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# Welcome to Physics Pages

This blog consists of my notes and solutions to problems in various areas of mainstream physics. An index to the topics covered is contained in the links in the sidebar on the right, or in the menu at the top of the page.

This isn’t a “popular science” site, in that most posts use a fair bit of mathematics to explain their concepts. Thus this blog aims mainly to help those who are learning or reviewing physics in depth. More details on what the site contains and how to use it are in the Welcome menu above.

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# Equipartition theorem: qualitative treatment

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.23 – 1.25.

Each translational degree of freedom in a molecule has an associated average kinetic energy of ${\frac{1}{2}kT}$, so for the molecules in a gas the total kinetic energy due to translational motion is ${\frac{3}{2}kT}$. It turns out that, in general, any energy component with a quadratic dependence on its associated position or velocity contributes ${\frac{1}{2}kT}$ to the total thermal energy. This includes rotational degrees of freedom, since the energy is ${\frac{1}{2}I\omega_{i}^{2}}$, where ${I}$ is the moment of inertia about axis ${i}$ and ${\omega_{i}}$ is the angular velocity about that same axis. Chemical bonds also act approximately like springs, so two bonded atoms can vibrate relative to each other, giving rise to another kinetic energy term and a potential energy term given by ${\frac{1}{2}k_{x}x^{2}}$ where ${k_{x}}$ is the spring constant and ${x}$ is the molecular separation.

The total number of degrees of freedom ${f}$ for a given molecule can be tricky to calculate for more complex molecules, but once ${f}$ is found, the equipartition theorem says that the thermal energy of a collection of ${N}$ molecules is

$\displaystyle U_{thermal}=\frac{1}{2}kT\times Nf \ \ \ \ \ (1)$

Example 1 A monatomic gas such as helium is the simplest case, as it has only the three translational degrees of freedom, so for a collection of ${N}$ atoms

$\displaystyle U=\frac{3}{2}NkT \ \ \ \ \ (2)$

A litre (${10^{-3}\mbox{ m}^{3}}$) of helium at ${T=293\mbox{ K}}$ (room temperature) and a pressure of ${1\mbox{ atm}=1.01325\times10^{5}\mbox{ Pa}}$ has a thermal energy of

 $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{3}{2}NkT\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}PV\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 152\mbox{ J} \ \ \ \ \ (5)$

For air, if we neglect argon (which accounts for about 1% of air), all the molecules are diatomic (${\mbox{N}_{2}}$ and ${\mbox{O}_{2}}$). At room temperature, the vibrational modes don’t contribute, since these diatomic molecules have to be hit very hard to get a vibration going, so the only extra degrees of freedom are the two rotational degrees of freedom that we get by rotating a dumbbell shape about the two axes that are perpendicular to the bond joining the two atoms. (A rotation about the axis of the bond isn’t counted as such a rotation doesn’t change the position of the molecule.) So in this case ${f=5}$ and we have

$\displaystyle U=\frac{5}{2}NkT=\frac{5}{2}PV=253\mbox{ J} \ \ \ \ \ (6)$

Example 2 In an elemental solid such as lead, each atom has 3 translation degrees of freedom and 6 vibrational degrees of freedom (3 kinetic and 3 potential, as described above). There are no rotational degrees of freedom for a single atom. For 1 gram of lead, we have

$\displaystyle N=\frac{10^{-3}}{\left(207.2\mbox{ amu}\right)\left(1.66\times10^{-27}\mbox{ kg amu}^{-1}\right)}=2.91\times10^{21} \ \ \ \ \ (7)$

The thermal energy at ${T=293\mbox{ K}}$ is

$\displaystyle U=\frac{9}{2}NkT=2.61\mbox{ J} \ \ \ \ \ (8)$

Example 3 In a water molecule, the two hydrogen atoms are not in a straight line with the oxygen atom, so water has 3 rotational degrees of freedom in addition to the 3 translational degrees of freedom. [A linear molecule such as carbon dioxide has only 2 rotational degrees of freedom since the rotation about the O-C-O axis doesn’t count.] Each O-H bond can vibrate like a spring, so that would add 4 more degrees of freedom. The total number is therefore ${f=3+3+4=10}$. [Actually, come to think of it, there could be other vibrational modes as well. The H-O-H angle could change without the length of either H-O bond changing, so there’s probably another vibrational mode there. However, that’s not really a spring-type vibration, so I’m not sure if it counts in the equipartition theorem.]

# Effusion: gas leaking through a small hole

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.21 – 1.22.

Here are a few examples of calculating the pressure on a surface due to the collisions of a number of molecules (or other small objects) with that surface.

Example 1 For a macroscopic example, suppose there’s a hailstorm that sends hailstones of mass 2 g (actually, these would be very big hailstones: larger than ${2\mbox{ cm}^{3}}$. They’d probably break the window, but never mind) with a speed of 15 m/s at a ${45^{\circ}}$ angle into a window at a rate of ${N=30}$ per second. What average pressure do they exert on the window?

