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
$$\frac{d\mathcal{P}}{dT} = \frac{1}{T}\frac{4}{3}\rho$$
$$= \frac{\rho + \mathcal{P}}{T}$$
Monday, May 12, 2014
Townsend - Problem 1.2
Chapter 1 - Stern Gerlach Experiments
Problem 1.2
Show for a solid spherical ball of mass $m$ rotating about an axis through its center with a charge $q$ uniformly distributed on the surface of the ball that the magnetic moment $\mu$ is related to the angular momentum $L$ by the relation
$$\mu = \frac{5q}{6mc}L$$
Solution
We are going to work in Gaussian units.
Recall that the magnetic moment is given by $$\mu = \frac{IA}{c}$$. Since we are dealing with uniformly charged sphere, we will cut up the sphere into infinitely many rings that are infinitely small in height. Without loss of generality we can cut them horizontally so that they are parallel to the x-axis.
Because the ball is rotating around the y-axis, there is an induced current. The current for each ring is given by $$I = \frac{\sigma dA}{T}$$, where $\sigma = \frac{q}{4\pi R^2}$ is the surface charge density for a sphere of radius $R$ and $T = \frac{2\pi}{\omega}$ is the period of the sphere's rotation.
Assuming the sphere has radius R, each ring has radius $r = Rsin\theta$ and thus $dA = 2\pi r Rd\theta$ is the area of each ring.
Thus $d\mu = \frac{dIA}{c} = \frac{dI}{c}\pi R^2sin^2\theta$.
Then to obtain $\mu$ of the whole sphere we need to integrate over all angles to get $$\mu = \int d\mu = \int \frac{\sigma dA}{Tc}\pi R^2sin^2\theta = \int\limits_0^{\pi} \frac{\sigma}{Tc}2{\pi}^2 R^4sin^3\theta d\theta \\ = \int\limits_0^{\pi} \frac{\pi q}{2Tc} R^2sin^3\theta d\theta \\ = \frac{q}{3c}\frac{2\pi}{T}R^2 \\ = \frac{q}{3c}\omega R^2 $$
Since the angular momentum of sphere is given by $$L = \frac{2}{5}mR^2\omega$$ we finally get $$\mu = \frac{5q}{6mc}L$$ as required.
Problem 1.2
Show for a solid spherical ball of mass $m$ rotating about an axis through its center with a charge $q$ uniformly distributed on the surface of the ball that the magnetic moment $\mu$ is related to the angular momentum $L$ by the relation
$$\mu = \frac{5q}{6mc}L$$
Solution
We are going to work in Gaussian units.
Recall that the magnetic moment is given by $$\mu = \frac{IA}{c}$$. Since we are dealing with uniformly charged sphere, we will cut up the sphere into infinitely many rings that are infinitely small in height. Without loss of generality we can cut them horizontally so that they are parallel to the x-axis.
Because the ball is rotating around the y-axis, there is an induced current. The current for each ring is given by $$I = \frac{\sigma dA}{T}$$, where $\sigma = \frac{q}{4\pi R^2}$ is the surface charge density for a sphere of radius $R$ and $T = \frac{2\pi}{\omega}$ is the period of the sphere's rotation.
Assuming the sphere has radius R, each ring has radius $r = Rsin\theta$ and thus $dA = 2\pi r Rd\theta$ is the area of each ring.
Thus $d\mu = \frac{dIA}{c} = \frac{dI}{c}\pi R^2sin^2\theta$.
Then to obtain $\mu$ of the whole sphere we need to integrate over all angles to get $$\mu = \int d\mu = \int \frac{\sigma dA}{Tc}\pi R^2sin^2\theta = \int\limits_0^{\pi} \frac{\sigma}{Tc}2{\pi}^2 R^4sin^3\theta d\theta \\ = \int\limits_0^{\pi} \frac{\pi q}{2Tc} R^2sin^3\theta d\theta \\ = \frac{q}{3c}\frac{2\pi}{T}R^2 \\ = \frac{q}{3c}\omega R^2 $$
Since the angular momentum of sphere is given by $$L = \frac{2}{5}mR^2\omega$$ we finally get $$\mu = \frac{5q}{6mc}L$$ as required.
Wednesday, May 7, 2014
Townsend - Problem 1.1
Chapter 1 - Stern Gerlach Experiments
Problem 1.1
Determine the field gradient of a 50-cm-long Stern Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at $T = 1500 K$. Assume the detector is located 50 cm from the magnet. Note that the emitted atoms have average kinetic energy $2k_bT$.
Solution
We are interested in the field gradient necessary to produce a given pattern on the detector, hence we need to find $\frac{\partial B}{\partial z}$.
From equation 1.4 we have that the force due to a Stern-Gerlach magnet is given by
$$F_z = \mu_z \frac{\partial B}{\partial z}$$,
where $\mu_z$ is the magnetic dipole moment of the particle given by
$$\mu_z = \frac{gq}{2mc}S_z$$ with $S_z$ being the intrinsic spin angular momentum of the particle in the z-direction.
For silver atoms, the magnetic dipole moment is due to the intrinsic spin of a single electron. Hence $g = 2$ and $q = e$, the charge of an electron.
