Understanding Stress Energy Tensor of Fluids

[Silviu]

Hello! I am reading about stress energy tensor of a perfect fluid and I don't understand the $T^{ij}$ terms. They are defined to be the flux of i-th momentum through the j-th surface. Now you take a fluid element and in its momentary comoving reference frame (MCRF) you calculate these $T^{ij}$. In the book they look at the case where fluid elements have just perpendicular forces on each other. And from here they get that $T^{ij}$ is equal to the pressure, if i=j and 0 otherwise. Although it makes sense unit-wise, I am not sure I understand physically. In MCRF, the fluid element doesn't move in space, so how can you have a flux of momentum? Can someone clarify all this stuff for me? I am sure I am missing something important but I don't know what. Thank you!

 

[PeterDonis]

In MCRF, the fluid element doesn't move in space, so how can you have a flux of momentum?

The average motion of the fluid in the MCRF is zero, but that does not mean the individual atoms/molecules of the fluid aren't moving. They are, and their motion is what produces the pressure. There is no net momentum flux through the boundary of the fluid element, which is why there are no nonzero terms $T^{0i}$ (i.e., one index 0 and one index 1, 2, or 3).

 

[Silviu]

The average motion of the fluid in the MCRF is zero, but that does not mean the individual atoms/molecules of the fluid aren't moving. They are, and their motion is what produces the pressure. There is no net momentum flux through the boundary of the fluid element, which is why there are no nonzero terms $T^{0i}$ (i.e., one index 0 and one index 1, 2, or 3).

Sorry, I am not sure I understand. In the book it says that $T^{11}=T^{22}=T^{33}=p$ (pressure). Doesn't this mean I have a net flux of momentum $p^1$ across the surface of constant $x^1$ (and same for 2 and 3)? The book I use is A first course in general relativity by Bernard Schutz and this is discussed in section 4.5 in the second edition.

 

[Orodruin]

Sorry, I am not sure I understand. In the book it says that $T^{11}=T^{22}=T^{33}=p$ (pressure). Doesn't this mean I have a net flux of momentum $p^1$ across the surface of constant $x^1$ (and same for 2 and 3)? The book I use is A first course in general relativity by Bernard Schutz and this is discussed in section 4.5 in the second edition.

Yes you have a net flux of momentum through a surface. This does not prevent the situation from being static. You also have a net flow of momentum through a stationary rope under tension.

 

[Silviu]

Yes you have a net flux of momentum through a surface. This does not prevent the situation from being static. You also have a net flow of momentum through a stationary rope under tension.

But if it is static, the force in a given direction, should't be 0? And thus $\frac{dp^i}{dx^i}=0$?

 

[Orodruin]

Also note that exactly this issue was recently discussed here.

 

[PeterDonis]

In the book it says that $T^{11}=T^{22}=T^{33}=p$ (pressure). Doesn't this mean I have a net flux of momentum $p^1$ across the surface of constant $x^1$ (and same for 2 and 3)?

I misstated my previous response a bit. We have to carefully distinguish between "flux" of momentum and "density" of momentum.

Momentum density is what I think you are thinking of when you say "flux" of momentum. In a frame in which the fluid is not at rest, there is a net momentum density in some direction--the direction in which the fluid, on average, is moving. (It's a momentum density, not just momentum, because we're treating the fluid as a continuous medium.)

Momentum flux is pressure or stress, and, as @Orodruin says, it can be present even if the fluid is static (i.e., in a frame in which the fluid is at rest). As I said before, it arises from the individual motions and collisions of atoms/molecules in the fluid. It is something different from average motion of the fluid as a whole.

 

[Orodruin]

But if it is static, the force in a given direction, should't be 0? And thus $\frac{dp^i}{dx^i}=0$?

I strongly suggest reading the thread I linked to in post #6. Pay particular attention to the last post (#10) in that thread.

Momentum transfer is just a different name for force across the surface.

 

[pervect]

Sorry, I am not sure I understand. In the book it says that $T^{11}=T^{22}=T^{33}=p$ (pressure). Doesn't this mean I have a net flux of momentum $p^1$ across the surface of constant $x^1$ (and same for 2 and 3)? The book I use is A first course in general relativity by Bernard Schutz and this is discussed in section 4.5 in the second edition.

I don't have Schutz. But the following might help. First, let's talk about what a flux is. We'll use wikipedia. <<link>>. The defintion I'll end up using is the general mathematical one, that the flux of a vector field is the surface integral of the vector field.