The horizontal velocity component is

$\displaystyle v_{x}=15\cos45^{\circ}=\frac{15}{\sqrt{2}}\mbox{m s}^{-1} \ \ \ \ \ (1)$

The horizontal momentum is reversed in each collision (assuming the collisions are elastic), so in one second, the total force exerted on the window is

$\displaystyle \Delta F=2mv_{x}N=2\times0.002\times\frac{15}{\sqrt{2}}\times30=1.275\mbox{ N} \ \ \ \ \ (2)$

The pressure on a window of area ${A=0.5\mbox{ m}^{2}}$ is therefore

$\displaystyle P=\frac{\Delta F}{A}=2.55\mbox{ N m}^{-2} \ \ \ \ \ (3)$

This is much smaller than atmospheric pressure, which is around ${10^{5}\mbox{ N m}^{-2}}$. Atmospheric pressure doesn’t break windows, of course, because it’s the same on both sides of the glass.

If there is a pressure difference on either side of a small hole, gas will leak out of the hole in a process called effusion. We can get an estimate of the time taken for gas to leak out of a hole using similar ideas to those above.

Suppose we have a container of volume ${V}$ that contains gas at initial pressure ${P}$ and temperature ${T}$. A small hole of area ${A}$ is made in the container. Suppose that the average ${x}$ component of the velocity of those molecules that are moving towards the hole is ${\overline{v_{x}}}$. Assuming elastic collisions, each molecule has its ${x}$ momentum of ${m\overline{v_{x}}}$ reversed, so in time ${\Delta t}$, the momentum change per unit time as a result of collisions with an area ${A}$ is the force, so

 $\displaystyle F$ $\displaystyle =$ $\displaystyle \frac{\Delta N\left(2m\overline{v_{x}}\right)}{\Delta t}=PA\ \ \ \ \ (4)$ $\displaystyle \Delta N$ $\displaystyle =$ $\displaystyle \frac{PA\Delta t}{2m\overline{v_{x}}} \ \ \ \ \ (5)$

We’ve seen from kinetic energy arguments that the mean square of ${v_{x}}$ is

$\displaystyle \overline{v_{x}^{2}}=\frac{kT}{m} \ \ \ \ \ (6)$

so we can use the rms of ${v_{x}}$ as an estimate for ${\overline{v_{x}}}$:

$\displaystyle \overline{v_{x}}\approx\sqrt{\overline{v_{x}^{2}}}=\sqrt{\frac{kT}{m}} \ \ \ \ \ (7)$

If the area ${A}$ is now made into the hole, then assuming that there is a vacuum on the other side of the hole, all the molecules that would have collided with the hole will now escape from the container, and no molecules will pass into the container. In that case, the change ${\Delta N}$ in the number of molecules in the container over time interval ${\Delta t}$ is (using the ideal gas law; the minus sign is because ${N}$ decreases)

 $\displaystyle \frac{\Delta N}{\Delta t}$ $\displaystyle =$ $\displaystyle -\frac{PA}{2m\overline{v_{x}}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\frac{PA}{2m}\sqrt{\frac{m}{kT}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{NkTA}{2mV}\sqrt{\frac{m}{kT}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{A}{2V}\sqrt{\frac{kT}{m}}N \ \ \ \ \ (11)$

Treating this as a differential equation, we can solve it to get

$\displaystyle N\left(t\right)=N\left(0\right)e^{-At\sqrt{kT}/2V\sqrt{m}} \ \ \ \ \ (12)$

so the characteristic time (after which ${N}$ drops to ${\frac{1}{e}}$ of its initial value) is

$\displaystyle \tau=\frac{2V}{A}\sqrt{\frac{m}{kT}} \ \ \ \ \ (13)$

Note that ${\tau}$ is independent of pressure, which may seem odd, since we would think that if the pressure inside the container is higher, the gas would escape faster. It is true that for a higher pressure, more gas will escape in a given time, but the fraction of the initial quantity of gas that escapes in a given time will be the same, regardless of the pressure.

Example 2 Suppose we have a 1 litre container at room temperature (in a vacuum) punctured by a ${1\mbox{ mm}^{2}}$ hole. The average mass of an air molecule is

$\displaystyle m=\frac{28.9697\times10^{-3}}{6.02\times10^{23}}=4.81\times10^{-26}\mbox{ kg} \ \ \ \ \ (14)$

The characteristic time is therefore

$\displaystyle \tau=\frac{2\times10^{-3}}{10^{-6}}\sqrt{\frac{4.81\times10^{-26}}{1.38\times10^{-23}\times293}}=6.9\mbox{ s} \ \ \ \ \ (15)$

Example 3 Puncture in a bicycle tire. A typical bicycle tire has a diameter of around 0.63 m and a cross-sectional diameter of about 0.03 m, so its volume is around

$\displaystyle V=2\pi\frac{0.63}{2}\times\pi\left(\frac{0.03}{2}\right)^{2}=1.4\times10^{-3}\mbox{ m}^{3}=1.4\mbox{ litres} \ \ \ \ \ (16)$

Suppose the tire goes flat after 1 hour, so that ${\tau=3600\mbox{ s}}$. How big is the hole?

