Richard Feynman, the bongo-playing, womanising Byron of 20th century physics, comes to mind, as does the apocryphal exam question about determining the height of a building with a barometer. Perhaps the best modern examples are xkcd and popular books on Fermi estimates. a goofy irreverence which is fine and good, but tends

January 6, 2020. Physics is awesomely powerful. Contrary to popular opinion, you don’t need years of advanced mathematics to taste this power! Simple physics hacks and pre-calculus mathematics are enough, as I copiously illustrate.

## 1. Introduction

How many calculus courses does it take to change a light bulb? The answer is none, of course. And how many does it take to work out why clouds float, the number of fish in the sea, the heart rate of a whale, or follow Einstein’s proof of the existence of atoms? Surprisingly, the answer is still none! As we’ll see below, high school math and some simple physics hacks are sufficient to answer these questions quantitatively. The goal of this post is to explain these hacks and convince you of their awesome power.

I study quantum gravity, so by necessity I am a user (and abuser!) of advanced mathematics. But preparing for a comprehensive exam and running a high school physics circle opened my eyes to the fact that you can do more with less. Without calculus or any deep physical lore, you can still discover black hole entropy, the dark energy density of the universe, gravitational lensing, and even the number of extra dimensions predicted by string theory. (See my physics circle problems if you want to learn how. They have very little overlap with the problems here.)

This post focuses on a set of generic problem-solving tools: dimensional analysis, Fermi approximation, and scaling laws. It is not oriented towards problem-solving heuristics (e.g. limiting arguments or symmetry) or surprisingly useful physical laws (e.g. Stefan-Boltzmann), though there is nontrivial overlap and I hope to say more about these in future. I have stuck to methods accessible to high school and first year students, though it is fair to say some scientific maturity is required of the former. The different sections can be read more or less independently.

## 5. Conclusion

It’s mysterious to me that these techniques are not usually taught in either high school or first year physics. Lack of mathematical background is not the problem.

#### References

Finally, I recommend Sanjoy Mahajan’s book Street-Fighting Mathematics, covering similar ground in greater depth.

https://web.archive.org/web/20160103040425/http://www.isoc.org/inet2000/cdproceedings/2a/2a_2.htm

http://www.fractalcities.org/book/Fractal%20Cities%20Introduction.pdf

https://users.math.yale.edu/~bbm3/web_pdfs/076stochasticModels.pdf

https://dogfishtacklecompany.com/blogs/news/112355910-first-post

https://www.ias.ac.in/article/fulltext/reso/011/01/0063-0078

https://arxiv.org/pdf/physics/0504201.pdf

http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1906_thesis.pdf

http://physics.bu.edu/~duffy/HTML5/brownian_motion.html

https://onlinelibrary.wiley.com/doi/pdf/10.1111/j.1478-4408.1983.tb03728.x

http://ruina.tam.cornell.edu/research/topics/locomotion_and_robotics/simplest_walking/simplest_walking.pdf

https://www.quora.com/Theoretically-how-tall-or-large-can-a-land-animal-evolve-What-would-happen-if-an-animal-exceeded-this-size

## 4. Scaling

Both dimensional analysis and Fermi approximation assume the answer to a question can be written as a product of powers of some other factors:

$\text{answer} \sim (\text{factor 1})^a \times (\text{factor 1})^b \times \cdots.$

Instead of trying to completely factorise the answer, we can zoom in on a single factor, leading to a proportionality relation:

$\text{answer} \propto \text{factor}^p.$

The symbol $\propto$ means we are ignoring all the other factors and not just dimensionless numbers. Sometime, we can nail down a proportionality relation, and in particular the scaling exponent $p$, without considering any other factors.

Exercise 11 (speed of sound). Sound travels through a fluid (such as air) as longitudinal waves, which wobble particles back and forth. The two factors determining how fast this wobbling can propagating are (1) the stiffness of the fluid, measured using the bulk modulus $K$, with the same units as pressure; and (2) the density $\rho$ of the fluid.

(a) Without performing a full dimensional analysis, show that the speed of sound $c_s$ scales with bulk modulus as

$c_s \propto \sqrt{K}.$

Hint. Find a dimensional factor in $c_s$ which only appears in $K$.

(b) Complete the dimensional analysis and figure out how $c_s$ depends on $\rho$. Given that the bulk modulus of air depends very weakly on temperature, would you expect sound to travel faster or slower on a hot day?

### 4.1. Allometry

The study of scaling laws in biology is called allometry. Here, we give a few simple examples. Tissue density is roughly the same for different species (up to the usual order of magnitude numbers). If an organism has length around $\mathcal{L}$, then the mass of the organism should scale as

$M \propto \mathcal{L}^3.$

The surface area, on the other hand, should scale as $\mathcal{L}^2$. (In both cases, we are assuming the organism has a simple shape.) The fact that these things scale differently is called the square-cube law, and was first observed by Galileo. It may seem trivial, but it has important biological consequences! Let’s do the most famous example.

Galileo realised that materials like bone and wood will break when subject to too much pressure. This maximum pressure $P_\text{max}$ does not scale with $\mathcal{L}$, but depends only on the material. Since mass goes as $\mathcal{L}^3$, the weight force also goes as $\mathcal{L}^3$. On the other hand, the cross-section of a weight-bearing element (like legs or a tree trunk) goes as $\mathcal{L}^2$. It follows that the pressure will scale with

$P = \frac{F}{A} \propto \frac{\mathcal{L}^3}{\mathcal{L}^2} = \mathcal{L}.$

An organism cannot get too large without exceeding this maximum pressure. Thus, there is no need to worry about giant ants from outer space: they will collapse under their own weight as soon as they try to invade earth!

Exercise 12 (walking speed). (a) Argue that walking speed scales as $\mathcal{L}^{1/2}$. Hint. Model the leg as a pendulum.

(b) Average human walking speed is $\sim 1.4$ m/s. Estimate the walking speed of a horse and a garden spider.

Exercise 13 (thickness). (a) Argue that the radius $r$ of a weight-bearing element should scale as $r \propto M^{1/2}$ in order to support the organism’s weight.

(b) General Sherman is a giant sequoia tree in California’s Sequoia National Park, probably the largest tree in the world. The diameter of the trunk is $7.7 \text{ m}$. Estimate its mass by comparing to the thickness of the human thighbone ($r = 2.3$ cm) and using the radius scaling.

### 4.2. Fractals

Box counting. If $\mathcal{L}$ is some linear measure of an organism’s size, it’s natural to expect that volume and surface area scale as $\mathcal{L}^3$ and $\mathcal{L}^2$, respectively, because volume and area have dimensions $L^3$ and $L^2$. For a cube and a square these dimensions are obvious, using the formulas $V_\text{cube} = s^3$ and $A_\text{square} = s^2$ for side length $s$. To argue that a general volume or area has these dimensions, you can imagine splitting the volume into many tiny cubes, and the area into many tiny squares. This may be hard in practice (it requires integral calculus!), but in principle it settles the matter.

