Add first problem set

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_posts/2019-02-13-pset1.md

----
-title: Problem Set 1
----
-
-This will be my first problem set.
-
-$$
-\begin{equation*}
-  \sum_{i=0}^3 \left( x^2 + i! \right)
-\end{equation*}
-$$

_posts/2019-02-14-pset1.md

+---
+title: Problem Set 1
+---
+
+## 2.1
+
+### (a)
+
+How many atoms are there in a yoctomole?
+
+Yocto is the SI prefix for $$10^{-24}$$. So there are
+$$
+\frac{N_a}{\si{\mol}} \cdot 10^{-24}
+= \frac{6.022 \cdot 10^{23}}{\si{\mol}} \cdot 10^{-24}
+= 0.6022
+$$
+atoms in a yoctomole.
+
+### (b)
+
+How many seconds are there in a nanocentury? Is the value near that of any important constants?
+
+Nano is the SI prefix for $$10^{-9}$$.
+
+$$
+\num{1e-9} \si{century}
+\cdot \frac{100 \si{years}}{1 \si{century}}
+\cdot \frac{365 \si{days}}{1 \si{year}}
+\cdot \frac{24 \si{hours}}{1 \si{day}}
+\cdot \frac{60 \si{minutes}}{1 \si{hour}}
+\cdot \frac{60 \si{\s}}{1 \si{minute}}
+= 3.15 \si{\s}
+$$
+
+This is relatively close to $$\pi$$.
+
+## 2.2
+
+A large data storage system holds on the order of a petabyte. How tall would a 1 petabyte stack of
+CDs be? How does that compare to the height of a tall building?
+
+CDs are nominally $$1.2 \si{\mm}$$ thick, and store 650 megabytes. (Some actually store more because
+they write data in regions the Orange Book standard reserves for manufacturing slop.)
+
+$$
+10^{15} \si{bytes} \cdot \frac{\num{1.2e-3} \si{\m}}{\num{6.5e8} \si{bytes}}
+\cdot \frac{\si{km}}{10^3 \si{m}}
+= 1.8 \si{\km}
+$$
+
+Currently the tallest building in the world is $$0.8 \si{km}$$ (the Burj Khalifa). The Jeddah Tower,
+currently under construction, is intended to be $$1 \si{km}$$. So for the foreseeable future the
+stack of CDs would be roughly twice as tall as any other human-built structure.
+
+## 2.3
+
+If all the atoms in our universe were used to write an enormous binary number, using one atom per
+bit, what would that number be (converted to base 10)?
+
+There are about $$10^{80}$$ atoms in the observable universe. So if each of these was used to encode
+one bit, we'd have $$10^{80}$$ bits of information. If all of this was used to store one big
+unsigned integer, the largest representable value would be $$2^{10^{80}} - 1$$ (not that the 1 makes
+any noticeable difference). Every decimal digit encodes $$\log_2(10) \approx 3.32$$ bits of
+information, so $$10^{80}$$ bits is equivalent to $$10^{80} / \log_2(10) \approx \num{3e79}$$
+decimal digits of information. Thus the largest representable unsigned integer can also be written
+as $$10^\num{3e79}$$.
+
+## 2.4
+
+Compare the gravitational acceleration due to the mass of the Earth at its surface to that produced
+by a 1 kg mass at a distance of 1 m. Express their ratio in decibels.
+
+The acceleration due to gravity at the surface of the Earth is $$9.8\si{m.s^{-2}}$$, while that of a
+$$1\si{kg}$$ mass at $$1\si{m}$$ is $$G \cdot 1\si{kg} / 1\si{m^2} = \num{6.67e-11} \si{m.s^{-2}}$$.
+In decibels this is
+
+$$
+20 \cdot \log_{10} \frac{9.8 \si{m.s^{-2}}}{\num{6.67e-11} \si{m.s^{-2}}} = 223 \si{db}
+$$
+
+## 2.5
+
+### (a)
+
+Approximately estimate the chemical energy in a ton of TNT. You can assume that nitrogen is the
+primary component; think about what kind of energy is released in a chemical reaction, where it is
+stored, and how much there is.
+
+TNT is a chemical explosive, meaning its potential energy is stored in chemical bonds. Chemical
+bonds are held together by the electromagnetic force. Covalent bonds in particular result from
+overlapping orbitals that enable multiple atoms to share electrons. I don't know what compounds
+result from the combustion of TNT, but clearly the atoms must end up in lower energy states on
+average. So for a rough estimate, I'll compute the number of nitrogen atoms per ton of TNT, and
+assume we free around 1 eV per atom during combustion, since this is a typical amount of energy for
+an atomic excitation.
