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

by ccnu
Tags: ##nmu, called, pressure, tmunu##
 P: 9 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.
 P: 1,045 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?
P: 9
 Quote by ChrisVer 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!

 P: 680 Why ##n^\mu T_{\mu\nu}## is called the pressure? 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.
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PF Gold
P: 2,026
 Quote by 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.
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}##.
P: 1,045
 Quote by 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.
oops sorry... yes the correct form would have to be:
$n^{i} T_{i \mu} \equiv n_{i} p$
P: 680
 Quote by ChrisVer 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.
 P: 1,045 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
 P: 680 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.

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