My problem is the following: I'm trying to model a dust (pressure-less relativistic gas) in the presence of electromagnetic field using colisioness vlasov-equation (relativistic version of boltzmann equation). Please note that I'm in flat minkowski spacetime with signature $(+,-,-,-)$.

So, I have the following:

Colision-less Vlasov Equation:

$ p^\mu \partial_\mu f_k + q_k\left(p^0\vec E +\vec p \times \vec B\right)\cdot \frac{\partial f_k}{\partial \vec p} = 0$

Everything is in the mass shell, so, $p^0=\sqrt{\vec p^2+m^2}$

Now, I know that the current and the stress tensor are given by:

$j^\mu = m \int \frac{d^3\vec p}{p^0} p^\mu f_k(x^\mu,\vec p)$

$T^{\mu\nu} = m \int \frac{d^3\vec p}{p^0} p^\mu p^\nu f_k(x^\mu,\vec p)$

Using the vlasov equation directly I arrive at:

$\partial_\mu j^\mu = 0$

$\partial_\mu T^{\mu\nu} = q_k F^{\nu\mu}j_\mu$

And I also know that the stress tensor for dust is:

$T^{\mu\nu}= \rho u^\mu u^\nu$

where $\rho$ would be the mass density of the dust. I wanted to find the $f_k$, the probability density in phase space that would give me the above stress energy tensor and also would satisfy the vlasov-equation.

My guess was:

$f_k = \frac{p^0}{m^2} n_k(x^\mu) \delta (\vec p - \vec p_k(x^\mu))$

With $n_k$ being a proper number density in space, because, when I plug this in the definition of stress tensor, I arrive at:

$T^{\mu\nu}_k = \frac{n_k}{m} p^\mu p^\nu$

Which is pretty much what I was looking for. The problem starts when I try to get this to satisfy the vlasov equation, what I do is to derivate the above equation, separate the term in $\delta$ and the one in $\delta'$, and whatever end in each term equals to zero by itself. Doing this I arrive here:

$p^\mu_k \partial_\mu \vec p_k = q_k (p^0_k \vec E + \vec p_k \times \vec B)$

$p^\mu_k \partial_\mu n_k = -n_k q_k \frac{\vec p}{p^0} \cdot \vec E$

So, the first equation is ok, I can convert it to $p^\mu_k \partial_\mu p_k^\nu = q_k F^{\nu\mu}p_{k\ \mu}$ and I get a nice covariant equation for $p^\mu_k$. My headache is on the second one. The right hand side, as far as I could imagine, is not Lorentz invariant, and so spoils all the merit of my former guess.

Just to complicate things even more, when I try to derive $T^{\mu\nu}_k$ and I use the equations for $n_k$ and for $p^\mu_k$, I get the following:

$\partial_\mu T^{\mu\nu} = q_k n_k F^{\nu\mu} p_{k\ \mu} + n_k \left[\partial_\mu p^\mu_k - n_k q_k \frac{\vec p}{p^0} \cdot \vec E \right] p^\nu_k$

Which is what I wanted plus a trash in the end which I haven't found how to to get rid of, thus messing the original equation that I had in the first place. Also I have a similar situation when I try to calculate $\partial_\mu j^\mu_k$.

So, my questions are the following:

1) Do anyone know the right probability density in phase-space to recover the dust stress-tensor? If not, is there anything obviously wrong with my guess?

2) If my guess is reasonable, have I done any miss calculation along the way that would make me have the above problems? (What bothers me is not only that ugly non-covariant term but also the $\partial_\mu p^\mu_k$ which I don't have any idea how to deal with).


As for you distribution, I think it should be correct, because you can note $p^0=\frac{m}{\sqrt{1-v^2}}$, so your distribution is actually $$ f_k=\frac{1}{m}n'_k\delta(p-p_k), $$ where $n'_k$ is the particle density in your lab frame, which is (up to your normalization $m$ which is unclear to me) really the phase-space distribution of dust.

I think that a possible source of errors can be the following.