+
+$$
+1 \si{ton} \cdot \frac{\num{907.18e3} \si{g}}{\si{ton}}
+\cdot \frac{\si{mol}}{14 \si{g}}
+\cdot \frac{\num{6.022e23} \si{atoms}}{\si{mol}}
+\cdot \frac{1 \si{eV}}{\si{atom}}
+\cdot \frac{\num{1.6e-19} \si{J}}{\si{eV}}
+= \num{6e9} \si{J}
+$$
+
+This is off by a factor of about 1.5, which isn't so bad for not knowing what the reaction is.
+
+### (b)
+
+Estimate how much uranium would be needed to make a nuclear explosion equal to the energy in a
+chemical explosion in 10,000 tons of TNT (once again, think about where the energy is stored).
+
+Nuclear explosions utilize nuclear energy instead of atomic (electromagnetic) energy. So we're
+dealing with a typical energy of $$10^6 \si{eV}$$ rather than $$1 \si{eV}$$. On top of that, we're
+getting about that much energy per proton or neutron, rather than per atom. Most uranium has 238
+nucleons, but fissionable uranium has only 235. However we're penalized for atomic mass: uranium is
+much heavier than nitrogen so there are less atoms of it per ton. So overall uranium should be about
+$$10^6 \cdot 235 \cdot 14/235 = \num{1.4e7}$$ times as potent as TNT per unit mass. Thus $$10^4$$
+tons of TNT is roughly equivalent to $$10^4 / (\num{1.4e7}) = \num{7e-4}$$ tons of fissionable
+uranium, which is about 650 grams.
+
+### (c)
+
+Compare this to the *rest mass energy* $$E = mc^2$$ of that amount of material (Chapter 15), which
+gives the maximum amount of energy that could be liberated from it.
+
+$$ E = mc^2 = 0.65 \si{kg} \cdot (\num{3e8} \si{m/s})^2 = \num{5.8e16} \si{J} $$
+
+This is roughly one quarter as powerful as Tsar Bomba, the largest single bomb ever built, which had
+a 50 megaton = $$\num{2e17} \si{J}$$ yield. It weighted 27 megagrams, though not all of that was
+fissile material. Still, if we had that much antimatter, we could get
+
+$$
+E = mc^2 = 2 \cdot \num{27e6} \si{kg} \cdot (\num{3e8} \si{m/s})^2 = \num{5e24} \si{J}
+$$
+
+I incude the factor of 2 since the antimatter would convert an equivalent amount of matter to energy
+on contact.
+
+Let's compare this to the gravitational binding energy of the planet, since this is the final
+frontier of any serious arms race within a uniplanetary species. We can calculate this by finding
+the energy required to remove a thin shell of matter from the surface of a sphere (moving it
+infinitely far away), and integrating it from the center of the Earth to the surface. For each
+shell, the spherical ball of matter inside behaves like a point mass, so the binding energy is just
+$$G M m / r$$. Assume a constant density $$\rho$$, so that the mass of a sphere of radius $$r$$ is
+$$(4/3) \pi r^3 \rho$$, and the mass of the shell on top of it is $$4 \pi r^2 \rho \mathrm{d}r$$.
+Thus the binding energy of the shell is $$G ((4/3) \pi r^3 \rho) (4 \pi r^2 \rho \mathrm{d}r) / r$$.
+The total energy needed to unbind the Earth is
+
+$$
+\begin{align*}
+E &= \int_0^R \frac{16}{3} G \pi^2 \rho^2 r^4 \mathrm{d}r \\
+&= \frac{16}{3} G \pi^2 \rho^2 \left[ \frac{1}{5} r^5 \right]_0^R \\
+&= \frac{16}{15} G \pi^2 \rho^2 R^5 \\
+\end{align*}
+$$
+
+Since $$\rho$$ must be $$3 M/(4 \pi R^3)$$, we have
+
+$$
+\begin{align*}
+E &= \frac{3 G M^2}{5 R} \\
+&= \frac{3 \cdot (\num{6.67e-11} \si{m^3.kg^{-1}.s^{-2}})
+   \cdot (\num{6e24} \si{kg})^2}{5 \cdot (\num{6.4e6} \si{m})} \\
+&= \num{2e32} \si{J}
+\end{align*}
+$$
+
+So you'd need a lot more than an antimatter Tsar Bomba to turn the Earth to dust. In fact, you'd
+need about 3 exagrams of the stuff. Wikipedia tells me this is about the mass of all coal deposits
+on Earth. Luckily, we can't currently produce (and catch) more than billions of antiprotons, which
+puts us about 14 orders of magnitude away from creating a single gram. (Coincidentally this also
+makes it the world's most valuable substance by mass.)
