Chapter 2 - The Smooth, Expanding Universe
Exercise 4
Find how the energy of a massive, nonrelativistic particle changes as the universe expands. Recall that in the massless case we used the fact that $g_{\mu\nu}P^{\mu}P^{\nu} = 0$. In this case, it is equal not to zero, but to $-m^2$.
Solution
We consider the zeroth component of the geodesic equation to obtain $$\frac{d^2x^{\mu}}{d\lambda ^2} = - \Gamma_{\alpha\beta}^{\mu}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}$$
Recall that we define $\lambda$ via the energy-momentum four vector: $$P^{\alpha} = \frac{dx^{\alpha}}{d\lambda}$$ Thus we obtain $$\frac{d}{d\lambda} = \frac{dx^{0}}{d\lambda}\frac{d}{dx^0} = E \frac{d}{dt}$$ This enables us to rewrite the zeroth component of the geodesic equation: $$E \frac{dE}{dt} = - \Gamma_{ij}^{0}P^{i}P^{j}$$ $$= -\delta_{ij}a^2P^{i}P^{j}$$ Another important fact that we are employing is that for a massive particle we have $$g_{\mu\nu}P^{\mu}P^{\nu} = -E^2 + \delta_{ij}a^2P^{i}P^{j} = -m^2$$ and so we can, again, rewrite the geodesic equation to read $$E \frac{dE}{dt} = - \frac{\dot{a}}{a}(E^2 - m^2)$$ $$\frac{E}{E^2 - m^2}\frac{dE}{dt} = -\frac{1}{a}\frac{da}{dt}$$ where we can cancel the dt terms and integrate making use of a substitution $u = m^2 + E^2$ to obtain $$\frac{1}{2}ln(m^2 + E^2) = - ln(a) + c$$ $$m^2 + E^2 \propto a^{-2}$$ and finally $$E \propto \frac{1}{a}$$.
Tuesday, July 15, 2014
Friday, July 11, 2014
Dodelson - Problem 2.2
Chapter 2 - The Smooth, Expanding Universe
Exercise 2
Show that the geodesic equation gets the correct equations of motion for a particle traveling freely in two dimensions using polar coordinates. You can get the Christoffel symbols one of two ways (or both!) and then proceed to (b).
(a) Get the Christoffel symbol either directly from the term in brackets in Eq. (2.17) or from the 2D metric $$g_{ij} = \left( \begin{array} {cc} 1&0 \\ 0&r^2\\ \end{array} \right)$$ using Eq. (2.19). Show that the only nonzero Christoffel symbols are $$\Gamma_{12}^{2} = \Gamma_{21}^{1} = \frac{1}{r} ; \Gamma_{22}^{1} = -r$$ with 1,2 corresponding to r,$\theta$.
(b) Write down the two components of the geodesic equation using these Christoffel symbols. Show that these give the proper equations of motion for a particle traveling in a plane.
Solution
(a) First note that $$g^{ij} = g_{ij}^{-1} = \left( \begin{array} {cc} 1&0 \\ 0&\frac{1}{r^2}\\ \end{array} \right)$$ Now we make use of Eq. (2.19): $$ \Gamma_{\alpha\beta}^{\mu} = \frac{g^{\mu\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ We have only two variables, r and $\theta$, labelled as 1 and 2 respectively so $\mu$ can only take on values of 1 and 2. First we consider the case $\mu = 1$. Then $$ \Gamma_{\alpha\beta}^{1} = \frac{g^{1\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ $$ = \frac{g^{11}}{2} \left[ \frac{\delta g_{\alpha 1}}{\delta x^{\beta}} + \frac{\delta g_{\beta 1}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{1}} \right] $$ since $g^{12} = 0$. Also notice that the only non zero derivative is $\frac{\delta g_{22}}{\delta x^{1}}$ and so the above reduces to: $$ \Gamma_{22}^{1} = \frac{g^{11}}{2} \left[ - \frac{\delta g_{22}}{\delta x^{1}} \right] $$ $$ = - \frac{1}{2} 2r = -r$$ as required.