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{2V}{\tau}\sqrt{\frac{m}{kT}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\times1.4\times10^{-3}}{3600}\sqrt{\frac{4.81\times10^{-26}}{1.38\times10^{-23}\times293}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.68\times10^{-9}\mbox{ m}^{2} \ \ \ \ \ (19)$

The hole has a diameter of

$\displaystyle d=2\sqrt{\frac{A}{\pi}}=0.058\mbox{ mm} \ \ \ \ \ (20)$

Example 4 Schroeder’s final example cites an episode from Jules Verne’s novel Around the Moon, in which space travelers eject a dead dog by opening a porthole in the spaceship, throwing the dog out and quickly closing the window again. This seems risky, but how much air would they actually lose in doing so?

Suppose it’s a small dog, so the porthole has a diameter of 30 cm, and the spaceship has a volume of ${V=100\mbox{ m}^{3}}$ (about the size of a smallish bedroom). The crew is efficient, so they can open the window, toss out the dog and close the window within 1 second. We get

 $\displaystyle A$ $\displaystyle =$ $\displaystyle \pi\left(\frac{0.3}{2}\right)^{2}=0.071\mbox{ m}^{2}\ \ \ \ \ (21)$ $\displaystyle \tau$ $\displaystyle =$ $\displaystyle \frac{2\times100}{0.071}\sqrt{\frac{4.81\times10^{-26}}{1.38\times10^{-23}\times293}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 9.72\mbox{ s} \ \ \ \ \ (23)$

With ${t=1\mbox{ s}}$, they would retain a fraction of their air given by

$\displaystyle \frac{N\left(1\right)}{N\left(0\right)}=e^{-1/9.72}=0.9 \ \ \ \ \ (24)$

so they’d lose about 10% of their air. They’d probably survive, but I think I’d rather use an airlock.

# Ideal gas: relation of average speed of molecules to temperature

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.18 – 1.20.

By using the ideal gas law and an analysis on the molecular scale, we can derive a relation between the speed of the molecules in an ideal gas and the temperature. Schroeder does this in detail in his section 1.2 so I won’t go through the whole derivation again; I’ll just summarize the main ideas.

The model places a single molecule inside a cylinder with a movable piston at one end. The molecule is moving at some speed ${v}$ and whenever it hits a wall or the piston it bounces off elastically. If the axis of the cylinder is the ${x}$ axis, then whenever the molecule hits the piston, the ${x}$ component of its velocity, ${v_{x}}$, is reversed, which means that its ${x}$ momentum changes by ${\Delta p_{x}=-2mv_{x}}$. The rate of change of momentum is the force exerted on the piston, and the force per unit area is the pressure, so we can relate ${v_{x}}$ to the pressure exerted on the piston. Since we’re considering only one molecule, the pressure is felt only at the moments when the molecule collides with the piston, so what we’re really interested in is the time average of the pressure. It turns out that this is

$\displaystyle \bar{P}=\frac{mv_{x}^{2}}{V} \ \ \ \ \ (1)$

where ${m}$ is the mass of the molecule and ${V}$ is the volume of the cylinder.

We can now extend the argument by putting a large number ${N}$ of molecules in the cylinder. In this case, the ${v_{x}^{2}}$ factor is replaced by the average of ${v_{x}^{2}}$ over all the molecules, so we get

$\displaystyle \bar{P}V=Nm\overline{v_{x}^{2}} \ \ \ \ \ (2)$

Comparing with the ideal gas law

$\displaystyle PV=NkT \ \ \ \ \ (3)$

we see that

$\displaystyle m\overline{v_{x}^{2}}=kT \ \ \ \ \ (4)$

or, since ${\frac{1}{2}m\overline{v_{x}^{2}}}$ is the kinetic energy from the ${x}$ motion of the molecules,

$\displaystyle \frac{1}{2}m\overline{v_{x}^{2}}=\frac{1}{2}kT \ \ \ \ \ (5)$

However, since the molecules are moving at random, there’s nothing special about the ${x}$ direction, so we’d expect the same contribution to the kinetic energy from the ${y}$ and ${z}$ directions, giving the relation

$\displaystyle \bar{K}_{trans}=\frac{1}{2}m\overline{v^{2}}=\frac{3}{2}kT \ \ \ \ \ (6)$

where ${\bar{K}_{trans}}$ is the average kinetic energy due to the translational motion of the molecules (if the molecules contain two or more atoms, then we can also have rotational and vibrational kinetic energy, so the total kinetic energy is greater than ${\frac{3}{2}kT}$).

A reasonable estimate of the average speed of molecules is the root mean square speed, defined as

$\displaystyle v_{rms}\equiv\sqrt{\overline{v^{2}}}=\sqrt{\frac{3kT}{m}} \ \ \ \ \ (7)$

Example 1 The molecules in a gas at room temperature are actually moving pretty fast. For example, for a nitrogen molecule (nitrogen makes up around 4/5 of the air) at room temperature (293 K), we have

 $\displaystyle m$ $\displaystyle =$ $\displaystyle 28.0134\mbox{ amu}=4.65\times10^{-26}\mbox{ kg}\ \ \ \ \ (8)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle 1.38\times10^{-23}\mbox{ m}^{2}\mbox{kg s}^{-2}\mbox{K}^{-1}\ \ \ \ \ (9)$ $\displaystyle v_{rms}$ $\displaystyle =$ $\displaystyle 510.7\mbox{ m s}^{-1} \ \ \ \ \ (10)$

Example 2 Consider a gas containing hydrogen and oxygen molecules in thermal equilibrium. The ratio of their rms speeds is

$\displaystyle \frac{v_{H}}{v_{O}}=\sqrt{\frac{m_{O}}{m_{H}}}=\sqrt{\frac{31.9988}{2.016}}=3.984 \ \ \ \ \ (11)$

where the masses are in amu. The hydrogen molecules are moving about 4 times faster than the oxygen molecules.

