Finding tangents, computing turning points and establishing extreme values of functions all seem like problems that require calculus for us to solve. But in fact there are often techniques which involve nothing more than some geometric thinking and elementary algebra. Imagine we were in a world where calculus hadn’t yet been invented (or that you haven’t learned any calculus yet)—how would we go about maximising and minimising functions, finding tangents and so on?

As a first problem, let’s find the minimum value of $x+y$ for points on the unit circle $x^{2}+y^{2}=1$.

One way of solving this problem is to consider lines with equation $x+y=c$ for different values of the constant $c$:

Since the circle lies to the right of $x+y=-2$, we can see that $x+y$ has got to be larger than $-2$, as otherwise $(x,y)$ cannot lie on the circle. It should also be clear that we need to choose a value of $c$ so that the line is a tangent to the circle.

We can do this by observing that if we try to find the value of $x$ (or $y$) for which the line is a tangent to the curve, there will only be one solution to the resulting equation (corresponding to the line ‘touching’ the circle at only one point). But this corresponds to the discriminant of the quadratic resulting from solving the two equations simultaneously being zero: substituting $y=c-x$ into $x^2+y^2=1$, then doing a little rearranging gives
\[2x^{2}-2cx+(c^{2}-1)=0.\]
The discriminant of this quadratic is
\begin{align*}
\text{“}b^{2}-4ac\text{”}&=(-2c)^{2}-4\times2\times(c^{2}-1)\\&=8-4c^{2}.
\end{align*}
This is zero when $c=\pm\sqrt{2}$, and so the minimum value of $x+y$ for points on the unit circle is $-\sqrt{2}$.

Tangents and turning points

We can use the idea of finding a single intersection point (corresponding to a line being tangent to a curve) to solve some problems where differentiation would be the standard approach: finding tangents to curves and turning points on curves.

As an example of the former, let’s suppose that we want to find the tangent to the curve $y=2x^{2}-5x+5$ at the point $(1,2)$. We know that a line with gradient $m$ through the point $(1,2)$ has equation $y-2=m(x-1)$. The diagram below shows the curve with several lines through the point $(1,2)$:

If we rearrange the equation of the general line we get $y=mx+2-m$, which we can then substitute into the equation of the curve to give
\[mx+2-m=2x^{2}-5x+5.\]
With a bit of rearranging, we can turn this into
\[2x^{2}-(5+m)x+(3+m)=0.\]
The discriminant of this quadratic is
\begin{align*}
(-(5+m))^{2}-4\times2\times(3+m)&=m^{2}+2m+1\\&=(m+1)^{2},
\end{align*}
and so the value of $m$ for which the discriminant is zero is $-1$. So the tangent to the curve at $(2,1)$ is $y=-x+3$:

As an example of finding turning points, let’s consider the curve $y=(x^{2}+3)/(x-1)$:

From the diagram, we can see that there are two turning points—shown by blue crosses—where the curve has a local maximum or minimum value. It might be worth noting at this point that we would require the quotient rule if we were to find these points by differentiation—the single intersection method here is certainly more elementary! We start by considering the graph intersecting a horizontal line; some examples are shown on the graph: one intersects twice, the other two do not intersect at all. If a horizontal line touches the curve just once—and so is a tangent—it will do so at one of the turning points. For a single intersection, we again require a discriminant to be zero. A horizontal line has equation $y=c$: solving this simultaneously with the equation of the curve gives $$c = \frac{x^2+3}{x-1},$$ which rearranges to \[x^{2}-cx+(3+c)=0.\]
The discriminant of this quadratic is
\begin{align*}(-c)^{2}-4(3+c)&=c^{2}-4c-12\\&=(c-6)(c+2),\end{align*}
so the values of $c$ that give a discriminant of zero are $-2$ and $6$. These correspond to the $y$-coordinates of the turning points, and it’s a quick task to find the corresponding $x$ values, giving the turning points as $(-1,-2)$ and $(3,6)$:

Cubics and beyond

So far, our problems have all boiled down to setting a quadratic’s discriminant to zero and working from there. Unfortunately, with many functions this approach won’t be possible. If we want to use our method above for finding a tangent to a cubic, we find it’s not so straightforward—when we substitute our straight line equation into our cubic we get another cubic rather than a quadratic, so there’s no way of setting the discriminant equal to zero. We can observe that when we reach this second cubic it must have a double factor at the point of tangency, so we can force it to be of the form $(x-a)^{2}(x-b)$ by equating coefficients.

There is, however, a lovely trick using polynomial division to find tangents to polynomial curves, which was brought to my attention by David T Williams on Twitter (and it was still Twitter then…), which is to use the fact that when a polynomial, $p(x)$, is divided by $(x-a)^2$, the remainder gives us the expression for the tangent at $x=a$.