Following the same procedure as above except letting $\mu = 2$ we obtain $$ \Gamma_{\alpha\beta}^{2} = \frac{g^{22}}{2} \left[ \frac{\delta g_{\alpha 2}}{\delta x^{\beta}} + \frac{\delta g_{\beta 2}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{2}} \right] $$ The third term in the bracket is always zero and we can only get non-zero values if either $\alpha = 1$ and $\beta = 2$ or $\alpha = 2$ and $\beta = 1$. In either case we will get $$ \Gamma_{12}^{2} = \Gamma_{21}^{2} = \frac{g^{22}}{2} \left[\frac{\delta g_{22}}{\delta x^{1}} \right] = \frac{1}{r}$$ again as we expected.
(b) Recall the geodesic equation is given by $$ \frac{d^2x^l}{dt^2} + \Gamma^{l}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0 $$ This time we have two choices for l, again because we have two independent variables. The first equation of motion is going to be for $l = 1$ and $x^1 = r$. $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0$$ $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{22}\frac{d\theta}{dt}\frac{d\theta}{dt} = 0$$ $$ \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 = 0$$ where we made use of the fact that $\Gamma^{1}_{22} = -r$ and $\Gamma^{1}_{12} = \Gamma^{1}_{21} = \Gamma^{1}_{11} = 0$. Thus we obtain the equation of motion for r: $$\ddot{r} - r\dot{\theta}^2 = 0$$ To obtain the second equation of motion we choose $l = 2$ from which $x^2 = \theta$ follows. Then we proceed as before to obtain: $$ \frac{d^2\theta}{dt^2} + \Gamma^{2}_{12}\frac{d\theta}{dt}\frac{dr}{dt} + \Gamma^{2}_{21}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ $$ \frac{d^2\theta}{dt^2} + \frac{2}{r}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ Thus we obtain the equation of motion for $\theta$: $$\ddot{\theta} + \frac{2}{r}\dot{\theta}\dot{r} = 0$$ The obtained equations are the equations of motion for a particle in a plane and our job is done.
Exercise 2
Show that the geodesic equation gets the correct equations of motion for a particle traveling freely in two dimensions using polar coordinates. You can get the Christoffel symbols one of two ways (or both!) and then proceed to (b).
(a) Get the Christoffel symbol either directly from the term in brackets in Eq. (2.17) or from the 2D metric $$g_{ij} = \left( \begin{array} {cc} 1&0 \\ 0&r^2\\ \end{array} \right)$$ using Eq. (2.19). Show that the only nonzero Christoffel symbols are $$\Gamma_{12}^{2} = \Gamma_{21}^{1} = \frac{1}{r} ; \Gamma_{22}^{1} = -r$$ with 1,2 corresponding to r,$\theta$.
(b) Write down the two components of the geodesic equation using these Christoffel symbols. Show that these give the proper equations of motion for a particle traveling in a plane.
Solution
(a) First note that $$g^{ij} = g_{ij}^{-1} = \left( \begin{array} {cc} 1&0 \\ 0&\frac{1}{r^2}\\ \end{array} \right)$$ Now we make use of Eq. (2.19): $$ \Gamma_{\alpha\beta}^{\mu} = \frac{g^{\mu\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ We have only two variables, r and $\theta$, labelled as 1 and 2 respectively so $\mu$ can only take on values of 1 and 2. First we consider the case $\mu = 1$. Then $$ \Gamma_{\alpha\beta}^{1} = \frac{g^{1\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ $$ = \frac{g^{11}}{2} \left[ \frac{\delta g_{\alpha 1}}{\delta x^{\beta}} + \frac{\delta g_{\beta 1}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{1}} \right] $$ since $g^{12} = 0$. Also notice that the only non zero derivative is $\frac{\delta g_{22}}{\delta x^{1}}$ and so the above reduces to: $$ \Gamma_{22}^{1} = \frac{g^{11}}{2} \left[ - \frac{\delta g_{22}}{\delta x^{1}} \right] $$ $$ = - \frac{1}{2} 2r = -r$$ as required.
