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Velocity is relative, which means kinetic energy is. Since, according to general relativity, energy bends spacetime around it, wouldn't this mean observers moving in different inertial frames measure different values of curvature?

I think this is slightly different from this question: Is space curvature relative?

Qmechanic
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gardenhead
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It depends on what you mean by “curvature”.

The most complete description of curvature is in terms of something called the Riemann curvature tensor, $R_{\mu\nu\lambda\kappa}$. In four dimensions, it has 256 components (only 20 of which are independent), and the value of these components is different in different reference frames. There is a transformation rule for how this tensor transforms between frames. In other words, the components of the Riemann curvature tensor are relative, i.e. observer-dependent.

However, from this tensor you can construct invariants such as the Ricci scalar $R$ which have the same value in all reference frames. These curvature invariants are absolute, i.e., observer-independent.

This is similar to how energy and momentum are relative — they are components of a four-vector which transforms between frames — but a particular combination of them, mass, is absolute.

G. Smith
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  • Taking into account symmetry the 256 reduce to 55. Still impressive. – my2cts Jan 07 '19 at 20:39
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    @my2cts No. For confirmation that 20 is correct, please see http://mathworld.wolfram.com/RiemannTensor.html. – G. Smith Jan 07 '19 at 20:45
  • Could you go into more detail on how we can combine energy and momentum to get mass? Also, isn't mass also relativistic? – gardenhead Jan 07 '19 at 21:05
  • In units where $c=1$, the invariant mass (which is sometimes called "rest mass") is $m=\sqrt{E^2-\mathbf{p}^2}$. (With the $c$'s, this is $m=\sqrt{E^2-\mathbf{p}^2 c^2}/c^2$.) The mass is the "length" of the energy-momentum vector $(E,\mathbf{p})$ in Minkowski space. Relativistic mass is just $E/c^2$ and is widely considered an obsolete concept. It is either proportional to the energy, or, in units with $c=1$, equal to the energy. – G. Smith Jan 07 '19 at 21:15
  • @my2cts If the downvote is from you, and is because you mistakenly think the number of independent components is 55, it would be polite for you to acknowledge your mistake and reverse your downvote. – G. Smith Jan 07 '19 at 21:21
  • I don't downvote, I prefer exchange of views. That is what I get out of this site. Have you already read my paper in enough detail to be able to judge it? I always believed that $R_{\mu\nu\lambda\kappa} = R_{\lambda\kappa\mu\nu}$ and $R_{\mu\nu\lambda\kappa} = - R_{\nu\mu\lambda\kappa}$. That led me to 55 elements, but I must have made a mistake as 20 is correct. – my2cts Jan 07 '19 at 22:09
  • @my2cts Thank you. Perhaps the mystery downvoter will come forward to explain why he or she found my answer objectionable. If you are talking about your paper on electromagnetism that you mentioned on another thread, I haven’t had time to read it. We should not discuss it here. – G. Smith Jan 07 '19 at 22:35
  • @my2cts I will try to find time to read your paper in the next few days and would be willing to chat about my reactions. However, I think feedback from someone like QMechanic would be more valuable, because I am less knowledgeable about QFT than I am about GR. – G. Smith Jan 07 '19 at 22:50
  • @G.Smith could you pls tell what is the "length" of Reimann tensor called... Like the length of energy momentum 4 vector is mass.... It will help to draw further analogy – Shashaank Jun 04 '20 at 14:37
  • @Shashaank The Ricci scalar curvature is the simplest invariant that can be formed from the Riemann curvature tensor, but I don't recommend thinking of it as a “length”. – G. Smith Jun 04 '20 at 16:14
  • @G.Smith Like there is a norm of a vector which is a scalar which denotes it's length if it is a vector in $R^3$... I understand that even abstract vectors like matrices cannot be denoted a length. But can a tensor be denoted a length... Something like norm of a tensor... even in just $R^3$ is it possible – Shashaank Jun 04 '20 at 16:30
  • @Shashaank If you have a tensor, say $T^i{}_j$, you can form scalars like $T^i{}_i$ and $T^i{}_jT^j{}_i$. I do not think it is useful to call any such invariant (formed from a tensor rather than a vector) a “length”. Comments are not an appropriate place to ask new questions. – G. Smith Jun 04 '20 at 16:37
  • @G.Smith ok thanks I understand. If I wish to draw your attention to a particular question ( since it might be linked with what you have answered), what would the best way in general. Will tagging you ( or someone) do the job or you suggest anything else. – Shashaank Jun 04 '20 at 16:42
  • @Shashaank As far as I know, tagging me in a comment to a question or answer in which I haven’t participated won’t notify me. I read basically every new question, so you don’t need to call my attention to one; I’ve probably already seen it, and if I didn’t participate it was intentional. – G. Smith Jun 04 '20 at 17:02
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No, it is absolute. The presence of curvature (non vanishing Riemann tensor) is mathematically equivalent to the existence of geodesical deviation of timelike geodesics. This physically means that there is a relative acceleration of free falling bodies. This fact is absolute, it does not depend on any reference frame as it is a relative phenomenon.

Valter Moretti
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Locally, yes. Globally no.

The Equivalence principle tells you that you can always find a local Inertial reference frame, which means that given a point in spacetime you can always find neighborhood of this point where spacetime is flat, and special relativity holds.

Globally you can't do that because spacetime is, mathematically speaking, a Riemannian manifold, and its curvature is an intrinsic property.

RenatoRenatoRenato
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