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Why ##n^\mu T_{\mu\nu}## is called the pressure? 
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#1
Aug1114, 09:39 PM

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 energymomentum tensor and ##n^\mu## is a four dimensional normal vector.



#2
Aug1214, 05:11 PM

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:
[itex]T_{\mu \nu} = diag( \rho, p , p , p ) [/itex] Then: [itex] n^{\mu}T_{\mu \nu} = n^{0}T_{00} + n^{ii}T_{ii} = n^{0} \rho + (n^{11}+n^{22}+n^{33}) p[/itex] Which doesn't have to be equal to the pressure... If though you choose [itex]n^{0}=0[/itex] (so it's not just any 4dim unit vector) things can get better. Can you give your reference? 


#3
Aug1214, 08:24 PM

P: 9




#4
Aug1314, 01:17 AM

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 4velocity of the fluid. 


#5
Aug1314, 09:53 AM

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P: 2,026

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}##. 


#6
Aug1314, 02:29 PM

P: 1,045

[itex]n^{i} T_{i \mu} \equiv n_{i} p[/itex] 


#7
Aug1314, 03:17 PM

P: 680

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


#8
Aug1314, 04:18 PM

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 


#9
Aug1314, 05:09 PM

P: 680

Even if the expression is nonzero 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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