Blake Test

\[\textrm{Heat loss} \propto \textrm{Area} \times T^4\]

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

\[\ket{00}\otimes\ket{0} \\ \ket{01}\otimes\ket{1} \\ \ket{10}\otimes\ket{0} \\ \ket{11}\otimes\ket{1} \]

\[\begin{align} + \ket{1000} & \rightarrow +\ket{1000} \quad \textrm{No alarm}\\ + \ket{0100} & \rightarrow +\ket{0100} \quad \textrm{No alarm}\\ + \ket{0010} & \rightarrow -\ket{0010} \quad \color{red} \textrm{Alarm!} \color{black}\\ + \ket{0001} & \rightarrow -\ket{0001}\quad \color{red} \textrm{Alarm!} \color{black}\\ \end{align}\]

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

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

\[\begin{align} + \ket{1000} & \rightarrow +\ket{1000} \quad \textrm{No alarm}\\ + \ket{0100} & \rightarrow -\ket{0100} \quad \color{red} \textrm{Alarm!} \color{black}\\ + \ket{0010} & \rightarrow -\ket{0010} \quad \color{red} \textrm{Alarm!} \color{black}\\ + \ket{0001} & \rightarrow +\ket{0001} \quad \textrm{No alarm} \end{align}\]

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

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

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:

\[ \begin{align} D(t = \Delta t) &= D(0) - D(0) k \Delta t \\ &= D(0) \left( 1 - k \Delta t \right) \end{align}\]

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:

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

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:

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

Substituting \(|v|\) as the norm of the two spatial components:

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

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 \(\frac{v}{c} \geq 0.1\).