Example 3 To separate the two naturally occurring isotopes of uranium ${^{235}\mbox{U}}$ and ${^{238}\mbox{U}}$, the uranium is combined with fluorine to make uranium hexafluoride gas ${\mbox{UF}_{6}}$. The two isotopes will result in different rms speeds for the two types of molecules. We have

 $\displaystyle m_{235}$ $\displaystyle =$ $\displaystyle 235.04+6\times18.998\mbox{ amu}=5.794\times10^{-25}\mbox{ kg}\ \ \ \ \ (12)$ $\displaystyle m_{238}$ $\displaystyle =$ $\displaystyle 238.02891+6\times18.998\mbox{ amu}=5.843\times10^{-25}\mbox{ kg}\ \ \ \ \ (13)$ $\displaystyle v_{235}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3k\times293}{m_{235}}}=144.692\mbox{ m s}^{-1}\ \ \ \ \ (14)$ $\displaystyle v_{238}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3k\times293}{m_{238}}}=144.084\mbox{ m s}^{-1} \ \ \ \ \ (15)$

Thus the speed difference is quite small.

# Virial expansion for a gas

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.17.

The ideal gas law isn’t entirely accurate for any real gas. For low density gases, one way of accounting for deviations from the ideal gas law is to use a virial expansion:

$\displaystyle PV=nRT\left[1+\frac{B\left(T\right)}{\left(V/n\right)}+\frac{C\left(T\right)}{\left(V/n\right)^{2}}+\ldots\right] \ \ \ \ \ (1)$

where ${B}$ and ${C}$ are the virial coefficients, and depend on the particular gas we’re modelling. For nitrogen molecules ${\mbox{N}_{2}}$ the measured values of ${B}$ are

 ${T\mbox{ (K)}}$ ${B\mbox{ (m}^{3}/\mbox{mol})}$ ${V/n\left(\mbox{ m}^{3}/\mbox{mol}\right)}$ ${B\left(T\right)/\left(V/n\right)}$ 100 ${-160\times10^{-6}}$ 0.0082 ${-0.0195}$ 200 ${-35\times10^{-6}}$ 0.0164 ${-0.00213}$ 300 ${-4.2\times10^{-6}}$ 0.0246 ${-0.00017}$ 400 ${9.0\times10^{-6}}$ 0.0328 ${0.000274}$ 500 ${16.9\times10^{-6}}$ 0.0410 ${0.000412}$ 600 ${21.3\times10^{-6}}$ 0.0492 ${0.000433}$

Using the ideal gas law and the gas constant ${R=8.31\mbox{ J/K}}$ to get values for ${V/n}$ at each temperature gives the third column in the table, and then we can use these values to calculate the ${B/\left(V/n\right)}$ terms in the fourth column. [Note that I’ve converted Schroeder’s values to SI units.] The corrections are very small so the ideal gas law should work well under these conditions.

As to why ${B}$ is negative for low temperatures and positive for high temperatures, it is known that gas molecules feel a weak attraction when fairly close to each other. At low temperatures, the molecular speed is lower, so this attraction would have a chance to be more strongly felt. Thus the molecules would tend to be closer to each other than if they didn’t interact, resulting in a slightly smaller volume. A negative value of ${B}$ (at a given ${P}$ and ${T}$) means a smaller volume.

For higher temperatures, the molecules are moving too fast for this attraction to have any effect, so molecules simply bounce off each other. Because the molecules have a non-zero volume (as opposed to the point molecules assumed by the ideal gas law), a slightly larger volume is needed at a given (high) temperature and pressure.

Another equation of state (that is, a relation between ${P}$, ${V}$ and ${T}$) is the van der Waals equation:

$\displaystyle \left(P+\frac{an^{2}}{V^{2}}\right)\left(V-nb\right)=nRT \ \ \ \ \ (2)$

where the parameters ${a}$ and ${b}$ are constant for a given gas. To compare this to the virial expansion above, we can write this as

 $\displaystyle \left(P+\frac{an^{2}}{V^{2}}\right)\left(V-nb\right)$ $\displaystyle =$ $\displaystyle \left(P+\frac{an^{2}}{V^{2}}\right)V\left(1-\frac{nb}{V}\right)\ \ \ \ \ (3)$ $\displaystyle PV$ $\displaystyle =$ $\displaystyle nRT\left(1-\frac{nb}{V}\right)^{-1}-\frac{an^{2}}{V}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle nRT\left[\left(1-\frac{b}{V/n}\right)^{-1}-\frac{a}{RT\left(V/n\right)}\right] \ \ \ \ \ (5)$

By Taylor-expanding the first term in brackets, assuming ${bn/V\ll1}$, we get

 $\displaystyle PV$ $\displaystyle \approx$ $\displaystyle nRT\left[1+\frac{b}{V/n}+\frac{b^{2}}{\left(V/n\right)^{2}}-\frac{a}{RT\left(V/n\right)}\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle nRT\left[1+\frac{1}{V/n}\left(b-\frac{a}{RT}\right)+\frac{b^{2}}{\left(V/n\right)^{2}}\right] \ \ \ \ \ (7)$

