Why ##n^\mu T_{\mu\nu}## is called the pressure?

[ccnu]

I read in the textbook which says that, according to the usual definition the absolute value of $n^\mu T_{\mu\nu}$ is just the pressure. $T_{\mu\nu}$ is the energy-momentum tensor and $n^\mu$ is a four dimensional normal vector.

 

[ChrisVer]

As given I don't see that this holds in general... For example take a perfect fluid for which it can be written in rest frame:
$T_{\mu \nu} = diag( \rho, p , p , p )$
Then:
$n^{\mu}T_{\mu \nu} = n^{0}T_{00} + n^{ii}T_{ii} = n^{0} \rho + (n^{11}+n^{22}+n^{33}) p$

Which doesn't have to be equal to the pressure... If though you choose $n^{0}=0$ (so it's not just any 4-dim unit vector) things can get better.

Can you give your reference?

 

[ccnu]

If though you choose $n^{0}=0$ (so it's not just any 4-dim unit vector) things can get better.

Oh~~ yes, I forgot to say that $n^0 = 0$, and $n^\mu$ is pure spatial. Thanks a lot!

 

[Orodruin]

Chris, your LHS has a free index and your RHS does not.

In a general frame you would have
$$
n^\mu T_{\mu\nu} = n^\mu \left((\rho+p) u_\mu u_\nu - p g_{\mu\nu}\right)
= (\rho+p) u_\nu (n \cdot u) - p n_\nu,
$$
where u is the 4-velocity of the fluid.

 

[Meir Achuz]

I read in the textbook which says that, according to the usual definition the absolute value of $n^\mu T_{\mu\nu}$ is just the pressure. $T_{\mu\nu}$ is the energy-momentum tensor and $n^\mu$ is a four dimensional normal vector.

The association of $n^\mu T_{\mu\nu}$ with the pressure at a point follows from the fact that its integral over a closed surface equals the rate of change of momentum within the closed surface.
There are simple examples where $n^\mu T_{\mu\nu}$ is NOT the pressure at a point on the surface. For instance the pressure on a dielectric slab due to a point charge a distance d from the slab is not $n^\mu T_{\mu\nu}$.

 

[ChrisVer]

Chris, your LHS has a free index and your RHS does not.

In a general frame you would have
$$
n^\mu T_{\mu\nu} = n^\mu \left((\rho+p) u_\mu u_\nu - p g_{\mu\nu}\right)
= (\rho+p) u_\nu (n \cdot u) - p n_\nu,
$$
where u is the 4-velocity of the fluid.

oops sorry... yes the correct form would have to be:
$n^{i} T_{i \mu} \equiv n_{i} p$

 

[Orodruin]

oops sorry... yes the correct form would have to be:
$n^{i} T_{i \mu} \equiv n_{i} p$

$n^{i} T_{i \mu} \equiv n_{\color{Red}\mu} p$



Assuming $n^0 = 0$ and that we are in the rest frame of the fluid.

 

[ChrisVer]

the n0 is zero
and also mu=i for the expression not to be zero...

For the rest frame yes, I just corrected the expression I gave in my previous post

 

[Orodruin]

Even if the expression is non-zero only for spatial $\mu$, you must still have the same free indices on both sides of your equality. In your case you have i as a summation index on one side and as a free index on the other. I can guess what you mean because I know what you are aiming for, but formally it does not make sense.