Or does it? In reality, a horse or a garden spider is made out of atoms, not tiny cubes, and atoms are point-like. When you zoom in, any animal resolves itself into a cloud of $0$-dimensional objects! It seems like everything is really $0$-dimensional, and should scale as $L^0$. This is obviously nonsense. The problem is that we are looking too close: if we zoom in enough, all we can see are the constituents of matter, losing sight of the forest for the trees. To reason about “coarse” properties like the dimension, we should make sure our resolution (the size of our cubes) is much larger than the length scale $\epsilon_\min$ associated with whatever constitutes a horse.

So far, so good: don’t look too close and you won’t see atoms. But the surface and volume of a horse are both made of atoms, and splitting a surface area into little squares, for instance, requires that we already know that it is a surface. That’s cheating! How can I figure out the scaling without assuming the answer? One elegant method is called box counting, where we measure everything with cubes. First, fix a resolution, $\epsilon$, and generate a pile of (imaginary) cubes of size $\epsilon$. Now count how many cubes you need to cover a spatial object of interest, for instance, the bulk of a horse, or its surface area.

Let’s do an example. If the horse has surface area $A$, and $\epsilon$ is small (but much larger than an atom), then the number of cubes needed to cover its skin is

$N(\epsilon) \approx \frac{A}{\epsilon^2} \propto \frac{1}{\epsilon^2},$

since a tiny cube will overlap the skin with area around $\epsilon^2$. (Actually, there are larger cross-sections if you tilt the cube, but the scaling $\propto 1/\epsilon^2$ remains the same.) Similarly, if you tried to replace the horse’s whole body with boxes, the number would scale as

$N(\epsilon) \approx \frac{V}{\epsilon^3} \propto \frac{1}{\epsilon^3}.$

Hopefully you see the pattern: the dimension of the object is the index of $\epsilon$! In general, a set has box counting or fractal dimension $d$ if

$N(\epsilon) \propto \epsilon^{-d}.$

Here, we assume that $\epsilon \gg \epsilon_\min$ (so we can’t see atoms, or whatever things constitute our object) and $\epsilon \ll \mathcal{L}$ (so the cubes are relatively small). Then the dimensional-analysis dimension will be $L^d$, and if we make the object bigger or smaller, it’s “generalised volume” will scale as $\mathcal{L}^d$. The appearance of the word “fractal” will be justified in a moment.

Interdimensional monsters. Box counting seems like a rather elaborate way to reproduce things we already know: volumes have dimension $L^3$, areas $L^2$, curves $L$ and points $L^0$. Usually, we can tell what’s going on immediately by drawing a picture! But there are interdimensional monsters lurking in the thickets between. Let’s catch one!

You can build a simple example with a pencil and eraser. Start by drawing a straight line segment, which we’ll label with the unit interval $[0, 1]$. Then erase the middle third, leaving two small intervals of width $1/3$, from $[0, 1/3]$ and $[2/3, 1]$. Let’s repeat this step for each remaining segment, erasing middle thirds and leaving $4 = 2^2$ segments of length $1/9 = 3^{-2}$. We can iterate this process of erasing middle thirds, and after the $n$th step, we will have $2^n$ segments of length $3^{-n}$.

In real life, the process will eventually terminate when the segments reach the atomic length scale. But in the mathematical realm, $\epsilon_\text{min} = 0$, and we can let the process run forever! The Cantor set is the resulting, infinitely holey pencil line that remains. We can use the construction to easily find the fractal dimension. Since the $n$th step is $2^n$ segments of length $3^{-n}$, and we only delete segments from then on, we can cover the Cantor set using these segments. Using log laws, we find that for $\epsilon = 3^{-n}$,

$N(\epsilon) = N(3^{-n}) = 2^{n} = 3^{n\log_3 2} = \epsilon^{-\log_3 2}.$

You may worry about other box sizes, but these smoothly interpolate between these points and don’t change the scaling law. Thus, the box-counting dimension is $d = \log_3 2 \approx 0.63$. This is neither a point nor a line, but somewhere in between!

Self-similarity. We have just discovered fractals: objects whose fractal dimension is different from the dimensionality of objects they are made out of. The Cantor set also exhibits a characteristic feature of fractals, namely that it is self-similar. By construction, the section of the Cantor set from $[0, 1/3]$ looks the same as the whole thing, but scaled down by $1/3$. This is not an accident, but rather, a feature of scaling laws in general! If a quantity $A$ scales with some quantity $x$ as

$A \propto x^p,$

then when we “zoom in” or “zoom out” by choosing a rescaled $\tilde{x} = x/\ell$, for instance, then we get

$A \propto x^p = \ell^p \tilde{x}^p \propto \tilde{x}^p.$

Using a different, rescaled variable $\tilde{x}$ does not change the exponent $p$. This is what self-similar means! It is therefore natural for objects with scalings $N \propto \epsilon^{-d}$ (including less exotic things like volumes and surface areas) to be self-similar.

Exercise 14 (more monsters). You can generalise the pencil-and-eraser construction above by iteratively removing a fraction $\gamma$ (instead of $1/3$) from the middle of each segment, where $0 < \gamma < 1$.

(a) Show that the box-counting dimension is

$d = -\frac{\log 2}{\log [(1-\gamma)/2]} = \log_{2/(1-\gamma)}2.$

(b) Conclude that any dimension in the range $0 \leq d \leq 1$ is possible.

Exercise 12 (fractional random walks). There is a generalisation of random walks called fractional random walks, where the average spread scales with the number of steps as

$d \propto n^{H},$

for some number $0 < H < 1$ called the Hurst index. Random walks have $H = 1/2$.

(a) Explain why $H > 1/2$ requires that steps be correlated, i.e. directions persist.

(b) What relation between steps does Hurst index $H < 1/2$ require?

This not just a theoretical exercise. The outlines of a coast are jagged, random curves, typically described by a fractional random walk with Hurst index $H \sim 0.8$.

(c) Why should a coastline consist of correlated random steps?

VERSION 2

There is a generalisation of random walks called fractional random walks, where

$d \sim \ell n^{H}$

for some number $0 < H < 1$ called the Hurst index. Random walks have $H = 1/2$.

(a) Explain why $H > 1/2$ requires that steps be correlated. What does $H < 1/2$ require?

In a fractional random walk, there is a small amount of alignment (or anti-alignment) between steps. Strangely, this depends on how many steps we are looking at!

(b) Suppose that consecutive steps $\vec{s}, \vec{s}’$ in a fractional random walk of $n$ steps have average alignment

$\vec{s}\cdot \vec{s}' = \ell^2 \alpha(n).$

Relate $\alpha(n)$ to $H$ and $n$, and confirm your answers to (a).

