Not, I don't bother doing anything to change the matrix. I just know the squared length of the e_t basis vector is "g_tt", so I divide by 1/sqrt(g_tt) to normalize.
I try to explain it quickly at 25:00. The steps are: 1. choose a path 2. Determine 4-velocity U for the path 3. Use U to calculate 4-acceleration A for the path 4. proper acceleration is the magnitude of A In our case: 1. the path is on observer at constant spatial position (constant r), but moving through time 2. Since the observe is at a constant position in schwarzschild coordinates, U points completely in the time direction, so U = c*e_t 3. Calculate A as shown on the slides 4. The magnitude of A gives the result
Massive respects for making these video series! At 7:46 Can you explain why this gives us the tangent vector of the lightbeams worldline? I would have tried to derive it from the equation at 5:42 , but I'm guessing that is just a more inconvenient way of doing it. But I don't really see how the 2 equations (5:42 and 8:23) are related.
The equation at 7:46 gives the generic equation of tangent vectors on any path parameterized by lambda. We found back at 5:42 the equation of a light beam between "ct" and "r". At 8:23 we decide to set the path parameter lambda to "r", so we're really calculating the tangent vectors d/dr.
just a suggestion for the future. i would really find it useful if there was a question sheet written in relation to each video. answering the questions as i go along would help me retain a lot of the information much better. chris, if you have any suggestions on what i could do, i would really appreciate it. thank you for all these amazing videos!
Realistically, I'm probably not going to make exercises for each of the relativity videos in the near future. But maybe I can consider it for my next series. One exercise you could try for this video is to calculate the frequency shift seen by a free-falling observer that has zero velocity at infinity. I might upload an "exercise-style" video on that later, where I let people pause and work on it before giving them the answer.
@@eigenchris I tried to calculate the frequency shift seen by a free-falling observer, but I am not sure what is the answer I am supposed to get. First of all, I tried to find the equation of the geodesic of the observer. I used the same method you used in video 108c and I got a differential equation that is a bit complicated (I think I got the right equation, because for r>>rs I got d^2(r)/d(tau)^2=-GM/r^2, just like in Newtonian Mechanics). I can try using the internet to solve and get the function r=r(tau), but where do I go from here? I can also try getting an expresion for tau in terms of ct and then use the frequency formula in this video to get f=f(tau) which would give us the frequency of light in terms of proper time of the observer. Is this the answer that I am supposed to get?
@eigenchris I have a question, at 6:00 you mentioned the log term goes to 0 when we are far from the Schwartzschild radius, but doesn't the log term goes to positive infinity when r approach positive infinity?
Hi chris, I have idea to explain the curl zero in 11:59. We can define scalar function "phase (phi)" on this plane. We can set in a certain light wave front phi=0, and in next wave front phi=2pi and so on. The wave number kr can be understanded as partial phi/partial r. The wave frequence kt can be understanded as partial phi/partial ct. I think we can assume phi has continuous second partial derivatives thus can exchange order of partial derivative. This can explain the curl zero. The other thing is about Doppler Effect for Rinderler coordinate. The formular in 27:36 looks different with video 106a in 30:39. If I use formular in this video, with increasing r, I got redshift. If I use formular in 106a video, I got blueshift. Do you know what's wrong.
If you took the 3D formula for curl, this is just 1 of the components of that formula. But in relativity, the 3D curl you learn in vector calculus is usually not used, since it only works in 3D. The "true" formula comes from the exterior derivative of kappa (you can see it in the first bullet point on the slide at 12:02). This gives a curl-like formula but it works in any dimension.
@@eigenchris do you have a derivation for it? I have understood almost everything in the playlist. Do you have any resources that could show an alternative derivation?
@@cinemaclips4497 You could try reading the first couple sections of this wikipedia article: en.wikipedia.org/wiki/Closed_and_exact_differential_forms All I'm really trying to say is that, if you drew tangent vectors along the paths of light beams, the trajectories should never "cross over" each other or form loops. This is equivalent to saying the curl of the vector field is zero. To make the equivalent statement for a covector field is to say the covector field is "exact", meaning its exterior derivative is zero. To fully understand this you'd have to take some time to study differential forms, possibly the wedge product, and the exterior derivative. Unfortunately I don't have videos on these. Saying "the curl of the vector field is zero" is an equivalent statement and results in the same formula.
The "proper" way of getting the formula is computing something called the "exterior derivative", which involves differential forms. I don't have videos on that, so I simplified the explanation a little.
@@eigenchris so in this case if both the covector components and partial derivatives are covariant would it be curl(κ)=εⁱ∂/∂xⁱ∧εʲκⱼ=1/2(∂κⱼ/∂xⁱ−∂κᵢ/∂xʲ)(εⁱ⊗εʲ−εʲ⊗εⁱ)=1/c(∂κₓ/∂t−∂ f/∂x)εᵗ⊗εˣ
how would you find the frequency from the perspective of falling into a black hole? would you instead put a unit length vector tangent to the objects wordline into the covector field at that point? also GREAT videos!