+
+## 2.6
+
+### (a)
+
+What is the approximate de Broglie wavelength of a thrown baseball?
+
+A baseball weighs $$1.45^{-1} \si{kg}$$. A major league fastball is around 100 miles per hour, so
+160 km per hour, but I'll just go with 100 km per hour since we're not all professional pitchers.
+
+$$
+\lambda = \frac{h}{p}
+= \frac{\num{6.626e-34} \si{J.s}}{\num{1.45e-1} \si{kg} \cdot 10^2 \si{km/hour}}
+\cdot \frac{\si{hour}}{60 \si{minutes}} \cdot \frac{\si{minute}}{60 \si{s}}
+= \num{1.3e-35} \si{m}
+$$
+
+### (b)
+
+Of a molecule of nitrogen gas at room temperature and pressure? (This requires either the result of
+Section 3.4.2, or dimensional analysis.)
+
+By the equipartition theorem, there is $$\frac{1}{2} k T$$ energy in the x, y, and z components of
+the molecule's velocity. We know $$E = \frac{1}{2} m v^2$$, and $$p = m v$$, so $$p = \sqrt{m k
+T}$$. Since the x, y, and z components add in quadrature, a typical total momentum will be $$\sqrt{3
+m k T}$$. The mass of a molecule of N2 is
+
+$$
+2 \cdot \frac{14 \si{g}}{\si{mol}} \cdot \frac{\si{mol}}{\num{6.022e23}}
+= \num{4.6e-23} \si{g}
+= \num{4.6e-26} \si{kg}
+$$
+
+Thus
+
+$$
+\lambda = \frac{h}{p}
+= \frac{h}{\sqrt{3 m k T}}
+= \frac{\num{6.626e-34} \si{J.s}}{\sqrt{3 \cdot \num{4.6e-26} \si{kg}
+  \cdot \num{1.38e-23} \si{J/K} \cdot 300 \si{K}}}
+= \num{2.7e-11} \si{m}
+$$
+
+### (c)
+
+What is the typical distance between the molecules in this gas?
+
+The typical distance can be estimated based on the number of particles per volume. If the volume is
+$$V$$, then we can arrange $$n$$ particles on a cubic lattice with a spacing of $$(V / n)^{1/3}$$.
+By the ideal gas law, $$V/n = RT/P$$, so the typical distance is
+
+$$
+\left( \frac{V}{n} \right) ^ \frac{1}{3}
+= \left( \frac{R T}{P} \right) ^ \frac{1}{3}
+= \left( \frac{8.3 \si{J/mol/K \cdot 300 K}}{10^5 \si{Pa}}
+  \cdot \frac{\si{mol}}{\num{6.022e23}} \right) ^ \frac{1}{3}
+= \num{3.5e-9} \si{m}
+$$
+
+### (d)
+
+If the volume of the gas is kept constant as it is cooled, at what temperature does the wavelength
+become comparable to the distance between the molecules?
+
+The typical spacing depends on the volume of the gas and number of molecules, both of which are
+fixed in this scenario. So we just need to solve
+
+$$
+\num{3.5e-9} \si{m} = \lambda = \frac{h}{\sqrt{3 m k T}}
+\implies T = \frac{h^2}{3 m k \lambda^2}
+= \frac{(\num{6.626e-34} \si{J.s})^2}{3 \cdot \num{4.6e-26} \si{kg}
+  \cdot \num{1.38e-23} \si{J/K} \cdot (\num{3.5e-9} \si{m})^2}
+= 0.02 K
+$$
+
+## (2.7)
+
+### (a)
+
+The potential energy of a mass m a distance r from a mass $$M$$ is $$−GMm/r$$. What is the escape
+velocity required to climb out of that potential?
+
+Assuming small and slow enough masses that relativistic corrections are negligible, we just need to
+solve
+
+$$
+E = \frac{1}{2} m v^2 = \frac{G M m}{r} \implies v = \sqrt{\frac{2 G M}{r}}
+$$
+
+### (b)
+
+Since nothing can travel faster than the speed of light (Chapter 15), what is the radius within
+which nothing can escape from the mass?
+
+$$
+c = \sqrt{\frac{2 G M}{r}} \implies r = \frac{2 G M}{c^2}
+$$
+
+This derivation is very much not rigorous, since neither the gravitational potential nor kinetic
+energy is expressed relativistically (in fact there's no such thing as a gravitational potential in
+general relativity, since gravity isn't a force). But remarkably we do end up with the real
+Schwarzschild radius.