(c)

This not just a theoretical exercise. The outlines of a coast are jagged, random curves, typically described by a fractional random walk with Hurst index $H \sim 0.8$.

(c) Why should a coastline consist of correlated random steps?

Fractals in nature. All this would be rather esoteric and pathological if fractals were not ubiquitous in nature. But it turns out they are everywhere! The most infamous example is coastlines. Naively, a coastline is just a curve bounding a land mass, and we would expect it to have dimension $L^1$. But in fact, coastlines get more jagged and intricate as you zoom in. The closer you look, the longer they get! The box-counting dimension is usually between $1$ and $2$, with the typical dimension around $d \approx 1.3$. We will give a (partial) explanation for this in the next section, but the basic idea is that coastlines are formed by random processes, and we can study how those random processes scale.

Fractals not only result from random physical process, but from evolution. Our bodies are full of them! This is not just a fun fact, but explains some puzzling biological scalings. One of the most famous is Kleiber’s law, which is the observation that metabolic rate $R$ (i.e. rate of energy consumption) scales as $R \propto M^{4/3}$. This is impossible to explain using the square-cube arguments we gave above. But it is explained by fractals!

The rough idea is as follows. Energy consumption is governed by the rate blood (which contains oxygen needed for combustion) is delivered to cells. We now assume the circulatory system is a self-similar fractal built out of ever smaller tubes, a bit like gluing the different stages in the construction of the Cantor set together. We start with a single big tube, the aorta, which branches into some child tubes, and these branch again. With a few physics and design requirements, we can deduce Kleiber’s law!

Let’s be more precise. Our first assumption is a very reaonable one: blood is conserved. Suppose each tube has $k$ children, and the children are narrower by a factor $\beta$ and shorter by a factor $\gamma$. The rate blood flows through a tube of cross-section $A$ is $A v$. There are $k$ children, and each has a cross-section of $\beta^2 A$, so the total cross-section is $k \beta^2 A$. Assuming the velocity doesn’t change, conservation of blood implies

$A v = k\beta^2 A v \quad \Longrightarrow \quad \beta = k^{-1/2}.$

The second assumption is also very plausible: the circulatory system spreads throughout the body. Since we’re assuming the structure is self-similar, we guess that each level approximately fills a volume, and hence the circulatory system has fractal dimension $d = 3$. If a tube has length $\ell$, it covers an approximate volume $\ell^3$ when we use cubes at that resolution. It has $k$ children of length $\ell \gamma$, and they should cover approximately the same volume at the lower resolution, so

$\ell^3 \approx k (\ell \gamma)^3 \quad \Longrightarrow \quad \gamma = k^{-1/3}.$

Our third assumption concerns the smallest tubes in system: capillary width and length are independent of organism. The basic idea here is that smallest tubes, which deliver material to the cells directly, have a cross-section $A_\text{capillary}$ and length $\ell_\text{capillary}$ dictated by the biological requirements of the cell and not the organism. Thus, if the capillaries are $c$ levels down, then the rate of delivery is

$R = A_\text{aorta}v = k^c A_\text{capillary} v \propto k^c,$

since $v$ and $A_\text{capillary}$ don’t depend on the organism’s size. We’re almost done!

Our last assumption is the plausible guess that the total volume of blood is proportional to the organism’s mass. Put differently, the average composition of a lump of flesh doesn’t depend on where it came from. Let’s also assume that each tube has many children $k$. Then a tube will carry much more blood than its children, since its volume $A \ell$ is much larger than the combined volume of its children:

$k (A \beta^2 \ell \gamma) = A \ell k^{-1/3} \ll A \ell.$

The total volume at the capillary level is then

$V_\text{capillary} = k^c \gamma A_\text{capillary} \ell_\text{capillary} \propto k^{c},$

since we are assuming $A_\text{capillary}$ and $\ell_\text{capillary}$ do not depend on organism size. Since most of the blood is actually carried in the aorta, we can “go up” $c$ levels by multiplying by the $(k^{-1/3})^{-c} = k^{c/3}$, to obtain

$M \propto V \propto k^{c} \cdot k^{c/3} = k^{4c/3}.$

Since $R \propto k^c$, we have finally arrived at Kleiber’s law:

$R \propto k^c = \left(k^{4c/3}\right)^{3/4} \propto M^{3/4}.$

Exercise 15 (tick tock). (a) The volume of the heart scales as $\mathcal{L}^3$. Using Kleiber’s law, deduce that heart rate $B$ (beats per unit time) scales as

$B \propto M^{-1/4}.$

(b) Assume that the size of a cell, and amount of blood it can process before dying, are independent of organism size. Argue that the lifetime of an organism scales as

$T_\text{organism} \propto M^{-1/4}.$

Hint. You can assume that $T_\text{organism}\propto T_\text{cell}$ in an organism-independent way.

(c) Combining (a) and (b), deduce that the number of hearbeats is roughly constant, independent of organism size! Since a normal human heart rate is around $60$ bpm, estimate how many tocks of the ticker are allotted to each creature on earth.

Exercise 16 (walking speed revisited). Above, we predicted a $\mathcal{L}^{1/2}$ scaling for walking speed with organism size. Here, we’re going to instead consider how human walking speed scales with city size. The analysis treats the city as a fractal for distributing people!

## Extra

Physics is often presented as the most arcane and mathematically challenging of natural sciences. But simple physical ideas, combined with pre-calculus mathematics, can be mind-bogglingly powerful. I give several examples.

##### 2.9. Relativity

fractals, quantisation, entropy

Our best theories of nature seem to involve hard mathematics. We have electromagnetism, formulated in the language of vector [ \begin{align} \nabla \cdot \mathbf{E} & = \frac{1}{\epsilon}\rho
\nabla \cdot \mathbf{B} & = 0
\nabla \times \mathbf{E} & = -\mu\dot{\mathbf{B}}
\nabla \times \mathbf{B} & = \mu\mathbf{J} + \epsilon\dot{\mathbf{E}} \end{align
} ] to Schrödinger [ i\frac{\partial}{\partial t}|\psi\rangle = \left(-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}\right)|\psi\rangle ] to Einstein [ R_{\mu\nu} - \frac{1}{2}\Lambda R g_{\mu\nu} = 8 \pi G T_{\mu\nu}, ] physics seems at its most powerful and exotic when expressed in mathematical language. The more opaque, the more powerful and exotic it seems.

January 6, 2020. Physics is often presented as arcane and mathematically challenging. In this post, I try to counter this notion by showing that simple physical ideas, combined with pre-calculus mathematics, can be mind-bogglingly powerful.

