Liouville operator in Robertson Walker metric

In summary, the Liouville operator in a Robertson Walker metric can be calculated using the general form of $$ \mathbb{L} = \dfrac{\text{d} x^\mu}{\text{d} \lambda} \dfrac{\partial}{\partial x^\mu} - \Gamma^{\mu}_{\nu \rho} p^{\nu} p^{\rho} \dfrac{\partial}{\partial p^\mu} $$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
  • #1
addaF
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0

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.
 
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  • #2
Welcome to PF
addaF said:
##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.
 
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Likes addaF
  • #3
TSny said:
Should there be a factor of ##m## in the right-hand side?

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

TSny said:
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.
 

Related to Liouville operator in Robertson Walker metric

1. What is the Liouville operator in Robertson Walker metric?

The Liouville operator in Robertson Walker metric is a mathematical operator used in quantum mechanics to describe the time evolution of a system in a curved spacetime. It is named after mathematician Joseph Liouville and physicist Arthur Geoffrey Walker.

2. How is the Liouville operator related to the Robertson Walker metric?

The Liouville operator is a key component of the Robertson Walker metric, which is a mathematical model used to describe the expansion of the universe in cosmology. The Liouville operator is used to calculate the time evolution of quantum mechanical systems within this expanding universe.

3. What is the significance of the Liouville operator in cosmology?

The Liouville operator plays a crucial role in understanding the behavior of quantum mechanical systems in an expanding universe. It helps us to understand how the expansion of the universe affects the time evolution of these systems and how they interact with the curved spacetime of the Robertson Walker metric.

4. How is the Liouville operator used in practical applications?

The Liouville operator is used in a variety of practical applications, such as in studying the behavior of quantum particles in the early universe, understanding the effects of cosmic inflation, and in developing quantum field theories in curved spacetime.

5. Are there any limitations or challenges associated with using the Liouville operator in Robertson Walker metric?

One limitation of using the Liouville operator in Robertson Walker metric is that it assumes a classical spacetime background, which may not be accurate in certain extreme situations, such as near black holes or during the very early stages of the universe. Additionally, the calculations involved in using the Liouville operator can be complex and require advanced mathematical techniques.

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