Liouville operator in Robertson Walker metric

[addaF]

Homework Statement

I'd like to calculate the form of Liouville operator in a Robertson Walker metric.

Homework Equations

The general form is
$$ \mathbb{L} = \dfrac{\text{d} x^\mu}{\text{d} \lambda} \dfrac{\partial}{\partial x^\mu} - \Gamma^{\mu}_{\nu \rho} p^{\nu} p^{\rho} \dfrac{\partial}{\partial p^\mu} $$
where $x^\mu$ are the coordinates, $\lambda$ is an affine parameter and $p^\mu = \frac{\text{d} x^\mu}{\text{d} \lambda}$ is the particle four momentum, such as $g_{\mu \nu} p^\mu p^\nu = m^2$.
Given the Robertson-Walker metric
$$
\text{ds}^2 = \text{d}t^2 - a^2(t) \left( \dfrac{\text{d}r^2}{1-kr^2} + r^2 \text{d} \theta^2 + r^2 \sin^2 \theta \text{d}\varphi^2 \right)
$$
I know that I should get the following expression, considering a mean distribution of the particles $f(E,t)$
$$ \mathbb{L}[f] = E \dfrac{\partial f}{\partial t} - \dfrac{\dot{a}}{a} \left| \vec{p} \right|^2 \dfrac{\partial f}{\partial E} $$
where $\frac{\dot{a}}{a} = H$ is the Hubble parameter and $\left| \vec{p} \right|$ is the modulus of the three momentum.

The Attempt at a Solution

I'm actually struggling a bit on this problem, but I do not really know where I'm making the mistake. Since the metric is diagonal, and the distribution $f$ do not depend on spatial components of $x^\mu$ and $p^\mu$, the only needed Christoffel connection components should be:
$$
\Gamma^{0}_{\nu \rho} = \dfrac{1}{2} g^{0 \sigma} \left( \partial_{\nu} g_{\rho \sigma} +\partial_{\rho} g_{\sigma \nu} - \partial_{\sigma} g_{\nu \rho} \right)
$$
again because the metric is diagonal, $\sigma = 0$ and the first two terms in the brackets are vanishing. Then I calculate
$$
\Gamma^{0}_{11} = - \dfrac{1}{2} g^{0 0} \partial_{0} g_{11} = \dfrac{a \dot{a}}{1 - k r^2}
$$
$$
\Gamma^{0}_{22} = - \dfrac{1}{2} g^{0 0} \partial_{0} g_{22} = a \dot{a} r^2
$$
$$
\Gamma^{0}_{33} = - \dfrac{1}{2} g^{0 0} \partial_{0} g_{33} = a \dot{a} r^2 \sin^2 \theta
$$
Now there are the problems: if i suppose that $p^\mu = \left(E,\vec{p} \right)$, the sum over $\nu, \rho$ gives
$$
\Gamma^{0}_{\nu \rho} p^\nu p^\rho= \dfrac{a \dot{a}}{1 - k r^2} p^1 p^1 + a \dot{a} r^2 p^2 p^2 + a \dot{a} r^2 \sin^2 \theta p^3 p^3 \neq H \left| \vec{p} \right|^2
$$
Since i cannot find the error, can anyone help me? Thanks in advance.

 

[TSny]

Welcome to PF

$p^\mu = \frac{\text{d} x^\mu}{\text{d} \lambda}$ is the particle four momentum

Should there be a factor of $m$ in the right-hand side?

Now there are the problems: if i suppose that $p^\mu = \left(E,\vec{p} \right)$, the sum over $\nu, \rho$ gives
$$
\Gamma^{0}_{\nu \rho} p^\nu p^\rho= \dfrac{a \dot{a}}{1 - k r^2} p^1 p^1 + a \dot{a} r^2 p^2 p^2 + a \dot{a} r^2 \sin^2 \theta p^3 p^3 \neq H \left| \vec{p} \right|^2
$$

I believe the middle expression will reduce to $H \left| \vec{p} \right|^2$. To see why, write out the expression $g_{\mu \nu} p^\mu p^\nu = m^2$ explicitly.

 

[addaF]

Should there be a factor of $m$ in the right-hand side?

Yes, of course I forgot the mass in the 4-momentum definition.

I believe the middle expression will reduce to $H \left| \vec{p} \right|^2$. To see why, write out the expression $g_{\mu \nu} p^\mu p^\nu = m^2$ explicitly.

I immediately realized it as soon as I read your suggestion. I was writing the 3-momentum modulus like in Minkowski metric.

Thank you very much for the reply.