photons. And all photons lose energy. So how is it conserved? Where does the energy go? Well, dE=-pdV. The energy goes towards the expansion of the universe. The photons slow the expansion down. All energy slows the expansion down. So, the photon energy isn't lost. The part of the photon energy does work -- by slowing the expansion down. As a result, the conservation of energy requires photon energy to decrease over time. And indeed, photons do have a non-vanishing energy tensor, so they do contribute to the expansion. They do curve spacetime. Now, if we look at the equation dE+pdV=0, we see that it looks a lot like a covariant derivative, and also the covariant derivative is zero, meaning the energy is covariantly conserved. These are the Bianchi identities. They conserve the energy covariantly. So the energy isn't lost. This way, the energy of the universe IS a conserved quantity. Otherwise GR would make no sense. The only exception is the gravitational field itself, because there is no energy tensor of the gravitational field. For gravitational field, t=0, or equivalently, R=0, the condition for empty space. Any attempt of constructing a non-trivial energy tensor of the gravitational field itself fails, and the next best thing one can do is the field pseudo-tensor. Since not a true tensor, the field energy is not conserved indeed. This is a hint at GR being possibly wrong. These questions are super important, since they are fundamental. If you don't have conservation of energy... you go against the zeroth law of therma. And that is bad! :D Bad on a very fundamental level. It is super disturbing indeed.
Sorry, I missed this comment a month ago. You might be interested in watching the last video in the series (Relativity 110f). The first law of thermodynamics does translate into a formula (I think I call it the "3rd Friedmann Equation because it doesn't have an official name) which dictates how energy densities change as the universe expands. You can leave additional comments on that video if you have questions.
@@eigenchris Right :D
chris is god.
The energy tensor for vacuum solution such as Kerr metric and Schwarzschild metric is actually not zero in 3-D submanifold.
Watch this youtube video for detail. ua-cam.com/video/jxDbulr9RDE/v-deo.html
Once upon a time Einstein took a balloon and observed it carefully. After a few minutes he realised that the air-molecules in the balloon had an average velocity and thus the inside of the balloon had to have a certain energy-density (Joule/Volume). He also realised that these molecules kept bumping into the surface of the balloon. Thus the surface experienced a pressure from within, called stress (Newton/Area). All these stresses can be decomposed in normal and shear stresses. That's what he had learned at the ETH in Zurich. But he wasn't satisfied yet, because all his balloons deflated after some time. Giving this some thought he came up with this great idea that there has to be a flow of particles through the surface! This kind of dynamics is introduced in physics through momentum and flux, he remembered. Well, that's easy, Einstein thought. Momentum-density = energy-density / velocity, and energy-flux = energy-density x velocity. Great, Einstein yelled. Now I've got all the components of my Stress-Energy-Tensor! Then he grabbed the balloon and popped it, lol.
translation in french please
What quality (i.s. such as a force), of 'curved space time' acts on electromagnetic energy to decrease its momentum on the macro scale? This appears to me to be utter nonsense. Just because there is a mathematical effect of translating the component vectors over a large scale that does not mean there is a real physical effect negating conservation of momentum, conservation of energy.
Energy on its own isn't an invariant concept and depends on the reference frame (it's just the time component of the 4-momentum vector). Since the universe is expanding, and light we receive on earth is not received in the same reference frame that it was sent, the energy measurement will be different. Basically, all light will be redshifted because the earth is constant traveling "away" from all other points in the universe. I admit I could have explained this better. I think I'll need to more carefully study the relationship between energy conservation and curved spacetime for my upcoming videos on cosmology.
@@eigenchris Thanks very much sincerely for this response, and I am really grateful for this entire series. Your comment seems to imply that for certain interactions and calculations there are preferred reference frames, I would expect conservation of momentum to be preserved in all reference frames.
@@leschwartz In both classical mechanics and relativity, when "conservation of momentum" is stated, it's basically a law of adding vectors together. But these vectors can have different components to different people. In classical mechanics, you might observe 2 particles collide along the x-axis, with 0 y-momentum, but if I'm rotated 90-degrees with respect to you, I might see the particles collide along the y-axis with 0 x-momentum. The vector addition rules are true in all frames, but the components are different in different frames. In relativity, energy is just another vector component, so it will be a different number in different frames. The way I describe conservation of 4-momentum is basically as a vector addition law. But again, the specific components will look different for different people. If I fire a photon from my galaxy to your galaxy, the sending and receiving frames are different due to the expansion of the universe. Neither frame is "correct" or "preferred". They are just different frames and will measure different numbers for the energy-component of 4-momentum.
@@eigenchris Right, thanks very much again, I get that for this set of examples, explanations. I was thinking about the comments regarding apparent lack of conservation of momentum for large scale curved space time.
I see. I guess all I can try to do is explain it better in the future in my cosmology videos. If it turns out I made a mistake in this video, all I can do is try to correct myself in the future.
This I lost my brain
Spacetime curvature is a useful tool for geometry but it does little for engineering purposes. I think a better way to interpret Einstein's equation is that, on the RHS we have the Energy Momentum Tensor that describes the forces, pressures, flows, etc... of energy and momentum acting on "matter" as discussed in the video. Logically then, the LHS should be interpreted as the geometry that is measured using rulers and clocks that have been deformed by the RHS of the equation. We have an accurate theory of quantized matter, and if gravity is interpreted as an effect on matter, not an effect on space-time, then the mystery of quantum gravity is solved. In fact, all the classical tests of GR can be replicated based on a simple radiative damping model of the harmonic oscillator. References available.
Damm bro you just solved quantum gravity. But you're not gonna publish a paper about it, hmmm.... i wonder why...
@@mastershooter64 I've already published multiple papers on it!
This video... Galileo was right... his theory should be implemented not special or general... galileo,s does both... with time...for time dilation ..and set time to zero and make time...spacetime then u can calculate curves with galileo,s theory...lol
what bruh
hi thanks for your great job i watched all your videos please can you make a video which explains loop quantum gravity
1- why the loop?
2- what is a loop?
3- why is space-time discrete?
(especially) 4- how can a loop give us general relativity, that is to say the metric tensor / riemann tensor? (how to reconstruct classical general relativity from these loops?)
And thank you in advance
(I saw a lot of books but I did not understand this theory and these questions
I have some knowledge in general relativity, quantum mechanics, quantum field theory)
Sorry, but I don't know anything about loop quantum gravity, other than the name.
@@eigenchris thank you any way