1
$\begingroup$

I learned that dust is a collection of particles at rest with repect to each other. Thus the four velocity field $U^\mu$ of dust will be a constant and it is easy to think of what the rest frame of dust means. It is simply the frame that every individual particle is at rest and is applicable to every point in space.

But what does the rest frame of a perfect fluid mean? A perfect fluid would not have a constant four velocity field $U^\mu$ so how do we choose the rest frame?

On pg. 34-35 of Sean Carroll's general relativity textbook, it was also written that to guess the energy momentum tensor for perfect fluid, we start with that of dust which is $$T^{\mu\nu}=\rho U^\mu U^\nu$$ where $\rho $ is the rest frame energy density. He then said that using this expression for a perfect fluid meant that $T^{ii}=0$, which I understand is because $U^i=0$.

That means that the four velocity of a perfect fluid in its rest frame is zero throughout space. But how is that possible if a perfect fluid does not have a constant four velocity in the first place?

$\endgroup$
2
  • $\begingroup$ You can see my answer here: physics.stackexchange.com/questions/659947/… . In particular, the fluid rest frame implies that the fluid particle velocity $U^{\mu}$ must be same as the unit timelike vector field $u^a$ (as mentioned in my answer). This is true for fluids with single species system $\endgroup$
    – KP99
    Commented Aug 22, 2021 at 14:46
  • $\begingroup$ Related (possibly duplicate) physics.stackexchange.com/q/694491/226902 $\endgroup$
    – Quillo
    Commented Feb 22, 2022 at 22:03

1 Answer 1

0
$\begingroup$

In the rest frame, $U^0 = 1$. So $T^{\mu \nu} = \rho$.

If you view the dust from another frame, the individual components will change, but the magnitude will not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.