Yeah, the procedure I show is pretty general. Just take the tangent vector for a geodesic along any worldline and you'll get the frequency that worldline observes.
So glad you made this video, the comparison btwn the Rindler and Schwarzschild frame was very illuminating. Does this mean though that all observers will see a time dilation gradient about a mass because of the Doppler effect? My understanding is that in the special relativity frames and the Rindler frame, all the time/spatial dilation effects are kinematical. Can time dilation still be considered to stem from a kinematical effect when spacetime is curved?
To me it seems a bit "wrong" to attribute time dilation to the Doppler effect. I'd think time dilation is more fundamental, since frequency shift is about making measurements in time and space. I've seen some people say you can view gravitational frequency shift as a series of kinematic doppler shifts across a chain of close observers, each sending a light signal from one to the next. I think this is basically approximating the curved spacetime near a mass as a series of flat tangent plane spacetimes. Time dilation could stem from a combination of gravitational and kinematic effects in GR. For example, instead of looking at two observers at constant radii above a planet and calculating their proper times, you could take one of the observers and make them "zig-zag" back and forth around a fixed position. This will add a kinematic aspect to the time dilation as well as the gravitational effects.
Awesome explanations as expected! Conceptually, it makes sense wavefronts passing by a stationary observer and getting denser near a massive object is a different situation than a spaceship passing through wavefronts in an accelarated frame, because those effects combine in a certain way since the spaceship have to stay still too, in a gravitational field. My only question is if it works the same way for objects in orbit, because accelaration vector would be tangential/oblique to some wavefronts of incoming light, for example, light from Andromeda coming to earth and being captured by tellescopes in orbit. So, do we have to consider two kinds of doppler effects (galilean/lorentzian and gravitational)?
I haven't done any of the calculations involving angular coordinates. Most of my schwarzschild videos focus on radial behaviour because angular + radial behaviour can get complicated (as seen in 108c). But in theory the covector approach should work. Covectors in 4D spacetime would be stacks of 3D hyperplanes. The formula for forcing them to be "exact" is more complicated, and you' probably need to learn what the "exterior derivative" is to get it. You don't need to worry about the source of frequency shift, whether it be doppler or gravitarional. Light will travel in whatever way the metric tells it to travel. After you have the wave covector, you just pick any observer and see how much their tangent vector sees in 1 unit of proper time.
Amazing video as always! Two questions : 10:14 - Is the Schwarzchild solution static over time for convenience or necessity? Is there an example of a solution to the EFE that is not static over time? 11:00 - The definition of 'exact' covector fields sounds very similar to the notion of flux in the FLRW metric. Such that worldlines and wavefronts can't simply pop into existence. Is there any way for a gravitational covector field to not be 'exact'?
The "static" criteria is one of the assumptions we made during the derivation of the Schwarzschild metric in 108a. So we just re-use that assumption again there. The specific case of covector fields for waves requires that the covector field be exact so that the wavefronts "join up". Mathematically, this is similar to how a gravitational field in Newtonian gravity must be conservative (otherwise we could get energy for free). Magnetic fields are examples of vector fields that are non-conservative (because they curl). Is there an example of a non-exact covector field in physics? Maybe, but I don't know of one. It would have to be an example where the "joining up" of the stack surfaces doesn't violate any physical laws. "Covector fields" are also called "differential forms". You can google "differential forms" and "exterior derivative" to learn more. I have a couple videos in my Tensor Calculus series that touch on this (I think 10-13) but I don't go into a ton of depth.
Sorry, I missed answering "is there a solution to the EFE that is not static overtime". Yes. Gravitational waves and the expanding universe (aka FLRW) metric are some examples that I discuss in the 109 and 110 videos. There's also rotating black holes (aka "Kerr metric"). I don't plan to do videos on the Kerr metric in the near future.
@@eigenchris I just want to say, that your videos are so refreshing, informative, and clear. There is a huge 'fight' right now about this video (ua-cam.com/video/PjT85AxTmI0/v-deo.html) which claims a bunch of other physics youtubers are 'wrong' about time dilation. And I'm like "...just go watch Eigenchris people. jeeeze." Anyway. Thanks for being an incredible teacher. I really appreciate your work and your patient responses to all of my weird questions.
@@DanSternofBeyer Yeah, I watched that video and read some of the chaos in the replies. I'll admit, after all my relativity studies so far, this "time dilation gradient" = "gravity" interpretation isn't something that's clear to me. It could just be that no one has explained it to me in the right way yet. Or maybe it's wrong. I can't really say one way or the other currently.
