Dodelson Cosmology 4.8 Temperature of Nonrelativistic Matter

[StuckPhysicsStudent]

Homework Statement

Show that the temperature of non-relativistic matter scales as $a^{-2}$ in the absence of interactions. Start from the zero-order part of Eq. (4.68) and assume a form $f_{dm} \propto e^{-E/T}=e^{-p^2/2mT}$. Note that his argument does not apply to electrons and protons: as long as they are couple to the photons, their temperature scales as $a^{-1}$.

Homework Equations

Eq. (4.68)
$$\frac {\partial f_{dm}} {\partial t}+\frac{\hat p^i}{a}\frac{p}{E}\frac {\partial f_{dm}} {\partial x^i}-\frac {\partial f_{dm}} {\partial E} \bigg[ \frac{da/dt}{a}\frac{p^2}{E}+\frac{p^2}{E}\frac {\partial \Phi} {\partial t}+\frac{\hat p^i p}{a}\frac {\partial \Psi} {\partial x^i} \bigg]=0$$

The Attempt at a Solution

So I will use overdots to represent partial derivative with respect to time and primes for derivatives with respect to $x^i$. I'm assuming since $e^{-E/T}=e^{-p^2/2mT}$ that $E=p^2/2m$ and that p is a function of both $x^i$ and $t$.

Since this is first order I ignore the terms with $\Phi$ and $\Psi$ because they are perturbations to the metric. So 4.68 is just

$$\frac {\partial f_{dm}} {\partial t}+\frac{\hat p^i}{a}\frac{p}{E}\frac {\partial f_{dm}} {\partial x^i}-\frac {\partial f_{dm}} {\partial E} \bigg[ \frac{da/dt}{a}\frac{p^2}{E}\bigg]=0$$

So taking the derivatives of $f_{dm}$ and just sticking with $e^{-E/T}$ for now, we have

$$\frac{\partial f_{dm}} {\partial t}=-\frac{\dot{E}}{T} e^{-E/T}$$
$$\frac{\partial f_{dm}} {\partial x^i}=-\frac{E'}{T} e^{-E/T}$$
$$\frac{\partial f_{dm}} {\partial E}=-\frac{1}{T}e^{-E/T}$$

where all the $-\frac{1}{T}e^{-E/T}$ go away and I am left with no term for temperature. So I messed up somewhere because I need a T to remain.

Thanks for the help in advance.

 

[nrqed]

Homework Statement

Show that the temperature of non-relativistic matter scales as $a^{-2}$ in the absence of interactions. Start from the zero-order part of Eq. (4.68) and assume a form $f_{dm} \propto e^{-E/T}=e^{-p^2/2mT}$. Note that his argument does not apply to electrons and protons: as long as they are couple to the photons, their temperature scales as $a^{-1}$.

Homework Equations

Eq. (4.68)
$$\frac {\partial f_{dm}} {\partial t}+\frac{\hat p^i}{a}\frac{p}{E}\frac {\partial f_{dm}} {\partial x^i}-\frac {\partial f_{dm}} {\partial E} \bigg[ \frac{da/dt}{a}\frac{p^2}{E}+\frac{p^2}{E}\frac {\partial \Phi} {\partial t}+\frac{\hat p^i p}{a}\frac {\partial \Psi} {\partial x^i} \bigg]=0$$

The Attempt at a Solution

So I will use overdots to represent partial derivative with respect to time and primes for derivatives with respect to $x^i$. I'm assuming since $e^{-E/T}=e^{-p^2/2mT}$ that $E=p^2/2m$ and that p is a function of both $x^i$ and $t$.