Following the same procedure as above except letting $\mu = 2$ we obtain $$ \Gamma_{\alpha\beta}^{2} = \frac{g^{22}}{2} \left[ \frac{\delta g_{\alpha 2}}{\delta x^{\beta}} + \frac{\delta g_{\beta 2}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{2}} \right] $$ The third term in the bracket is always zero and we can only get non-zero values if either $\alpha = 1$ and $\beta = 2$ or $\alpha = 2$ and $\beta = 1$. In either case we will get $$ \Gamma_{12}^{2} = \Gamma_{21}^{2} = \frac{g^{22}}{2} \left[\frac{\delta g_{22}}{\delta x^{1}} \right] = \frac{1}{r}$$ again as we expected.
(b) Recall the geodesic equation is given by $$ \frac{d^2x^l}{dt^2} + \Gamma^{l}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0 $$ This time we have two choices for l, again because we have two independent variables. The first equation of motion is going to be for $l = 1$ and $x^1 = r$. $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0$$ $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{22}\frac{d\theta}{dt}\frac{d\theta}{dt} = 0$$ $$ \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 = 0$$ where we made use of the fact that $\Gamma^{1}_{22} = -r$ and $\Gamma^{1}_{12} = \Gamma^{1}_{21} = \Gamma^{1}_{11} = 0$. Thus we obtain the equation of motion for r: $$\ddot{r} - r\dot{\theta}^2 = 0$$ To obtain the second equation of motion we choose $l = 2$ from which $x^2 = \theta$ follows. Then we proceed as before to obtain: $$ \frac{d^2\theta}{dt^2} + \Gamma^{2}_{12}\frac{d\theta}{dt}\frac{dr}{dt} + \Gamma^{2}_{21}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ $$ \frac{d^2\theta}{dt^2} + \frac{2}{r}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ Thus we obtain the equation of motion for $\theta$: $$\ddot{\theta} + \frac{2}{r}\dot{\theta}\dot{r} = 0$$ The obtained equations are the equations of motion for a particle in a plane and our job is done.
Tuesday, July 8, 2014
Dodelson - Modern Cosmology
Here is an index of my solutions to problems in Dodelson's "Modern Cosmology". Please note that some
solutions might be incorrect. If you do spot an error, let me know about it
in the comments.
Chapter 1 - The Standard Model and Beyond
Chapter 2 - The Smooth, Expanding Universe
2.1 2.2 2.3 2.4 2.5 2.6 2.13 2.14 2.15
Chapter 3 - Beyond Equilibrium
3.1 3.2
Chapter 1 - The Standard Model and Beyond
Chapter 2 - The Smooth, Expanding Universe
2.1 2.2 2.3 2.4 2.5 2.6 2.13 2.14 2.15
Chapter 3 - Beyond Equilibrium
3.1 3.2
Monday, July 7, 2014
Dodelson - Problem 2.14
Chapter 2 - The Smooth, Expanding Universe
Exercise 14
(a) Compute the pressure of a relativistic species in equilibrium with temperature $T$. Show that $\mathcal{P} = \frac{\rho}{3}$ for both Fermi-Dirac and Bose-Einstein statistics.
(b) Suppose the distribution function depends only on E/T as it does in equilibrium. Find $d\mathcal{P}/dT$. A simple way to do this is to rewrite df/dT in the integral as -(E/T)df/dE and then integrate Eq. (2.62) by parts.
Solution
(a) Recall from Special Relativity that $pc = \sqrt{E^2 - m^2c^4}$. Thus in the relativistic limit where $E >> mc^2$ we have that $pc \approx E$. Since we are employing units where $c = 1$, we have that $$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{p^2}{3E(p)}$$
$$ = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{E(p)}{3}$$
$$ = \frac{\rho}{3}$$
as required.