Physics is embedded in mathematics. Our best theories of nature seem to require it, and are often stated in the pithy but impenetrable form of equations. For instance, we can summarise gravity with Einstein’s field equations: [ R_{\mu\nu} -\frac{1}{2}R g_{\mu\nu} = 8\pi G T_{\mu\nu}. ] This is a startingly beautiful result, but requires years of training to appreciate mathematically.

https://web.archive.org/web/20160129142844/http://www.eftaylor.com/exploringblackholes/GravWaves150909v1.pdf

Suppose you want to build a pendulum-driven grandfather clock, with the mass suspended on a light rod $\ell = 1.5$ metres long. Approximately how heavy should the pendulum itself be if you want it to swing back and forth with a period of $t = 1$ second? We can answer this with dimensional analysis. We first identify a target quantity: the thing we are trying to control, predict or measure. In this case, it is the mass of the pendulum $m$, which has the dimension $M$:

$[m] = M.$

We then list the relevant parameters which physically determine our target quantity. In this case, they are the period of the pendulum $t$, the length of the rod $\ell$, and the strength of gravity $g = 9.8 \text{ m/s}^2$, without which the pendulum will not oscillate! From the units, this has dimension

$[g] = [9.8 \text{ m/s}^2] = \frac{[\text{m}]}{[\text{s}^2]} = \frac{L}{T^2}.$

In summary, the relevant parameters and dimensions are:

• period, $[t] = T$;
• length, $[\ell] = L$;
• gravitational acceleration, $[g] = L/T^2$. We now guess that the target quantity is some product of powers of the relevant parameters,
$m = t^a \ell^b g^c,$

and by analysing the dimensions on both sides, figure out what those powers are. On the LHS, we have dimension $m$. On the RHS, we have dimensions

\begin{align*} [t^a \ell^b g^c] & = T^a L^b \cdot \frac{L^c}{T^{2c}} = \frac{T}{} \end{align*}

#### 2.2. Example: spring-mass system

Suppose you want to build a spring-driven clock, with time marked off by the oscillations of a mass stuck to the spring. When the spring is compressed or extended, you measure the restoring force $F$ and notice that it is proportional to the displacement $x$ from the spring’s equilbrium position (Hooke’s law):

$F = kx.$

This constant $k$ is called the stiffness, and it has dimensions

$[k] = \frac{[F]}{[x]} = \frac{[ma]}{[x]} = \frac{ML}{T^2L} = \frac{M}{T^2}.$

To make our clock, we want the mass to oscillate with a period of $t_\text{period} = 1$ second. If you are familiar with harmonic motion, you know that a nicer quantity than period is angular frequency:

$\omega = \frac{2\pi}{t_\text{period}},$

with dimensions $1/T$. To help design the clock, we want to know how the angular frequency $\omega$ depends on the stiffness $k$ of the spring and mass $m$.

The general procedure for dimensional analysis is as follows. We first identify a target quantity: the thing we are trying to control, predict or measure. In this case, it is the angular frequency. We then list the relevant parameters which our target quantity should depend on. In this case, these are the stiffness $k$ and mass $m$, with dimensions

• stiffness, $[k] = M/T^2$;
• mass $[m] = M$.

We now guess that the target quantity is a product of powers of the relevant parameters, taking the form:

$\omega = k^a m^b.$

(I’ll comment on how this assumption can break down in the usage notes.) We can find the powers $a$ and $b$ from the requirement that the dimensions on both sides are equal:

\begin{align*} [k^a m^b] &= \frac{M^a}{T^{2a}}\cdot M^b = \frac{M^{a+b}}{T^{2a}} \\ [\omega] &= \frac{1}{T}. \end{align*}

Since there are no factors of mass for the RHS, we have $b = -a$, and hence $T^{2a} = T$, or $a = 1/2$. Thus, dimensional analysis gives

$\omega \sim \sqrt{\frac{k}{m}},$

where $\sim$ indicates that some numbers may have gone astray. In fact, our sneaky choice of angular frequency $\omega$ instead of period $t_\text{period}$ means this is exactly correct, numbers and all. (If we stuck with period we would be off by a factor of $2\pi$, which is not ideal if we want to design a precision timepiece.) We didn’t need to analyse any forces, solve a differential equation, or even deal with numbers. Dimensional analysis let us skip straight to the answer!

So, if your high school physics lab has springs of stiffness $k = 10^2 \text{ N/m}$ ($100$ Newtons for every meter displaced), then to obtain an oscillation period of $t_\text{period} =1$ second, you should attach a mass

$m = \frac{k t_\text{period}^2}{4\pi^2} = \frac{10^2 \text{ kg}}{4\pi^2} \approx 2.5 \text{ kg}.$

MAKE MORE INTERESTING

1. Using dimensional analysis, show that the angular frequency does not depend on the initial displacement $x$ of the mass.

#### 2.3. Example: the wobbling pupm

The spring-mass example is neat, but not particularly exciting. We can make it more interesting by adding some quantum mechanics into the mix. The fundamental constant appearing in quantum mechanics is Planck’s constant, with SI value

$\hbar = 1.05 \times 10^{-34} \text{ J s}.$

(Technically, this is the “reduced” Planck constant favoured by theoretical physicists. The original constant is $h = 2\pi \hbar$.)

Pushing an object through a fluid takes work, since as we plough through we must push the fluid aside. In realistic fluids, adjacent layers like to stick together, so they resist the shearing; a phenomenon called viscosity. This is measured by a number $\mu$,

with the simplest being to the terminal velocity of a sphere falling slowly through a fluid. This may sound rather artificial, but it lets us explain why clouds stay in the air, despite being made of droplets much denser than air.

The hacks are in no particular order except for the “fundamental hacks” of dimensional analysis and Fermi estimates. Everything else can be read independently. For each hack, I quickly outline the physics, do some examples, and finish with usage notes.

The mother of all Fermi estimates was is due to Enrico Fermi himself, who estimated the strength of the Trinity nuclear test by dropping a few pieces of paper. Fermi doesn’t explain his precise reasoning, but states that he dropped the paper from a height of around $6$ feet ($1.8$ m) and observed a displacement of $2.5$ m as the blast wave passed. Fermi was $16$ km away from the detonation point.

What is the key physical mechanism? The paper moved because the air around it was displaced.

Exercise 1 (spring-driven clock). You can build a clock out of a spring and an old pumpkin by attaching the pumpkin to one end and fixing the other, compressing the spring, and letting it wobble back and forth. (If you do this horizontally, you can ignore gravity.) When the spring is moved from equilibrium a distance $x$, there is a restoring force proportional to displacement, $F = kx$, where $k$ is the stiffness. If the hardware store only stocks springs of stiffness $k = 100 \text{ N/m}$, and you want a period of $t = 2$ s, how heavy should your pumpkin be?

We’ve already seen examples of KISS when performing dimensional analysis. When choosing a list of relevant parameters, for instance, we neglected things that we did not expect , our answers are usually only correct to within an order of magnitude because we cannot determine the numbers out front.

If you are estimating “in real life” (rather than in a test) check Google.

But if for whatever reason, act like a crowd, and try to intuit some over- and underestimates you can average.

Here’s a simple example.

