I Flux of i-momentum: Is it Correct?

[epovo]

I am trying to understand the stress-energy tensor. Say you have a gas of particles moving with random direction, but all with mass $m$ and velocity $v$ in a frame where the centre of mass is at rest. The contribution to the flux of x-momentum of a particle moving with $\vec p = p \vec e_x$ would be the same as that of another with $\vec p = - p \vec e_x$ and the total would be $2pv / A$ , where A is the Area of the considered element of the gas. Is this correct?

 

[Pencilvester]

I am trying to understand the stress-energy tensor. Say you have a gas of particles moving with random direction, but all with mass $m$ and velocity $v$ in a frame where the centre of mass is at rest. The contribution to the flux of x-momentum of a particle moving with $\vec p = p \vec e_x$ would be the same as that of another with $\vec p = - p \vec e_x$ and the total would be $2pv / A$ , where A is the Area of the considered element of the gas. Is this correct?

If the center of mass of the gas is at rest in this frame, then there is no net momentum flux in any spatial direction. There is only pressure and net flux in the time direction, i.e. energy density.

 

[epovo]

I don't think that is true, Pencilvester

 

[PeterDonis]

I don't think that is true, Pencilvester

What don't you think is true? Everything @Pencilvester said looks correct to me.

The contribution to the flux of x-momentum of a particle moving with $\vec p = p \vec e_x$ would be the same as that of another with $\vec p = - p \vec e_x$

This seems obviously wrong since momentum is a vector quantity and vectors pointing in opposite directions cancel out.

the total would be $2pv / A$ , where A is the Area of the considered element of the gas.

What stress-energy tensor component do you think this corresponds to?

 

[epovo]

I am probably wrong (that's why I am asking!). But I thought $T^{11} = T^{22} = T^{33} = p$ where p is the pressure in a perfect fluid in the MCRF. The $2pv/A$ would correspond to $T^{11}$

 

[PeterDonis]

I thought $T^{11} = T^{22} = T^{33} = p$ where p is the pressure in a perfect fluid in the MCRF.

That is correct. But the momentum-based argument you used is not.

 

[epovo]

If the momentum cancels out, where does the pressure come from, if not from adding up the contribution from the particles traversing the x=constant surface?

 

[PeterDonis]

If the momentum cancels out, where does the pressure come from, if not from adding up the contribution from the particles traversing the x=constant surface?

Pressure does not come from the free motion of particles, which is what you were calculating. It comes from the momentum exchanged in collisions between particles. The usual way is to imagine a small parcel of fluid enclosed in a box that is held stationary, and figure out the momentum flux into the box walls as particles of fluid bounce off them. In an actual fluid the role of the box is played by the particles in neighboring parcels of fluid.

 

[epovo]

But if the two particles in my example collide and bounce back, isn't that situation indistinguishable from the one where each particle goes on its own way? Both represent the exact same transfer of momentum between the two boxes you alluded to.

 

[Pencilvester]

But if the two particles in my example collide and bounce back, isn't that situation indistinguishable from the one where each particle goes on its own way? Both represent the exact same transfer of momentum between the two boxes you alluded to.

Sure, but again, that describes pressure, which is something you can have with zero net momentum flux. Of course you can have individual particles passing through a given surface, but flux involves taking a total sum of vectors of particles passing through a given surface, so if you have two momentum vectors of equal magnitude pointing in opposite directions, then the sum is zero, i.e. no flux.

 

[Dale]

But if the two particles in my example collide and bounce back, isn't that situation indistinguishable from the one where each particle goes on its own way?

In the collide and bounce situation each particle has a change of momentum. In the go on its own way situation the particles do not change momentum. The net momentum is the same, but the change in momentum is different.

 

[epovo]

If the net momentum is zero for my two particles, bouncing or not, how is there a momentum flux to account for the non-zero component of the tensor? Because it's obviously not zero. $T^{11}$ is by definition x-momentum flux. According to Pencilvester it should be zero.

