Relativity 105e: Acceleration - Covariant Derivative in Flat Spacetime (Rindler Coordinates)

[@sagarjc5423]

At 25:57 At Bonus challenge: Pause and check if eigenchris made any typos. Bonus Prize: pain, sadness, fear. Made my day Lol.

[@eigenchris]

I think you're the first lerson to comment on that since I uploaded this video.

[@ibrahimalsamak5268]

Man you are a legend, as a physics student I really appreciate your work. Thank you very much

[@davidwagner6116]

As a 4th year undergrad, Christoffel symbols were very confusing. 40 years later, not confusing at all. And thanks for your very clear explanations!

[@leonardp8517]

The only channel I actually keep the bell rung for

[@Mikey-mike]

I used to teach this in the late 80s. This is my favorite part. Good one

[@HighWycombe]

Thank you... I've wondered for some time what Christoffel symbols were!! btw, I love it when you start a section with the words : "It turns out that.." It means you are saving me a lot of brain work to follow some difficult maths!!.

[@jurgenanklam3226]

If teaching is an art, you are really a great artist! I have also been following your tensor calculus series, which is a big advantage for this one here. In any case your explanation of the Christoffel symbols together with the visualization in this video here is the best one I ever came around. Thanks a lot!

[@aclearlight]

Lovely work, thank you. Very comprehensive and detailed...I keep having to stop and ponder the very well-deliniated steps and then take painful note of just how rusty my calculus is!

[@tensorific]

Tiny typo at 11:56 - It should say $U^{x} = \frac{dx}{d\tau}$ Otherwise a brilliant series - the only channel I have notifications on for!

[@reinerwilhelms-tricarico344]

At 5:00 you show the two maps from (ct,x) to Rindler coordinates and back. But I think you got the arrows pointing between the two domains in reverse direction. I simply tried this with an example (using Octave), choosing a/c^2 = 1: [xtd,ctd]=meshgrid(linspace(1/8, 1.25, 30),linspace(-1,1,40)); ct=xtd.*sinh(ctd); x=xtd.*cosh(ctd); plot(x,ct,x',ct'); axis('equal'); That gives a grid just like on the right side with the Rindler coordinates The inverse: xi = sqrt(x.^2-ct.^2); ti = atanh(ct./x); plot(xi,ti,xi',ti');axis('equal'); then gives me the rectangular mesh grid back. Your image may make sense if I assume that the labels ct, and x mean actually the bases: e_t, e_x. ?

[@DmAlmazov]

If we will take into account the symmetry of the Christoffel symbols than there will be n*(n^2+n)/2=4*(16+4)/2=40 Christoffel symbols to calculate for 4-dimentional space-time. We need to calculate 64 symbols exactly for 4-dimentional space-time when we are taking torsion into account where the Christoffel symbols are not symmetric. But in relativity theory the torsion vanishes and we need to calculate 40 symbols only as i mentioned above.

[@beagle1008]

What a star eigenchris is!

[@jvillas171]

I think I missed something when you say at 6:10 that we can build the 4-position vector S completely out of the ex tilda basis vector. Since ex tilda can point to many directions, how can we not also need et tilda in order to depict the specific direction of S?

[@eigenchris]

ex-tilde always points outward from the origin, so it's always parallel with S, which also always points out from the origin.

[@jvillas171]

@@eigenchris Yes, just like R in polar coordinates is paralel to any position vector, and we still need Theta to specify the direction of these vectors

[@phugoidoscillations]

My life forever changed at 20:00.

[@doaamohamedz9z9z9]

Thank you so so much 💙💙💙💙💙

[@guilhermebahia7843]

Interesting point about "covariant" having two meanings. Now I'm struggling to decide which one applies best to the extra covariant rank gained by tensors acted upon by the covariant derivative😁

[@lameretmer126]

I have some enquirements please, like what's the academic position you hold? And based on any university? Please if we could contact you that would be so helpful. Your lectures are so informative and detailed. It's for master student, who's asking to know more about you to confirm the resourse of the information, the references and so on... Thank you so muck for such this massive helping in relativity...we appreciate it. All support from Syria.🌷

[@eigenchris]

Hello. I have an undergrad degree in engineering/physics and a master's degree in computer engineering. I'm not an academic right now... I just work. I don't have any particular textbook I used to make these videos.

[@lameretmer126]

@@eigenchris thank you so much. About these links, you've provided in the "About" section of your channel; can we trust them to be as references?

[@eigenchris]

@@lameretmer126 They are good for learning but they are probably not acceptable to use as references in an academic paper.

[@lameretmer126]

@@eigenchris you're right, we will consider this. Thank you so much, it's really so helpful and please continue to general relativity because there where the effort comes up. Thank you...all support 🌷.

