Deriving perfect fluid energy tensor from point particles

[mjordan2nd]

Homework Statement

For a system of discrete point particles the energy momentum takes the form

$$T_{\mu \nu} = \sum_a \frac{p_\mu^{(a)}p_\nu^{(a)}}{p^{0(a)}} \delta^{(3)}(\vec{x}-\vec{x}^{(a)}),$$

where the index a labels the different particles. Show that, for a dense collection of particles with isotropically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor.

Homework Equations

Energy-momentum tensor of a perfect fluid:

$$T^{\mu \nu} = (\rho + p)U^\mu U^\nu + p \eta^{\mu \nu}$$

Here rho is the rest-frame energy density, p the isotropic rest-frame pressure, and U the four-velocity.

The Attempt at a Solution

[/B]
I'm not really sure how to approach this problem. I would assume "smooth over" means average, so the only thing I can think of trying is

$$\Delta s = \int \sqrt{\eta_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d \! \lambda,$$

$$T_{\mu \nu} = \frac{1}{\Delta s} \int \sum_a \frac{p_\mu^{(a)}p_\nu^{(a)}}{p^{0(a)}} \delta^{(3)}(\vec{x}-\vec{x}^{(a)}) \sqrt{\eta_{\mu \nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d \! \lambda,$$

but I'm not sure how to proceed with this integral, or if this is even the right approach. Can someone help me figure out how to approach this problem?

 

[George Keeling]

I don't suppose @mjordan2nd is still worrying about this. But, just in case, I found a solution from ~2012 at https://physics.stackexchange.com/questions/21278/discrete-point-particles-stress-energy-tensor. I think it is a bit flakey towards the end and I can put my (slightly less flakey) solution up here if anybody is still interested. (When I edit physics.stackexchange I get rejected!)

 

[vanhees71]

What's missing is the fact that you have to average the microscopic expression over some "Gibbs ensemble". To get the perfect-fluid equations you have to assume that you are always in local thermal equilibrium, i.e., neglecting all dissipation effects like viscosity and/or heat conduction.

 

[George Keeling]

I'm not sure how

some "Gibbs ensemble"

helps...

First I restate the question, then I have reproduced Bobrick's solution along with a critique, then I suggest improvements but some parts are dubious so I would love some help!

Question
"For a system of discrete point particles the energy-momentum tensor takes the form
\begin{align}
T^{\mu\nu}=\sum_{a}{\frac{p^{\mu\left(a\right)}p^{\nu\left(a\right)}}{p^{0\left(a\right)}}\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\left(a\right)}\right)}&\phantom {10000}(1)\nonumber
\end{align}where the index $a$ labels the different particles. Show that, for a dense collection of particles with isotropically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor 1.114."

'Answer'
1.114 was
\begin{align}
X^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}&\phantom {10000}(2)\nonumber
\end{align}where $\rho$ is the energy density of the fluid measured in its rest frame (rest-frame energy density) and $p$ is the isotropic rest-frame pressure. $X^{\mu\nu}$, which Carroll wrote $T^{\mu\nu}$, is the energy-momentum tensor of a fluid element with a four velocity $U^\mu$. Alexey Bobrick wrote it as $\mathbf{T}^{\mu\nu}$ which is hard in latex, so I use $X$.