If you know that $$ \delta(p-p_k)f(p)+\delta'(p-p_k)g(p)=0 $$ it implies that \begin{align} g(p_k)=0\\ f(p_k)-g'(p_k)=0 \end{align} This is because $g(p)\delta'(p-p_k)\neq g(p_k)\delta'(p-p_k)$, but rather \begin{align} g(p)\delta'(p-p_k)=&\left(g(p)\delta(p-p_k))\right)'-g'(p)\delta(p-p_k)=\\ =&\left(g(p_k)\delta(p-p_k))\right)'-g'(p_k)\delta(p-p_k)=\\ =&g(p_k)\delta'(p-p_k)-g'(p_k)\delta(p-p_k) \end{align} where you now have the coefficients before linearly-independent functions as real numbers, not functions (this is what you need). The above is just the Leibniz rule, if you recall the definition of distributions (functions like $\delta$), and the definition of 'a distribution is zero'.

When you do your derivation, you should be carefull. You should keep the distinction between $p$ and $p_k$ up to the very end. Look. You have derived the first equation by using the composition rule, and then just set the coefficient before the $\delta'$ to zero. It is ok, but you either have to assume that $p_0$ in your definition of $f$ is a function of $p_k$ or add the $-g'(p_k)$ term to the second equation. I gather that you have not added this term. Then, in deriving the second equation we should consider $p_0$ to be a function of $p_k$ and thus to have derivatives wrt $x$. Lets see, lhs comes from differentiating $n_k$ wrt $x$, and rhs comes from differentiating $p_0$ wrt $p$. So, you do not follow one of the possibilities, hence the problems.

  • $\begingroup$ @user23873 let me know if I should be more specific $\endgroup$ May 19 '13 at 18:51
  • $\begingroup$ I'm redoing my calculations right now, you probably already helped me enough with this detail that I must admit I totally forgot. I'm having some problems with writing explicitly the vectorial $\delta'$ since it introduces a derivative that I believe that it's the divergent, but the calculation is getting very messy with it. If I need more help, I'll post here in 1h $\endgroup$
    – Hydro Guy
    May 19 '13 at 20:38
  • $\begingroup$ @user23873, there should be no divergences -- the definition of $\frac{\partial}{\partial p_i}\delta^3(\vec{p})$ is that $\int f(\vec{p})\frac{\partial}{\partial p_i}\delta^3(\vec{p})d^3p=-\frac{\partial}{\partial p_i}f(0)$, it is motivated by integration by parts. $\endgroup$ May 19 '13 at 20:59
  • $\begingroup$ but what I need is $\frac{\partial}{\partial \vec p}\delta^3(\vec p-\vec p_k)$. My question is how to properly evaluate $p^\mu \delta'(\vec p - \vec p_k)$ $\endgroup$
    – Hydro Guy
    May 19 '13 at 21:02
  • $\begingroup$ @user23873 yes, and $\frac{\partial}{\partial\vec{p}}$ of smth is a vector whose components are $\frac{\partial}{\partial p^i}$, so it is essentially what I wrote. In the answer I write delta prime just for brevity. $\endgroup$ May 19 '13 at 21:38

Relativistic kinetic theory is not that easy because you can't define what "t" stands for. One knows how to write the interaction between two particles in terms of retarded times and proper time, but beyond two interacting particles the question arises of what time do we have to use ? Poulain if I correctly remember has adressed this problem by extending the phase space dimension and by considering the proper time of each particle as a dynamical variable. Regarding the use of Vlasov's equation do not forget that E and B are both generated by the particles and thus are functional of fk. Vlasov equation is thus highly non linear. If you consider a neutral dust which does not generate any emf there is no need to bother with a Vlasov equation, just use Hamiltonian dynamics with delta distribution for each non interacting particles .

  • $\begingroup$ Thanks for the clarification, but still you haven't answered any of my questions. Please remember that I may (in this case I do) want to have other things also generating EM field. $\endgroup$
    – Hydro Guy
    May 14 '13 at 22:01

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