@@eigenchris The main idea behind the interpretation, if I understood correctly, is that newtonian gravity is explained to a very good approximation(in a certain coordinate system) by the 00 component of the metric alone. And in that coordinate system that component presents as a simple decreasing function of r. I think it's fair to call that a time dilation gradient but there is some subtlety to it that isn't really addressed in most presentations of the analogy. Also please tell me if I'm overstepping by invading your comments. This is something that's I'm really trying to understand both sides of right now.
Hi Chris. One question: When two black holes merge and send out gravitational waves, is there a redshift in frequency of the measured gravitational waves, at least at times their Schwarzschild surfaces come close ?
That's not something I'd thought about before, so I had to google it. The two stack exchange posts I've linked below seem to think the answer is "yes". Light and gravitational waves both travel at the speed "c", so I guess it makes sense that they behave the same way? Interesting question. astronomy.stackexchange.com/questions/13326/redshift-for-gravitational-waves physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted
Can you explain how a variable that parametrized a curve can be used to make a tangent vector? The same variable can be used to parametrize another curve so there’s confusion. And what if the coordinate curves aren’t functions of the specific curve? (Like when the curve is a line parallel to a coordinate axis).
A worldline parameterized by λ would be a collection of functions for each coordinate. For example: (ct(λ), r(λ), θ(λ), φ(λ)). If the curve is parallel to a coordinate axis, it just means one or more of the above functions is constant, so the derivative with respect to λ for that coordinate will be zero.
@@eigenchris how would the parametrization work in terms of how close it is to being arc-length parametrized? I’d imagine the parametrization depends on the velocity, but how so? Also, I meant perpendicular to a coordinate axis.
@@biblebot3947 Not sure I understand the question. For timelike paths, the "arc length" would be proper time. For spacelike paths, the arc length is actual arc length. I use lambda as an arbitrary parameter that isn't necessarily equal to either of these. But you can always get the "arc length" by integrating the lengths of tangent over the curve, regardless of what the parameter used is. You'll get the same result either way.
@@eigenchris My questions were: 1) What happens when a curve can’t be expressed as a function of a coordinate curve (like a straight line perpendicular to the x-axis can’t be written as a function of x)? 2) A curve can be parametrized in multiple ways. How would the choice of parameter affect the tangent vector?
@@biblebot3947 For 1), I'd figure pretty much any smooth curve can be parameterized. An example of a curve perpendicular to the x-axis is: (ct(λ)=λ, x(λ)=5, y(λ)=0, z(λ)=0) For 2), basically, if you have "less λ" between two points on the curve, the size of the tangent vector will increase. You can loosely think of "λ" as being like a "time" parameter, telling you how long it takes to move from point to point. If λ goes from 0 to 100 between two points, you spend a lot of "time" between those two points so the tangent velocity vector is small. If λ goes from 0 to 1 between the same two points, the tangent velocity vectors will be large since you travel across very fast. I realize "time" is a confusing term to use though... since there is already a time coordinate in relativity. It's more like a mathematical "time" and not a physical time.
I have one question that is bothering me. Does LIGO detect gravitational waves the way we predict in weak gravity limit (the + and X polarization)? This bothers me because on Earth we can't use the weak gravity limit, so that means that gravitational waves should take different forms.
Wavefronts (drawns as planes in 3D) are perpendicular to the direction of wave travel in space. But we're looking at spacetime, where each individual wavefront will travel along a lightlike worldline as it moves across the spacetime diagram. In this 2D spacetime diagram, wavefronts are just points, and they trace out a set of lines on the spacetime diagram as they move.
It's so that the covector surfaces don't "split" into multiple light beams. It's similar to saying the divergence of a vector field is zero. If you draw a spacetime box and 3 light beams enter the box, then 3 light beams must leave.
@@eigenchris thank u. I will try to get spoiled and get another answer to this question: if i have a satellite orbiting earth if have a time dilation due to its velocity relative to an observer on earth (linked to sr and gamma) and one due to gravitational effect (linked to r and r_s), how one can "sum" the 2 and get the proper time dilation?
@@domenicobianchi8When it comes to general relativity, there isn't really a "simple" time dilation formula you can use. If you have 2 spacetime curves, you can calculate the total proper time along each by doing an integral along dτ. You can then compare the results to see which path has longer proper and and which has less.
@@eigenchris i see. Lets say i do the same path in a curve spacetime, it makes sense saying i could do the same path going slow, and the doing it at at speed say half the speed of light? How the proper time is affected? Could you point me to some calculations of this kind
@@domenicobianchi8 These would be different paths on a spacetime diagram, since going "fast" and going "slow" would have different slopes on the spacetime diagram. I do some proper time integrals in video 105d I believe, when I cover the twin paradox.
Do infalling reference frames see the change in frequency? I guess not exactly, but I am curious about what happens if every second of the fall a beam is generated inside the rocket, what would happen when the rocket crosses the event horizon? Would the beam of light change a lot its frequency just by crossing the rocket's lenght?