To see why this is the case, write
\[p(x)=q(x)(x-a)^{2}+r(x).\]
When $x$ is close to $a$ we can see that $x-a$ is small and so $(x-a)^{2}$ is very small—in other words $p(x)\approx r(x)$. Since $(x-a)^{2}$ is a quadratic, the remainder will be a linear function, so we’re approximating $p(x)$ by a linear function $r(x)$ at $x=a$. But this is exactly what a tangent is! (For those who have studied Taylor series, our remainder $r(x)$ represents the first two terms of the Taylor series for $p(x)$ about $x=a$.) This method works nicely for polynomial functions only, where we can use long division to find the appropriate linear remainder.

As an example, let’s find the tangent to $y=x^{4}-7x^{3}+20x^{2}-37x+42$ at $x=3$.

All we need to do is divide the polynomial by $(x-3)^{2}=x^{2}-6x+9$:

And so the required tangent is $y=2x-3$:

So it turns out there’s a lot we can do with curves without resorting to calculus. In the example above we don’t even need to calculate the $y$-value at $x=3$. if you’re fluent with polynomial division it’s quicker than using differentiation.

As a final teaser, try the following problem using (a) a discriminant method and (b) calculus:

If $x^{2}-2xy-4x+4y^{2}=12$, what’s the maximum value of $y-2x$?

The first Fermi problem I tackled was in an introductory astrophysics lecture. One of the questions on the first problem sheet was: Are there more mosquitoes in the world than stars in our galaxy? Questions like these are intriguing, easy to understand, and have a definite answer. However, with only limited information available, we need to use a creative way of guesstimating to answer them. In principle a mighty (but hypothetical) being like Maxwell’s demon or the flying spaghetti monster would be able to count all mosquitoes and stars at one instance in time (or, as physicists might put it, in one specific frame of reference), but this brute force approach might not be necessary to answer the question: educated guesses can do the job just as well.

Total Eclipse of the Hair-t. Bonnie Tyler would be proud.

The legendary Italian physicist Enrico Fermi was the master of these questions. He was a key player in the Manhattan project to build the first nuclear weapon—you might spot him in the background of Christopher Nolan’s Oppenheimer, which unfortunately doesn’t show how he estimated how much energy the bomb would release. During the first detonation of the bomb, the Trinity test on 16 July 1945, Fermi threw a piece of crumpled paper in the air and observed how far it was blown away by the shock waves released in the explosion. He came up with an instant estimate of the energy released by the bomb, which was later confirmed after the analysis of all the data obtained during the test. Fermi solved a Fermi problem in the most elegant way!

Proposing and solving Fermi questions is an essential skill not only for physicists. Companies use questions like How many golf balls would fit in a car? or How many leaves are on a tree? in job interviews to test applicants’ creativity and capability for abstraction.

For me, Fermi problems lurk around every corner and I think they make life more fun… like the other day when I was waiting at the hairdressers for a new hair cut. This is usually the time when you can think about the universe, and stumble across deep questions of major importance for humankind. Questions like: How quickly is my hair growing and how much hair do I have? As an astrophysicist, I wanted to think about them on a cosmic scale. Next time you’re struggling for small talk at the hairdressers, maybe you can ask them…

Are there more hairs or stars in the universe?

If we want to make an educated estimate, it’s often best to start with simple observations in our daily life. Looking in the mirror, my head is roughly a sphere, with diameter 16cm. About half of this sphere is covered with hair, so the total hair-ea on my head is \[A_\text{head} = \frac12 \times 4\pi \times \frac{(16\,\text{cm})^2}{4} \approx 400\,\text{cm}^2\] or, to use traditional German units, $5.6 \times 10^{-5}$ standard football fields.

Hairy Earps: About half of the surface hair-ea of Mary’s head is covered in hair. Image: James Boyes, CC BY 2.0

If all the hairs were squashed up on top of each other, how many could fit on a head? Each hair is between 50 and $120\,\mu\text{m}$ wide—we’ll call it $80\,\mu\text{m}$—so the average cross section of a single hair is \[\overline{A}_\text{hair} = \pi \times \frac{(80\,\mu\text{m})^2}{4} = 5000\,\mu\text{m}^2.\] This is about $7\times 10^{-13}$ standard football fields. So if all the hairs were tightly packed, we’d get an estimate of $A_\text{head}/\overline{A}_\text{hair} \approx \text{8,000,000}$ hairs on our heads. However, scalp hair is not densely packed—between two single hairs there is a distance. I had another look in the mirror, and observed that the average distance between two hairs is on the order of 1mm. That means between one hair and the next one, about 1mm $/$80$\mu$m $\approx$ 13 hairs could fit.

Rubeus Hair-grid: it’s helpful to assume that hairs grow in a grid pattern with 1mm gaps between hairs. That’s a gap of around 13 hair-widths, so only 1 out of 196 `hair spaces’ are filled.