Comparing with 1 we see that the van der Waals model predicts

 $\displaystyle B\left(T\right)$ $\displaystyle =$ $\displaystyle b-\frac{a}{RT}\ \ \ \ \ (8)$ $\displaystyle C\left(T\right)$ $\displaystyle =$ $\displaystyle b^{2} \ \ \ \ \ (9)$

By fitting the curve 8 to the data in the table above, we can get estimates for ${a}$ and ${b}$. I used Maple’s Fit function (which does a least squares fit), with the result:

 $\displaystyle a$ $\displaystyle =$ $\displaystyle 0.0219\mbox{ J m}^{3}\mbox{mol}^{-2}\ \ \ \ \ (10)$ $\displaystyle b$ $\displaystyle =$ $\displaystyle 6.42\times10^{-5}\mbox{m}^{3}\mbox{mol}^{-1} \ \ \ \ \ (11)$

Comparing the value of ${b}$ with the values of ${V/n}$ in the above table, we see that our assumption of ${bn/V\ll1}$ is consistent, so we’re safe.

The following plot illustrates how good the fit is:

The green curve is the van der Waals fit 8 and the red crosses are the data from the table above. [Note that ${B}$ is plotted using Schroeder’s units, which are just SI units multiplied by ${10^{6}}$.] The fit is actually fairly good, so the van der Waals equation is a decent model for these data.

# Barometric equation: the exponential atmosphere

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.16.

A model for the pressure in the atmosphere as a function of height can be estimated from the ideal gas law. Suppose we have a horizontal slab of air with thickness ${dz}$ and a unit cross sectional area. The density of air is a function ${\rho\left(z\right)}$ of height ${z}$ above sea level. The pressure of the air at height ${z}$ must be equal to the pressure of the air above it plus the weight of the air in the slab. In other words, the change in pressure as we go from height ${z}$ to ${z+dz}$ is just the weight of the air in the slab of thickness ${dz}$, so since pressure is force per unit area

$\displaystyle dP=-\rho g\; dz \ \ \ \ \ (1)$

where ${g}$ is the acceleration of gravity. The minus sign accounts for the fact that pressure decreases as we go higher.

The density of air can be written using the ideal gas law as

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \frac{Nm}{V}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{kT}P \ \ \ \ \ (3)$

where ${m}$ is the average mass of an air molecule, ${N}$ is the number of molecules in volume ${V}$ and ${T}$ is the temperature. We therefore get a differential equation for the pressure:

 $\displaystyle \frac{dP}{dz}$ $\displaystyle =$ $\displaystyle -\rho g\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{mg}{kT}P \ \ \ \ \ (5)$

This is called the barometric equation. In a realistic model, both ${g}$ and ${T}$ would depend on ${z}$, although since the atmosphere’s thickness isn’t really large enough to affect ${g}$ all that much, we can safely take it to be the usual ${g=9.8\mbox{ m s}^{-2}}$. The temperature does decrease substantially as we get higher, but if we want a crude model we can take it to be roughly constant. In that case, we can integrate the equation to get

 $\displaystyle \frac{dP}{P}$ $\displaystyle =$ $\displaystyle -\frac{mg}{kT}dz\ \ \ \ \ (6)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle P\left(0\right)e^{-mgz/kT} \ \ \ \ \ (7)$

so the pressure decreases exponentially with altitude. Further, from 3, we have

$\displaystyle \rho\left(z\right)=\frac{m}{kT}P\left(z\right)=\frac{m}{kT}P\left(0\right)e^{-mgz/kT}\equiv\rho\left(0\right)e^{-mgz/kT} \ \ \ \ \ (8)$

Assuming a temperature of ${288\mbox{ K}}$ (${15^{\circ}\mbox{ C}}$), and taking ${P\left(0\right)=1\mbox{ atm}}$, we get the following values for pressure at various altitudes:

 Altitude ${z}$ (m) ${P\left(z\right)}$ (atm) 1430 0.844 3090 0.693 4420 0.592 8850 0.350

The last value is the altitude at the top of Mount Everest and compares with the measured value of 0.333 atm (33.7 kilopascals), so the model isn’t too bad.

A graph of pressure versus altitude is as follows:

# Quantum versus classical mechanics in solids and gases

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.18.

The de Broglie wavelength of a particle, which is the wavelength of an idealized ‘free particle’ which has a precise momentum ${p}$ and thus a completely indeterminate position, is

$\displaystyle \lambda=\frac{h}{p} \ \ \ \ \ (1)$

In general, quantum mechanics is needed to describe systems in which the de Broglie wavelength of the constituent particles is larger than some characteristic size of the system itself. For example, if the wavelength of a free electron (that is, an electron not bound to a particular atom) in a solid is greater than the average spacing between atoms, then quantum mechanics is needed to describe these electrons. If the wavelength is much smaller than the size of the system, the wave nature of a particle isn’t noticeable and we can get away with using classical mechanics.

