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

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In summary, the textbook states that the absolute value of ##n^\mu T_{\mu\nu}## is typically defined as the pressure, where ##T_{\mu\nu}## is the energy-momentum tensor and ##n^\mu## is a four-dimensional normal vector. However, this does not hold in general, as shown by the example of a perfect fluid in its rest frame. The correct expression would be ##n^\mu T_{\mu\nu} = (\rho+p)u_\nu(n\cdot u) - pn_\nu##, where u is the 4-velocity of the fluid.
  • #1
ccnu
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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.
 
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  • #2
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 4-dim unit vector) things can get better.

Can you give your reference?
 
Last edited:
  • #3
ChrisVer said:
If though you choose [itex]n^{0}=0[/itex] (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!:smile:
 
  • #4
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.
 
  • #5
ccnu said:
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}##.
 
  • #6
Orodruin said:
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:
[itex]n^{i} T_{i \mu} \equiv n_{i} p[/itex]
 
  • #7
ChrisVer said:
oops sorry... yes the correct form would have to be:
[itex]n^{i} T_{i \mu} \equiv n_{i} p[/itex]

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

:smile:

Assuming ##n^0 = 0## and that we are in the rest frame of the fluid.
 
  • #8
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
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.
 

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

The quantity ##n^\mu T_{\mu\nu}## is called the pressure because it represents the force per unit area that a fluid exerts on its container. This is similar to how pressure is defined in classical mechanics, where it is the force per unit area exerted by a gas or liquid on its surroundings.

2. How is ##n^\mu T_{\mu\nu}## related to thermodynamics?

In thermodynamics, pressure is one of the fundamental quantities used to describe the state of a system. ##n^\mu T_{\mu\nu}## is related to thermodynamic pressure through the thermodynamic identity, which relates the pressure, energy density, and entropy density of a system.

3. Why is ##n^\mu T_{\mu\nu}## used in the context of relativistic fluids?

In the context of relativistic fluids, ##n^\mu T_{\mu\nu}## is used to describe the pressure because it takes into account the effects of special relativity. This is important when dealing with high-speed fluids, such as those found in astrophysical scenarios.

4. How does the pressure affect the behavior of a fluid?

The pressure of a fluid affects its behavior in several ways. First, it determines the direction and magnitude of the fluid's flow. It also plays a role in determining the fluid's density, temperature, and other thermodynamic properties. Additionally, the pressure can cause changes in the fluid's volume and shape.

5. Can ##n^\mu T_{\mu\nu}## be negative?

Yes, ##n^\mu T_{\mu\nu}## can be negative. This would indicate that the fluid is exerting a force in the opposite direction of its motion, which can occur in certain situations such as shockwaves. In general, however, we expect ##n^\mu T_{\mu\nu}## to be positive, as it represents the pressure of the fluid.

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