There’s only one hair spot occupied in each square of $14\times 14$ hair diameters, so we should divide our estimate by 196. This means that there are about to get $\text{8,000,000} / 196 \approx \text{40,000}$ hairs on each person’s head, at a density of roughly 100 hairs per $\text{cm}^{2}$. In reality, hairs don’t grow in a grid pattern, so the actual number is a bit higher: the average person has 100,000 hairs on their head. We’re out by a factor of 2: pretty good going for a guesstimate!

We just need two more facts to be able to answer our question. Firstly, how many stars are there? In the observable universe we believe there are $10^{11}$ galaxies, with $10^{11}$ stars in each galaxy (on average)—so there are about $10^{22}$ stars in total. Secondly, there are about $8\times 10^{9}$ humans in our solar system. We’ve just worked out that each one has about $10^5$ hairs on their heads, adding up to $8\times 10^{14}$ hairs, which we’ll round up to $10^{15}$. Thus we have more hairs in our galaxy than stars but there are more stars in the observable universe than hairs!

How long would it take for a single hair to grow to the moon?

All this talk of stars is lovely, but I wanted to focus on something a bit closer to home. I wondered: how long would it take for a single hair to grow long enough to reach the moon?

Hair-ston, we have a problem. Image: Gregory H Revera, CC BY-SA 3.0

Let’s start with the speed of hair growth. As a child I learned that my scalp hair grows at a rate of about $v_\text{hair} = 1\,\text{cm}$ per month (As a grown-up I can check Wikipedia: the range is more or less $v_\text{hair} = 0.6$–$3.35\,\text{cm}/\text{month}$.). This seems like a natural unit to use for hair growth, but unfortunately it is not a SI unit, nor does it relate to football fields. As an astronomer, I’d rather work in more familiar units: \[v_\text{hair} = 3.86\times 10^{-12}\,\text{km}/\text{s}.\] In order to have a better feeling of how fast this is, we can compare the speed of hair growth with the speed of light, $c_\text{light} = \text{299,792.458}\,\text{km}/\text{s}$. It turns out that photons travel through space by a factor of $c_\text{light}/v_\text{hair}\approx 10^{17}$ times faster than hair grows!

The distance between the Earth and the moon is \[D_\text{moon} =\text{384,400}\,\text{km},\] which is about one light-second. Since hair grows $10^{17}$ times more slowly than light moves, our total growth time is $t=D_\text{moon}/v_\text{hair}\approx 10^{17}\,\text{s}$. In human-readable units this time is about $3\times 10^{9}\,\text{yrs}$—that’s pretty close to the age of Earth, or $4.5\times 10^{9}\,\text{yrs}$. Sending humans to the moon via spaceships based on human hair growth is thus not a very smart idea—not to mention the amount we’d be spending on shampoo.

Ah, but what if we all teamed up? Since we have about $8\times 10^{9}$ contributing people on the planet, the global human scalp hair production rate (or GHSHPR for short) is about $ 8\times 10^{9} \times \text{100,000} \times v_\text{hair} \approx 3000\,\text{km}/\text{s}$. This means that, if we all pitched in, we could grow enough hair to reach the moon in about two minutes!

On the other hand, with 8 billion people and $100\,000$ hairs each, sticking all those hairs together into one mega-hair might be a problem. We’d need $8 \times 10^{14}$ pieces of tape, and even if we could stick ten hairs per second, it would take us $8\times 10^{14}\times 0.1\,\text{s}$, or about two and a half million years, to glue it all together.

Total Eclipse of the Hair-t. Bonnie Tyler would be proud.

What if we wanted to cover the Earth in hair?

Let’s admit it—one big downside of our planet is that it’s not very fluffy. But we can make it fluffy (like the Tribbles in Star Trek) by covering it with hair!

The Earth’s radius is $R_\text{Earth} = 6371\,\text{km}$, so its surface area is $4 \pi \times (6371\,\text{km})^2 \approx \text{40,000,000}\,\text{km}^2$. To make the Earth satisfyingly hairy, we’d need $4\pi R^2_\text{Earth} /\overline{A}_\text{hair} \approx 10^{22}$ hairs. This is about a tenth of a mole of hair, but it’s several orders of magnitude more than the hair we’ve got available. Even if we all teamed up, it would take absolutely ages to grow enough hair. According to the American Academy of Dermatologists, most people lose about 50–100 hairs per day, and since we seem to have the same number of hairs from one day to the next, let’s assume we’re growing 100 hairs a day each. At 8 billion people on Earth, that’s $8 \times 10^{11}$ new hairs every day. It would take us $10^{10}$ days—or 27,000 years—to reach our goal. This is probably the main reason why our proposed plan of creating the fluffiest planet in the universe will never get funding. On the other hand, we do have enough hair to turn Manchester into Mane-chester. If anyone wants to get in on my grant application, please do let me know.

Engineering: without it, we’d have a hard time applying all the brilliant maths people do to the real world, and an even harder time pronouncing “Stem”.

But what does a career in engineering really look like? How much maths does it involve? Is it really just emails?