In statistical mechanics, the average energy of each particle in a system is ${\frac{1}{2}k_{B}T}$ per degree of freedom of the particle, where ${k_{B}}$ is Boltzmann’s constant and ${T}$ is the temperature in kelvins. For a single particle such as an electron, there are three degrees of freedom (one per coordinate direction) so its average energy is

$\displaystyle E=\frac{p^{2}}{2m}=\frac{3}{2}k_{B}T \ \ \ \ \ (2)$

Combining this with the definition of the de Broglie wavelength above, we get

$\displaystyle \lambda=\frac{h}{\sqrt{3mk_{B}T}} \ \ \ \ \ (3)$

so the condition for quantum mechanics to apply is that ${\lambda>d}$ where ${d}$ is the size of the system.

Example 1 Solids. Using the typical lattice spacing of ${d=3\times10^{-10}\mbox{ m}}$ for a solid, what is the maximum temperature at which we need to use quantum mechanics to describe free electrons in such a solid? For quantum mechanics to apply, we need

 $\displaystyle d$ $\displaystyle <$ $\displaystyle \frac{h}{\sqrt{3mk_{B}T}}\ \ \ \ \ (4)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle \frac{h^{2}}{3mk_{B}d^{2}} \ \ \ \ \ (5)$

We have the values (in SI units)

 $\displaystyle h$ $\displaystyle =$ $\displaystyle 6.626\times10^{-34}\ \ \ \ \ (6)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 9.1\times10^{-31}\ \ \ \ \ (7)$ $\displaystyle k_{B}$ $\displaystyle =$ $\displaystyle 1.38\times10^{-23}\ \ \ \ \ (8)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle 3\times10^{-10} \ \ \ \ \ (9)$

so we get

$\displaystyle T<1.29\times10^{5}\mbox{ K} \ \ \ \ \ (10)$

so electrons most definitely need to be described quantum mechanically.

For the atomic nuclei in a solid, the critical temperature is much lower. For sodium, with an atomic mass of about 23 atomic mass units (where ${1\mbox{ amu}=1.66\times10^{-27}\mbox{ kg}}$), we have

$\displaystyle T<3.1\mbox{ K} \ \ \ \ \ (11)$

Example 2 The ideal gas. The ideal gas law is

$\displaystyle PV=Nk_{B}T \ \ \ \ \ (12)$

where ${P}$ is the pressure, ${V}$ is the volume and ${N}$ is the number of gas molecules. We can get an estimate of ${d}$ by calculating the average volume per molecule ${v}$:

 $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{V}{N}\ \ \ \ \ (13)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle v^{1/3}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{k_{B}T}{P}\right)^{1/3} \ \ \ \ \ (15)$

Therefore, the condition for quantum mechanics to apply to an ideal gas is

 $\displaystyle \left(\frac{k_{B}T}{P}\right)^{1/3}$ $\displaystyle <$ $\displaystyle \frac{h}{\sqrt{3mk_{B}T}}\ \ \ \ \ (16)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle \frac{h^{6/5}P^{2/5}}{k_{B}\left(3m\right)^{3/5}} \ \ \ \ \ (17)$

For helium at atmospheric pressure we have

 $\displaystyle P$ $\displaystyle =$ $\displaystyle 10^{5}\mbox{N m}^{-2}\ \ \ \ \ (18)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 4\times\left(1.66\times10^{-27}\right)\mbox{ kg}\ \ \ \ \ (19)$ $\displaystyle T$ $\displaystyle <$ $\displaystyle 2.92\mbox{ K} \ \ \ \ \ (20)$

This is actually below the boiling point of helium (${4.55\mbox{ K}}$) so whenever helium is a gas, we don’t need quantum mechanics to describe it.

For hydrogen atoms (protons) in outer space, ${d=1\mbox{ cm}}$ and ${T=3\mbox{ K}}$. In this case, the critical temperature is given by 5 with ${m=1.66\times10^{-27}\mbox{ kg}}$:

$\displaystyle T<\frac{h^{2}}{3mk_{B}d^{2}}=6.4\times10^{-14}\mbox{ K} \ \ \ \ \ (21)$

Definitely no quantum mechanics needed here.

# Uncertainty principle: an example

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.17.

Here’s another example of calculating the uncertainty principle. We have a wave function defined as

$\displaystyle \Psi\left(x,0\right)=\begin{cases} A\left(a^{2}-x^{2}\right) & -a\le x\le a\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)$

The constant ${A}$ is determined by normalization in the usual way:

 $\displaystyle \int_{-a}^{a}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\left.\left(\frac{x^{5}}{5}-\frac{2}{3}a^{2}x^{3}+a^{4}x\right)\right|_{-a}^{a}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\frac{16a^{5}}{15}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\sqrt{15}}{4a^{5/2}} \ \ \ \ \ (5)$

The expectation value of ${x}$ is ${\left\langle x\right\rangle =0}$ from the symmetry of the wave function. The expectation value of ${p}$ is

 $\displaystyle \left\langle p\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\Psi^*\frac{\partial}{\partial x}\Psi dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\left(-2Ax\right)A\left(a^{2}-x^{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

[We can’t calculate ${\left\langle p\right\rangle =\frac{d}{dt}\left(m\left\langle x\right\rangle \right)}$ in this case, because we know the value of ${\left\langle x\right\rangle }$ only at one specific time (${t=0}$), so we don’t have enough information to calculate its derivative.]