Now it's a bit awkward to talk about the flux of $p^i$, so I will be more specific. We will let $x^1=x, x^2=y, x^3=z, and x^0=t$. Then instead of talking about the flux of $p^i$, we can talk about the specific example and generalize. The specific example we'll use is the flux of $p^1$=, which in our new notation we can say is the flux of the x component of the momentum, $p^x$.

The rate of transport of x-momentum is just $d(p^x)/dt$, where $p^x$ is the x-component of the momentum. But this is just the x-component of the force $F^x$ on the fluid element. So the rate of transport of x-momentum is just the x-component of the force.

In the integral formulation (see the second defintion in wiki) we say that the flux $\Phi$ = $\int \int_A F \cdot dA$, where A is an orientable surface represented by a vector. The orientability is important. On the left surface at $x=-\delta$, we have a negative force F and a negatrive vector-valued area element dA as a boundary. If this seems confusing, try doing a similar example from electromagnetism, and think about current as being "a flux of charge".

The product of the two negative numbers is a positive flux. On the right surface at $x=+\delta$, we have a postive force F and a positive vector-valued area element. So again we have a positive flux. The total flux is the sum of the flux across the left surface and the right surface, and since both numbers are positive, they add together.

So, quick summary.

The transport of momentum dp/dt is physically just a force.
We can review the formalism of flux from EM to help us with this application in GR. I suggest using charge as a flux of current, but there are other useful EM analogies.

The result is as the textbooks say, when careful attention is paid to signs. It's important that the surface be orientable, and it's particularly important to get the sign of the vector representation of the area correct.

The treatment above is really a 3-dimensional one, hopefully that will be more familiar to you than a 4-dimensional treatment, and sufficient to answer your question.

 

[Silviu]

I don't have Schutz. But the following might help. First, let's talk about what a flux is. We'll use wikipedia. <<link>>. The defintion I'll end up using is the general mathematical one, that the flux of a vector field is the surface integral of the vector field.

Now it's a bit awkward to talk about the flux of $p^i$, so I will be more specific. We will let $x^1=x, x^2=y, x^3=z, and x^0=t$. Then instead of talking about the flux of $p^i$, we can talk about the specific example and generalize. The specific example we'll use is the flux of $p^1$=, which in our new notation we can say is the flux of the x component of the momentum, $p^x$.

The rate of transport of x-momentum is just $d(p^x)/dt$, where $p^x$ is the x-component of the momentum. But this is just the x-component of the force $F^x$ on the fluid element. So the rate of transport of x-momentum is just the x-component of the force.

In the integral formulation (see the second defintion in wiki) we say that the flux $\Phi$ = $\int \int_A F \cdot dA$, where A is an orientable surface represented by a vector. The orientability is important. On the left surface at $x=-\delta$, we have a negative force F and a negatrive vector-valued area element dA as a boundary. If this seems confusing, try doing a similar example from electromagnetism, and think about current as being "a flux of charge".

The product of the two negative numbers is a positive flux. On the right surface at $x=+\delta$, we have a postive force F and a positive vector-valued area element. So again we have a positive flux. The total flux is the sum of the flux across the left surface and the right surface, and since both numbers are positive, they add together.

So, quick summary.

The transport of momentum dp/dt is physically just a force.
We can review the formalism of flux from EM to help us with this application in GR. I suggest using charge as a flux of current, but there are other useful EM analogies.

The result is as the textbooks say, when careful attention is paid to signs. It's important that the surface be orientable, and it's particularly important to get the sign of the vector representation of the area correct.

The treatment above is really a 3-dimensional one, hopefully that will be more familiar to you than a 4-dimensional treatment, and sufficient to answer your question.

Thank you for your detailed answer. It looks a bit more clear now, but I still have some little confusions. If you add up the forces on both sides, shouldn't the diagonal (in the case of only perpendicular forces) have 2p instead of p (I assume p is the pressure of only one of the walls of fluid element)? Also, I understand that the fluid element exerts a force and thus a net flow of momentum. But at the same time the neighbour fluid element will exert an equal opposite force. Thus the amount of momentum coming in is equal to the one going out. So in the end shouldn't it be 0? Mathematically speaking, in the MCRF, the energy 4-vector is $(mc^2,0,0,0)$ and the flux 4-vector is $n_0 U = (n_0c, 0, 0, 0)$. How can I get the pressure from these 2 vectors? Based on what you said, the flux of $p^x$ would be $\int \int_A p^x dA = \int \int_A p^x dydz$. But in MCRF, $p^x$ is 0, so how can the flux be non-zero. I guess I missunderstand a definition somewhere, but I am not sure where.