Since this is first order I ignore the terms with $\Phi$ and $\Psi$ because they are perturbations to the metric. So 4.68 is just

$$\frac {\partial f_{dm}} {\partial t}+\frac{\hat p^i}{a}\frac{p}{E}\frac {\partial f_{dm}} {\partial x^i}-\frac {\partial f_{dm}} {\partial E} \bigg[ \frac{da/dt}{a}\frac{p^2}{E}\bigg]=0$$

So taking the derivatives of $f_{dm}$ and just sticking with $e^{-E/T}$ for now, we have

$$\frac{\partial f_{dm}} {\partial t}=-\frac{\dot{E}}{T} e^{-E/T}$$
$$\frac{\partial f_{dm}} {\partial x^i}=-\frac{E'}{T} e^{-E/T}$$
$$\frac{\partial f_{dm}} {\partial E}=-\frac{1}{T}e^{-E/T}$$

where all the $-\frac{1}{T}e^{-E/T}$ go away and I am left with no term for temperature. So I messed up somewhere because I need a T to remain.

Thanks for the help in advance.

You are assuming that T does not depend on time or on space. Why?

 

[StuckPhysicsStudent]

You are assuming that T does not depend on time or on space. Why?

Well, it must depend on time since it is dependent on a. I am unsure exactly how it would depend on space, but regardless if I assume T does depend on both I get

$$\frac{\partial f_{dm}}{\partial t}=-\frac{\dot{E}T-\dot{T}E}{T^2}e^{-E/T}$$
$$\frac{\partial f_{dm}}{\partial x^i}=-\frac{E'T-T'E}{T^2}e^{-E/T}$$

So that the equation becomes

$$0=-\frac{\dot{E}T-\dot{T}E}{T^2}e^{-E/T}-\frac{E'T-T'E}{T^2}e^{-E/T}\frac{p\hat{p^i}}{aE}+\frac{1}{T}e^{-E/T}\frac{\dot{a}}{a}\frac{p^2}{E}$$

Then dividing through by $e^{-E/T}$ and multipling by $T^2$ we get

$$0=-(\dot{E}T-\dot{T}E)-(E'T-T'E)\frac{p\hat{p^i}}{aE}+T\frac{\dot{a}}{a}\frac{p^2}{E}$$
$$=-\dot{E}T+\dot{T}E-E'T+T'E\frac{p\hat{p^i}}{aE}+T\frac{\dot{a}}{a}\frac{p^2}{E}$$
$$=T(\frac{\dot{a}}{a}\frac{p^2}{E}-\dot{E}-E') +E(\dot{T}+T'\frac{p\hat{p^i}}{aE})$$

Then using $E=p^2/2m$, $\dot{E}=2p\dot{p}/2m=p\dot{p}/m$, and $E'=2pp'/2m=pp'/m$

$$0=T(\frac{\dot{a}}{a}2m-\frac{p\dot{p}}{m}-\frac{pp'}{m}) +\frac{p^2}{2m}(\dot{T}+T'\frac{2m\hat{p^i}}{ap})$$

Now this is where I feel comfortable so far, the following I don't feel like it's right but I am going to post it anyways.

While I still not sure on the space dependence of T I am going to assume that it really doesn't matter for the following because it's a(t) not a(x,t) so we get

$$0=T(\frac{\dot{a}}{a}2m-\frac{p\dot{p}}{m}-\frac{pp'}{m}) +\frac{p^2}{2m}(\dot{T})$$

$$T(\frac{\dot{a}}{a}2m-\frac{p\dot{p}}{m}-\frac{pp'}{m}) =-\frac{p^2}{2m}\dot{T}$$

and using the knowledge (which we are suppose to be finding in the first place) that $T\propto a^{-2}$ so that $\dot{T}=-2\dot{a}/a^3$ so the above goes to

$$T(\frac{\dot{a}}{a}2m-\frac{p\dot{p}}{m}-\frac{pp'}{m}) =\frac{\dot{a}p^2}{a^3 m}$$

And then just ignoring everything beside T, a, and $\dot{a}$

$$T\frac{\dot{a}}{a}=\frac{\dot{a}}{a^3}\rightarrow T=a^{-2}$$

Which I don't think is the right method because I just set $-\frac{p\dot{p}}{m}-\frac{pp'}{m}=0$ and $2m=1$ (which I guess could work if $p\dot{p}$ and $pp'$ both equal zero, but I doubt that's the case).Thanks for the help so far, the time dependence was quite obvious once you pointed it out.