The remaining statistics are (the integrals are all just integrals of polynomials, so nothing complicated):

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-a}^{a}\left|\Psi\right|^{2}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15}{16a^{5}}\left.\left(\frac{x^{7}}{7}-\frac{2}{5}a^{2}x^{5}+\frac{1}{3}a^{4}x^{3}\right)\right|_{-a}^{a}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a^{2}}{7}\ \ \ \ \ (11)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-a}^{a}\Psi^*\frac{\partial^{2}}{\partial x^{2}}\Psi dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15\hbar^{2}}{8a^{5}}\left.\left(a^{2}-x^{2}\right)\right|_{-a}^{a}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5\hbar^{2}}{2a^{2}}\ \ \ \ \ (14)$ $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{\sqrt{7}}\ \ \ \ \ (16)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{2}}\frac{\hbar}{a}\ \ \ \ \ (18)$ $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{14}}\hbar\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle 0.598\hbar>\frac{\hbar}{2} \ \ \ \ \ (20)$

Thus the uncertainty principle is satisfied in this case.

# Inner product of two wave functions is constant in time

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.16.

The fact that the normalization of the wave function is constant over time is actually a special case of a more general theorem, which is

$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}\Psi_{1}^*\Psi_{2}dx=0 \ \ \ \ \ (1)$

for any two normalizable solutions to the Schrödinger equation (with the same potential). The proof of this follows a similar derivation to that in section 1.4 of Griffiths’s book.

The derivative in the integrand is (where we’re using a subscript ${t}$ or ${x}$ to denote a derivative with respect that variable):

 $\displaystyle \frac{\partial}{\partial t}\left(\Psi_{1}^*\Psi_{2}\right)$ $\displaystyle =$ $\displaystyle \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t} \ \ \ \ \ (2)$

From the Schrödinger equation

 $\displaystyle \Psi_{2t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\Psi_{2xx}-\frac{i}{\hbar}V\Psi_{2}\ \ \ \ \ (3)$ $\displaystyle \Psi_{1t}^*$ $\displaystyle =$ $\displaystyle -i\frac{\hbar}{2m}\Psi_{1xx}^*+\frac{i}{\hbar}V\Psi_{1}^*\ \ \ \ \ (4)$ $\displaystyle \Psi_{1t}^*\Psi_{2}+\Psi_{1}^*\Psi_{2t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\left(-\Psi_{1xx}^*\Psi_{2}+\Psi_{2xx}\Psi_{1}^*\right)+\frac{i}{\hbar}V\left(\Psi_{1}^*\Psi_{2}-\Psi_{1}^*\Psi_{2}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi_{2x}\Psi_{1}^*-\Psi_{1x}^*\Psi_{2}\right) \ \ \ \ \ (6)$

Inserting this into 1 and integrating gives zero because all wave functions go to zero at infinity. [Of course, the theorem doesn’t hold if ${\Psi_{1}}$ and ${\Psi_{2}}$ are solutions for different potentials, because in that case the potential term wouldn’t cancel out in 5.]

# Unstable particles: a crude model

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.15.

A rather unrealistic way of modelling an unstable particle is to introduce an imaginary component to the potential. We can see this by modifying the proof given in Griffiths’s section 1.4 that, for a real potential, the normalization of the wave function is constant in time. We propose that

$\displaystyle V\left(x\right)=V_{0}\left(x\right)-i\Gamma \ \ \ \ \ (1)$

where ${V_{0}}$ is the ‘true’ potential and ${\Gamma}$ is a positive real constant.

The Schrödinger equation then says (where we’re using a subscript ${t}$ or ${x}$ to denote a derivative with respect that variable):

 $\displaystyle i\hbar\Psi_{t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\Psi_{xx}+V_{0}\Psi-i\Gamma\Psi\ \ \ \ \ (2)$ $\displaystyle -i\hbar\Psi_{t}^*$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\Psi_{xx}^*+V_{0}\Psi^*+i\Gamma\Psi^* \ \ \ \ \ (3)$

where the second line is the complex conjugate of the first.

Retaining the interpretation of the wave function as a probability of finding the particle at a given place and time, we can calculate the time derivative of the total probability of finding the particle anywhere:

 $\displaystyle \frac{dP}{dt}\equiv\frac{d}{dt}\int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\frac{\partial}{\partial t}\left|\Psi\right|^{2}dx \ \ \ \ \ (4)$

The derivative in the integrand is

 $\displaystyle \frac{\partial}{\partial t}\left|\Psi\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial t}\left(\Psi^*\Psi\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Psi_{t}^*\Psi+\Psi^*\Psi_{t} \ \ \ \ \ (6)$

From 2 and 3 we have

 $\displaystyle \Psi_{t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\Psi_{xx}-\frac{i}{\hbar}V_{0}\Psi-\frac{\Gamma}{\hbar}\Psi\ \ \ \ \ (7)$ $\displaystyle \Psi_{t}^*$ $\displaystyle =$ $\displaystyle -i\frac{\hbar}{2m}\Psi_{xx}^*+\frac{i}{\hbar}V_{0}\Psi^*-\frac{\Gamma}{\hbar}\Psi^*\ \ \ \ \ (8)$ $\displaystyle \Psi_{t}^*\Psi+\Psi^*\Psi_{t}$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\left(\Psi_{xx}\Psi^*-\Psi_{xx}^*\Psi\right)-2\frac{\Gamma}{\hbar}\Psi^*\Psi\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\frac{\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi_{x}\Psi^*-\Psi_{x}^*\Psi\right)-2\frac{\Gamma}{\hbar}\Psi^*\Psi \ \ \ \ \ (10)$