Notice that, using symmetry arguments, the displacement in the z-direction of the upper particle is the same as the displacement of the lower particle. Let this displacement be $z$. Then $d = 1 mm = 2z$.
Therefore, we can focus on the displacement in the z-direction of the upper particle.
While the particle is traveling through the magnet it experiences a force and thus accelerates in the z-direction. We have
$$a_z = \frac{F_z}/{M} = \frac{\mu_z}/{M} \frac{\partial B}{\partial z}$$ where $M$ is the mass of the silver atom.
When the magnet is not traveling through the magnet it does not accelerate in the z-direction.
Initially there is no velocity in the z-direction and we have that the displacement in the z-direction of the upper particle while it is traveling through the magnet is given by
$$z_1 = \frac{1}{2}a_zt^2 = \frac{1}{2}a_z \left(\frac{l}{v_x}\right)^2$$, where $v_x$ is the initial velocity of the particle in the x-direction.
Outside the magnet the displacement in the z-direction of the upper particle is given by:
$$z_2 = v_zt = v_z\frac{l}{v_x} = a_z\left(\frac{l}{v_x}\right)^2$$
because v_z is the velocity of the particle when it just left the magnet.
The total displacement is thus
$$z = z_1 + z_2 = \frac{3}{2}a_z\left(\frac{l}{v_x}\right)^2$$
We can solve for $v_x$ by making use of the fact that the particles are ejected with average kinetic energy $2k_bT$.
$$\frac{1}{2}M{v_x}^2 = 2k_bT$$
Thus ${v_x}^2 = \frac{4k_bT}{M}$ and we can solve
$$d = 2z = 3a_zl^2\frac{M}{4k_bT} = \frac{3}{4}\frac{\mu_z}{M} \frac{\partial B}{\partial z}\frac{4k_bTl^2}{M}$$
Finally, we rearrange for $\frac{\partial B}{\partial z}$ and plug in the values:
$$\frac{\partial B}{\partial z} = 1200 G/cm$$
Problem 1.1
Determine the field gradient of a 50-cm-long Stern Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at $T = 1500 K$. Assume the detector is located 50 cm from the magnet. Note that the emitted atoms have average kinetic energy $2k_bT$.
Solution
We are interested in the field gradient necessary to produce a given pattern on the detector, hence we need to find $\frac{\partial B}{\partial z}$.
From equation 1.4 we have that the force due to a Stern-Gerlach magnet is given by
$$F_z = \mu_z \frac{\partial B}{\partial z}$$,
where $\mu_z$ is the magnetic dipole moment of the particle given by
$$\mu_z = \frac{gq}{2mc}S_z$$ with $S_z$ being the intrinsic spin angular momentum of the particle in the z-direction.
For silver atoms, the magnetic dipole moment is due to the intrinsic spin of a single electron. Hence $g = 2$ and $q = e$, the charge of an electron.
Notice that, using symmetry arguments, the displacement in the z-direction of the upper particle is the same as the displacement of the lower particle. Let this displacement be $z$. Then $d = 1 mm = 2z$.
Therefore, we can focus on the displacement in the z-direction of the upper particle.
While the particle is traveling through the magnet it experiences a force and thus accelerates in the z-direction. We have
$$a_z = \frac{F_z}/{M} = \frac{\mu_z}/{M} \frac{\partial B}{\partial z}$$ where $M$ is the mass of the silver atom.
When the magnet is not traveling through the magnet it does not accelerate in the z-direction.
Initially there is no velocity in the z-direction and we have that the displacement in the z-direction of the upper particle while it is traveling through the magnet is given by
$$z_1 = \frac{1}{2}a_zt^2 = \frac{1}{2}a_z \left(\frac{l}{v_x}\right)^2$$, where $v_x$ is the initial velocity of the particle in the x-direction.
Outside the magnet the displacement in the z-direction of the upper particle is given by:
$$z_2 = v_zt = v_z\frac{l}{v_x} = a_z\left(\frac{l}{v_x}\right)^2$$
because v_z is the velocity of the particle when it just left the magnet.
The total displacement is thus
$$z = z_1 + z_2 = \frac{3}{2}a_z\left(\frac{l}{v_x}\right)^2$$
We can solve for $v_x$ by making use of the fact that the particles are ejected with average kinetic energy $2k_bT$.
$$\frac{1}{2}M{v_x}^2 = 2k_bT$$
Thus ${v_x}^2 = \frac{4k_bT}{M}$ and we can solve
$$d = 2z = 3a_zl^2\frac{M}{4k_bT} = \frac{3}{4}\frac{\mu_z}{M} \frac{\partial B}{\partial z}\frac{4k_bTl^2}{M}$$
Finally, we rearrange for $\frac{\partial B}{\partial z}$ and plug in the values:
- d = 0.1 cm
- l = 50 cm
- T = 1500
$$\frac{\partial B}{\partial z} = 1200 G/cm$$
Townsend - A Modern Approach to Quantum Mechanics
Here is an index of my solutions to problems in Townsend's "A Modern
Approach to Quantum Mechanics", 2nd edition. Please note that some
solutions might be incorrect. If you do spot an error, let me know about it
in the comments.
Chapter 1 - Stern Gerlach Experiments
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15
Chapter 1 - Stern Gerlach Experiments
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15
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