(b) If the distribution function is function of E/T only, we get
$$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(E/T) \frac{p^2}{3E(p)}$$.
Since there is no angular dependence we can easily transform to spherical coordinates and write
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{p^4}{3E(p)}dp$$,
where we are integrating over all space. Changing the variable of integration to the energy
$$E^2 = p^2 +m^2$$
$$dp = \frac{E}{p}dE$$
and noting that we are in the relativistic limit where $p \approx E$, we obtain
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{E^3}{3}dE$$.
Differentiating with respect to temperature under the integral sign yields
$$\frac{d\mathcal{P}}{dT} = g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta T} \frac{E^3}{3}dE$$
$$ = -g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta E} \frac{E^4}{3T}dE$$
where we made use of the fact that $\frac{df}{dT} = -(E/T)\frac{df}{dE}$.
Now we integrate by parts. Letting
$$u = E^4 -> du = 4E^3 dE$$
$$dv = \frac{df}{dE} dE -> v = f$$
we obtain
$$\frac{d\mathcal{P}}{dT} = \frac{g_i}{3T} \frac{4\pi}{(2\pi)^3} \left( E^4f \big{|}_{0}^{\infty} - \int_{0}^{\infty} E^4 f dE \right)$$
Notice that $E^4f \big{|}_{0}^{\infty} = 0$ and so we simply get
$$\frac{d\mathcal{P}}{dT} = \frac{1}{T}\frac{4}{3}\rho$$
$$= \frac{\rho + \mathcal{P}}{T}$$
Exercise 14
(a) Compute the pressure of a relativistic species in equilibrium with temperature $T$. Show that $\mathcal{P} = \frac{\rho}{3}$ for both Fermi-Dirac and Bose-Einstein statistics.
(b) Suppose the distribution function depends only on E/T as it does in equilibrium. Find $d\mathcal{P}/dT$. A simple way to do this is to rewrite df/dT in the integral as -(E/T)df/dE and then integrate Eq. (2.62) by parts.
Solution
(a) Recall from Special Relativity that $pc = \sqrt{E^2 - m^2c^4}$. Thus in the relativistic limit where $E >> mc^2$ we have that $pc \approx E$. Since we are employing units where $c = 1$, we have that $$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{p^2}{3E(p)}$$
$$ = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{E(p)}{3}$$
$$ = \frac{\rho}{3}$$
as required.
(b) If the distribution function is function of E/T only, we get
$$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(E/T) \frac{p^2}{3E(p)}$$.
Since there is no angular dependence we can easily transform to spherical coordinates and write
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{p^4}{3E(p)}dp$$,
where we are integrating over all space. Changing the variable of integration to the energy
$$E^2 = p^2 +m^2$$
$$dp = \frac{E}{p}dE$$
and noting that we are in the relativistic limit where $p \approx E$, we obtain
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{E^3}{3}dE$$.
Differentiating with respect to temperature under the integral sign yields
$$\frac{d\mathcal{P}}{dT} = g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta T} \frac{E^3}{3}dE$$
$$ = -g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta E} \frac{E^4}{3T}dE$$
where we made use of the fact that $\frac{df}{dT} = -(E/T)\frac{df}{dE}$.
Now we integrate by parts. Letting
$$u = E^4 -> du = 4E^3 dE$$
$$dv = \frac{df}{dE} dE -> v = f$$
we obtain
$$\frac{d\mathcal{P}}{dT} = \frac{g_i}{3T} \frac{4\pi}{(2\pi)^3} \left( E^4f \big{|}_{0}^{\infty} - \int_{0}^{\infty} E^4 f dE \right)$$
Notice that $E^4f \big{|}_{0}^{\infty} = 0$ and so we simply get
$$\frac{d\mathcal{P}}{dT} = \frac{1}{T}\frac{4}{3}\rho$$
$$= \frac{\rho + \mathcal{P}}{T}$$
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