When we write our guess as a product of subestimates, we are assuming the final answer is proportional to each subestimate. A very prosaic instance of nonlinearity is when we should use the usual arithmetic mean, $(a+b)/2$. For instance, with power, there may be economies of scale which reduce the average power usage when there are more people. Or maybe larger cities will have more amenities, and average power usage will increase. Who knows? Whatever the answer, it is probably a fairly mild power law nonlinearity. We’ll discuss power laws below, but I’ll finish with the much more severe example of exponential nonlinearity.

When we write our guess as a product of subestimates, we are assuming the final answer is proportional to each subestimate, e.g. that the power usage in Vancouver is proportional to the number of people. But this may not be true! For instance, with power usage, there may be economies of scale which reduce the average power usage when there are more people. Conversely, maybe larger cities have more amenities, and average power usage increases. Examples like this usually obey power laws, where

$\text{final guess} \propto (\text{factor})^\gamma,$

for some number $\gamma$.

With this sort of dependence, you can use Fermi estimate techniques to make a guess, but the answer isn’t really well-defined to an order of magnitude. Different fists will differ by more than two quarters! (Our estimate of $n$ will probably be accurate up to an order of magnitude, i.e. it could be around $3n$ or $n/3$. So our estimate of the probability is likely to be good up to a cube or cube root! We could view this as ticks on a doubly logarithmic ruler, but it’s a bit mind-boggling.)

### 1. Introduction

For the last year or so, I’ve been heavily involved in running a high school physics circle. I’ve learned a lot about logistics, leadership, and leaning towers of pizza, but surprisingly, the biggest lesson has been about physics itself. In a physics circle, the mandate is to write problems that high school students find challenging but fair given their limited background knowledge. In particular, problems should avoid calculus, which is not part of Canada’s high school curriculum. Judging from competition problems, this restricts us to tediously intricate mechanics problems and a other elementary topics.

But preparing for my PhD comprehensive exam, I realised that it was possible to do more with less.

I decided to view this as a challenge: do more with less. I am not above writing tediously intricate mechanics problems, but it was much more instructive to master some simple physics hacks and use them to discover new things. I’m not sure about the students, but it has certainly made me a better physicist!

The goal of this post is to share a few of these hacks and convince you of their awesome power. To keep things interesting, I avoid overlap with my physics circle problems though I encourage you to have a look if you want more examples. I also recommend Sanjoy Mahajan’s book Street-Fighting Mathematics, covering similar ground in greater depth.

January 6, 2020. Physics is often presented as the most arcane and mathematical of natural sciences. Here, we show how simple physical ideas, with a dash of pre-calculus mathematics, can still be awesomely powerful.

, and we can actually use dimensional analysis to account for some factors of $\pi$ (see Extra dimensions).

We say that the answer scales as $\text{factor}^p$, or with power $p$.

Suppose we have some random process, like tossing a coin, whose outcome we label by $X = \text{H, T}$ (heads and tails). The expectation of a function $f(X)$, denoted $\langle f(X)\rangle$, is the average over outcomes, weighted by their probability. For a fair coin, this is just

$\langle f(X)\rangle = \frac{1}{2}[f(\text{H}) + f(\text{T})],$

but if the coin is biased, and heads has probability $p$, then

$\langle f(X)\rangle = pf(\text{H}) + (1-p)f(\text{T}).$

Suppose we toss two coins independently, labelling the first outcome $X$ and the second $Y$. “Independent” means that the probability of $X$ and $Y$ is always a product of the single-coin probability of $X$, and the single-coin probability of $Y$:

$P(X, Y) = P(X)P(Y).$

For instance, if you drop dye into water, it will initially spread quickly before slowing, since it consists of many random walkers (dye molecules) obeying the $d \propto \sqrt{t}$ scaling.

World Wide Web conference-2006, Bar-Yossef et a

Exercise 14 (mid-air collision). (a) Show that the ideal gas law (Exercise 3) implies an ideal gas has number density

$n = \frac{\mathcal{N}}{V} = \frac{P}{k_B\mathcal{T}},$

where $P$ is the pressure, $k_B$ is Boltzmann’s constant, and $\mathcal{T}$ the temperature (in Kelvin).

(b) The average air molecule has size $r = 4 \times 10^{-10} \text{ m}$. Using this data, estimate the density of air molecules around you right now.

(c) Find the average distance between collisions of air molecules.

Collision cylinders. Let’s now turn back to random walks of colliding particles. There is a nice way to approximate the step length $\ell$ of the walk in terms of the size and density of colliding particles. Suppose our particles are spherical, with radius $r$. We assume the number of particles per unit volume is $n$, or equivalently, there is on average one particle in a volume $U = 1/n$.

If another particle of the same size comes within $2r$ of the first, the particles will collide, since the edges just touch at that distance. So, as a particle moves, sweep out a collision cylinder of radius $2r$ around it. For a distance $\ell$, the volume of this cylinder is $V = \pi(2r)^2 \ell$. To guess the average distance $\ell$ between collisions, we assume that when the cylinder volume $V$ is large enough to contain one particle on average, a collision will occur. Since $U = 1/n$ is the volume per particle, we find that

$V = 4\pi r^2 \ell = U = \frac{1}{n} \quad \Longrightarrow \quad \ell = \frac{1}{4\pi r^2 n}.$

More generally, particles are more interesting than rigid billiard balls, bouncing off each other elastically. They can attract and repel one another, and particle size is no longer the relevant parameter. Instead, this will be given by an effective area or cross-section $\sigma$:

$\sigma \ell = U \quad \Longrightarrow \quad \sigma = \frac{1}{n\ell}.$

By definition, a collision cylinder of length $\ell$ and cross-section $\sigma$ will contain one particle on average.

Exercise 15 (chain). A chain lies

Exercise 14 (taking the air). Heat is just the kinetic energy of particles. More precisely, the the average kinetic energy per particle, $\epsilon$, is proportional to temperature,

$\epsilon \sim k_B \mathcal{T}.$

This is called the equipartition theorem. I think it gives the most sensible way to think about heat.

(a) Show that if our particles have mass $m$, the average speed is

$v_\text{avg} \sim \sqrt{\frac{k_B \mathcal{T}}{m}}.$

(b) Show using Exercise 3 that an ideal gas has number density

$n = \frac{\mathcal{N}}{V} = \frac{P}{k_B\mathcal{T}}.$

(c) Using the collision cylinder method, show that particles of mass $m$ and size $r$, in a gas with pressure $P$ and temperature $\mathcal{T}$, will spread out according to

$d \sim \left[\frac{(k_B \mathcal{T})^3}{16\pi^2 P^2 r^4 m}\right]^{1/4}\sqrt{t}.$

(d) Finally, the average mass and radius of an air molecule is $m = 5 \times 10^{-26}$ kg and $r = 4 \times 10^{-10}$ m. Estimate how long it takes an air molecule starting in the middle of a room to reach a wall.