I haven't done the calculation of looking at frequencies measured by an infalling timelike geodesic, but according to the equivalence principle, I would have to conclude a freely falling observer would not observe any change in frequency.
@@viniciusaraujoritzmann Sorry, but thinking about this further... I tried going through the calculations and I did not get the result that a freely falling observer would observe the same frequency all throughout the fall. I may have assumed too quickly that the equivalence principle applies globally to a free-falling observer's entire journey--this is not correct, as the E.P. only applies locally, as stated in my video. I'm trying to find some sources to back up my calculations. This Physics Stack Exchange question seems similar to what you're asking, although I'm not yet sure if the answer they give is algebraically equivalent to the one I calculated... I'll keep looking into this. physics.stackexchange.com/questions/685802/what-is-the-frequency-of-the-light-seen-by-an-observer-free-falling-into-a-black
@@eigenchris Thank you Chris I was reading online about this and wasn't able to find an answer yet, but I started to think what you said, if the E.P would be valid since the curvature changes so rapdly close to the black hole ( I mean, it seems the more close you get the smaller it is the time and space where aproximations work). Thak you again I'll keep researching as well.
To add a little more information... Alfred's answer at the end states: "whatever distance from the black hole you start your jump in free-fall, you can tell that you crossed the horizon precisely when the light emitted from your starting point reaches you with exactly half the frequency it is emitted." This seems a bit counter-intuitive, because it actually means the light you see while freely-falling into the black hole will be redshifted compared to where it is emitted, but this is consistent with the result I calculated. If you want to try calculating this yourself, I got that an observer starting at rest from infinity (Ut=c, Ur=0) will have the following 4-velocity components while freely-falling into a black hole: ( Ut = c/(1-rs/r), Ur = -c*sqrt(rs/r) ). You can divide by "c" to normalize to a length of 1 (since 4-velocity vector always has a length of "c") and then feed it into the Kappa covector seen in this video. After some algebra, I get f/c = a* 1/(1 + sqrt(rs/r)). As you can see, when plugging in r=rs, I get an answer of f/c = a/2, meaning the frequency "a" at infinity will get divided by 2 for an observer freely falling across the event horizon. This is a surprising result to me... and I am not totally certain my calculations are correct, but they match up with someone else who did them independently, so this may be true.
d/dλ is a vector tangent to light worldlines. The covector is a "stack" that's parallel to light beams. When a covector acts on a vector parallel to it, you get zero.
For gravitational redshift it's the frequency seen at infinity. For the accelerati g ships in flat spacetime,it' the frequency seen by the first ship (at x~ = D). Mumtiplication by "c" is needed if you want the right units.
@@federicopagano6590 Since (wavelength)^-1 = frequency / c, it could also be viewed as the (wavelength)^-1 at infinity. If you want to talk about "shifting", you might want to look at the formula for taking the ratio of frequencies at two different positions.
The formula for proper acceleration for hovering above a planet was missing the square root due to an algebra error I made. I also notice this version has an unfortunate audio glitch near the end, but youtube auto-uploaded it sooner than I wanted. Must have misclicked something.
@@NoNameAtAll2 Yes, I did that maybe 4 times, through various iterations of finding small things that are wrong. The audio glitch didn't appear in any of the previous versions, or the video editor, so I didn't bother checking that timestamp. Alas.
Last two videos it is not very clear. Rather we can say it it illusion more provocative way to tell the a unstable field is being fixed . Weak gravity is in question? Geometry is overpowerd the space-time. Couldn't understand.
These relativity videos are hidden gems. So well put and concise. Thank you!!!
The normalization that you speak of at 14:17, is that accomplished by the elementary row operation of scale?
Not, I don't bother doing anything to change the matrix. I just know the squared length of the e_t basis vector is "g_tt", so I divide by 1/sqrt(g_tt) to normalize.
@@eigenchris Thank you for the knowledge and for taking the time.
I think that at 25:30 the metric tensor component should be g_rr instead of g_tt even though the formula is correct downstair.
Hello Chris ! Can you explain where does formula for proper acceleration in gravity in GR come from at 23:50
I try to explain it quickly at 25:00. The steps are:
1. choose a path
2. Determine 4-velocity U for the path
3. Use U to calculate 4-acceleration A for the path
4. proper acceleration is the magnitude of A
In our case:
1. the path is on observer at constant spatial position (constant r), but moving through time
2. Since the observe is at a constant position in schwarzschild coordinates, U points completely in the time direction, so U = c*e_t
3. Calculate A as shown on the slides
4. The magnitude of A gives the result
Thanks a lot for such quick response . Love from India 🙏🙏🙏🙏🙏
Massive respects for making these video series!