Fear not: Chalkdust has the answers. In this issue’s edition of our day in the life series, we hear from three people who put engineering mathematics into practice in their daily lives:

• Isobel Voysey is a PhD student at the Edinburgh Centre for Robotics.
• David Fairbairn is a mathematician at Tharsus, and a PhD student in Durham University’s department of mathematical sciences
• Kirsten Ross is a design engineer, specialising in building service design.

From designing heating systems for schools to helping robots avoid crashing into each other, they’re all working on real-world problems every day. Read on to find your new dream job…

Continue reading →

A group of children stand around in a circle. Each one puts a foot into the centre. One of the children starts tapping the feet around the circle in a clockwise direction and reciting one word per foot:

Ip dip dip,
My little ship,
Sailing on the water,
Like a cup and saucer,
But you are not in…
it!

The child whose foot is tapped on the final it is eliminated, then the rhyme starts again. Eventually, there is only one child left: the winner. The winner might get some chocolate, or might get to be it in the next game of tag. Imagine there are 13 children to start with. They begin by standing in a circle:

In the first round, they count round starting from 1 until number 8 is eliminated:

They then continue from the next number 9,until eventually number 4 is eliminated:In the next few rounds, 2, 3, 7, 13, 12, 6, 9, 10 and 5 are eliminated (in that order), leaving just 1 and 11. The next ‘ip’ starts on 11; 1 is dip’; 11 is the second ‘dip’… this continues until finally 11 is ‘it’. The winner is 1. If you were one of the children, you might want to save a lot of time and work out who was in the winning position before the eliminations started. This is often called ‘the Josephus problem’.

Generalisation

In the example above, we worked out who would win if we started with 13 people and counted out the whole ‘ip dip dip’ rhyme. But we’d like to be able to work out the winner if we start with any number of people. To make it a little easier, we’re going to use a shorter rhyme with just two words (‘Ip it!’). We can start by doing some examples and looking for a pattern. We can choose some values for the number of children, $n$, and by following the rules of the Josephus problem find the location of the winner, $w$. Here is the table for the winner depending on the number of children:

From the table, we can notice a couple of patterns: the winner is always odd; if we start with $2^k$ players, the winner is always 1. We can prove that both of these patterns will also hold true for larger numbers. To show that the winner is always odd, we simply need to notice that all the even positions lose during the first set of rounds. We can show that the winner is always 1 when we start with $2^k$ players by induction: in this case, the $2^{k-1}$ even numbers are eliminated first, leaving $2^{k-1}$ numbers. We can then renumber the remaining players, and in the subsequent round of eliminations, the even numbers are eliminated again. The number 1 is never removed in these rounds of eliminations, so remains as the winner.

Modular arithmetic

Thinking about everyone standing in a circle suggests that modular arithmetic might be the light of our problem. But what is modular arithmetic? Modular arithmetic uses the modulo function—a function that generates the remainder when dividing by a number $k$. For example, when 7 is divided by 2, the remainder is 1, and when 8 is divided by 2, the remainder is 0. We can write these facts as
\begin{align*} 7\text{ mod }{2}&= 1,\\ 8\text{ mod }{2}&=0. \end{align*}
Notice that when we divide by 2, the remainder is either 0 or 1. We can write \[ a\text{ mod }{2}=b, \] where $b$ is 0 or 1. In general, the remainder of $a\div k$ can be from 0 as a minimum to $k-1$ as a maximum. Modular arithmetic is often called clock arithmetic, as a 12-hour clock counts in mod 12. This way of thinking about counting around a circle like on a clock suggests that it might be useful for our problem. Let’s see if it can help us work out what happens if we start with a number of players that isn’t a power of 2.

Finding a solution

We start by guessing that the solution for $n=2^k+r$ (with $r<2^k$) might be \[w = c + a(n\text{ mod }{2^k}),\] where $c$ and $a$ are positive constants. When $n=2^k$, we know from earlier that $w$ is equal to 1. We can deduce that \begin{align*} 1 &= c + a(2^k\text{ mod }{2^k})\\ &=c, \end{align*} and so if our guess was right, the solution is \[w = 1 + a(n \text{ mod} {2^k}).\] We can now try this formula using some values of $n \not={2^k}$. For example, when $n=11$ we know from the table that $w=7$. The nearest power of 2 less than 11 is 8, which is $2^3$. So $k=3$ and our solution formula becomes \begin{align*} 7 &= 1 + a(11\text{ mod }{2^3})\\ &= 1 + a(11\text{ mod }{8})\\ &= 1 + 3a. \end{align*} Our formula works if $a=2$, and so our solution is \[ w=1+2(n\text{ mod }{2^k}). \] If you try this formula out on the other numbers in the table, you’ll see that it always gives the right answer. But checking all of these examples isn’t enough to know that this formula will work for all values of $n$. For that, we need a proof. Luckily finding a proof isn’t too hard. First, we know from earlier that if we start with $2^k$ players then player 1 will win. From this we also know that if we’re in the middle of a game and have reached a point where there are $2^k$ players left, then the player who we are starting with next will win. So if we start with $n=2^k+r$ players, we can find the winner by working out who we’ll be pointing to once the extra $r$ players are eliminated. Noticing that $r\equiv n\mod 2^k$, and that we count out two people for each elimination tells us that in eliminating these $r$ players, we’ll have counted on $2(n\text{ mod }2^k)$ places from the 1 we started at. We have a general formula connecting the number of players, $n$, the winning position, $w$, and we’re not certain this it’s right. But maybe you’re wondering if there’s a nice observation we could use to save ourselves time? For that, we’re going to need binary numbers.

