Blake Responses

Happy to help John!

So this substitution is aiming to change a somewhat tricky integrand $\frac{1}{\sqrt{x^2+a^2}}$ into one that the course has already talked about -- specifically one of the sides of these triangles

So the substitution wants to change $x^2+a^2$ into $u^2 + 1$ times a constant factor. The constant factor is just factored out of the integral and more-or-less ignored because it's not essential to understanding the physics at play. Also if you're just trying to solve an integral, the constant factor you factor out can just be multiplied by the integrated function at the end. 

The way to turn $x^2+a^2$ into $u^2+1$ times a constant factor is to be able to factor $a^2$ out. Let's try to do that.

$\begin{aligned} x^2 + a^2 =& a^2 \left( \frac{x^2}{a^2} + 1 \right) \\ &= a^2 \left( \left( \frac{x}{a} \right) ^2 + 1 \right) \\ & = a^2 \left( u^2 + 1 \right) \, \textrm{ when } u = \frac{x}{a} \end{aligned}$

OK so now we've made it into the form we wanted, $u^2+1$ times a constant factor, and we can also see what $u$ has to be -- it's $\frac{x}{a}$.

So to finish the substitution we need to apply it to $dx$ and the limits as well:

First the $dx$, this one's easy because the substitution is just a constant factor, so if

$u = \frac{x}{a}$

then

$du = \frac{dx}{a}$

and

$dx = a du$

Now the limits, which are $x=L$ and $x=-L$. To substitute these with $u$, we ask -- what is $u$ when $x=L$? Well, since $u = \frac{x}{a}$, these limits are:

When $x = L$, $u = \frac{L}{a}$, and when $x=-L$, $u = \frac{-L}{a}$.

Now we put all that together into the original integral:

$\int^{x=L}_{x=-L} \frac{dx}{\sqrt{x^2 + a^2}} = \int^{u=\frac{L}{a}}_{u=-\frac{L}{a}} \frac{a du}{\sqrt{ a^2 \left( u^2 + 1 \right) }}$

We can pull the \(a^2) out of the square root and it cancels out with the (a) on top!

\[ \int^{x=L}_{x=-L} \frac{dx}{\sqrt{x^2 + a^2}} = \int^{u=\frac{L}{a}}_{u=-\frac{L}{a}} \frac{du}{\sqrt{ u^2 + 1 }}\]

This was a walkthrough of how to figure out how to solve this type of integral, but doing constant factor substitutions is a really common way to make an integrand that looks kinda like one you know, into exactly one you know.

This all above is not what I thought about when you asked what the substitution was, though. That can be figured out just by looking at the points you circled. The limits are when $x$ is $L$ and $-L$, and the transformed limits are when $u$ is $\frac{L}{a}$ and $-\frac{L}{a}$, so then $u$ must just be whatever $x$ is divided by $a$; $u = \frac{x}{a}$. You can verify that this is actually the substitution done by putting $\frac{x}{a}$ in for $u$ in the denominator and seeing you get the original expression back when you multiply through by $a^2$.

This is a lot to go through, and hopefully you were able to follow everything above. The truth is the calculus you're playing with is pretty advanced and it seems to me like you need some more practice on the foundations of differential and integral calculus before lessons like "Integration in Physics II" are going to completely make sense. The part that comes next in the lesson is meant to combine everything you've learned in the entire course into one problem, it's tricky and the solutions will assume you're familiar with everything that came before.

Good luck!

Your wish is my command.

I'm going to be playing fast and loose with physics here, so bear with me. To start, everyone is correct in saying that once the teaspoon of neutron matter shows up on Earth, it's now longer got the gravitational pressure to hold all the neutrons together, so suddenly all the energy required to pack the neutrons nearly as close as a nucleus is suddenly going to be released. So let's find this energy.

Even a nucleus can be treated in a cartoon model as a non-interacting fermion gas, so I think this will be enough to at least approximate the confinement energy in neutron star matter, since it's really nothing more than a big neutron-rich nucleus held together by gravity. This brings with it a big caveat though: at the extremely high densities we're talking about here, the short-range nuclear interactions will begin to come into play, and that is a little bit above my pay grade.

Particle(s) in a box

We'll start with a particle in a box – a concept that's probably familiar to those who have taken introductory quantum mechanics. Disregarding the ground state energy, the kinetic energy of a single particle in a one-dimensional box is:

$E = \frac{\pi^2 n^2 h^2}{2 m L^2}$

Where $n$ is a positive integer (an indicator of a possible quantize energy level the particle can occupy – this is quantum, remember!), $h$ is Planck's constant, $m$ is the mass of the particle, and $L$ is the length scale of the box. We can already tell that as the box we squeeze the particle in gets smaller, the energy gets higher. This is the degeneracy pressure we're looking for! But we must go deeper.

Let's bump up this box to three dimensions, meaning we have to introduce two more positive integer energy level indices; one for each dimension ($n_x, n_y, n_z$). Each possible state has a unique combination of quantum numbers (the indices).

$E = \frac{\pi^2 h^2}{2 m L^2} (n_x^2 + n_y^2 + n_z^2)$

Okay so since we're dealing with neutrons (which are fermions), they follow Pauli's exclusion principle, and they can't occupy the same state. If there's only a single particle in this box, it'll be in the lowest energy state $n_x = n_y = n_z = 1$, but then when we put another one in it'll have to take a higher energy state (maybe $n_x = 2, n_y = n_z = 1$). As we put more particles in, the energy level quantum numbers are going to go up, and with it, the energy $E$. If we have to put in $N$ particles, how high is the energy going to go if we fill up the available energy levels starting at $n_x = n_y = n_z = 1$?

Energy level space

The way to model this as $N$ gets big is to treat the three quantum numbers as positions in some kind of "energy level space". The energy levels fill up, and the particles' energies depend on the norm $|n| = \sqrt{n_x^2 + n_y^2 + n_z^2}$. This can be treated as the Euclidean distance between the origin ($n_x = n_y = n_z = 0$) and the outer filled levels. But the quantum numbers can only be positive, so the available levels are kind of like the volume of the positive eighth of a sphere in "energy level space".