A drop of dye in water consists of many random walkers, spreading quickly before the $\sqrt{t}$ scaling slows it down.

Dye particles are much larger than water molecules, so a cloud of dye spreading in a glass of water is Brownian. A drop of dye takes around The viscosity of water at room temperature is $\mu \approx 10^{-3}$ kg/m s.

In 1908, Jean Perrin measured the dance of pollen grains in water. Perrin (1926) and Einstein (1921) picked up Nobel prizes for their efforts.

This is also the reason we use geometric means. If the true value is $x$, and $a$ underestimates by a factor $c$, while $b$ overestimates by a factor $c$, then $\sqrt{ab} = \sqrt{x^2} = x$ is the true value. Note that more subestimates will tend to increase the variance of our answers, so once again, we should KISS.

Exercise 12 (randomness is slow). If you drop some dye into a glass of water, it will bloom out and eventually discolour the whole glass. Dye particles have size $\sim 1\, \mu$m, and water at room temperature has viscosity $\eta = 10^{-3}$ kg/m s.

(a) If the spreading was due to Brownian motion alone, how long would it take?

Exericse 13 (hypercubes). We can regard the random steps $s_i = \pm 1$, $i = 1, \ldots, n$, as consecutive left-right steps on a line. But there is an alternative interpration: we can view them as random coordinates of a point in $n$-dimensional space,

$\vec{x} = (s_1, s_2, \ldots, s_n).$

There are $2^n$ such vectors, forming the corners of an $n$-dimensional hypercube of side length $2$:

$I_n = \{(x_1, x_2, \ldots, x_n) : -1 \leq x_i \leq 1\}.$

The length squared of one of these random corners (see Exercise 18) is

$d^2 = |\vec{x}|^2 = s_1^2 + s_2^2 + \cdots + s_n^2.$

(a)

For more on hypercubes and random process, see my earlier blog post.

If you find this unconvincing, the details of the proof are spelt out in an optional section below.

Model humans and elephants as spheres. There are around $4 \times 10^5$ elephants left in the world. Assume they are distributed at random, and you move in a straight line without deviating when you spot an elephant. Roughly how many elephants would you expect to collide with in a lifetime?

All this assumes that the colliding objects are like billiard balls, and we won’t need anything fancier. But $\sigma$ is more interesting when objects interact, undergoing attraction or repulsion as a function of distance. In fact, particle physicists spend most of their time calculating cross-sections to understand what happens in the Large Hadron Collider!

where $\sigma$ is your cross-section in the direction of motion. Put differently, it is your surface area, viewed head on. (We are ignoring the “bumps” in volume at the start and end of the cylinder.)

If a fluid particle comes within a distance $r$ of the pollen, a collision will happen, so to help us keep track of possible collisions we draw a collision cylinder of cross-section $\sigma = \pi r^2$ around the grain as it moves. On average, there is a single fluid particle in each volume $U = V/\mathcal{N}$ of fluid We can determine $\ell$ by assuming that when the collision cylinder has length $\ell$, it has volume $U$, and hence on average contains a fluid particle. This leads to

$\sigma \ell = \frac{V}{\mathcal{N}} \quad \Longrightarrow \quad \ell = \frac{V}{\pi r^2 \mathcal{N}}.$

Exercise 13 (midair collision). Two spheres of radius $r$ collide when their centres come within $2r$ of each other, since the edges touch. To find the average distance between colliding particles of the same size, we should draw our collision cylinder with cross-section $\sigma = \pi(2r)^2$.

The average air molecule has size $r = 4 \times 10^{-10} \text{ m}$. Use this to estimate the average distance between collisions of air molecules in the room around you, $\ell = U/\sigma$.

Hint. The ideal gas law lets you rewrite $U = V/\mathcal{N}$ in terms of temperature and pressure. Atmospheric pressure is $101$ kPa.

Exercise 13 (zombies). A group of $500$ zombies shambles around a postapocalyptic car park of size $\sim 50$ m. One of the zombies decides to call it a day, moving in a straight line at walking speed until they hit another zombie, at which point they mumble “brains…” and randomly change direction. Roughly how long will it take them to escape, and how many “brains” will they mutter?

Hint. Use a collision rectangle of width $d$, where $d$ is the diameter of a zombie.

Exercise 14 (mirrorball madness). An eccentric billionaire decides to have a disco in space. The dance floor lies at the centre of a huge glass sphere, with a “gas” of $\mathcal{N}$ mirrorballs floating around it. The mirrorballs are illuminated by laser light shot out from pointers on the dance floor. The giant glass sphere has radius $R$, while the $\mathcal{N}$ mirror balls have radius $r$. Let’s follow the trajectory of a single photon.

(a) Argue that the collision cylinder around the photon has radius $r$.

(b) Compute the approximate time it will take the photon to escape the giant disco ball. You should find

$t_\text{esc} \sim \frac{3r^2 \mathcal{N}}{4 Rc},$

where $c$ is the speed of light.

If you are colliding with objects much smaller than you, then the cross-section is just your surface area viewed head-on, $A_\text{head-on}$. Similarly, if you are colliding with objects much larger than you, then $\sigma$ is related to their cross-section only. If they are randomly oriented with respect to you, this will be the average surface area they present, $A_\text{avg}$. If you and the objects are comparable sizes, things can get even more complicated, so we will just Keep It Spherical. If the colliding objects are spheres, they look the same from any angle, and no averaging is required.

So far, we’re assuming that the objects we’re colliding with don’t move, or at least, don’t move on average. This means our estimates will work for stationary elephants, or a bunch of elephants moving in random directions, but not a herd of elephants charging towards you! Our main interest will be particles in a gas, where this condition holds true. But it’s not too hard to take movement into account, as we now illustrate.

Exercise 15 (zombies). A group of $500$ zombies shambles around a post-apocalyptic car park of size $\sim 50$ m. One of the zombies decides to call it a day, moving in a straight line at walking speed until they hit another zombie, at which point they mumble “brains…” and randomly change direction. Roughly how long will it take them to escape, and how many “brains” will they mutter?

Hint. Use a collision rectangle of width $d$, where $d$ is the diameter of a zombie.

(c) Plug in reasonable values for a box model of a human, and compare the size of the collision cylinders associated with the front and top of the box. You can determine the speed $v$ using Stokes’ law and the fact that raindrops are typically $1$ mm in diameter. On these grounds, can you explain why MythBusters might have had a hard time testing this myth?

(c) If you like trigonometry, determine how many raindrops you encounter for arbitrary horizontal velocity $u$. If you really like trig, do it for box people!

Exercise 2 (falling balls). (a) The Bathysphere was a hollow ball of steel designed for deep-sea exploration. It weighed $2.25$ tons (above water), had a diameter of $1.45$ m, and held the world record for deepest dive until 1949, sinking almost $1$ km ($923$ m to be precise) into the ocean. Roughly how long did it take to reach that depth? The viscosity of cold water is $\mu \approx 0.0016$ kg/m s.