Binary numbers

In our usual decimal numbers, each digit represents the next power of 10. If instead, we use each digit to represent a power of 2, we have binary numbers. There are 2 digits in this number system: 0 and 1. For example, $1101_2$ is a binary number (we include the $_2$ to make it obvious that we’re using binary): it is equal to the decimal number 13 as it has ones in the 1, 4, and 8 positions. Every decimal number can be represented as a unique binary number: this is the key to our trick. Let’s try writing some values of $n$ and $w$ from the table in binary. First up, let’s try $n=10$ and $w=5$: \begin{align*} n &= 1010_2 & w_1 &= 0101_2. \end{align*} Another example: $n=14$ and $w=13$ becomes \begin{align*} n &= 1110_2 & w_1 &= 1101_2. \end{align*} Notice that if we take the 1 at the front of the binary number $n$, then move it to the end, we get $w$. This works for the other numbers too, for example for $n=8$:

This gives the correct value for $w$! You can prove that this version of the solution works for every value of $n$ using our formula from the previous section. I’ll leave you to think about how to do this. We’ve found two ways to work out who the winner is if we eliminate the second person. While I go away and enjoy all the chocolate I’ve won, you can try to work out the solution when we use a longer rhyme…

Do you ever find yourself wondering how the maths you learn helps us to understand and model small things we can’t even see? Yes, I thought so. I know you might hate how physicists mess with maths but bear with me on this one. Hold on tight as we take a joyride through the fusion of mathematics and subatomic wizardry—it’s time to unlock the secrets of perturbative unitarity and those elusive gravitons you may have heard about!

The gravity of the situation

Alright then, imagine this: particle physicists are the ultimate puzzle-solvers, working to understand the tiniest building blocks of the universe. But guess what? It’s like they’ve been handed a humongous jigsaw puzzle with no picture to refer to. The jigsaw pieces? Those are our fundamental particles, such as electrons, quarks and photons. The picture? Well, we aren’t fully sure yet, but the so-called standard model was a decent start and I will introduce it to you soon.

The standard model has some flaws in the sense that it does not include gravity, neutrino masses or dark matter. Among other issues! The standard model features the fundamental particles that exist in the universe. These are the smallest building blocks of matter, light and gravity. The standard model describes how atomic nuclei are held together via the strong force and it also describes what matter consists of. These particles are our building blocks for almost everything else in the universe.

Quarks come in six different flavours and these quarks combine in different ways to form other larger particles. The so called bosons in the standard model mediate the fundamental forces, photons which carry electromagnetism, which we see in light. The W and Z bosons are responsible for the weak nuclear force that is responsible for radioactive decay. The standard model is basically the particle physics version of the periodic table.

The different regimes of physics

So far so particle-like, but I heard particles are waves too, I hear you cry. Imagine you’re looking at a nice pond, and you throw a stone into it; ripples will spread out from where the stone landed and they interact with each other. These ripples create waves on the water’s surface. Now, think of this pond as the universe, and those ripples as our particles.

Quantum field theory, or QFT for short, is a way physicists try to understand how these tiny particles work and interact with each other. Instead of treating particles like billiard balls, QFT treats all these different particles as vibrations or disturbances in quantum fields. Think of this field as an invisible substance filling all of space, and particles are like the ripples or waves in this field. Quantum field theory can successfully describe three of the four fundamental interactions: the weak interaction, electromagnetism and the strong interaction. However, it needs some extra work to describe quantum gravity.

Gravity at large distances

When we place a bowling ball in the centre of a trampoline, it creates a dent in the trampoline’s surface. This is analogous to how massive objects, like planets, warp spacetime according to Einstein’s theory of general relativity. Now imagine our trampoline as spacetime and some kids bouncing around chaotically as our particles. They move around and interact with the dent created by the bowling ball, which is like how gravity acts. This interaction between the kids and the trampoline is what we usually describe using general relativity. Typically, when we think of gravity what comes to mind is the force that pulls things with mass together. It is what causes apples to fall from trees and keeps the planets in our solar system orbiting around the sun, but how do we describe gravity at very small length scales?