Imagine every possible combination of positive integers $n_x, n_y, n_z$ is a point in this corner of a sphere. As the sphere gets big, we can estimate the number of points by calculating its volume.

The volume of this corner of the "energy level sphere" is the number of possible energy states $N$ that could fit in a certain radius |n| = \sqrt{nx^2 + ny^2 + n_z^2}).

$N = \frac{1}{8} \frac{4}{3} \pi |n|^3$

We're forgetting one thing: spin. Fermions can occupy the same energy state as long as they've got different spins. Since neutrons are spin $\frac{1}{2}$ particles, they have two different spin states, so we can add a factor of $2$ to the total number of fermion states $N$:

$N = 2 \frac{1}{8} \frac{4}{3} \pi |n|^3$

We can isolate $|n|$:

$|n| = \left( \frac{3N}{\pi} \right)^{\frac{1}{3}}$

This is the magnitude of the quantum numbers $n_x, n_y, n_z$ that we've reached if we put $N$ fermions into our box. Since each fermion needs its own state, the norm $|n|$ can give us an estimate for the energy scale of the most energetic particles:

$\begin{aligned} E &= \frac{\pi^2 h^2}{m L^2} |n|^2 \\ &= \frac{\pi^2 h^2}{m L^2} \left( \frac{3N}{\pi} \right)^{\frac{2}{3}} \end{aligned}$

The Fermi Energy

Alright, so now we've got an equation for the energy of the particles when we've squeezed $N$ into a box of length $L$. We can switch out $L$ for $V^{\frac{1}{3}}$ to yield a number density, $\rho = \frac{N}{V}$:

$\begin{aligned} E &= \frac{h^2}{2m} \left( \frac{3 \pi ^2 N}{V} \right)^{\frac{2}{3}} \\ &= \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \end{aligned}$

We've got a pretty great estimate for the energy scale (per particle) of a bunch of fermions trapped in a box with density $\rho$, this is called the Fermi Energy $E_F$. To keep all our ducks in a row we could integrate over $N$ from low energy initial particles (with low quantum numbers) to super high energy final particles (with high quantum numbers) which would yield an energy per particle of $E = \frac{3}{5} E_F$ (thanks to the factor of N^{\frac{2}{3}. But what's a factor of $\frac{3}{5}$ between friends.

Will it destroy the Earth?

Alright! Now we can just do some googling to find the density of a neutron star: looks like $\approx \SI{10^{17}}{\kilo \gram \per \meter \cubed}$. The mass of a neutron is about $m_n = 1.66 \times 10^{-27} \textrm{kg}$, so the number density in a neutron star is $\rho \approx 10^{43} \frac{\textrm{neutrons}}{\textrm{m}^3}$. Wow, that's a lot.

Let's plug that into our energy:

$E = \frac{3}{5} \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \approx 3.5 \times 10^{-11} \textrm{J}$

So per neutron, we're talking about $3.5 \times 10^{-11} \textrm{J}$ of energy. That's about the energy released per Uranium atom during a nuclear explosion. If we give the neutron this much kinetic energy (via $E = \frac{1}{2} m v^2$), it ends up going at about $0.35 c$. This means the entire model we used above is probably a bit off, and needs a relativistic correction (ie. we need to drop a $E = mc^2$ in there).

So to finally reveal about how much energy we can get from a teaspoon worth of neutrons, we can use the density again to find the number of neutrons that would fit in a teaspoon, and multiply it by our per-particle energy above. From the density of $\approx 10^{43}$ neutrons per cubic meter, we'd get about $5 \times 10^{37}$ neutrons per teaspoon. This would yield a total energy of

$E = 3.5 \times 10^{-11} \textrm{J} \times (5 \times 10^{37}) = 1.7 \times 10^{27} \textrm{joules}$

With such a big number, we can turn to Wikipedia and Wolfram Alpha for some energy comparisons. This is about equal to the total energy output of the sun in a second, and about ten thousand times the energy release of the asteroid that killed the dinosaurs. So would it destroy the Earth? Probably not – the gravitational binding energy of the Earth is over a thousand times this energy.

A cursory look around the neutron radiation therapy and nuclear reactor neutron moderation literature indicates that fast neutrons like the ones we'd have here have a relatively large mean free path... something like $\approx \SI{20}{\centi \meter}$ with dense matter like water or concrete (1).

This large mean free path compared to charged alpha and beta particles (which can be blocked with wax paper and tin foil, respectively) makes some sense, since a fast uncharged neutron will breeze through the electron clouds around atoms like they're nothing -- it'll only interact with matter if it hits a nucleus (which has a tiny cross-section compared to the whole atom).

Unless the neutron hits head on a nucleus of identical mass (like a proton/Hydrogen nucleus), an elastic collision will only transfer a fraction of its energy. This fraction depends on the mass difference between the particles colliding: a fast neutron will take many collisions (well over 100) with heavier nuclei in rock to dissipate all its energy. Vaporized rock/water several meters deep around the Earth's surface sounds about right.

(1) One source for the mean free path of neutrons.

@David Herman asked for an estimate of the energy release here, which got me going down a physics rabbit hole. Hope you enjoy.

I'm going to be playing fast and loose with physics here, so bear with me. To start, everyone is correct in saying that once the teaspoon of neutron matter shows up on Earth, it's now longer got the gravitational pressure to hold all the neutrons together, so suddenly all the energy required to pack the neutrons nearly as close as a nucleus is suddenly going to be released, blasting the neutrons out in all directions. So let's find this energy.

Even a nucleus can be treated in a cartoon model as a non-interacting fermion gas, so I think this will be enough to at least approximate the confinement energy in neutron star matter, since it's really nothing more than a big neutron-rich nucleus held together by gravity. This brings with it a caveat though: at the extremely high densities we're talking about here, the short-range nuclear interactions will begin to come into play, and that is a little bit above my pay grade. We'll also encounter another caveat about relativity a bit later on.