[@canyadigit6274]

Congrats on 30k mate. Can you make a Q+A vid?

[@eigenchris]

I'm not sure if I want to make that type of content right now. You can feel free to ask me questions in the comments if you want.

[@tw5718]

These videos are great. I was actually wondering if you would be ok with me taking some screen caps and using them in my undergrad senior thesis presentation. I'll obviously cite them.

[@eigenchris]

Thanks! I don't have a problem with you using my images or slides. But I'm guessing your profs would not consider youtube a reputable source to cite for math or physics.

[@tw5718]

@@eigenchris it's less a citation as in it's true because eigenchris said so, more as in I am very slow at latex Also a lot of the visuals are very nice

[@eigenchris]

@@tw5718 Ah, fair enough. Go ahead, then. I made this in the microsoft powerpoint equation editor. So that's an alternative to LaTeX if you need to make something quick.

[@MudahnyaFizik]

New video! Thanks!

[@partakerofbread]

Are we saying that tau is equal to t tilde in Rindler coordinates?

[@partakerofbread]

I think that's just on the hyperbolic worldline...

[@RakeshKumar-ec6wf]

Plz suggest book for tensor algebra tensor calculus and for reletivity plz plz plz

[@adityanadgir3769]

Hey Chris, coordinates like ct tilda are basically vector components, right? Sometimes I get confused.

[@eigenchris]

There are two things to consider: position vector components (which measure a vector) and point coordinates (which mark a point in spacetime). In cartesian coordinates these are the same thing. But in Rindler coorsinates they are different. In Rindler coordinates, the spacetime point (ct~, x~) is given by the position vector S = x~ e_x~ (the time component is always zero since the basis vector e_x~ always points in the correct direction). Sorry for the confusion.

[@gguevaramu]

Dea Chris. In minute 11:01 you derive [ ct (tilde) wrt tau and your result is (c). But in the former video, you got the relation between [ ct (tilde) and tau ]. It is [ ct (tau) = module to the square of vector [e t (tilde)] times [ ct (tilde)] so, if you derive this last expression wrt tau you get [ c divided by the module to the square of vector e t (tilde)] isn't it?

[@eigenchris]

In the 105d video? This was because I was taking the limit of the magnitude of a vector times detla tau, so the in the limit of a small tau, I got the magnitude of the tangent vector times d tau.

[@ProfeARios]

Great video!!! I loved it!!!!

[@safdarsafdari7289]

it would be great if you add some examples of using this in real situation and solving them, thanks. I mean whole relativity problems.

[@Physics_PI]

Comprehensive 👍👍

[@demr04]

If I can ever achieve to use tensor calculus in plant physiology, I would give you credit in my work 🥺.

[@angeld23]

can i please get my pain, sadness, and fear

[@se7964]

Sorry to keep trolling your channel, but the “covariant vs. invariant” derivative is an interesting topic, thought I’d add my two cents. I take the origin of the name to come from the fact that derivatives along manifolds lie along the tangent / basis vectors, and since basis vectors transform covariantly with respect to a coordinate transformation, even though something like 4-velocity is ultimately an invariant geometric object, the “covariant” label still got attached to it. But I think one of your earlier videos offered another unique insight. You said somewhere earlier in this series that the laws of physics are invariant in inertial frames, but when we transform to non-inertial frames the laws become covariant, which essentially means we have to add additional terms for additional forces, etc. I think in this sense of term “covariant”, which implies the laws of physics and certain transformations “co-vary” or change together, can be applied to the 4-derivative, and four tensors in general, since in non-inertial frames these objects are always distinguished by the appearance of christoffel symbols in their expanded expressions.

[@eigenchris]

I think I remember saying "the laws of physics are invariant in inertial frames" in an early video... I'm questioning the value of that statement now. 3-vectors like Newtonian force, acceleration and velocity are not truly "invariant" because you can make the disappear to zero by changing reference frames (possibly to a non-inertial frame). But you can never make 4-velocity or 4-acceleration disappear by changing frames... you can only change its components. I think what I was trying to say in that earlier video is that the form of Newton's 2nd law, when written using 3-vectors, doesn't change when we switch to another inertial frame. Part of the problem when my making these videos is that (to some extent) I'm understanding things better as I make them, and so I sometimes look back on earlier videos and think "hmm that's not the best explanation". My current point of view is that there are two ways to write equations in physics: (1) is to write them using 4-vectors or related tensors, and (2) is to write them using the COMPONENTS of 4-vectors and tensors. In (1), invariance in all reference frames is guaranteed because 4-vectors and tensors don't depend on coordinate systems by definition. In (2), when we change coordinates, terms can disappear and reappear based on the properties of the coordinate system. Sometimes in the case of (2), we isolate special transformation types that don't introduce new terms (Galilean transformations in Galilean relativity, or Lorentz transformations in Special relativity). I think that (1) is truly the best option for writing the laws of physics and so I have been trying to do that. This is why I write the geodesics equation in 105f as "d^2 S / d lambda^2" (which uses true tensor only), instead of the "traditional" way with the Christoffel terms (which uses only components, which is why some parts of the equation can appear or disappear, depending on the coordinates).