Hint. Assume it is travelling at terminal velocity, and Stokes’ law applies. You should also take buoyancy forces into account!

(b) Keeping with our theme of whimsical rulers, devise a practical scheme to measure the size of small spheres of known density by dropping them into water.

So, imagine a sphere moving through a fluid, e.g. a Bathysphere sinking to the ocean floor. (You can explore this example in Exercise 2.)

Exercise 14 (pachyderm pileup). (a) Estimate how far you will walk in your lifetime.

(b) There are around $4 \times 10^5$ elephants left in the world. Assuming they are distributed at random over the landmass of the globe, and you do not deviate when you spot one, give a rough estimate of the number of elephants you are likely to collide with. (Model yourself and the elephant as spheres.)

The estimates in the previous section work for elephants at rest, but not a herd of elephants charging towards you!

Exercise 2 (falling balls). Keeping with our theme of whimsical rulers, devise a scheme to measure the size of small spheres of known density by dropping them into water.

# Random walks

Dimensional analysis and Fermi estimates are conventional napkin algorithms, though in my opinion, both are ripe for hacking. In contrast, our last algorithm of random walks is not usually viewed as a back of the napkin tool.

Imagine an atom jiggling around randomly in a hot gas. On average, it will travel some distance $\ell$ between collisions. Surprisingly, after $n$ collisions, the approximate distance $d$ from where it started is proportional to the square root of the number of steps:

$d \sim \ell \sqrt{n}.$

How is this possible? The basic trick is to consider the displacement, a vector we label $\vec{x}$, which is made up of $n$ steps $\vec{s}_i$:

$\vec{x} = \vec{s}_1 + \cdots + \vec{s}_n.$

The distance squared is just the length of the displacement square, $d^2 = |\vec{x}|^2$. We can expand the displacement into steps as

\begin{align*} d^2 = |\vec{x}|^2 & = (\vec{s}_1 + \vec{s}_2 + \cdots + \vec{s}_n)^2 \\ & = (s_1^2 + s_2^2 + \cdots + s_n^2) + \text{cross-terms}. \end{align*}

This is just a generalisation of the familiar algebraic fact that

$(x + y)^2 = x^2 + y^2 + 2xy = x^2 + y^2 + \text{cross-terms}.$

If the steps have length $\ell$, then each $s_i^2 = \ell^2$. If different steps are independent and have no preferred direction, then on average, the cross-terms are zero, since different steps have no preferred orientation with respect to each other. For instance, if $x$ is chosen to be $\pm 1$ with probability $1/2$, and $y$ is independently chosen the same way, then $xy = +1$ half the time, and $xy = -1$ the other half. On average, that gives zero. The conclusion is that

$d^2 \sim s_1^2 + s_2^2 + \cdots + s_n^2 = n\ell^2 \quad \Longrightarrow \quad d \sim \ell \sqrt{n},$

as claimed above.

This $\sqrt{n}$ scaling is the defining feature of a random walk. Remarkably, nothing depends on the number of dimensions the displacement $\vec{x}$ lives in. It is just as true for an atom jiggling in three dimensions, a drunkard wandering in two dimensions, or a virtual bacterium foraging in a 216-dimensional simulation.

If a random walker moves with speed $v$, a step takes time $\tau = \ell/v$. After time $t$, the random walker will tend to wander a distance

$d \sim \ell \sqrt{n} = \ell \sqrt{\frac{t}{\tau}} = \sqrt{\ell v}\cdot \sqrt{t} = \sqrt{Dt},$

where we will call $D = \ell v$ the diffusion coefficient. Even though the walker moves at constant speed, the average distance from the origin scales as $d \propto \sqrt{t}$! It’s important to note that “average distance” is a bit of a misnomer. We really mean the average spread of distance travelled. In time $t$, an individual walker will explore a region of size $\propto \sqrt{t}$, while a batch of walkers released from the same point will fan out to cover that region.

We haven’t talked about how to deal with colliding objects all moving in the same direction, and here’s the nice thing: we don’t need to! Instead, let’s think about everything in the reference frame of the rain. Imagine it extends infinitely in all directions, with density of drops $n$. You, the sphere, travel in a vertical direction at speed $v$, and horizontally towards the shelter (represented as a vertical line a distance $d$ away) at some speed of your choice, $u$. The question is: does the number of drops you collide with on your way to shelter depend on $u$? If it doesn’t, you may as well walk.

So, you are a sphere of radius $R$ caught in a rainstorm, and shelter is a distance $d$ away. The rain consists of tiny balls of water, falling directly down (no wind) at some speed $v$. Viewed in the reference frame of the falling drops, there are $n$ drops per unit volume. The speed $v$ and density $n$ can change with height, but we’re only interested in these quantities near the ground, so we are at liberty to imagine that they are constant everywhere.

Exercise 12 (flipping coins). Our discussion so far has been a little abstract, but we can make random walks concrete with coin flips. So, take a fair coin and start flipping it, with the outcome of the $i$th flip $s_i = \pm 1$, with $+1$ for tails and $-1$ for heads. The sum of $n$ flips is $x$, and really do just multiply and expand, with

$x^2 = (s_1 + \cdots + s_n)^2 = s_1^2 + \cdots s_n^2 + 2 \left[s_1 s_2 + \cdots + s_{n-1}s_n\right].$

We can think of $x$ as describing the position of a random walk on the number line. Let’s check the assertions we made above work!

(a) For a fair coin, show that on average, $s^2 = 1$.

(b) Consider two fair coin flips, $s$ and $s’$. Show that $ss’ = +1$ with probability $1/2$ and $ss’ = -1$ with probability $1/2$. Conclude that, on average, $ss’$ vanishes.

(c) Combining the last two arguments, conclude that after $n$ coin flips, you tend to spread a distance $\sqrt{n}$ around the origin, with $x \sim \sqrt{n}$.

Let’s now briefly consider two ways for the random walk description to fail: bias and correlation. A coin has bias when it has probability $p \neq 1/2$ of giving $+1$, and probability $q = 1- p$ of giving heads. The average value of a coin flip $s$ is then $p - q$. Finally, coin flips $s, s’$ have correlation $c$ if the average value of $ss’$ is

$(ss')_\text{avg} = c + (p-q)^2,$

i.e. there is a tendency $c$ to align, in addition to the bias.

(d) Show that $s^2 = 1$ even if the coin is biased or flips are correlated.

(e) Suppose that coin flips are uncorrelated, with $c = 0$ for all pairs of distinct flips. Argue that

$x^2 \sim n + 2n(n-1)(p-q)^2.$

(f) Show that, for large $n$, biased coins execute a walk obeying

$x \sim \sqrt{2}(p-q)n.$

(g) Now consider an unbiased coin. Show that if only successive coin flips are correlated, the walk still obeys $d \propto \sqrt{n}$.