Particle(s) in a box

We'll start with a particle in a box – a concept that's probably familiar to those who have taken introductory quantum mechanics. We could also start with Heisenberg's uncertainty principle, which would end up in more or less the same equation in a few steps. Disregarding the ground state energy, the kinetic energy of a single particle in a one-dimensional box is:

$E = \frac{\pi^2 n^2 h^2}{2 m L^2}$

Where $n$ is a positive integer (an indicator of a possible quantize energy level the particle can occupy – this is quantum, remember!), $h$ is Planck's constant, $m$ is the mass of the particle, and $L$ is the length scale of the box. We can already tell that as the box we squeeze the particle in gets smaller, the energy gets higher. This is the degeneracy pressure we're looking for! But we must go deeper.

Let's bump up this box to three dimensions, meaning we have to introduce two more positive integer energy level indices; one for each dimension ($n_x, n_y, n_z$). Each possible state has a unique combination of quantum numbers (the indices).

$E = \frac{\pi^2 h^2}{2 m L^2} (n_x^2 + n_y^2 + n_z^2)$

Okay so since we're dealing with neutrons (which are fermions), they follow Pauli's exclusion principle, and they can't occupy the same state. If there's only a single particle in this box, it'll be in the lowest energy state $n_x = n_y = n_z = 1$, but then when we put another one in it'll have to take a higher energy state (maybe $n_x = 2, n_y = n_z = 1$). As we put more particles in, the energy level quantum numbers are going to go up, and with it, the energy $E$. If we have to put in $N$ particles, how high is the energy going to go if we fill up the available energy levels starting at $n_x = n_y = n_z = 1$?

Energy level space

The way to model this as $N$ gets big is to treat the three quantum numbers as positions in some kind of "energy level space". The energy levels fill up, and the particles' energies depend on the norm $|n| = \sqrt{n_x^2 + n_y^2 + n_z^2}$. This can be treated as the Euclidean distance between the origin ($n_x = n_y = n_z = 0$) and the outer filled levels. But the quantum numbers can only be positive, so the available levels are kind of like the volume of the positive eighth of a sphere in "energy level space".

Imagine every possible combination of positive integers $n_x, n_y, n_z$ is a point in this corner of a sphere. As the sphere gets big, we can estimate the number of points by calculating its volume.

The volume of this corner of the "energy level sphere" is the number of possible energy states $N$ that could fit in a certain radius $|n| = \sqrt{n_x^2 + n_y^2 + n_z^2}$.

$N = \frac{1}{8} \frac{4}{3} \pi |n|^3$

We're forgetting one thing: spin. Fermions can occupy the same energy state as long as they've got different spins. Since neutrons are spin $\frac{1}{2}$ particles, they have two different spin states, so we can add a factor of $2$ to the total number of fermion states $N$:

$N = 2 \frac{1}{8} \frac{4}{3} \pi |n|^3$

We can isolate $|n|$:

$|n| = \left( \frac{3N}{\pi} \right)^{\frac{1}{3}}$

This is the magnitude of the quantum numbers $n_x, n_y, n_z$ that we've reached if we put $N$ fermions into our box. Since each fermion needs its own state, the norm $|n|$ can give us an estimate for the energy scale of the most energetic particles:

$\begin{aligned} E &= \frac{\pi^2 h^2}{m L^2} |n|^2 \\ &= \frac{\pi^2 h^2}{m L^2} \left( \frac{3N}{\pi} \right)^{\frac{2}{3}} \end{aligned}$

The Fermi Energy

Alright, so now we've got an equation for the energy of the particles when we've squeezed $N$ into a box of length $L$. We can switch out $L$ for $V^{\frac{1}{3}}$ to yield a number density, $\rho = \frac{N}{V}$:

$\begin{aligned} E &= \frac{h^2}{2m} \left( \frac{3 \pi ^2 N}{V} \right)^{\frac{2}{3}} \\ &= \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \end{aligned}$

We've got a pretty great estimate for the energy scale (per particle) of a bunch of fermions trapped in a box with density $\rho$, this is called the Fermi Energy $E_F$. To keep all our ducks in a row we could integrate over $N$ from low energy initial particles (with low quantum numbers) to super high energy final particles (with high quantum numbers) which would yield an energy per particle of $E = \frac{3}{5} E_F$ (thanks to the factor of $N^{\frac{2}{3}}$). But what's a factor of $\frac{3}{5}$ between friends.

Will it destroy the Earth?

Alright! Now we can just do some googling to find the density of a neutron star: looks like $\approx \SI{10^{17}}{\kilo \gram \per \meter \cubed}$. The mass of a neutron is about $m_n = 1.66 \times 10^{-27} \textrm{kg}$, so the number density in a neutron star is $\rho \approx 10^{43} \frac{\textrm{neutrons}}{\textrm{m}^3}$. Wow, that's a lot.

Let's plug that into our energy:

$E = \frac{3}{5} \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \approx 3.5 \times 10^{-11} \textrm{J}$

So per neutron, we're talking about $3.5 \times 10^{-11} \textrm{J}$ of energy. That's about the energy released per Uranium atom during a nuclear explosion. If we give the neutron this much kinetic energy (via $E = \frac{1}{2} m v^2$), it ends up going at about $0.35 c$. This means the entire model we used above is probably a bit off, and needs a relativistic correction (ie. we need to drop a $E = mc^2$ in one of the energy equations above. Bonus points if someone can perform this correction in a response).

So to finally reveal about how much energy we can get from a teaspoon worth of neutrons, we can use the density again to find the number of neutrons that would fit in a teaspoon, and multiply it by our per-particle energy above. From the density of $\approx 10^{43}$ neutrons per cubic meter, we'd get about $5 \times 10^{37}$ neutrons per teaspoon. This would yield a total energy of

$E = 3.5 \times 10^{-11} \textrm{J} \times (5 \times 10^{37}) = 1.7 \times 10^{27} \textrm{joules}$

With such a big number, we can turn to Wikipedia and Wolfram Alpha for some energy comparisons. This is about equal to the total energy output of the sun in a second, and about ten thousand times the energy release of the asteroid that killed the dinosaurs. So would it destroy the Earth? Probably not – the gravitational binding energy of the Earth is over a thousand times this energy.

