Pressure term in the energy stress tensor

[Ron19932017]

Hi all, I am reading Bernard Schutz's a first course in general relativity. In Chapter 4 it introduced the energy stress tensor in two ways: 1.) Dust grain 2.) Perfect fluid.

The book defined the energy stress tensor for dust grain to be $p⊗N$, where $p$ is the 4 momentum for a single dust grain, which writes $(m,0,0,0)$ in the rest frame (m is mass of a dust grain), and $N$ is the number density flux, which writes $(n,0,0,0)$ (n is scalar number density) in the rest frame. Thus
$T$ writes $((mn,0,0,0),(0,0,0,0),(0,0,0,0),(0,0,0,0))$ in the rest frame.
Such definition of energy stress tensor $T$ has a physical meaning that: $T^{ij}$ is the i-th component of 4 momentum density flux into j-th surface.

When we are dealing with fluid, which has random motion thus pressure, the book claim the energy stress tensor in the rest frame to be $((\rho,0,0,0),(0,P,0,0),(0,0,P,0),(0,0,0,P))$ where $\rho$ is energy density and P is the pressure. I understand the $\rho$ term but not the $P$ term.
I understand that the unit of a component of energy stress tensor is same as pressure, and the idea of momentum density flux matches the idea of pressure. However, using such expression, when we try to compute the x-momentum in x-direction, i.e. $T^{1,1}=P$, it does not have the meaning of "x-momentum density flux in x-direction", because such pressure will not lead to net bulk flow of momentum. The net flow of momentum is 0 as we are assuming the fluid in its equilibrium.

Can someone explain the "inconsistency"? Or the energy stress tensor has simply different physical definition for dust grain and fluid?

 

[Orodruin]

it does not have the meaning of "x-momentum density flux in x-direction",

It is exactly the meaning it has. That the fluid is in equilibrium just means the net change in momentum is zero. There are two ways momentum can transfer between volume through a surface. A flux carrying momentum with them or a force between the material on either side. (A force by definition is a change in momentum.)

 

[Ron19932017]

It is exactly the meaning it has. That the fluid is in equilibrium just means the net change in momentum is zero. There are two ways momentum can transfer between volume through a surface. A flux carrying momentum with them or a force between the material on either side. (A force by definition is a change in momentum.)

If the net momentum flux is zero, why do we write $T^{1,1}= P$, but not zero? For example an observer placing a plate in $dx$ orientation in a tank of perfect fluid will agree that momentum density flux measured is zero. I think it is natural to claim $T^{1,1}=0$.

 

[Dale]

Or the energy stress tensor has simply different physical definition for dust grain and fluid?

The stress energy tensor is physically different. A fluid has non zero pressure, a dust has zero pressure.

 

[Orodruin]

The stress energy tensor is physically different.

I would be careful with the statement here. The stress energy tensor still has the same physical interpretation. It contains the energy and momentum currents. Of course, a fluid is (generally) different from dust and so the stress energy tensor will take a different form and in that sense is physically different. It depends on what meaning the OP puts into "physically different". My interpretation was that the components of the stress energy tensor would have a different physical interpretation in the different cases.

If the net momentum flux is zero, why do we write $T^{1,1}= P$, but not zero? For example an observer placing a plate in $dx$ orientation in a tank of perfect fluid will agree that momentum density flux measured is zero. I think it is natural to claim $T^{1,1}=0$.

Because in a fluid under pressure you have transfer of momentum between different parts of the fluid and therefore a pressure. Note that there being transfer of momentum through a given surface does not mean that the momentum of the fluid changes as long as for any given volume the flux into the volume is the same as that out of the volume. Compare with a steady state heat flux through a metal bar between two reservoirs at different temperature. That the bar will take a stationary state does not mean that heat is not flowing through it, it just means that for any part of the bar the heat flow into it is the same as the heat flow out of it.

It is definitely not "natural" to claim $T^{11} = 0$ for a general fluid under pressure. This would mean that there is no momentum transfer, i.e., force, between the fluid on either side of the surface - which is kind of the definition of what pressure is.

Note that dust is just a special case of a perfect fluid where the pressure is zero. The approximation is that nothing is moving and that the dust particles are non-interacting. This leads to the momentum flux being zero.

 

[Ron19932017]

Because in a fluid under pressure you have transfer of momentum between different parts of the fluid and therefore a pressure. Note that there being transfer of momentum through a given surface does not mean that the momentum of the fluid changes as long as for any given volume the flux into the volume is the same as that out of the volume. Compare with a steady state heat flux through a metal bar between two reservoirs at different temperature. That the bar will take a stationary state does not mean that heat is not flowing through it, it just means that for any part of the bar the heat flow into it is the same as the heat flow out of it.

Consider an interval $(x,x+dx)$ in a 1-D tube of fluid in equilibrium. In your picture do you mean there is a momentum flux flow in at surface $x$, meanwhile there is a flow out at surface $x+dx$. Or do you mean at any surface (say $x$), there is both a flow in and flow out such that net flow in any surface is zero?

 

[Orodruin]

In your picture do you mean there is a momentum flux flow in at surface xxx, meanwhile there is a flow out at surface x+dxx+dxx+dx.

This is the case (assuming you are looking at the 1D case now). You need to look at the flow of momentum in and out from a volume (not only through a given surface) and consider the equilibrium conditions for that volume. In essence, the infinitesimal version of that equilibrium condition will involve the divergence of the stress tensor and the volume force.

Consider the case when you are hanging something from a rope and there is a tension in the rope. There will then be a continuous transfer of momentum from the roof, through the rope, to the hanging object. This momentum transfer exactly cancels the momentum transfer due to gravity on the object.

 

[Dale]

It depends on what meaning the OP puts into "physically different".

Yes, I agree. I meant it in the sense that the pressure in an inflated tire is different than the pressure in a deflated tire. They have the same meaning but different values.

 

[Ron19932017]

This is the case (assuming you are looking at the 1D case now). You need to look at the flow of momentum in and out from a volume (not only through a given surface) and consider the equilibrium conditions for that volume.
.

If the x-momentum is flowing in at surface $x$ and flowing out at surface $x+dx$, then there will be energy flowing in and out through these two surfaces too. The two energy flow balances each other and results in no change in energy density in the interval $(x,x+dx)$. Why does the stress energy tensor have the $T^{01}$ component zero?

 

[Orodruin]

then there will be energy flowing in and out through these two surfaces too.

No, this is incorrect. There is no necessity for these two to be related, momentum and energy are not the same thing. For example, if you look on your bookshelf, there is a transfer of momentum (i.e., force) from the shelf to the books that exactly balances the gravitational force on the books. This transfer of momentum does no work and therefore there is no transfer of energy from the bookshelf to the books.

When it comes to particles carrying momentum crossing the border between two volumes, consider a particle crossing in the positive direction. This particle will carry positive momentum in the positive direction, resulting in a net flux of momentum in the positive direction. If a particle crosses in the negative direction it carries negative momentum in the negative direction, resulting in the net flux being in the positive direction. If you look at the energy of these particles instead, both particles carry positive energy in opposite directions and therefore cancel out (assuming the same number of particles cross in each direction, which happens in the fluid rest frame - in other frames the net energy flux is not zero).