We are going to average (smooth over world lines) (1) over a 4-volume $\Delta V_4$ which is small enough to be treated as a single 'fluid element', which is lots of actual particles. We get
\begin{align}
X^{\mu\nu}=\frac{1}{\Delta V_4}\int_{\Delta V_4}^{\ }{T^{\mu\nu}dV}=\frac{1}{\sqrt{-g}d^3x^idx^0}\int_{\Delta V_4}^{\ }{T^{\mu\nu}\sqrt{-g}d^3x^idx^0}&\phantom {10000}(3)\nonumber
\end{align}I believe that $dV=\sqrt{-g}d^3x^idx^0$ where $g$ is the determinant of the metric is a well known formula and in this case $\sqrt{-g}=1$. (The solution just states that $g$ is constant over the volume, which is small). Then, putting in the $T^{\mu\nu}$ from (1), we get
\begin{align}
X^{\mu\nu}=\frac{1}{d^3x^idx^0}\int_{\Delta V_4}^{\ }{\sum_{a}{\frac{p^{\mu\left(a\right)}p^{\nu\left(a\right)}}{p^{0\left(a\right)}}\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\left(a\right)}\right)}d^3x^idx^0}&\phantom {10000}(4)\nonumber
\end{align}now the summation over $a$ is only over particles in $\Delta V_4$ which we consider just having the same four momentum throughout. So we can take it out of the integration. This is questionable. Then there are no $x^0$ terms in the integration and we can take out the $dx^0$ part and cancel it. So we have
\begin{align}
X^{\mu\nu}=\frac{1}{d^3x^i}\sum_{a}\frac{p^{\mu\left(a\right)}p^{\nu\left(a\right)}}{p^{0\left(a\right)}}\int_{\Delta V_3}^{\ }{\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\left(a\right)}\right)d^3x^i}&\phantom {10000}(5)\nonumber
\end{align}the integral is now over the three volume. By definition $\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\left(a\right)}\right)=1$ at one point in the volume, so the whole integral is 1 and we simply have
\begin{align}
X^{\mu\nu}=\frac{1}{d^3x^i}\sum_{a}\frac{p^{\mu\left(a\right)}p^{\nu\left(a\right)}}{p^{0\left(a\right)}}&\phantom {10000}(6)\nonumber
\end{align}which we evaluate component by component. We'll also change $d^3x^i$ to $d^3x^k$ and remember that both just mean $dxdydz$ the volume of the fluid element.
\begin{align}
X^{00}&=\frac{1}{d^3x^k}\sum_{a} p^{0\left(a\right)}\equiv\rho&\phantom {10000}(7)\nonumber\\
X^{i0}&=\frac{1}{d^3x^k}\sum_{a} p^{i\left(a\right)}=0&\phantom {10000}(8)\nonumber
\end{align}"As the sum $\sum_{a} p^{i\left(a\right)}$ of 3-vectors is taken over a macroscopic [fluid element] volume, the result should result in a macroscopic 3-vector. However, if this vector was not zero, it would violate isotropy, which states that there exists no preferable direction. Hence $X^{i0}=0$."
\begin{align}
X^{ij}&=\frac{1}{d^3x^k}\sum_{a}\frac{p^{i\left(a\right)}p^{j\left(a\right)}}{p^{0\left(a\right)}}&\phantom {10000}(9)\nonumber
\end{align}"As just before, the sum should produce a symmetric macroscopic 3-tensor of second order. But all symmetric 3-tensors are defined by 3 eigenvectors. If eigenvalues are non-degenerate, then there exists 3 preferred directions (3 eigenvectors), if eigenvalues are single degenerate, then there are 2 preferred directions etc. No preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to Kronecker delta. The coefficient of proportionality is the pressure so: $X^{ij}\equiv p\delta^{ij}$.

Expressing $\delta^{ij}$ as $\eta^{ij}+U^0U^0$ ($U^i$ is zero by symmetry considerations, and $U^0$ is hence equal to unity), and $X^{00}$ as $\rho U^0U^0$ , one arrives at the final expression for $T$."

I have copied Alexey Bobrick's answer for those three equations and it should be clear why we have used $d^3x^k$

(7) is OK: if we consider the rest frame of the fluid element. $p^{0\left(a\right)}$ is the energy of each particle so the sum is the total energy and when we divide that by the volume we get $\rho$ the rest energy density. The fluid element is at rest, the real particles are not.

(8) is OK in the fluid rest frame at least if we assume that all the particles have the same mass. We know that there is no net movement so the sum of the movement of the particles must cancel out and be zero. I'm not sure how that works if the particles have different masses. The expression for $X^{0i}$ is the same as that for $X^{i0}$ so we do not even need to invoke symmetry, which was manifest from (6).