At 7:46 Can you explain why this gives us the tangent vector of the lightbeams worldline? I would have tried to derive it from the equation at 5:42 , but I'm guessing that is just a more inconvenient way of doing it. But I don't really see how the 2 equations (5:42 and 8:23) are related.
The equation at 7:46 gives the generic equation of tangent vectors on any path parameterized by lambda. We found back at 5:42 the equation of a light beam between "ct" and "r". At 8:23 we decide to set the path parameter lambda to "r", so we're really calculating the tangent vectors d/dr.
just a suggestion for the future. i would really find it useful if there was a question sheet written in relation to each video. answering the questions as i go along would help me retain a lot of the information much better.
chris, if you have any suggestions on what i could do, i would really appreciate it. thank you for all these amazing videos!
Realistically, I'm probably not going to make exercises for each of the relativity videos in the near future. But maybe I can consider it for my next series. One exercise you could try for this video is to calculate the frequency shift seen by a free-falling observer that has zero velocity at infinity. I might upload an "exercise-style" video on that later, where I let people pause and work on it before giving them the answer.
@@eigenchris I tried to calculate the frequency shift seen by a free-falling observer, but I am not sure what is the answer I am supposed to get. First of all, I tried to find the equation of the geodesic of the observer. I used the same method you used in video 108c and I got a differential equation that is a bit complicated (I think I got the right equation, because for r>>rs I got d^2(r)/d(tau)^2=-GM/r^2, just like in Newtonian Mechanics). I can try using the internet to solve and get the function r=r(tau), but where do I go from here? I can also try getting an expresion for tau in terms of ct and then use the frequency formula in this video to get f=f(tau) which would give us the frequency of light in terms of proper time of the observer. Is this the answer that I am supposed to get?
@eigenchris I have a question, at 6:00 you mentioned the log term goes to 0 when we are far from the Schwartzschild radius, but doesn't the log term goes to positive infinity when r approach positive infinity?
I think the correct description should be the slope (del r) of the log term goes to 0 when r approach positive infinity?
Hi chris, I have idea to explain the curl zero in 11:59. We can define scalar function "phase (phi)" on this plane. We can set in a certain light wave front phi=0, and in next wave front phi=2pi and so on. The wave number kr can be understanded as partial phi/partial r. The wave frequence kt can be understanded as partial phi/partial ct. I think we can assume phi has continuous second partial derivatives thus can exchange order of partial derivative. This can explain the curl zero.
The other thing is about Doppler Effect for Rinderler coordinate. The formular in 27:36 looks different with video 106a in 30:39. If I use formular in this video, with increasing r, I got redshift. If I use formular in 106a video, I got blueshift. Do you know what's wrong.
Where does the relation at 12:02 come from?
I know how to do curl so you can direct me to a resource where it is derived.
If you took the 3D formula for curl, this is just 1 of the components of that formula. But in relativity, the 3D curl you learn in vector calculus is usually not used, since it only works in 3D. The "true" formula comes from the exterior derivative of kappa (you can see it in the first bullet point on the slide at 12:02). This gives a curl-like formula but it works in any dimension.
@@eigenchris do you have a derivation for it? I have understood almost everything in the playlist. Do you have any resources that could show an alternative derivation?
@@cinemaclips4497 You could try reading the first couple sections of this wikipedia article: en.wikipedia.org/wiki/Closed_and_exact_differential_forms
All I'm really trying to say is that, if you drew tangent vectors along the paths of light beams, the trajectories should never "cross over" each other or form loops. This is equivalent to saying the curl of the vector field is zero. To make the equivalent statement for a covector field is to say the covector field is "exact", meaning its exterior derivative is zero. To fully understand this you'd have to take some time to study differential forms, possibly the wedge product, and the exterior derivative. Unfortunately I don't have videos on these. Saying "the curl of the vector field is zero" is an equivalent statement and results in the same formula.
11:53 how do we know that’s the curl: why doesn’t it depend on the metric?
The "proper" way of getting the formula is computing something called the "exterior derivative", which involves differential forms. I don't have videos on that, so I simplified the explanation a little.
@@eigenchris so in this case if both the covector components and partial derivatives are covariant would it be
curl(κ)=εⁱ∂/∂xⁱ∧εʲκⱼ=1/2(∂κⱼ/∂xⁱ−∂κᵢ/∂xʲ)(εⁱ⊗εʲ−εʲ⊗εⁱ)=1/c(∂κₓ/∂t−∂ f/∂x)εᵗ⊗εˣ
how would you find the frequency from the perspective of falling into a black hole? would you instead put a unit length vector tangent to the objects wordline into the covector field at that point? also GREAT videos!
Yeah, the procedure I show is pretty general. Just take the tangent vector for a geodesic along any worldline and you'll get the frequency that worldline observes.