 

[Orodruin]

Thus the amount of momentum coming in is equal to the one going out.

This is ultimately the same argument that would lead to a horse not being able to pull a wagon because of Newton’s third law. The point is that you are computing either the force from the fluid to the left on the fluid to the right or vice versa depending on surface orientation (it is important that the surface is oriented!). These forces form a third law pair and are of equal magnitude and opposite sign.

You need to integrate the pressure times the appropriate area element. Not the momentum density.

 

[Silviu]

This is ultimately the same argument that would lead to a horse not being able to pull a wagon because of Newton’s third law. The point is that you are computing either the force from the fluid to the left on the fluid to the right or vice versa depending on surface orientation (it is important that the surface is oriented!). These forces form a third law pair and are of equal magnitude and opposite sign.

You need to integrate the pressure times the appropriate area element. Not the momentum density.

Could you please write it down formally please... Or at least tell me where I am wrong? As I see it, the 4 momentum is $(E_0,0,0,0)$ in MCRF. So $p_x=0$. When you do the flux and integrate $p_x$ over a surface, you should get 0, regardless of the surface. It is like finding the electric flux, in a region where E is 0. Of course, I am wrong somewhere. Is not the 4-momentum that? Or where exactly am I wrong? Could you please write down for me the 4-momentum and the 4 flux and extract from there the flux of $p_x$, let's say (in the case of perpendicular forces for simplicity)? I think I would understand it better if I see the calculations done like this. Thank you!

 

[Orodruin]

That's the point. You are not integrating $p_x$, you are integrating the pressure, not the momentum density.

The force across a surface element $d\vec S$ is given by $dF_i = \sigma_{ij} dS_j$, where $\sigma$ is the stress tensor, which is the spatial-spatial part of the stress energy tensor. In the case of a perfect fluid in its rest frame, the stress tensor is proportional to identity with the diagonal elements given by the pressure $P$ (not the momentum density!). The force across a surface $S$ is given by
$$
F_i =\int_S \sigma_{ij} dS_j = -\int_S P\, dS_i
$$
(note that this is the force from the side that the surface normal is pointing to on the side that the surface normal is pointing from, therefore the sign).

Now in 1 spatial dimension consider a static fluid of pressure $P$ and a surface element in the $x$-direction. For any surface (in the 1D case, point), the force on the fluid to the left of the point from the fluid to the right of the point is given by
$$
F = -P
$$
(note that the integration disappears, an N-1-dimensional surface for N = 1 is zero dimensional). However, if you consider a finite region between $x_1$ and $x_2$, then the total force on the region is the flux of momentum to the right at $x_2$, which is equal to -P, and the flux of momentum to the left at $x_1$, which is equal to P. All in all, there is a total non-zero momentum current but the fluid remains in equilibrium. For the N-dimensional static fluid, this follows from
$$
\vec F = -\oint_S P\, d\vec S = -\int_V \nabla P \, dV = 0,
$$
where $V$ is the volume being considered and the gradient of $P$ is zero (uniform pressure so it does not depend on the position).

 

[Orodruin]

To put the first part slightly differently: $p_x$ is the momentum density. It is the density in the continuity for momentum in the $x$-direction. Any conserved quantity follows the continuity equation
$$
\partial_t \rho + \partial_i j^i = \partial_\mu j^\mu = 0,
$$
where $\rho$ is the density of the quantity, $\vec j$ is its current, and $j^\mu$ is the 4-current $(\rho,\vec j)$. Now, the stress-energy tensor is actually a tensor containing four different 4-currents since it relates to the conserved 4-momentum. For fixed $\nu$, $T^{\nu\mu}$ is the 4-current of momentum in the $\nu$-direction. In particular, for $\nu = 0$ the 4-current contains the energy density in the 0-component and the energy current in the spatial components.

For momentum in the $x$-direction, the continuity equation becomes
$$
\partial_t p_x - \partial_i \sigma_{xi} = 0.
$$
Note that the momentum density appears as the density of momentum in the $x$-direction - not as a current. For a static situation, the density should not depend on time and therefore $\partial_t p_x = 0$. This leads to the requirement $\partial_i \sigma_{ix} = 0$ - or in other terms when $\sigma_{ij} = -P\delta_{ij}$, $\nabla P = 0$. Therefore, as long as the pressure is not position dependent, the momentum density will remain constant. In particular, if it is zero it will remain zero.

Edit: Also note that I have chosen to use $P$ for the pressure instead of $p$ in order to separate it from the momentum density.