Edit: See my response to Xinxin below for some guesses about what ultrafast neutrons might do when they hit the Earth's surface.

I'm going to be playing fast and loose with physics here, so bear with me. To start, everyone is correct in saying that once the teaspoon of neutron matter shows up on Earth, it's no longer got the gravitational pressure to hold all the neutrons together, so suddenly all the energy required to pack the neutrons nearly as close as a nucleus is suddenly going to be released. So let's find this energy.

Even a nucleus can be treated in a cartoon model as a non-interacting fermion gas, so I think this will be enough to at least approximate the confinement energy in neutron star matter, since it's really nothing more than a big neutron-rich nucleus held together by gravity. This brings with it a big caveat though: at the extremely high densities we're talking about here, the short-range nuclear interactions will begin to come into play, and that is a little bit above my pay grade.

Particle(s) in a box

We'll start with a particle in a box – a concept that's probably familiar to those who have taken introductory quantum mechanics. Disregarding the ground state energy, the kinetic energy of a single particle in a one-dimensional box is:

$E = \frac{\pi^2 n^2 h^2}{2 m L^2}$

Where $n$ is a positive integer (an indicator of a possible quantize energy level the particle can occupy – this is quantum, remember!), $h$ is Planck's constant, $m$ is the mass of the particle, and $L$ is the length scale of the box. We can already tell that as the box we squeeze the particle in gets smaller, the energy gets higher. This is the degeneracy pressure we're looking for! But we must go deeper.

Let's bump up this box to three dimensions, meaning we have to introduce two more positive integer energy level indices; one for each dimension ($n_x, n_y, n_z$). Each possible state has a unique combination of quantum numbers (the indices).

$E = \frac{\pi^2 h^2}{2 m L^2} (n_x^2 + n_y^2 + n_z^2)$

Okay so since we're dealing with neutrons (which are fermions), they follow Pauli's exclusion principle, and they can't occupy the same state. If there's only a single particle in this box, it'll be in the lowest energy state $n_x = n_y = n_z = 1$, but then when we put another one in it'll have to take a higher energy state (maybe $n_x = 2, n_y = n_z = 1$). As we put more particles in, the energy level quantum numbers are going to go up, and with it, the energy $E$. If we have to put in $N$ particles, how high is the energy going to go if we fill up the available energy levels starting at $n_x = n_y = n_z = 1$?

Energy level space

The way to model this as $N$ gets big is to treat the three quantum numbers as positions in some kind of "energy level space". The energy levels fill up, and the particles' energies depend on the norm $|n| = \sqrt{n_x^2 + n_y^2 + n_z^2}$. This can be treated as the Euclidean distance between the origin ($n_x = n_y = n_z = 0$) and the outer filled levels. But the quantum numbers can only be positive, so the available levels are kind of like the volume of the positive eighth of a sphere in "energy level space".

Imagine every possible combination of positive integers $n_x, n_y, n_z$ is a point in this corner of a sphere. As the sphere gets big, we can estimate the number of points by calculating its volume.

The volume of this corner of the "energy level sphere" is the number of possible energy states $N$ that could fit in a certain radius |n| = \sqrt{nx^2 + ny^2 + n_z^2}).

$N = \frac{1}{8} \frac{4}{3} \pi |n|^3$

We're forgetting one thing: spin. Fermions can occupy the same energy state as long as they've got different spins. Since neutrons are spin $\frac{1}{2}$ particles, they have two different spin states, so we can add a factor of $2$ to the total number of fermion states $N$:

$N = 2 \frac{1}{8} \frac{4}{3} \pi |n|^3$

We can isolate $|n|$:

$|n| = \left( \frac{3N}{\pi} \right)^{\frac{1}{3}}$

This is the magnitude of the quantum numbers $n_x, n_y, n_z$ that we've reached if we put $N$ fermions into our box. Since each fermion needs its own state, the norm $|n|$ can give us an estimate for the energy scale of the most energetic particles:

$\begin{aligned} E &= \frac{\pi^2 h^2}{m L^2} |n|^2 \\ &= \frac{\pi^2 h^2}{m L^2} \left( \frac{3N}{\pi} \right)^{\frac{2}{3}} \end{aligned}$

The Fermi Energy

Alright, so now we've got an equation for the energy of the particles when we've squeezed $N$ into a box of length $L$. We can switch out $L$ for $V^{\frac{1}{3}}$ to yield a number density, $\rho = \frac{N}{V}$:

$\begin{aligned} E &= \frac{h^2}{2m} \left( \frac{3 \pi ^2 N}{V} \right)^{\frac{2}{3}} \\ &= \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \end{aligned}$

We've got a pretty great estimate for the energy scale (per particle) of a bunch of fermions trapped in a box with density $\rho$, this is called the Fermi Energy $E_F$. To keep all our ducks in a row we could integrate over $N$ from low energy initial particles (with low quantum numbers) to super high energy final particles (with high quantum numbers) which would yield an energy per particle of $E = \frac{3}{5} E_F$ (thanks to the factor of N^{\frac{2}{3}. But what's a factor of $\frac{3}{5}$ between friends.

Will it destroy the Earth?

Alright! Now we can just do some googling to find the density of a neutron star: looks like $\approx \SI{10^{17}}{\kilo \gram \per \meter \cubed}$. The mass of a neutron is about $m_n = 1.66 \times 10^{-27} \textrm{kg}$, so the number density in a neutron star is $\rho \approx 10^{43} \frac{\textrm{neutrons}}{\textrm{m}^3}$. Wow, that's a lot.