So glad you made this video, the comparison btwn the Rindler and Schwarzschild frame was very illuminating. Does this mean though that all observers will see a time dilation gradient about a mass because of the Doppler effect? My understanding is that in the special relativity frames and the Rindler frame, all the time/spatial dilation effects are kinematical. Can time dilation still be considered to stem from a kinematical effect when spacetime is curved?
To me it seems a bit "wrong" to attribute time dilation to the Doppler effect. I'd think time dilation is more fundamental, since frequency shift is about making measurements in time and space. I've seen some people say you can view gravitational frequency shift as a series of kinematic doppler shifts across a chain of close observers, each sending a light signal from one to the next. I think this is basically approximating the curved spacetime near a mass as a series of flat tangent plane spacetimes. Time dilation could stem from a combination of gravitational and kinematic effects in GR. For example, instead of looking at two observers at constant radii above a planet and calculating their proper times, you could take one of the observers and make them "zig-zag" back and forth around a fixed position. This will add a kinematic aspect to the time dilation as well as the gravitational effects.
Awesome explanation!
In 25:29, at third line there should be g(rr) instead of g(tt). Am I right Chris?
Yeah. My mistake.
Awesome explanations as expected!
Conceptually, it makes sense wavefronts passing by a stationary observer and getting denser near a massive object is a different situation than a spaceship passing through wavefronts in an accelarated frame, because those effects combine in a certain way since the spaceship have to stay still too, in a gravitational field.
My only question is if it works the same way for objects in orbit, because accelaration vector would be tangential/oblique to some wavefronts of incoming light, for example, light from Andromeda coming to earth and being captured by tellescopes in orbit. So, do we have to consider two kinds of doppler effects (galilean/lorentzian and gravitational)?
I haven't done any of the calculations involving angular coordinates. Most of my schwarzschild videos focus on radial behaviour because angular + radial behaviour can get complicated (as seen in 108c). But in theory the covector approach should work. Covectors in 4D spacetime would be stacks of 3D hyperplanes. The formula for forcing them to be "exact" is more complicated, and you' probably need to learn what the "exterior derivative" is to get it. You don't need to worry about the source of frequency shift, whether it be doppler or gravitarional. Light will travel in whatever way the metric tells it to travel. After you have the wave covector, you just pick any observer and see how much their tangent vector sees in 1 unit of proper time.
@@eigenchris Thank you.
How may I find the gravity of the black hole exactly on the event horizon?
Amazing video as always!
Two questions :
10:14 - Is the Schwarzchild solution static over time for convenience or necessity? Is there an example of a solution to the EFE that is not static over time?
11:00 - The definition of 'exact' covector fields sounds very similar to the notion of flux in the FLRW metric. Such that worldlines and wavefronts can't simply pop into existence. Is there any way for a gravitational covector field to not be 'exact'?
The "static" criteria is one of the assumptions we made during the derivation of the Schwarzschild metric in 108a. So we just re-use that assumption again there.
The specific case of covector fields for waves requires that the covector field be exact so that the wavefronts "join up". Mathematically, this is similar to how a gravitational field in Newtonian gravity must be conservative (otherwise we could get energy for free). Magnetic fields are examples of vector fields that are non-conservative (because they curl). Is there an example of a non-exact covector field in physics? Maybe, but I don't know of one. It would have to be an example where the "joining up" of the stack surfaces doesn't violate any physical laws.
"Covector fields" are also called "differential forms". You can google "differential forms" and "exterior derivative" to learn more. I have a couple videos in my Tensor Calculus series that touch on this (I think 10-13) but I don't go into a ton of depth.
Sorry, I missed answering "is there a solution to the EFE that is not static overtime". Yes. Gravitational waves and the expanding universe (aka FLRW) metric are some examples that I discuss in the 109 and 110 videos. There's also rotating black holes (aka "Kerr metric"). I don't plan to do videos on the Kerr metric in the near future.
@@eigenchris I just want to say, that your videos are so refreshing, informative, and clear.
There is a huge 'fight' right now about this video (ua-cam.com/video/PjT85AxTmI0/v-deo.html) which claims a bunch of other physics youtubers are 'wrong' about time dilation.
And I'm like "...just go watch Eigenchris people. jeeeze."
Anyway. Thanks for being an incredible teacher. I really appreciate your work and your patient responses to all of my weird questions.
@@DanSternofBeyer Yeah, I watched that video and read some of the chaos in the replies. I'll admit, after all my relativity studies so far, this "time dilation gradient" = "gravity" interpretation isn't something that's clear to me. It could just be that no one has explained it to me in the right way yet. Or maybe it's wrong. I can't really say one way or the other currently.
@@eigenchris The main idea behind the interpretation, if I understood correctly, is that newtonian gravity is explained to a very good approximation(in a certain coordinate system) by the 00 component of the metric alone. And in that coordinate system that component presents as a simple decreasing function of r. I think it's fair to call that a time dilation gradient but there is some subtlety to it that isn't really addressed in most presentations of the analogy.
