### Why Can't We Reach the Speed of Light By Boosting?

If you have ever come across someone talking about special relativity, there’s a good chance you will be able to tell me one of the two fundamental axioms in the subject: the fact that the speed of light is constant in all frames of reference.

After thinking about this for a few moments, one might come up with a thought experiment that *looks* like a counterexample. Imagine (for the sake of the experiment) that there’s a train moving along at a constant speed (we’ll call it *v*) along the ground with respect to you, the unmoving person. Then, imagine that there’s another train on top of the first train, and its moving at a speed *v*, but with respect to the first train. As such, you would undoubtedly agree that the second train seems to be moving faster than the first train, since it has the speed of the first train *and* its own speed.

Following this argument, it seems reasonably straightforward that if you keep on stacking trains on top of each other such that they all are moving relative to the last one with speed *v*, at some point, no matter how slow the speed *v* is, one of the trains should exceed the speed of light. Bingo, we’ve done it!

Unfortunately, this is not so. But what is the actual problem here? Why can’t we get past the speed of light?

To answer this, I’ll have to introduce a few things. But first, some notes. One, we are trying to see why the speed of light acts as a “cosmic speed limit”, so we aren’t going to simply say, “Let the speed of the first train be faster than the speed of light.” Instead, we want to see why we can’t build *past* it. Second, I want to note that this isn’t just a strange situation that has no applications. If you don’t like the scenario with the trains, imagine continually boosting oneself to a faster and faster speed. Of course, this has to still be done with the right Lorentz transformation, but you *will* hit a barrier.

With that out of the way, let’s dig into the problem. We could try and do successive Lorentz transformations in order to get the relative velocity between the observer on the ground and the *n*^{th} train. However, the fact is that this isn’t the “natural” or easy way to do the problem. In special relativity, velocities don’t add like we are used to. Instead, there’s a fancy factor γ and several other differences when transforming velocities. However, there is a quantity that does simply add when two velocities are composed. It is called the rapidity, and is denoted with the letter φ. It is defined as:

Here, β is just the ratio of the speed (that any train is moving at with respect to the one underneath it) and the speed of light. For our problem, this value is constant. We want to know how fast the *n*^{th} train is moving, so we just have to keep on adding the rapidity factors. But they’re all the same! This means we get the rather simple result that the rapidity for the *n*^{th} cart is *n* times the first train’s rapidity factor. We then use the definition for φ from above in order to get:

We could stop here since everything in the above expression is a constant that we would input, but we don’t want to know what happens for the *n*^{th} cart. We want to keep boosting forever, until we get past the speed of light! This means we need to take a limit as *n* tends to infinity. Therefore, it would be nice to have our above expression in a form that we can take a limit much more cleanly. I’ll spare you the gory details, but basically we can use the identities for hyperbolic functions that allow us to go from them to exponential and logarithmic functions. After doing this, you should be able to get something like the following:

This is actually a nice expression, because its limit is very easy to evaluate. Before this though, let’s look at plot of this function as a function of *n*. In other words, we want to see the behaviour of this function as *n* gets larger.

As you can see, the function gets closer and closer to one as *n* increases. Why isn’t it the speed of light *c*? Remember what β is. It’s the speed of the train divided by the speed of light, so having the plot asymptotically go towards one is what we should expect. Also, if you’re curious, the value for β_{1} is 0.5, which is quite a large value. This is why the function asymptotes to one very fast.

### Intuition and Rigour

I’ve always been a bit wary about intuition. For a long time, I would avoid using the term, because I found it was creating a dangerous habit in terms of thinking clearly while solving problems. Your intuition isn’t always right. As such, instead of trying to guess when your intuition is correct, my advice was simply to throw it out, and work with what you have. This seemed like a much easier and straightforward way to go about reasoning, particularly within mathematics.

However, as I’ve reflected on this matter further, I’ve come to realize that I actually *do* like using intuition, but not in the way that some people do. To see exactly what I mean, consider the classic example from calculus. We’ve all heard professors (particularly, if they aren’t teaching mathematics, but subjects in the sciences instead) say things like we should just “divide by $dt$ on both sides of the equation”. The problem with saying this isn’t that it works, but that it hides the reason *why* it works. There’s a reason our notation allows for this kind of manipulation, and it’s built right into the definition of a limit and how the derivative functions. It’s not a happy coincidence that we can then “divide” by $dt$ and get the correct answer. It really has to do with the way our mathematics is set up. I feel like the danger with this kind of “intuitive” manipulation is that students won’t remember the *actual* reason, and this causes problems down the road.