Let's plug that into our energy:

$E = \frac{3}{5} \frac{h^2}{2m} \left( 3 \pi \rho \right)^{\frac{2}{3}} \approx 3.5 \times 10^{-11} \textrm{J}$

So per neutron, we're talking about $3.5 \times 10^{-11} \textrm{J}$ of energy. That's about the energy released per Uranium atom during a nuclear explosion. If we give the neutron this much kinetic energy (via $E = \frac{1}{2} m v^2$), it ends up going at about $0.35 c$. This means the entire model we used above is probably a bit off, and needs a relativistic correction (ie. we need to drop a $E = mc^2$ in there).

So to finally reveal about how much energy we can get from a teaspoon worth of neutrons, we can use the density again to find the number of neutrons that would fit in a teaspoon, and multiply it by our per-particle energy above. From the density of $\approx 10^{43}$ neutrons per cubic meter, we'd get about $5 \times 10^{37}$ neutrons per teaspoon. This would yield a total energy of

$E = 3.5 \times 10^{-11} \textrm{J} \times (5 \times 10^{37}) = 1.7 \times 10^{27} \textrm{joules}$

With such a big number, we can turn to Wikipedia and Wolfram Alpha for some energy comparisons. This is about equal to the total energy output of the sun in a second, and about ten thousand times the energy release of the asteroid that killed the dinosaurs. So would it destroy the Earth? Probably not – the gravitational binding energy of the Earth is over a thousand times this energy.

Varsha-

Thanks so much for your thoughtful comment/report. We have features incoming to introduce new ways of responding to problems, including challenges or extensions like this one. Sorry those aren't live quite yet, but I appreciate you leaving a "solution" anyway 😊.

I've tweaked the text of the problem a bit (to make it clear we're not considering a single word, and that the information tends to go down, but not necessarily monotonically), but I realize it's just a bandaid on the issue that you're bringing up. If I added a caveat: "For the first letter in unknown English text, there are $26$ options, so revealing that letter gives $\log_2 {26} = 4.7$ bits of information if every possible first letter is equally likely." do you think the spirit of the problem would still shine through?

For those interested, one way to do this problem justice would be to treat an English string as a sequence of random variables $X$ which can assume a character value $x$ with probability $P(X=x)$. According to Shannon's definition, the entropy (or uncertainty) of a given variable in the sequence can thus be given by

$S(X)=-\sum_{x} P(X=x)\log{(P(X=x))}$

Generalizing to a sequence of characters, the conditional probability of the next character having value $x$ given the previous characters is defined as

$P(X_{n+1}=x|X_{1}=u,X_{2}=v,X_{3}=w,...,X_{n}=z)=P(X_{n+1}=x|X_1,...,X_n)$

So we can define the information content of each new character as

$S(X_{n+1)}|X_{1},...,X_{n})=-\sum P(X_{1},...,X_{n},X_{n+1})\log{P(X_{n+1}|X_{1},...,X_{n})}$

To determine these probabilities, we'd ideally need a large body of English text to analyze letter by letter, which is a bit computationally difficult.

An easier way to do this would be to use a different definition of uncertainty (or entropy) given by

$S = \log_2{\Omega}$

where $\Omega$ is the number of possible outcomes. Using this definition, we can use an English dictionary and just consider the body of English words. The one in Python has about $235 000$ words, yielding an uncertainty of

$S = \log_2{235 000} \approx 18 \textrm{ bits}$

If we reveal the first letter (let's say it is "e"), then the number of possible words goes way down, to about $8700$, leaving us with

$\log_2{8700} \approx 13 \textrm{ bits}$

of uncertainty. That means revealing the first letter gave us about 5 bits of information. If we reveal the next letter (let's say its a "n"), then the number of possible words goes down again, to $1700$ words, or $11 \textrm{ bits}$ of uncertainty. This means the second letter gave us about $2 \textrm{ bits}$.

The information content of each letter can be calculated in this way and graphed:

If you were to look at the information content of each new character over many words, the information content of each letter does tend to go down as you move deeper into a word. Though of course there are some notable exceptions.

In the word azureous, most of the information is contained in second letter, "z":

While in the word quarantine, basically no information is contained in the second letter, "u":

Averaged over a large dataset of words, the trend towards lower information content revealed as you proceed deeper into a word or phrase remains fairly consistent. Claude Shannon himself did a fun experiment to collect this averaged data by asking study participants to guess the next letter given previous ones for many different phrases and words. His estimated upper and lower limit is plotted here, and agrees roughly with data that we could calculate today.

This is a really fun problem to do computationally or experimentally, but is somewhat outside the scope of this track of daily problems. Keep on the lookout for more advanced content in the future though.

thanks again,

Blake

$\textrm{F}$ $\textrm{Heat loss} \propto \textrm{Surface Area} \times T^4$

$\ket{\psi_1\psi_2\psi_3\psi_4\ldots}$ $\ket{\psi^\prime_1\psi^\prime_2\psi^\prime_3\psi^\prime_4\ldots}$

$\ket{00}\otimes\ket{0} \\ \ket{01}\otimes\ket{1} \\ \ket{10}\otimes\ket{0} \\ \ket{11}\otimes\ket{1}$

$\begin{aligned} + \ket{1000} & \rightarrow +\ket{1000} \quad \textrm{No alarm}\\ + \ket{0100} & \rightarrow +\ket{0100} \quad \textrm{No alarm}\\ + \ket{0010} & \rightarrow -\ket{0010} \quad \color{#D61F06} \textrm{Alarm!} \color{#333333}\\ + \ket{0001} & \rightarrow -\ket{0001}\quad \color{#D61F06} \textrm{Alarm!} \color{#333333}\\ \end{aligned}$