Also please tell me if I'm overstepping by invading your comments. This is something that's I'm really trying to understand both sides of right now.
Hi Chris. One question: When two black holes merge and send out gravitational waves, is there a redshift in frequency of the measured gravitational waves, at least at times their Schwarzschild surfaces come close ?
That's not something I'd thought about before, so I had to google it. The two stack exchange posts I've linked below seem to think the answer is "yes". Light and gravitational waves both travel at the speed "c", so I guess it makes sense that they behave the same way? Interesting question.
astronomy.stackexchange.com/questions/13326/redshift-for-gravitational-waves
physics.stackexchange.com/questions/137292/can-gravitational-waves-be-red-shifted
How acceleration due to gravity is created is in which video?
Can you explain how a variable that parametrized a curve can be used to make a tangent vector? The same variable can be used to parametrize another curve so there’s confusion. And what if the coordinate curves aren’t functions of the specific curve? (Like when the curve is a line parallel to a coordinate axis).
A worldline parameterized by λ would be a collection of functions for each coordinate. For example: (ct(λ), r(λ), θ(λ), φ(λ)).
If the curve is parallel to a coordinate axis, it just means one or more of the above functions is constant, so the derivative with respect to λ for that coordinate will be zero.
@@eigenchris how would the parametrization work in terms of how close it is to being arc-length parametrized? I’d imagine the parametrization depends on the velocity, but how so? Also, I meant perpendicular to a coordinate axis.
@@biblebot3947 Not sure I understand the question. For timelike paths, the "arc length" would be proper time. For spacelike paths, the arc length is actual arc length. I use lambda as an arbitrary parameter that isn't necessarily equal to either of these. But you can always get the "arc length" by integrating the lengths of tangent over the curve, regardless of what the parameter used is. You'll get the same result either way.
@@eigenchris
My questions were:
1) What happens when a curve can’t be expressed as a function of a coordinate curve (like a straight line perpendicular to the x-axis can’t be written as a function of x)?
2) A curve can be parametrized in multiple ways. How would the choice of parameter affect the tangent vector?
@@biblebot3947 For 1), I'd figure pretty much any smooth curve can be parameterized. An example of a curve perpendicular to the x-axis is: (ct(λ)=λ, x(λ)=5, y(λ)=0, z(λ)=0)
For 2), basically, if you have "less λ" between two points on the curve, the size of the tangent vector will increase. You can loosely think of "λ" as being like a "time" parameter, telling you how long it takes to move from point to point. If λ goes from 0 to 100 between two points, you spend a lot of "time" between those two points so the tangent velocity vector is small. If λ goes from 0 to 1 between the same two points, the tangent velocity vectors will be large since you travel across very fast. I realize "time" is a confusing term to use though... since there is already a time coordinate in relativity. It's more like a mathematical "time" and not a physical time.
I have one question that is bothering me. Does LIGO detect gravitational waves the way we predict in weak gravity limit (the + and X polarization)? This bothers me because on Earth we can't use the weak gravity limit, so that means that gravitational waves should take different forms.
Brilliant educator
i dont get if sr effects are to be summed to gr effects or they are implied in them?
26:05 shouldnt that be f/c?
Yes, that's right. kappa(e_t_norm) = f/c
Why are the wavefronts parallel to the world lines and not orthogonal?
Wavefronts (drawns as planes in 3D) are perpendicular to the direction of wave travel in space. But we're looking at spacetime, where each individual wavefront will travel along a lightlike worldline as it moves across the spacetime diagram. In this 2D spacetime diagram, wavefronts are just points, and they trace out a set of lines on the spacetime diagram as they move.
Hi Chris Can you tell me which tools along with power point do you use to make these videos?
I use Audacity for the audio and an old copy of Windows Movie Maker for basic video editing.
Excellent! I want to discuss an alternative approach with you privately. How can I send you a direct message?
@@eigenchris Sent
can you explain why the covector field has to be exact phisically?
It's so that the covector surfaces don't "split" into multiple light beams. It's similar to saying the divergence of a vector field is zero. If you draw a spacetime box and 3 light beams enter the box, then 3 light beams must leave.
@@eigenchris thank u. I will try to get spoiled and get another answer to this question: if i have a satellite orbiting earth if have a time dilation due to its velocity relative to an observer on earth (linked to sr and gamma) and one due to gravitational effect (linked to r and r_s), how one can "sum" the 2 and get the proper time dilation?
@@domenicobianchi8When it comes to general relativity, there isn't really a "simple" time dilation formula you can use. If you have 2 spacetime curves, you can calculate the total proper time along each by doing an integral along dτ. You can then compare the results to see which path has longer proper and and which has less.