In everyday life, we experience gravity in the continuous sense, but on small scales gravity acts like a particle. Individual particles are responsible for these forces: we quantise continuous waves into particles. Quantising is like turning something continuous and smooth into tiny, discrete chunks. It’s like you are filling a glass with water, but instead of a smooth stream, the water comes out in little drops—each drop is a quantum of water. You can’t have half a drop; it’s the smallest possible amount. Quantum gravity works in the same way. We try to break up this continuous force of gravity which acts everywhere in the universe by introducing the quanta of gravity: the graviton particles.

From particles to waves

Many people think that quantum mechanics and general relativity are incompatible, and that the solution is to find some form of a grand unified theory that can account for all the fundamental forces simultaneously. A grand unified theory, often abbreviated as GUT, is like a big puzzle-solving idea in the world of physics such that maybe, there’s one big theory that explains everything: how electrons and quarks behave, how gravity works, and why the universe is the way it is. But this is a bit of an outdated idea. We all accept special relativity and classical mechanics to be different theories but valid in their own regimes (high speeds for SR). For speeds much less than that of light, we arrive at the same results as Newtonian mechanics. The same situation is applied to general relativity and quantum mechanics, we take quantum gravity as the high-energy short-distance limit of general relativity. The overall point of this is because quantum mechanics and general relativity are not fully compatible when dealing with more extreme limits, we introduce a bridge between them, which we call effective field theories (EFTs).

The standard model in all its glory

The actual description of the universe is like a big orchestra performing some very complex music at a busy stadium with lots of musicians (the particles). But when this orchestra plays at a smaller venue, they might use a smaller group of musicians, and the music is simplified. This is how EFTs work, they simplify our theories of the universe to focus on just the important pieces at a time. With these EFTs we do not need to choose, per se, a specific theory for quantum gravity—string theory or loop quantum gravity, for example—but merely just assume one exists. In the low-energy large-distance limit, the behaviour of quantum gravity is then described by general relativity, which is a classical theory, where spacetime curvature and gravity can be calculated using Einstein’s equations (as seen in Chalkdust‘s iconic hall of fail). EFTs are only valid within their energy regime, but this is OK because you wouldn’t expect to model a ball rolling down a hill with special relativity (unless you are crazy or the ball is extremely fast).

Some HEFTy techniques: perturbative unitarity to predict the Higgs boson

Maybe you have seen the standard model Lagrangian, where there are so many terms just to describe how particles move, their masses and how they interact. EFTs take terms in the Lagrangian densities which are relevant at the required energy scale to form approximations on particle interactions and scattering amplitudes. Think about a magnetic material: in principle we could describe it in terms of individual atoms and their spins, but this would be incredibly complicated. Instead we zoom out and describe the collective behaviour of the atoms through the magnetisation, a measure of how the material responds to external magnetic fields. This is an effective field theory, the effective part means averaging out unnecessary details. The aim is to average over short length scales to remove some of the intricacies of the theory, which the EFT does in terms of an action and a Lagrangian density.

In much of physics, the Lagrangian, like the energy, is a bookkeeping tool that helps us describe the behaviour of a system mathematically. Think of a mechanics problem like a ball rolling down a hill. The Lagrangian describing the ball’s motion is the difference between its kinetic and potential energies. Extremising this quantity leads to the familiar, Newtonian, equations describing the ball’s motion.The action is an integral over space and time of the Lagrangian density, a version of the Lagrangian for field theories, describing the motion of particles along different paths. Extremising the action leads to the equations of motion; it is an optimisation problem to find a particle’s most efficient trajectory.

In the quantum world all the paths that a particle can travel between two points, $A$ and $B$, are valid, those which extremise the action are just the most probable. The sum of all these probabilities must add up to $1$, which is encoded in the principle of unitarity.

Perturbations and probabilities

Unfortunately for many QFTs, the description in terms of a Lagrangian is still very complicated, and to extract any meaningful information we need an approximation scheme. However, this approximation scheme, known as perturbation theory, can break unitarity as it involves throwing information away. We use a technique called perturbative unitarity to ensure the total probability of all possible outcomes in a particle interaction or collision—like scattering in different directions—sums to 1.

Imagine trying to solve a quadratic equation like $\varepsilon x^{2}-2x+1=0$ for $\varepsilon \ll 1$, but you didn’t know the quadratic formula. How would you go about solving this equation? We could perhaps take a guess at an expansion in powers of $\varepsilon$, and then neglect terms of higher order, $x=x_{0}+\varepsilon x_{1} +\varepsilon^{2}x_{2}+\cdots$. We can then approximate a solution of this equation by comparing coefficients of $\varepsilon$, however we will only get one root of the equation, but it’s still a good way to approximate a solution. This is the essence of perturbation theory, we pick a small parameter and expand around it.