$\begin{aligned} \ket{0000} &- \ket{0001} - \ket{0010} + \ket{0011} \\ + \ket{0100} &- \ket{0101} - \ket{0110} + \ket{0111} \\ + \ket{1000} &- \ket{1001} - \ket{1010} + \ket{1011} \\ + \ket{1100} &- \ket{1101} - \ket{1110} + \ket{1111} \end{aligned}$

\begin{aligned} & a_1 \ket{000000} + a_2 \ket{000001} + a_3 \ket{000010} + a_4 \ket{000011} +\\ & a_5 \ket{000100} +a_6 \ket{000101} + a_7 \ket{000110} + a_8 \ket{000111} +
\\ & a_9 \ket{001000} + a_{10} \ket{001001} + a_{11} \ket{001010} + a_{12} \ket{001011} + \\ & a_{13} \ket{001100} + a_{14} \ket{001101} + a_{15} \ket{001110} + a_{16} \ket{001111} +
\\ & a_{17} \ket{010000} + a_{18} \ket{010001} + a_{19} \ket{010010} + a_{20} \ket{010011} + \\ & a_{21} \ket{010100} + a_{22} \ket{010101} + a_{23} \ket{010110} + a_{24} \ket{010111} +\\ & a_{25} \ket{011000} + a_{26} \ket{011001} + a_{27} \ket{011010} +a_{28} \ket{011011} + \\ & a_{29} \ket{011100} + a_{30} \ket{011101} + a_{32} \ket{011110} + a_{32} \ket{011111} +\\ &a_{33} \ket{100000} + a_{34} \ket{100001} + a_{35} \ket{100010} + a_{36} \ket{100011} + \\ & a_{37} \ket{100100} + a_{38} \ket{100101} + a_{39} \ket{100110} + a_{40} \ket{100111} +
\\ & a_{41} \ket{101000} + a_{42} \ket{101001} + a_{43} \ket{101010} +a_{44} \ket{101011} + \\ & a_{45} \ket{101100} + a_{46} \ket{101101} + a_{47} \ket{101110} +a_{48} \ket{101111} +
\\ & a_{49} \ket{110000} + a_{50} \ket{110001} + a_{51} \ket{110010} + a_{52} \ket{110011} + \\ & a_{53} \ket{110100} + a_{54} \ket{110101} + a_{55} \ket{110110} + a_{56} \ket{110111} +\\ & a_{57} \ket{111000} + a_{58} \ket{111001} +a_{59} \ket{111010} + a_{60} \ket{111011} + \\ & a_{61} \ket{111100} + a_{62} \ket{111101} + a_{63} \ket{111110} + a_{64} \ket{111111} \end{aligned}

$\begin{aligned} + \ket{1000} & \rightarrow +\ket{1000} \quad \textrm{No alarm}\\ + \ket{0100} & \rightarrow -\ket{0100} \quad \color{#D61F06} \textrm{Alarm!} \color{#333333}\\ + \ket{0010} & \rightarrow -\ket{0010} \quad \color{#D61F06} \textrm{Alarm!} \color{#333333}\\ + \ket{0001} & \rightarrow +\ket{0001} \quad \textrm{No alarm} \end{aligned}$

$0000 \quad 0001 \quad 0010 \quad 0011 \\ 0100 \quad 0101 \quad 0110 \quad 0111 \\ 1000 \quad 1001 \quad 1010 \quad 1011 \\ 1100 \quad 1101 \quad 1101 \quad 1111$

$\begin{aligned} \ket{0000} &+ \ket{0001} + \ket{0010} + \ket{0011} \\ + \ket{0100} &+ \ket{0101} + \ket{0110} + \ket{0111} \\ + \ket{1000} &+ \ket{1001} + \ket{1010} + \ket{1011} \\ + \ket{1100} &+ \ket{1101} + \ket{1101} + \ket{1111} \end{aligned}$

P1

Chemistry is the study of matter, its interactions, and how it can transform. And there is no more familiar example of matter in motion than the wondrous process of life.

Inside of every cell, the laws of chemistry work together to animate molecules. Sometimes life’s reactions build up grand molecular structures built of proteins and DNA, and other times they rapidly break down living matter into its barest constituents.

[[Figure of DNA monomers building up a genome and a cell (show cytoskeleton), and then that cell breaking down a log into CO2, or similar]]

The number of different molecules interacting in one cell is an intimidatingly large quantity, but even the most complex systems often arise from a combination of simple rules. In this chapter, we’ll seek to extract the simple rules governing chemical reactions by picking apart a biological process.

Q2

Most cells are little more than fluid-filled sacs, with very little structural integrity. For example, a white blood cell can twist and contort itself into any shape while chasing a bacterial invader. But sometimes, cells need to hold their shape. To do this, they build a sort of cellular skeleton.

[[figure of a cytoskeleton]]

The skeleton of a cell is made up of structures called microtubules that act like beams or supports to hold together the structure of a nerve cell or a skin cell. The microtubules are little more than stacked protein bricks called tubulins. The tubulins stack two at a time, so they’re called dimers.

[[zoom in on a dimer that builds up the cytoskeleton]]

If you were to represent this process in the language of a chemical reaction, how would you write it?

A: $\ce{2 Tubulin -> 1 Dimer}$ (Maybe N Dimer -> Microtubule?)

Q3

When a cell dies, its contents spills out into the surrounding environment. All the proteins, DNA and skeletal structures that were once animated in the cell spread out.

[[Figure of this sad event happening in a scientific flask]]

In this “dead” state, the chemical reactions that once occurred in the living cell have changed, but they have not stopped. If a cell’s contents are released in a flask otherwise filled with water, we can study the transformations that continue to occur as a way to predict the mechanisms inside of a cell.

What would you expect to happen to the cell’s microtubules once they’re released from the cell into the surrounding water?

A: * Eventually they would begin to break down into the tubulins that make them up. They would grow larger and larger once released from confinement in the cell ...etc

P4

The association reaction is used to build up microtubules in the cell, but once removed from the cell, we observe that there is an opposite reaction that can happen as well; the dimers can dissociate into their component pieces. This process will eventually break down most of the microtubules.

$\ce{1 Dimer -> 2 Tubulin}$

Inside the cell, both the association and dissociation reactions are acting on the tubulin dimers. But somehow the cell manages to build and maintain an orderly skeleton despite the reality of the dissociation reaction constantly working to break the dimers apart.

[[some illustration of both of these reactions occurring inside of a cell]]

In this quiz, we will first try to understand the breakdown that happens to isolated microtubules and their dimers. We’ll then use the governing equation for this transformation to predict how the dynamics microtubules in a cell.