@@eigenchris i see. Lets say i do the same path in a curve spacetime, it makes sense saying i could do the same path going slow, and the doing it at at speed say half the speed of light? How the proper time is affected? Could you point me to some calculations of this kind
@@domenicobianchi8 These would be different paths on a spacetime diagram, since going "fast" and going "slow" would have different slopes on the spacetime diagram. I do some proper time integrals in video 105d I believe, when I cover the twin paradox.
Do infalling reference frames see the change in frequency? I guess not exactly, but I am curious about what happens if every second of the fall a beam is generated inside the rocket, what would happen when the rocket crosses the event horizon? Would the beam of light change a lot its frequency just by crossing the rocket's lenght?
I haven't done the calculation of looking at frequencies measured by an infalling timelike geodesic, but according to the equivalence principle, I would have to conclude a freely falling observer would not observe any change in frequency.
@@eigenchris I think the same. Thank you
@@viniciusaraujoritzmann Sorry, but thinking about this further... I tried going through the calculations and I did not get the result that a freely falling observer would observe the same frequency all throughout the fall. I may have assumed too quickly that the equivalence principle applies globally to a free-falling observer's entire journey--this is not correct, as the E.P. only applies locally, as stated in my video. I'm trying to find some sources to back up my calculations. This Physics Stack Exchange question seems similar to what you're asking, although I'm not yet sure if the answer they give is algebraically equivalent to the one I calculated... I'll keep looking into this. physics.stackexchange.com/questions/685802/what-is-the-frequency-of-the-light-seen-by-an-observer-free-falling-into-a-black
@@eigenchris Thank you Chris I was reading online about this and wasn't able to find an answer yet, but I started to think what you said, if the E.P would be valid since the curvature changes so rapdly close to the black hole ( I mean, it seems the more close you get the smaller it is the time and space where aproximations work). Thak you again I'll keep researching as well.
To add a little more information... Alfred's answer at the end states: "whatever distance from the black hole you start your jump in free-fall, you can tell that you crossed the horizon precisely when the light emitted from your starting point reaches you with exactly half the frequency it is emitted." This seems a bit counter-intuitive, because it actually means the light you see while freely-falling into the black hole will be redshifted compared to where it is emitted, but this is consistent with the result I calculated. If you want to try calculating this yourself, I got that an observer starting at rest from infinity (Ut=c, Ur=0) will have the following 4-velocity components while freely-falling into a black hole: ( Ut = c/(1-rs/r), Ur = -c*sqrt(rs/r) ). You can divide by "c" to normalize to a length of 1 (since 4-velocity vector always has a length of "c") and then feed it into the Kappa covector seen in this video. After some algebra, I get f/c = a* 1/(1 + sqrt(rs/r)). As you can see, when plugging in r=rs, I get an answer of f/c = a/2, meaning the frequency "a" at infinity will get divided by 2 for an observer freely falling across the event horizon. This is a surprising result to me... and I am not totally certain my calculations are correct, but they match up with someone else who did them independently, so this may be true.
Excuse me teacher! Why covector act on D over D Lamda equal to zero?
d/dλ is a vector tangent to light worldlines. The covector is a "stack" that's parallel to light beams. When a covector acts on a vector parallel to it, you get zero.
Thanks
Ok, but what is the physical interpretation of the constant a?
For gravitational redshift it's the frequency seen at infinity. For the accelerati g ships in flat spacetime,it' the frequency seen by the first ship (at x~ = D). Mumtiplication by "c" is needed if you want the right units.
@@eigenchris could it be interpreted a=(shifted lambda)^-1?
@@federicopagano6590 Since (wavelength)^-1 = frequency / c, it could also be viewed as the (wavelength)^-1 at infinity. If you want to talk about "shifting", you might want to look at the formula for taking the ratio of frequencies at two different positions.
Do you know about higgs field ..!!
I don't know much. Only some very superficial facts.
Can you make a video explaining how you make your videos?
And then a video explaining how he makes a video explaining how he makes his videos.
I could maybe do something like that after I finish the relativity series. It's mostly just powerpoint.
was the worry about mistake justified?
what changed?
The formula for proper acceleration for hovering above a planet was missing the square root due to an algebra error I made. I also notice this version has an unfortunate audio glitch near the end, but youtube auto-uploaded it sooner than I wanted. Must have misclicked something.
@@eigenchris I know some channels upload video unlisted first, to check that nothing glitched
then making it public notifies everyone as normal
@@NoNameAtAll2 Yes, I did that maybe 4 times, through various iterations of finding small things that are wrong. The audio glitch didn't appear in any of the previous versions, or the video editor, so I didn't bother checking that timestamp. Alas.
great video, appreciate ya
أحسنت واصل
so colourful ships !
Last two videos it is not very clear. Rather we can say it it illusion more provocative way to tell the a unstable field is being fixed .
Weak gravity is in question?
Geometry is overpowerd the space-time.
Couldn't understand.
nice
Absolutely .... whatever!