Source tracelessness, divergencelessness at invariant speed

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Is the local propagation of an entity at invariant speed a sufficient condition for its stress-energy tensor, independently of its explicit mathematical form, to be trace-free, or to have null covariant divergence, or both in curved space-time?
In the book “An Introduction To Mechanics”, by Daniel Kleppner and Robert J Kolenkow, at page 326 says that, the electromagnetic stress-energy tensor, giving its exact mathematical form, is trace-free because electromagnetism propagates at the invariant speed. Until now I understood that, the traceless character of this tensor, as also its vanishing covariant divergence were a direct result of its mathematical form (symmetries), which results from the application of the least action principle for writing it. Is it that propagation at invariant speed is enough for tracelessness and divergencelessness?
I was not able to work out an answer using tensor calculus, or intuition. I will very much appreciate any help.
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EagleH
 

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bcrowell
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The reason it has to be divergence free is the same as the reason that any other stress-energy tensor has to be divergence-free: local conservation of mass-energy.

Yes, the trace-free property does relate to the velocity of propagation. The momentum of a beam of light is related to its energy by p=E (in units with c=1). This fact holds only because v=c. I find it more convenient to think in terms of a photon gas, because then you can go into the frame where it's at rest on average, and the [itex]T^\mu_\nu[/itex] form of the stress-energy tensor looks like a diagonal matrix with elements [itex](\rho,-P,-P,-P)[/itex]. The average momentum along each axis is proportional to E/3, so it's traceless.

Until now I understood that, the traceless character of this tensor, as also its vanishing covariant divergence were a direct result of its mathematical form (symmetries), which results from the application of the least action principle for writing it.
Tracelessness isn't a result of its symmetry, it's simply the word we use to describe its symmetry. Since all of physics can come from writing down Lagrangians, it's certainly true that a specific fact like this must come from the Lagrangian.

Is the local propagation of an entity at invariant speed a sufficient condition for its stress-energy tensor, independently of its explicit mathematical form, to be trace-free, or to have null covariant divergence, or both in curved space-time?
Yes, it's a sufficient condition for it to be trace-free. No, it's not a sufficient condition for it to be divergenceless. For example, if I say that photons don't redshift due to cosmological expansion, then their stress-energy tensor would have a nonvanishing divergence.

In the book “An Introduction To Mechanics”, by Daniel Kleppner and Robert J Kolenkow, at page 326 says that, the electromagnetic stress-energy tensor, giving its exact mathematical form, is trace-free because electromagnetism propagates at the invariant speed.
Are you sure it's that book and that page? I don't have my copy handy, but using amazon's "look inside" function, that doesn't seem right. Relativity comes much later than p. 326 in that book.
 
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bcrowell
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In the book “An Introduction To Mechanics”, by Daniel Kleppner and Robert J Kolenkow, at page 326 says that, the electromagnetic stress-energy tensor, giving its exact mathematical form, is trace-free because electromagnetism propagates at the invariant speed.
What edition do you have? I have one with a 1973 copyright, and p. 326 is about rigid body motion (section 7.7).
 

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