+
+### (c)
+
+If the rest energy of a mass $$M$$ is converted into a photon, what is its wavelength?
+
+$$
+E = M c^2 = \frac{h c}{\lambda} \implies \lambda = \frac{h}{M c}
+$$
+
+### (d)
+
+For what mass does its equivalent wavelength equal the size within which light cannot escape?
+
+$$
+\frac{h}{M c} = \frac{2 G M}{c^2} \implies M = \sqrt{\frac{h c}{2 G}} \approx \num{3.9e-8} \si{kg}
+$$
+
+### (e)
+
+What is the corresponding size?
+
+Plugging the above formula for $$M$$ into $$\lambda = h/(M c)$$, we find that
+
+$$
+\lambda = \sqrt{\frac{2 h G}{c^3}} \approx \num{5.7e-35} \si{m}
+$$
+
+### (f)
+
+What is the energy?
+
+$$
+E = \frac{hc}{\lambda} = \sqrt{\frac{h c^5}{2 G}} \approx \num{3.5e9} \si{J}
+$$
+
+### (g)
+
+What is the period?
+
+$$\tau = 1/f$$ and $$\lambda = c / f = c \tau$$, so
+
+$$
+\tau = \lambda / c = \sqrt{\frac{2 h G}{c ^ 5}} \approx \num{1.9e-43} \si{s}
+$$
+
+
+## 2.8
+
+Consider a pyramid of height H and a square base of side length L. A sphere is placed so that its
+center is at the center of the square at the base of the pyramid, and so that it is tangent to all
+of the edges of the pyramid (intersecting each edge at just one point).
+
+### (a)
+
+How high is the pyramid in terms of L?
+
+To fit within the square base, the radius of the sphere must be $$L / 2$$. Now imagine rotating the
+square base 90 degrees about one of its diagonals. Since the axis of rotation goes through the
+center of the sphere, the edges of the base remain tangent to the sphere. So the height of the
+pyramid is the same as the distance from the center of the sphere to any corner of the square,
+namely $$L / \sqrt{2}$$. (This could also be calculated using triangle similarities, but that's a
+lot more tedious.)
+
+### (b)
+
+What is the volume of the space common to the sphere and the pyramid?
+
+The sphere is tangent to three points on each wall of the pyramid (the midpoints of each edge). The
+intersection of each of these planes with the sphere is a circle, and the spherical caps on top of
+these circles are the portions outside the pyramid. So we can compute the volume of the intersection
+of the sphere and the pyramid by subtracting the volume of these caps from the volume of the (half)
+sphere.
+
+The volume of the caps can be computed via an integral over perpendicular distance from the planes,
+but to do so we need to know how far these planes are from the center of the sphere. If we align two
+of the base square's corners with the X and Y axes (so the base square is the set of points with L1
+norm of $$L/\sqrt{2}$$), and the tip of the pyramid along Z, then the normal vector of the planar
+wall in the first octant is $$(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$$. The vector from the center of
+the sphere to the X base corner is $$(L/\sqrt{2}, 0, 0)$$, so the distance from the center of the
+sphere to the plane is $$(L/\sqrt{2}, 0, 0) \cdot (1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}) =
+L/\sqrt{6}$$. (Again this could also be derived from triangle similarities.)
+
+Thus the volume of a single cap is
+
+$$
+\begin{align*}
+V_\text{cap} &= \int_\frac{L}{\sqrt{6}}^\frac{L}{2} \pi r^2 \mathrm{d}x \\
+&= \int_\frac{L}{\sqrt{6}}^\frac{L}{2} \pi \left( \frac{L^2}{4} - x^2 \right) \mathrm{d}x \\
+&= \left[ \frac{\pi L^2}{4} x \right]_\frac{L}{\sqrt{6}}^\frac{L}{2} - \left[ \frac{\pi x^3}{3} \right]_\frac{L}{\sqrt{6}}^\frac{L}{2} \\
+&= \pi L^3 \left( \frac{1}{12} - \frac{7}{36 \sqrt{6}}\right)
+\end{align*}
+$$
+
+And the volume shared by the sphere and pyramid is
+
+$$
+\frac{1}{2} \cdot V_\text{sphere} - 4 \cdot V_\text{cap} = \pi L^3 \left( \frac{7}{9 \sqrt{6}} - \frac{1}{4} \right)
+$$

psets.html

@@ -6,7 +6,7 @@ title: Problem Sets
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+      <h2><a href="{{ post.url | real_relative_url }}">{{ post.title }}</a></h2>
       <p>{{ post.excerpt }}</p>
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