Q5

We’ve already seen that the dimers which make up microtubules can break down into their component pieces through the dissociation reaction below:

\ce{1 Dimer} -> 2 Tubulin} <- This equation in a figure of a flask.

Every dimer contained in a microtubule is identical, and they have no “memory”. That is, in every given moment, each tubulin dimer might break apart into its component pieces with some given probability. We associate a number with this probability of breaking apart per unit time, and call it $k$, the rate constant. A higher rate constant means the probability of each dimer breaking up in a given moment is higher, and the rate of dissociation will be higher.

If there are $D(0)$ dimers in a flask at an initial time $t=0$, approximately how many will fall apart in a short time $\Delta t$?

A: $D(0) k \Delta t$

Q6

The rate constant $k$ has units of “per second” $(1 / \si{\second})$, because a probability itself has no units. To transform it into a rate of change of $D$, we need to multiply it by how many dimers are present in that moment:

$\textrm{rate of change of } D = D(0) k \Delta t$

Even if the probability of a single tubulin dimer breaking apart in a short time $\Delta t$ is very small, there are likely a large number of tubulin dimers initially in the flask that are liable to break apart. The number of dimers that remain after the reaction has proceeded for some time $\Delta t$ is given by:

$\begin{aligned} D(t = \Delta t) &= D(0) - D(0) k \Delta t \\ &= D(0) \left( 1 - k \Delta t \right) \end{aligned}$

If we measure the number of dimers in the flask in the first few moments of the reaction, approximately what do you expect it to look like?

A: Downward linear plot <- graph showing this

Q7

As the dissociation reaction continues, there will be fewer and fewer intact tubulin dimers left in the flask.

[[A gif of the dissociation reaction occuring in a flask]]

If we continue to measure the number of dimers in the flask as the reaction continues, how would you expect your measurements to change as time goes on?

A: The number of dimers will keep going down, but slower and slower over time. ie. it will flatten out. <- graph showing this.

P8

The probability of flipping heads on a coin doesn’t change after flipping tails several times, and a person’s probability of winning the lottery doesn’t change after playing fruitlessly for 20 years. Similarly, in this chemical reaction, the probability that a dimer breaks apart is the same from one moment to the next.

Whether it is in the first moment after the reaction begins:

$\frac{D(\Delta t)}{D(0)} = k \Delta t ,$

or after some unknown time $t$:

$\frac{D(t + \Delta t)}{D(t)} = k \Delta t ,$

the fraction of dimers which dissociate will always be proportional to the rate constant, $k$.

We can use this simple pattern to get a full quantitative picture of dimer dissociation by using a bit of math.

Q9

As you showed before, in the first moment $\Delta t$ of the reaction, the number of dimers changes according to

$D(\Delta t) = D(0) (1 - k \Delta t)$

And by the pattern we just saw, the next moment looks largely the same:

$D(2 \Delta t) = D(\Delta t) (1 - k \Delta t)$

The linear drop in dimers can't hold up as the reaction keeps going, because the breakup has to slow down as the dimers become fewer and fewer. After two periods of time $\Delta t$, how does the number of dimers compare to the initial number $D(0)$?

A: $D(2 \Delta t) = D(0) (1 - k \Delta t)^2$ <- found by substituting the first equation in the second.

Q10

We can make progress by keeping track of how $D$ changes from moment to moment over many $(n)$ short periods of time $\Delta T$.

In much the same way that we substituted the result of the first moment into the second, we can chain together these moments:

$\begin{aligned} D(n \Delta t) &= D(0) ( 1 - k \Delta t) ( 1 - k \Delta t) … ( 1 - k \Delta t) \quad (N \textrm{ times.}) \\ &= D(0) (1-k \Delta t)^n \end{aligned}$

If we break the time $t$ we’re letting the reaction go for into $n$ moments $\Delta t$, each moment is really just $\Delta t = t / n$:

$D(t) = D(0) (1-k t / n)^n$

This might be a familiar equation for those who have studied exponential functions, but for those who haven’t, it’s clear that the limit form of the exponential function $e^x$ could play a roll here:

$\lim_{n \to \infty} (1-x / n)^n = e^x$

What is the general formula for the number of dimers at any point in the reaction, $D(t)$?

A: $D(t) = D(0) e^{-kt}$

P11

The simple expression that we’ve derived here checks all the boxes that we expected for this reaction:

$D(t) = D(0) e^{-kt}$

[[Show this graph as an interactive one, with a tangent line that could be moved around the function showing the slope decreasing]]

The dissociation rate begins very high at the beginning of the reaction, when dimers are abundant, but as the dimers dissociate, their dissociation rate begins to slow. The exponential function isn’t just a few lectures that you might dimly recall from math class; it will show up again and again in the study of chemical reactions as we study their dynamics and rates of change.

Q12

In this quiz, we've pretended like the association reaction just doesn't happen.

$\ce{2 Tubulin -> 1 Dimer}$

In reality this isn't true — tubulin dimers are constantly breaking up, but the cell keeps things ordered by the continual production of new tubulins, which drives the reaction in the other direction. The cell is a ceaseless balance of breakup and formation, which we'll study in the next quiz.

For now, how would you expect the complication of this association reaction to change the graph of $D(t)$ from the simplified picture we used here?

A: Level out at a non-zero amount of dimers. <- graph of this

P13

The interaction between two competing reactions, the ceaseless break up of dimers along with their continual assembly tends to yield a middle ground result. Not all of the dimers break up into their component tubulins, and not all of the tubulins assemble into dimers.

Instead, the system will tend to settle on a certain number of dimers that remains largely unchanged, even though the chemical reactions have not stopped. This gives us the first clue for how unending reactions going on inside of a cell can still create a stable result, like the microtubules that make up the cell’s skeleton.