Putting this into 4 we can integrate the first term and get zero because ${\Psi\rightarrow0}$ as ${x\rightarrow\pm\infty}$ so we’re left with

$\displaystyle \frac{dP}{dt}=-2\frac{\Gamma}{\hbar}\int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx=-2\frac{\Gamma}{\hbar}P \ \ \ \ \ (11)$

We can solve this differential equation to get

 $\displaystyle \frac{dP}{P}$ $\displaystyle =$ $\displaystyle -2\frac{\Gamma}{\hbar}dt\ \ \ \ \ (12)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle P_{0}e^{-2\Gamma t/\hbar} \ \ \ \ \ (13)$

where ${P_{0}}$ is the probability of finding the particle at ${t=0}$. If we know the particle hasn’t decayed at ${t=0}$ then ${P_{0}=1}$.

The half-life of the particle is the time it takes for ${P}$ to be reduced to ${P_{0}/2}$, so

 $\displaystyle \ln0.5$ $\displaystyle =$ $\displaystyle -2\frac{\Gamma}{\hbar}t_{1/2}\ \ \ \ \ (14)$ $\displaystyle t_{1/2}$ $\displaystyle =$ $\displaystyle 0.347\frac{\Gamma}{\hbar} \ \ \ \ \ (15)$

# Continuous probability distribution: needle on a pivot

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problems 1.11-12.

This exercise in continuous probability distributions actually precedes the problem on Buffon’s needle, so it uses the same logic.

Suppose we have a needle mounted on a pivot so that the needle is free to swing anywhere in the top semicircle, so that when it comes to rest, its angular coordinate is equally likely to be any value between 0 and ${\pi}$. In that case, the probability density ${\rho\left(\theta\right)}$ is a constant in this range, and zero outside it. That is

$\displaystyle \rho\left(\theta\right)=\begin{cases} A & 0\le\theta\le\pi\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)$

From normalization, we must have

 $\displaystyle \int_{0}^{\pi}\rho\left(\theta\right)d\theta$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{\pi} \ \ \ \ \ (3)$

The statistics of the distribution are

 $\displaystyle \left\langle \theta\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\theta d\theta=\frac{\pi}{2}\ \ \ \ \ (4)$ $\displaystyle \left\langle \theta^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\theta^{2}d\theta=\frac{\pi^{2}}{3}\ \ \ \ \ (5)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle \theta^{2}\right\rangle -\left\langle \theta\right\rangle ^{2}}=\frac{\pi}{2\sqrt{3}}\ \ \ \ \ (6)$ $\displaystyle \left\langle \sin\theta\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\sin\theta d\theta=\frac{2}{\pi}\ \ \ \ \ (7)$ $\displaystyle \left\langle \cos\theta\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\cos\theta d\theta=0\ \ \ \ \ (8)$ $\displaystyle \left\langle \cos^{2}\theta\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\cos^{2}\theta d\theta=\frac{1}{2} \ \ \ \ \ (9)$

We now want the probability that the projection of the needle onto the ${x}$ axis lies between ${x}$ and ${x+dx}$. If the needle is at angle ${\theta}$, then its ${x}$ coordinate is ${r\cos\theta}$ (where ${r}$ is the length of the needle). If the angle changes by ${d\theta}$, its ${x}$ coordinate changes by ${dx=-r\sin\theta\; d\theta}$ so for the probability density, we take absolute values and get

 $\displaystyle \rho\left(\theta\right)d\theta$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\frac{dx}{r\sin\theta}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dx}{\pi y}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dx}{\pi\sqrt{r^{2}-x^{2}}}\ \ \ \ \ (12)$ $\displaystyle \rho\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\pi\sqrt{r^{2}-x^{2}}} \ \ \ \ \ (13)$

As a check:

 $\displaystyle \int_{-r}^{r}\rho\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{-r}^{r}\frac{dx}{\sqrt{r^{2}-x^{2}}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\left.\arctan\frac{x}{\sqrt{r^{2}-x^{2}}}\right|_{-r}^{r}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (16)$

Since ${x=r\cos\theta}$, we can get ${\left\langle x\right\rangle }$ and ${\left\langle x^{2}\right\rangle }$ from 8 and 9, but we can also calculate it the hard way, using ${\rho\left(x\right)}$:

 $\displaystyle \left\langle x\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-r}^{r}x\rho\left(x\right)dx\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{-r}^{r}\frac{x\; dx}{\sqrt{r^{2}-x^{2}}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-r}^{r}x^{2}\rho\left(x\right)dx\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{-r}^{r}\frac{x^{2}\; dx}{\sqrt{r^{2}-x^{2}}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\left.r^{2}\arctan\frac{x}{\sqrt{r^{2}-x^{2}}}-x\sqrt{r^{2}-x^{2}}\right|_{-r}^{r}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{r^{2}}{2} \ \ \ \ \ (23)$