 

[epovo]

I found this exercise from Schutz:

4.22 Many physical systems may be idealized as collections of *noncolliding particles* (for example, black-body radiation, rarified plasmas, galaxies and globular clusters). By assuming that such a system has a random distribution of velocities at every point, with no bias in any direction in the MCRF, prove that the stress-energy tensor is that of a perfect fluid. If all particles have the same speed $v$ and mass $m$, express $p$ and $\rho$ as functions of $m$, $v$ and $n$. Show that a photon gas has $p=1/3\rho$.

 

[Pencilvester]

I found this exercise from Schutz:

4.22 Many physical systems may be idealized as collections of *noncolliding particles* (for example, black-body radiation, rarified plasmas, galaxies and globular clusters). By assuming that such a system has a random distribution of velocities at every point, with no bias in any direction in the MCRF, prove that the stress-energy tensor is that of a perfect fluid. If all particles have the same speed $v$ and mass $m$, express $p$ and $\rho$ as functions of $m$, $v$ and $n$. Show that a photon gas has $p=1/3\rho$.

Okay, now I’m confused. Schutz does say (pg. 96 of First Course in GR) that $T^{ij}$ is the flux of the i component of momentum across a surface of constant $x^j$, and seems to justify this by giving the example of a fluid element exerting a force on its adjacent fluid element.

I don’t see how the fact that there’s a force indicates that there is flux. If i=j, Schutz would say that’s the flux of the i component of momentum across a surface of constant $x^i$, which is essentially the same as the flux of the entire momentum vector across that surface, but as we’ve been saying, you can have pressure, or force, with no flux. Is Schutz just using the word “flux” very flippantly?

 

[epovo]

It does seem however that you can have pressure with no collisions of particles.

 

[Ibix]

It does seem however that you can have pressure with no collisions of particles.

Well - without mutual collisions in the bulk of the gas, certainly. In operational terms, however, pressure is the result of the collision of gas/liquid particles with some kind of sensor, and those most certainly do occur even in a collisionless gas. And one can generalise this to a notion of pressure at some random location in the bulk of the gas by imagining a hypothetical pressure sensor there.

My (limited) understanding is that, as Peter mentions in #8, if you cut a small cuboid out of the matter which is $dx^\mu$ on a side (I'm assuming we're using some orthonormal coordinate system here), the stress-energy tensor is what forces per unit area you have to provide on its faces so that the bulk of the matter doesn't change.

That works well for the space-space components. You can imagine putting a microscopic evacuated box into a room full of gas. The wall perpendicular to $dx^1$ must provide a force per unit area of $p$ in the $dx^1$ direction, or else the box shrinks and the volume of the gas increases slightly. A force through the $i$th face in the $i$th direction gives your $T^{ii}$ component. There can't be shear stresses in a gas, but you can imagine cutting a small cuboidal cavity into a stretched rubber band. Typically that would be deformed into a rhombohedron of some kind, and the $dx^1$ face would need to exert forces in the $dx^2$ and $dx^3$ directions in order to revert to its unstressed cuboid shape - and those are your $T^{ij}$, $i\neq j$, components.

This picture doesn't work so well for the $T^{\mu 0}$ components. It would include cutting a small break into the worldline of any matter passing though a small volume, which doesn't make sense. This is where the momentum flux explanation comes in, for me. The $T^{i0}$ components describe the amount of momentum crossing a surface due to the bulk motion of the matter in the $dx^i$ direction. In the rest frame of a material, this is zero - even in a gas where particles cross a surface, they carry equal momentum in opposite directions. But if the wind is blowing in the $dx^i$ direction, we shall have snow then momentum flows in that direction. And finally, the $T^{00}$ component is the energy density in the volume - remember that this is a 4d volume, so it's the energy associated with particles (all on timelike trajectories, of course) in the purely timelike direction.

I think that's all correct. I must say that GR sources (at least the ones I've looked at, and I'm conscious I haven't looked in MTW yet) seem to go into far more detail on the subject of curved spacetime than on the stress-energy tensor, so this is a bit of a personal view patched together over time. Do take with a slight pinch of salt...

 

[PeterDonis]

I don’t see how the fact that there’s a force indicates that there is flux.