(9) is trickier. First Bobrick concludes by stating that $\delta^{ij}=\eta^{ij}+U^0U^0$ which is clearly false. Perhaps they meant $\delta^{ij}=\eta^{ij}+U^iU^j$. I think that makes the rest of it work. The parenthetic statement "$U^i$ is zero by symmetry considerations, and $U^0$ is hence equal to unity" is also not true. Fluid elements which all satisfy that are not very interesting! It is true in the fluid element rest frame. Even more problematic is the sudden announcement that pressure $p$ is the constant of proportionality. Why?

Once again we must be thinking of the fluid element rest frame, then we have a chance of extracting $p$ which is the rest frame pressure and is caused by all the real particles jiggling about inside the fluid element. If we do it for the rest frame, it must be true in other frames so we are done.
If we think of just one (moving) particle for the moment (9) becomes
\begin{align}
\frac{1}{d^3x^k}\frac{p^ip^j}{p^0}=\frac{1}{d^3x^k}\frac{m\frac{dx^i}{d\tau}m\frac{dx^j}{d\tau}}{m\frac{dt}{d\tau}}=\frac{1}{d^3x^k}m\frac{dx^idx^j}{dtd\tau}=\frac{dx^j}{dx^i}\frac{1}{dx^kdx^l}m\frac{dx^i}{d\tau dt}&\phantom {10000}(10)\nonumber
\end{align}no summation over $i$. The last part of that is (almost) the pressure $q^i$ exerted in the $i$ direction (which has an area ${dx}^k{dx}^l$ where $k,l$ are the other two spatial coordinates) divided by the. So we might write that
\begin{align}
\frac{dx^j}{{dx}^i}q^i&\phantom {10000}(11)\nonumber
\end{align}if we could only say that the $d$ is like the partial derivative operator $\partial$ then we would have
\begin{align}
\frac{dx^j}{{dx}^i}q^i=\frac{\partial x^j}{\partial x^i}q^i=\delta_i^jq^i&\phantom {10000}(12)\nonumber
\end{align}(still no summation over $i$) which would be perfect. We could add up the pressures of all particles in the fluid element to get the total pressure so
\begin{align}
X^{ij}=\delta_i^jQ^i&\phantom {10000}(13)\nonumber
\end{align}where $Q^i$ is the pressure in the $i$ direction. But by assumption the pressure is uniform so it must be the same in all directions and we can call it $p$. Further assuming that that was done in the fluid element rest frame, we have the rest frame energy momentum tensor
\begin{align}
X^{\mu\nu}=\left[\begin{matrix}\rho&0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\\\end{matrix}\right]&\phantom {10000}(14)\nonumber
\end{align}and that of course, in the rest frame, is the same as
\begin{align}
X^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}&\phantom {10000}(15)\nonumber
\end{align}and since that is covariant it must also be true in any frame. We are done.

 

[vanhees71]

I think the entire problem formulation is dubious. Why should you get perfect-fluid dynamics from a very general formulation? It depends on the material, what you get. It could even be an elastic solid. Then you get something completely different.

Another very simple solution is simply "dust", i.e., matter where interactions are completely negligible. Then you get simply $T^{\mu \nu} = m c^2 n u^{\mu} u^{\nu}$, where $n$ is the particle-number density in the local rest frame (i.e., a scalar field) and $u^{\mu}$ the (normalized) four-velocity field of the medium.

The energy-momentum tensor for an ideal fluid can be derived by, e.g., considering an ideal gas in local thermal equilibrium, where
$$f(x,\vec{p})=\frac{g}{(2 \pi \hbar)^3} \exp[-u^{\mu}(x) p_{\mu}(x)/T(x)+\alpha(x)].$$
Then the energy-momentum tensor is evaluated as
$$T^{\mu \nu}(x)= \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{p_{\mu} p_{\nu}}{p^0} f(x,\vec{p}),$$
where $p^0=\sqrt{m^2 c^2+\vec{p}^2}$ is the on-shell energy.

 

[George Keeling]

I think the entire problem formulation is dubious.

I will let Carroll know and not spend any more time on this