[[back to the pretty picture of the cell’s cytoskeleton]]

$1 / \sqrt{2} \ket{\psi }$ $\frac{1}{\sqrt{2}}$

There has been a great deal of discussion about this problem in the solutions. Many people have invoked an experiment performed by the Mythbusters several years back which showed that a fired bullet and a dropped bullet hit the ground at the same time. We here at Brilliant love the Mythbusters, but we love physics more. Let's break this problem down a bit:

Quadratic Drag

As some of the solutions have indicated, the interesting bit of physics in this problem comes from the fact that the drag force from air resistance is proportional to the squared velocity of the bullet. If we break down the velocity into its $v_x$ and $v_y$ components, that means the drag term is something like:

$F_d = -k|v|^2,$

where $|v| = \sqrt{v_x^2 + v_y^2}$.

Breaking down the force

Like any other force, we can break down the drag force into components along the $x$ and $y$ axes. If the bullet's trajectory is off from horizontal by an angle $\theta$, then the $y$ component of the force is $F_d \sin{\theta}$ and the $x$ component of the force is $F_d \cos{\theta}.$ To avoid any unnecessary variables, we can express these trigonometric functions using the bullet's $x$ and $y$ components. Using the right triangle above, we can observe that:

$\sin{\theta} = \frac{v_y}{|v|}, \\ \cos{\theta} = \frac{v_x}{|v|}.$

When substituted into the $x$ and $y$ components of the drag force, the divided factor of $|v|$ simplifies and yields $x$ and $y$ components of $-k |v| v_x$ and $-k |v| v_y$ respectively. Even though these components of the force are orthogonal, notice that they both still depend on the total velocity $|v|$ in addition to their velocity component.

Equations of motion

Now that we've broken down the drag force into the $x$ and $y$ components, we can construct the equations of motion according to Newton's second law:

$\begin{aligned} m \, v_x'(t) &= -k |v| v_x \\ m \, v_y'(t) &= -mg - k |v| v_y \end{aligned}$

Substituting $|v|$ as the norm of the two spatial components:

$\begin{aligned} m \, v_x'(t) &= -k v_x \sqrt{v_x^2 + v_y^2} \\ m \, v_y'(t) &= -mg - k v_y \sqrt{v_x^2 + v_y^2} \end{aligned}$

You might notice now that these equations are coupled through $|v|$, and likely can't be solved explicitly. Even without solving, we can consider what effect the total velocity of the bullet $|v| = \sqrt{v_x^2 + v_y^2}$ contributes to the drag term in both components of the bullet's velocity. If the bullet is fired from a gun, a large horizontal velocity will increase the vertical drag force that resists the pull of gravity. If a bullet is simply dropped, the horizontal velocity will be zero, and the total velocity will be much lower. This would yield a weaker drag force to resist gravity, and explains why a dropped bullet hits the ground quicker

Finding the trajectories

Though explicitly solving these coupled differential equations is nasty, we can use a numerical solver in Mathematica (or your computational tool of choice). Picking some physical constants for the drag coefficient $k$ and the mass $m$ yields the following trajectories for a dropped bullet and one fired at $\SI{500}{\meter \per \second}$.

Fired bullet trajectory and time dependence of height (Quadratic drag)

Dropped bullet trajectory and time dependence of height (Quadratic drag)

The fired bullet hits the ground after about 0.60 seconds, while the dropped one hits after only 0.45 seconds.

What about linear drag?

If we were to use a simpler linear approximation of air drag, the total velocity would no longer factor in to the drag of both components, and the equations would no longer be coupled. I leave that to the reader to construct similar equations of motion using a linear drag term like $F_d = -k |v|$. The trajectory of such a bullet looks something like the one below, and hits the ground just like the dropped one, after about 0.45 seconds.

Fired bullet trajectory and time dependence of height (Linear drag)

The wavelength of green light is about $\SI{500}{\nano \meter}$. As the wavelength of light gets smaller, its energy gets larger. Green light is about $200 \times$ smaller than the width of a hair, and about $2000 \times$ larger than most atoms.

We use visible light in microscopes to observe forensic samples of hair, and even to visualize the inside of cells, which are only about $2 \times$ larger than the wavelength of green light. But we can't observe individual atoms with a light microscope: In fact as a rule, you can't really observe objects much smaller than the wavelength you're using to look at them. It would be like trying to track a speck of dust in the sky with a radar station.

What does this imply about subatomic particle physics?

1. Subatomic particles can only be studied with very long wavelength light.
2. As we probe more deeply into subatomic particles, we'll need to build higher energy accelerators.
3. To observe individual subatomic particles, the experiment must be extremely cold.

Hydrogen is the simplest atom, with only one proton and one electron, interacting with a Coulomb potential:

$U = \frac{e^2}{4 \pi \epsilon_0 r}$

where $e$ is the elementary charge and $\epsilon_0$ is the permitivity of free space.

If the electron were a classical charged particle, its orbit around the nucleus would decay until it came to rest at the nucleus. Since the electron is a quantum object, this doesn't happen. The spread in position and momentum of the electron is limited by the uncertainty principle:

$\Delta x \Delta p \geq \frac{\hbar}{2}$

where $\hbar$ is the reduced Planck's constant.

The radius of the electron orbit must be larger than the spread in position, likewise the momentum of the electron must be larger than this spread in momentum. Otherwise, the Hydrogen atom would be violating this fundamental limit.

Treat the Hydrogen atom as a two body problem where the mass of the proton is much greater than the electron. Estimate the minimum radius of the electron orbit.

The founding fathers of quantum mechanics largely ignored relativity when they first applied quantum theory to chemistry:

"The general theory of quantum mechanics is now almost complete ... these give rise to difficulties only when high-speed particles are involved, and are therefore of no importance in the consideration of atomic and molecular structure and ordinary chemical reactions" - Paul Dirac

Treat an atom as a two body problem with one electron and a central nucleus with a charge $Z$ and a mass much greater than the electron. Find the characteristic length scale of a ground state electron orbit. At what element in the periodic table would relativistic effects become important for an electron in a ground state?

Note: Significant errors start to appear between classical and relativistic dynamics when $\frac{v}{c} \geq 0.1$.