[@se7964]

@@eigenchris You can actually make Newtonian 3-force and other classical quantities into invariant four vectors and tensor quantities via the Newton-Cartan formulation. To me this signals that four-tensors are more about generalizations of mathematical framework rather than fundamentally reconceptualizing what constitutes invariant or “real” things. As you state, there’s a special class of transformations that don’t introduce new terms, which corresponds to inertial systems, where the form of the laws are equivalent from frame to frame. This is the basis relative to which all other transformations are made, and the basis from which things like proper acceleration gain their objectivity. Tensor laws seem to be about generalizing how these inertial laws and objects change with corresponding changes in coordinates, but without the inertial frames as the reference class the tensors would be ultimately meaningless . You wouldn’t be able to define invariant objects like “proper time” or “proper acceleration” in the first place, and you would only end up with a vague theory which would tell you only how transformations and objects would relate to each other once defined. Thus it can’t really be said that GR puts all frames of reference on equal footing, which is the impression a lot of people take away after a cursory introduction to the topic. So I guess I think saying the laws of GR are generally covariant as opposed to invariant is a better distinction ultimately to make.

[@eigenchris]

I plan on covering Newton-Cartan in part D of my next set of videos, but I'm still researching it for now. I haven't come across 4 vectors in it yet. I'm not sure what to make of your last paragraph. I guess free-fall/inertial frames will always be different than non-inertial frames. But I'm not sure what you mean by "covariant as opposed to invariant is a better distinction".

[@se7964]

@@eigenchris I guess there’s enough leeway in terminology that ultimately it’s going to come down to personal preference... however I would strongly side with generally covariant over invariant for the following reasons: First off, when we say “a vector is invariant” we have to remember a vector is just a collection of numbers, and so in fact that statement is meaningless in-and-of-itself. What we mean to say is that certain quantities, e.g. direction & magnitude, associated with that vector will retain their same value after after a coordinate transformation. Similarly since tensors are also lists of numbers, they can neither be invariant nor variant with respect to a coordinate transformation. Rather they have certain invariant quantities associated with them (there are 3 invariants for a rank-two tensor I believe), quantities that do not change with coordinate transformations. Tensors CAN however be covariant or contravariant (we will come back to that point) Now if we form laws out of these vectors and tensors, we can express those laws in a way that emphasizes the relationships between the invariants: the relation expressing the sum of two vectors for instance A + B = C is true in any coordinate system. But should we say the “law” that A + B = C is itself invariant? Again this would be imprecise, because it’s only certain quantities associated with those vectors that can actually meet the definition of invariant. Here you could choose to liberally expand the definition of “invariant” to include that of laws expressed without relation to a particular coordinate system. But I actually that’s more confusing, because people learning the topic will be unable to make the distinction between the vector/tensor as an abstract object with certain invariant quantities associated to it and vector/tensor laws as merely symbolic relations meant to emphasize the importance of those invariants. Where does this come into play with GR? Well first we have to consider the fact that a coordinate transformation doesn’t “truly” preserve any quantities whatsoever. Scale all your basis vectors by two, and you don’t have to shrink your components by half to preserve length: rather you could simply choose to say that the length of your vector doubled with the transformation. In physics however we require - a priori - that certain quantities such as length and direction (or those quantities found in inertial frames) be left unchanged by coordinate transformations. This gives us our particular covariant/contravariant relationships of the quantities that then do change with the transformation. Tensors are all either covariant or contravariant, because their components transform in different ways with different transformations. Thus if we form a law from tensors alone, we could say this law is “covariant” or “contravariant” or mixed, because the quantities of interest will change in a certain way once we have chosen our coordinate system. Again, this won’t be wholly precise because there will also be invariant quantities involved in the tensor expression as well that won’t change with our choice of coordinates. So if we look at Einstein’s law for gravitational, we can notice all the tensors are covariant, and that it is a sort of “covariant law”. Again, it would be imprecise to call the law itself “covariant” since there are also invariant and probably contra-variant quantities involved. So for that particular reason we slap the world “general” in front of it and there you have it: “general covariance”. Not perfect obviously but better than invariant.

[@menjolno]

no face reveal pls. :)