Force is a flux of momentum. That's its definition.

More precisely, force is the time rate of change of momentum, so force per unit area is the flux (time rate of change per unit area) of momentum.

 

[Ibix]

More precisely, force is the time rate of change of momentum, so force per unit area is the flux (time rate of change per unit area) of momentum.

I meant to say something of this sort at the end of my last post - that the space-space stresses and strains are also momentum fluxes. In an orthonormal coordinate system like I was positing you are free to make a (physically meaningless but possibly mentally useful) distinction between space-space stresses/forces on the one hand and space-time and time-time momentum fluxes on the other. However, if you aren't working in a coordinate system where space and time are orthogonal then you can't really justify segregating the components that way, so not thinking of them in such segregated terms (even when you could) is probably a worthwhile goal in its own right.

 

[epovo]

imagining a hypothetical pressure sensor

I agree. So, going back to my original particle pair, a sensor would be hit by both particles, each contributing to the pressure in equal measure.

 

[Ibix]

I agree. So, going back to my original particle pair, a sensor would be hit by both particles, each contributing to the pressure in equal measure.

No, although I see where you are coming from. Thinking of my little excised cuboid, the second particle strikes the opposite face, so doesn't contribute to pressure on the face in the $dx^i$ direction.

Furthermore, if you adopted your reasoning, imagine a microscopic volume at the edge of the volume of gas. No particles strike the opposite face, so the pressure on the walls would be half the pressure in the bulk. That doesn't make an awful lot of sense - especially since any pressure sensor is effectively a boundary to the gas anyway.

 

[epovo]

the second particle strikes the opposite face

Why? Forget the cuboid. Just imagine a pressure sensor (which is placed perpendicular to the x-axis at any point inside the gas) being hit by both particles.
I don't think that the idealization works at the edges of the volume of gas. In fact, a gas of particles with random velocities would naturally expand.

 

[PeterDonis]

Just imagine a pressure sensor (which is placed perpendicular to the x-axis at any point inside the gas) being hit by both particles.

Yes, which means both particles have a collision and there is momentum transfer. That's what pressure comes from.

 

[Pencilvester]

Force is a flux of momentum. That's its definition.

More precisely, force is the time rate of change of momentum, so force per unit area is the flux (time rate of change per unit area) of momentum.

I thought that momentum flux looked at the momentum vectors of all particles passing through a given surface (from either side). I didn’t realize that the common definition of flux involves only looking at the particles coming from one side of it, as you would if you were considering a force on that surface. So without the qualifier “net” attached to it, the term momentum flux is really just pressure?

 

[PeterDonis]

I thought that momentum flux looked at the momentum vectors of all particles passing through a given surface (from either side).

Yes, but momentum flux is a different stress-energy tensor component from pressure. It is the components $T_{01}$, $T_{02}$, $T_{03}$. In the fluid's rest frame the momentum flux is zero, because, as has been pointed out, momentum vectors pointed in opposite directions cancel out.

If you want to look at pressure, then you have to look at momentum being exchanged in collisions, not just momentum that's flowing freely.

I didn’t realize that the common definition of flux involves only looking at the particles coming from one side of it

It doesn't. See above.

So without the qualifier “net” attached to it, the term momentum flux is really just pressure?

No. See above.

 

[epovo]

Yes, which means both particles have a collision and there is momentum transfer. That's what pressure comes from.

It does not mean that. The pressure is acted upon the (imaginary) pressure detector of unit magnitude. The particles do not have to collide. Pressure in a gas does not come from particles colliding with one another. Statistically, there will be collisions in a real situation, but that is not the source of pressure. Pressure comes from particles flying around. The more there are and the fastest they go, the more pressure you get.
You can have pressure and momentum flux without any collisions whatsoever.

 

[PeterDonis]

It does not mean that.

Well, since you already know what pressure is, we can just close this thread since you obviously don't have a question that needs to be answered, and you are not going to listen to those of us who have a different understanding of what pressure is anyway, even if we happen to be correct and you happen to be wrong.

Thread closed.