In classical mechanics, we can think of a collision between billiard balls as being alike to the scattering of fundamental particles. Consider two balls colliding with initial momenta $\boldsymbol{p}_{1}$ and $\boldsymbol{p}_{2}$ respectively. After the collision the momenta of the individual particles will have changed; they have final momenta $\boldsymbol{p}_{3}$ and $\boldsymbol{p}_{4}$. We can use Newton’s second law to find the forces acting on these balls and predict where they will travel to after the collision. However, in quantum mechanics our particles are not modelled by solid billiard balls, and instead we describe them by wavefunctions that represent probabilities of finding the particle in different positions and states. Instead of predicting the exact path of a particle like you would with billiard balls, quantum mechanics provides probabilities. We can calculate the likelihood of a particle scattering in a certain direction or having a specific outcome, but we can’t predict precisely where it will go.

You can use classical mechanics to work out where the cue ball’s going

We can think of the incoming and outgoing particles as the billiard balls, and where the billiard balls collide and fuse is the part where the graviton or Higgs particle enters existence. The billiard balls then move away from each other, becoming the outgoing particles.

S is for scattering

When particles collide, we model them using a tool called the S-matrix—the scattering matrix. Essentially we know what goes in to the collision and what comes out but we treat the collision itself as a mystery box where we do not know all the details. The S-matrix is a way of describing what we do know about the interaction. The S-matrix is such that $\boldsymbol{\mathsf{SS}}^{\dagger}=\boldsymbol{\mathsf{I}}$, where $\boldsymbol{\mathsf{S}}^{\dagger}$ is the Hermitian conjugate (complex conjugate transpose) of the S-matrix. This property gives us an equation for the scattering amplitude $\mathcal{A}$, which describes how particles interact during a collision. This matrix helped us predict the existence of the Higgs boson particle, postulated in 1964, which was eventually detected at the Large Hadron Collider in 2012. The scattering amplitude encodes the probability of a scattering interaction occurring.

The compact muon solenoid (CMS) detector played a part in detecting the Higgs boson. Image: Wikimedia commons user Tighef, CC BY-SA 3.0

Imagine that you’ve got a particle, maybe a tiny electron, and it’s moving from left to right along one direction. But suddenly, it encounters a localised potential, we’ll call it $V(x)$, at $x=0$. Our electron wants to get through this barrier and enter the right hand side where the party is kicking off. Our particle has what we call a wavefunction, $\Psi$, which is determined by the Schrödinger equation,
$$-\frac{{\hbar}^{2}}{2m}\nabla^{2}\Psi+V(x)\Psi=\mathrm{i}\hbar \frac{\partial\Psi}{\partial t}.$$ What’s the deal with this wavefunction, you ask? This wavefunction is like our particle’s ID card, holding all the information about where the particle might be hanging out and how it’s vibing with its momentum. A wavefunction encodes information about a particle’s position and momentum and allows us to calculate the probability of finding the particle in different states or locations. Solving the Schrödinger equation we obtain
$$\Psi(x)=
\begin{cases}
A\mathrm{e}^{\mathrm{i}kx} + B\mathrm{e}^{-\mathrm{i}kx} & \text{if } x<0 \\ C\mathrm{e}^{\mathrm{i}kx} + D\mathrm{e}^{-\mathrm{i}kx} & \text{if } x>0.
\end{cases}$$
If our particle is on the left side $x<0$, it's all about the coefficients $A$ and $B$, with $\Psi(x) = A\mathrm{e}^{\mathrm{i}kx} + B\mathrm{e}^{-\mathrm{i}kx}.$ $A$ and $B$ are coefficients telling us how much of the wave is coming and going. $A$ represents the incoming wave and $B$ stands for the reflected wave. But if our particle is on the right side of the potential $x>0$, it’s more about the coefficients $C$ and $D$, and we’ve got $\Psi(x) = C\mathrm{e}^{\mathrm{i}kx} + D\mathrm{e}^{-\mathrm{i}kx}.$ With $C$ representing the outgoing wave and $D$ the wave coming in from the right. Focusing on a particle starting to the left of the potential barrier and moving to the right, we can set $D$ to zero. Then $C$ tells us how much of the wave is transmitted through the barrier, while $B$ tells us how much has been reflected. We also have a wavevector $k=\sqrt{2mE}/\hbar$, which is like the DJ at our party. Our $k$ is the tempo based on the particle’s energy and mass; if $k$ is large, the music is fast, and the particle is zipping along at high energy. On the flip side, when $k$ is small, it’s a slow jam, and the particle is moving at lower energy.

Wave function scattering off a localised potential

The S-matrix is like the party host that connects the vibes from the incoming guests to the outgoing ones. Then $\boldsymbol{\Psi}_{\mathrm{out}}$ is the guest list for the party, with $B$ and $C$ rocking the party. There is also $\boldsymbol{\Psi}_{\mathrm{in}}$ which is on the left side trying to enter our party; $A$ makes it in but $D$ is still chilling at home.
These are related by
$$\boldsymbol{\Psi}_{\mathrm{out}}=\boldsymbol{\mathsf{S}}\boldsymbol{\Psi}_{\mathrm{in}},$$
with
$$\boldsymbol{\Psi}_{\mathrm{out}}=\begin{pmatrix} B \\ C \end{pmatrix},\; \boldsymbol{\Psi}_{\mathrm{in}}=\begin{pmatrix} A \\ D \end{pmatrix}.$$
We use the unitarity of the S-matrix to form an upper bound on the real and imaginary parts of a scattering amplitude, which we can then convert to an upper bound on the energy, with the potential to violate unitarity for collisions above this energy. To preserve unitarity a new particle must be introduced to our theory!