Thus,

### 1.1. Random walks

Prerequisites: basic probability theory; vectors.

We will prove the square root scaling of random walks, first in 1D, and then extend almost immediately to many dimensions. The only prerequisite is a little probability theory and knowledge of vectors.

Proof (1D). Suppose we toss a coin, and move a counter left or right one unit depending on whether we gets heads or tails. Label the outcome of the $n$th toss $s_n$, where $s_n = -1$ for heads and $s_n = +1$ for tails. If we start at $0$, the position $x$ after $n$ tosses is the sum of steps:

$x = s_1 + s_2 + \cdots + s_n.$

This is a random process, so what do we expect to happen on average? We can calculate this with an expectation: a weighted average over all the things that can happen, where the weights are probabilities. We denote this by $\langle f(s_1, \ldots, s_n)\rangle$, where $f$ is any function of the steps. For instance, if the coin is fair, then on average an individual step is zero:

$\langle s\rangle = P(s=-1)(-1) + P(s=+1)(+1) = \frac{1}{2}-\frac{1}{2} = 0.$

A very similar calculation shows that $x$ vanishes on average:

\begin{align*} \langle x\rangle &= \langle s_1 + \cdots + s_n\rangle\\ & = \langle s_1\rangle + \cdots + \langle s_n\rangle\\ & = 0 + \cdots + 0 = 0, \end{align*}

where we used the fact that expectation is linear, $\langle f + g\rangle = \langle f\rangle + \langle g\rangle$. This makes sense, since if the coin is unbiased, it has no preference between heads and tails. If $\langle x\rangle >0$, for instance, then the coin is exhibiting a bias towards tails. The story is different for the square of a step:

$\langle s^2\rangle = P(s=-1)(-1)^2 + P(s=+1)(+1)^2 = \frac{1}{2}+\frac{1}{2} = 1.$

What about a product of different coin flips? Each possible outcome $–$, $-+$, $+-$, $++$ has chance $1/4$, and hence

$\langle X_i \cdot X_j\rangle = \frac{1}{4}\left[(-1)^2 + (-1)(+1) + (+1)(-1) + (+1)^2\right] = 0.$

Combining these two facts, we can calculate the variance:

\begin{align*} \langle x^2\rangle & = \langle (s_1 + \cdots + s_n)^2\rangle \\ & = \langle s_1^2\rangle + \cdots + \langle s_n^2\rangle + 2\left\{\langle s_1\cdot s_2\rangle + \cdots + \langle s_{n-1}\cdot s_n\rangle\right\} \\ & = n \langle s^2\rangle = n. \end{align*}

The counter moves back and forth, and on average $\langle x\rangle = 0$. But the size of the region explored as it moves back and forth is the square root of the variance, also called the root mean square (rms) displacement, $\sqrt{\langle x^2\rangle} = \sqrt{n}$. This is the distance from the origin the counter will tend to wander in the first $n$ steps. If instead of steps $\pm 1$, we have steps $\pm \ell$, then $\langle s^2\rangle = \ell^2$ and the rms displacement becomes

$\sqrt{\langle x^2\rangle} = \ell\sqrt{n}$

as claimed above.

Proof (many D). The proof is almost identical in $d$ dimensions, where a step can be written as a vector $\vec{s}$ with $d$ components:

$\vec{s} = (s^1, s^2\, \ldots, s^d).$

The length of $\vec{s}$ is given by a generalisation of Pythagoras’ theorem (Exercise A.1):

$|\vec{s}| = \sqrt{(s^1)^2 + (s^2)^2 + \cdots + (s^d)^2}.$

Let’s assume that the average step length is $\ell$, and steps are unbiased, so

$\langle \vec{s} \rangle = \vec{0} = (0, 0, \ldots, 0), \quad \langle |\vec{s}|^2 \rangle = \ell^2.$

Finally, we assume that steps are independent, so any two components of distinct steps satisfy

$\langle s_i^a s^b_j \rangle = 0,$

where $\vec{s}_i, \vec{s}_j$ are distinct steps, and $a, b$ labels components. Then, if $\vec{x} = \vec{s}_1 + \cdots + \vec{s}_n$, the random walk has variance

\begin{align*} \langle |\vec{x}|^2\rangle & = \langle (s_1^1 + s_2^1 + \cdots s_n^1)^2 + \cdots + (s_1^n + s_2^n + \cdots s_n^n)^2 \rangle \\ & = \langle [(s_1^1)^2 + (s_2^1)^2 + \cdots (s_n^1)^2] + \cdots + [(s_1^n)^2 + (s_2^n)^2 + \cdots +(s_n^n)^2] \rangle \\ & \qquad \qquad + 2 \langle [s_1^1 \cdot s_2^1 + \cdots + s_{n-1}^1\cdot s_n^1]+\cdots + [s_1^n \cdot s_2^n + \cdots + s_{n-1}^n\cdot s_n^n] \rangle\\ & = \langle [(s_1^1)^2 + (s_1^2)^2 + \cdots (s_1^n)^2] + \cdots + [(s_n^1)^2 + (s_n^2)^2 + \cdots +(s_n^n)^2] \rangle \\ & = \langle [(s_1^1)^2 + (s_1^2)^2 + \cdots (s_1^n)^2] + \cdots + [(s_n^1)^2 + (s_n^2)^2 + \cdots +(s_n^n)^2]\\ & = \langle |\vec{X}_1|^2\rangle + \cdots +\langle |\vec{X}_n|^2\rangle\\ & = n \ell^2, \end{align*}

where on the third line we used the fact that the components of different steps are independent, on the fourth line we reorganised the terms $(s_i^a)^2$ into individual steps, on the fifth line we used the definition of length and the linearity of expectation, and on the last line we used the assumption about average step length. Taking square roots, we find the rms displacement

$\sqrt{\langle |\vec{x}|^2\rangle} = \ell\sqrt{n},$

in any number of dimensions. Notice also that we do not need to assume the steps are the same length or live on lattice.

Exercise A.1 (Pythagoras in higher dimensions). Consider a vector $\vec{C} = (C^1, \ldots, C^d)$ in $d$ dimensions.

(a) We can write the vector as a sum

$\vec{C} = (C^1, \ldots, C^{d-1}, 0) + (0, \ldots, 0, C^d) = \vec{A}+\vec{B}.$

Argue that these two summands are at right angles, and use Pythagoras’ theorem (in the plane spanned by $\vec{A}$ and $\vec{B}$) to argue that

$|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 = |\vec{A}|^2 + (C^d)^2.$

(b) Now iteratively repeat this argument for $\vec{A}$, and deduce that, as we claimed above, length squared in $d$ dimensions is

$|\vec{C}|^2 = (C^1)^2 + \cdots + (C^d)^2.$