Find Perfect Fluid & EM Tensor in Rest Frame

[victorvmotti]

we use perfect fluid which is characterized by a energy density and isotropic pressure for general forms of matter.

When guessing the values of energy momentum tensor indices we can use the physical insight that they are the flux of four momentum in a constant surface of spacetime.

The shortcut to find the general tensorial equation is making physical arguments in the "rest" frame.

Clearly the sheer stresses vanish and therefore only pressure remains in three orthogonal directions and the Energy Momentum matrix is diagonal.

What is not clear to me is how a rest frame makes sense in the case of a perfect fluid. Are we assuming an infinitesimal element of the fluid? Otherwise for the bulk of the fluid a rest frame is not possible given the random motions of particles.

Also related is that what does mean pressure for the elements of fluid when that element is on the boundary as opposed to be inside the fluid.

Put it other way, what does pressure, the spatial diagonal elements in the EM tensor, in the rest frame means, on the surface of a perfect fluid, say a star?

Does it mean that in a thought experiment if we put a rod close enough to the surface of a star we will sense a pushing force against us?

 

[PeterDonis]

for the bulk of the fluid a rest frame is not possible given the random motions of particles.

The random motions of particles average out to zero; the rest frame describes the average bulk motion once the random motions have been averaged out. (The effect of the random motions is to increase the energy density of the fluid, by giving it a finite temperature instead of zero temperature.)

what does pressure, the spatial diagonal elements in the EM tensor, in the rest frame means, on the surface of a perfect fluid, say a star?

If the star is surrounded by vacuum, then the pressure will be zero at its surface.

 

[pervect]

we use perfect fluid which is characterized by a energy density and isotropic pressure for general forms of matter.

When guessing the values of energy momentum tensor indices we can use the physical insight that they are the flux of four momentum in a constant surface of spacetime.

The shortcut to find the general tensorial equation is making physical arguments in the "rest" frame.

Clearly the sheer stresses vanish and therefore only pressure remains in three orthogonal directions and the Energy Momentum matrix is diagonal.

What is not clear to me is how a rest frame makes sense in the case of a perfect fluid. Are we assuming an infinitesimal element of the fluid? Otherwise for the bulk of the fluid a rest frame is not possible given the random motions of particles.

In the context of cosmology, where radiation is treated as a perfect fluid, the rest frame is the frame in which the universe is isotropic, i.e. it's the frame in which the cosmic microwave backround radiation (or it's more energetic equivalent earlier in the universe) is the same in all directions.

So the stress-energy tensor is $(\rho + p) u_a u_b + p g_{ab}$, and $u_a$ defines the "rest frame" of the fluid - you can't have a perfect fluid without a rest frame. $\rho$ is the density and $p$ is the pressure.

Also related is that what does mean pressure for the elements of fluid when that element is on the boundary as opposed to be inside the fluid.

Put it other way, what does pressure, the spatial diagonal elements in the EM tensor, in the rest frame means, on the surface of a perfect fluid, say a star?

Does it mean that in a thought experiment if we put a rod close enough to the surface of a star we will sense a pushing force against us?

The metric of a radiating star is not isotropic, it would be given by the Vaidya metric. Wiki https://en.wikipedia.org/wiki/Vaidya_metric gives the stress-energy tensor in that case as proportional to $l_a \, l_b$, where $l_a$ and $l_b$ are null vectors. I believe they're outgoing radial null vectors, but I could be mistaken. This is not a perfect fluid stress energy tensor.

 

[victorvmotti]

Can you explain what happens for a rotating perfect fluid?

What a rest frame could mean here?

 

[pervect]

Can you explain what happens for a rotating perfect fluid?

What a rest frame could mean here?

[/quote]"Frame" is a rather ambiguous term, but in the context of your question, it means is basically is that if you have some angular rotation frequency $\omega$, which we represent in the 3-vector form by a 3-vector $\vec{\omega}$, the 3-velocity of a fluid element at some location $\vec{r}$ will be $\vec{v} = \vec{r} \times \vec{\omega}$

Just imagine a sold rotating cylinder, and ask what the velocity of a mark on the cylinder at some location $\vec{r}$ is.

The tensor equation I gave will give you the stess-energy tensor in terms of the 4-velocity. I'm getting the sense that you may be asking about the stress-energy tensor without knowing what a tensor is. I'm not sure how much I can explain in that case, I'll do my best. Please let me know if I'm wrong and if you think you know what a tensor is.

There isn't any global inertial frame for a rotating cylinder (or rotating fluid), but assuming for the moment we are in flat space-time, at any point in the fluid, there is a "local frame" that you can think of as an inertial frame of reference co-moving with the fluid element with a velocity equal to the product of r and omega.

 

[victorvmotti]

Thank you, this local frame was what I asking, back to the first post, I see that you are referring to that infinitesimal element of the fluid.

 

[victorvmotti]

Thank you, this local frame was what I asking, back to the first post, I see that you are referring to that infinitesimal element of the fluid.