Perturbative bound for gravity

For gravity, we now take on the challenge of finding the energy scale at which unitarity breaks down, indicating the existence of some new physics in the same way as for the Higgs. We need this new particle to stop unitarity from being violated at high energies. We use a trick called partial wave decomposition. Essentially we know that the scattering amplitude solves some equation, and we also know that this equation has a basis of solutions which we can expand in. Think of this as a fancy cousin of the Fourier series known as a partial wave decomposition. This expansion gives
$$\mathcal{A}(s,\theta)=16\pi\sum\limits_{n=0}^{\infty}(2n+1)a_{n}(s)P_{n}(\cos{\theta}).$$ Where $a(s)$ are partial wave amplitudes, which is analogous to the transmission component of a wave in 1D scattering, determined by $A/C$ in our quantum mechanics example.

The same idea as what solved the Higgs case works again for gravity where we integrate the S-matrix unitarity constraint to find a bound for distinguishable (different) particles with scattering mediated by a graviton. In terms of the partial waves, the bound is
$$|a(s)|^{2}+\mathrm{Im}(a(s))=0.$$ This bound is not satisfied for every possible energy, there are values of energy $s$ where this bound is violated. But we can’t have unitarity violated, so physicists introduced correction terms and the Higgs boson to mediate the scattering process. If we apply this bound to the amplitude that can be calculated for $W^{+}W^{+}\rightarrow W^{+}W^{+}$ scattering, we find that without the Higgs particle, unitarity would be violated at an energy around 1.7TeV. This bound means at energies above 1.7TeV, our physics would stop working, and probabilities would go above 1. Now, we see the same problem with quantum gravity, except this time it is more difficult as gravity requires an infinite number of correction terms, whereas the Higgs case only needed finitely many.

We can visualise the bound by displaying it on an Argand diagram in order to see where unitarity breaks down and hence where we need to introduce a new particle. Let $y=\mathrm{Im}(a(s))$ and $x=\mathrm{Re}(a(s))$ and using the properties of complex numbers that $|a(s)|^{2}=x^{2}+y^{2}$, we obtain a circle in the complex plane which bounds the scattering amplitude. In other words for energies such that we are inside the circle unitarity is preserved, while for other energies it is violated. Our circle in this case is $x^{2}+(y+1/2)^{2}=1/4$, and the bound is $$|\mathrm{Re}(a(s))|\leq \frac{1}{2}.$$

Adding new particles to our theory changes the scattering amplitude as there are now more ways that the particles can interact which changes the bound on energies where unitarity is preserved. If this new particle has the right mass then the energy, $s$, can grow arbitrarily large without violating unitarity. The new physics that we need to preserve unitarity is the introduction of the Higgs and the graviton respectively for each case. The new particle gives a new way, or channel in particle physics language, for particles to interact.

Argand diagram with a circle of radius $1/2$ and centre $(0, -i/2)$, bounding the real part of the amplitude and partial waves as $|\mathrm{Re}\{a_n(s)\}|\leq1/2$.

Strings

For quantum gravity, one new channel is not enough to preserve unitarity at every energy scale. Instead we will get a new bound and at energies greater than the new bound unitarity would still be violated. To fix this requires us to add more resonance terms to our amplitudes so that we do not violate unitarity. The difference between the Higgs case and the graviton case is that gravitons require an infinite number of correction terms, while the Higgs does not. Physicists are taking what was learnt about the Higgs case and applying it to scattering mediated by gravity.

How do we go about adding an infinite number of correction terms? A similar area of physics called string theory offers a perspective on the amplitudes of gravitons, as there we can use a specific amplitude—among others—known as the Veneziano amplitude to model graviton scattering. While at first our EFT did not assume any specific theory of quantum gravity, string theory becomes helpful in modelling graviton scattering amplitudes. Investigating these amplitudes and what they can tell us about new physics at higher energies is a current area of research.

To sum it all up, the standard model of particle physics has guided us to analyse the interactions of elementary particles, while effective field theories have opened doorways to explore the gaps in the standard model. We marvelled at the magic of perturbative unitarity, the safety net that constrains our probabilities, and the S-matrix which describes the interactions of particles, revealing the hidden secrets of the Higgs boson and paving the way for experimental discoveries at the LHC. This led onto gravitons whose scattering amplitudes help us to bridge the gap between the quantum and the cosmic scales. The next step is to actually detect these gravitons (easier said than done)… Maybe check back in 100 years?