In my mind, the crux of the problem is captured by the following question. Is your argument for why something is true based upon your intuition? If so, then that’s a problem. It’s not the intuition itself that’s problematic. It’s the fact that you’re using that intuition as a pseudo-proof for why you think something is correct. If we want students to be more confident in their ability to prove things are true, then I think it’s worth it to highlight this distinction and make it implicit.

Furthermore, this change in perspective allows me to capture what I wanted to get out of intuition. In my mind, intuition is what lets you get a *foothold* into a new topic. By itself, you only get a rough idea of what is going on, and perhaps the direction an idea will take you. But the intuition is only there as a starting point. It’s not supposed to carry you through the whole concept. I like to think of it as the “one-sentence summary” of an idea, without the jargon.

This doesn’t mean that intuition is useless. In fact, it can be *difficult* to come up with these ideas, even if you know the topic well. That’s a challenge that I’ve been trying to work on, because I want to be able to explain an idea at its essence, without the extra baggage. *That’s* what I would argue intuition is for. Use it as a foothold into a new topic, but never forget that it’s only that. Don’t be lulled into thinking your intuition will carry you through, because mathematics has a lot of unintuitive surprises!

### When Are Two Things the Same?

Take a look at the following two graphs, and tell me what you think of them:

One thing you might say is that they look the “same”, up to the fact that the graphs are coloured differently, and are somewhat rotated with respect to each other. While some may say that these are important distinctions, for the present essay, let’s pretend that this is merely a trivial difference. I think we can agree that these two graphs are essentially the same. Not only do they have the same number of nodes (vertices), there’s a sense that you can “match” up the nodes between the graphs. It’s just like when you were in elementary school and had to find out if two shapes were congruent, similar, or completely different.

But what if I told you that the graph on the left represented the roadways of a small town, and the graph on the right represented a network of people? Suddenly, these graphs don’t seem so similar anymore, practically speaking. And yet, we can still agree that there’s something about their *structure* that’s the same. What’s great about this is that, if we study properties of the graph dealing with the roads, we can be sure that these same properties carry over to that of the network of people! This seems obvious when looking at the two graphs, because they look just the same, but it’s quite neat that the properties of two completely different kinds of systems can be applied from one to the other.

Formally, what I’m talking about in the case above with the two graphs is called an *isomorphism*. For graphs, this essentially boils down to showing that one can “relabel” the vertices of the graph on the left such that it has the same labels as the graph on the right, while still preserving the same edges between them. If two vertices are joined by an edge in the right graph, then our isomorphism better associate these two vertices in the graph on the left which have an edge joining them.

Isomorphisms are special functions. You can think of them as “sameness” functions. They provide a link between two mathematical objects that show how they can be thought of as two instances of the same thing. I love isomorphisms, because they showcase one of my favourite parts of mathematics: connecting things that seem totally unrelated. In this essay, I want to demonstrate to you the power of isomorphisms. After introducing the concept, we’ll look at an easy example to get a flavour of things, before moving on to the main course.

(Also, can you find an isomorphism for the two graphs? Check the end of the essay for one solution.)

## Preserving Structure

As I said above, isomorphisms are special functions. Typically, we denote them as $\phi$, and they are functions from one set $X$ to another set $Y$, with the following properties. First, they are bijective. Another name for this is one-to-one and onto. One-to-one means that any time time you take an element $x$ in $X$ and move it to an element $y$ in $Y$, the *only* element that gets mapped to $y$ is $x$. Essentially, each element in $X$ gets its own element in $Y$. Then, onto means that any element in $Y$ can be “reached” by using the function $\phi$ on a certain element of $X$. In other words, we have inverses. This is all summarized in the figure below.

An immediate consequence of a bijection is that the two sets in question have the same cardinality, the number of elements in the set. This is true because each element in $X$ gets sent to an element in $Y$, and nothing in $Y$ is “missed”. That’s a bit of handwaving, but the important point is that these two sets will have the same cardinality if you can establish a bijection between them (which is an exciting topic in and of itself, but will be saved for another time!).

This gives us a bijection between the two sets, but an isomorphism gives us a bit more than that. Not only do we have a direct “link” between the elements in $X$ and $Y$, we also have the fact that the *structure* of $X$ is preserved in $Y$. If we go back to our example with graphs, not only did we want to send every vertex in the left graph to a unique vertex on the right, but we also want their edges to coincide accordingly.

To encode this mathematically, we want the following property to be true. If $\phi$ is our isomorphism, then if we take $a,b \in X$, $\phi(ab) = \phi(a) \phi(b)$. Let’s untangle this a bit, because the equation is a bit subtle.

On the left hand side, we have $\phi(ab)$. The term $ab$ means you are doing some kind of operation that sticks $a$ and $b$ together. Despite what it may look like here, I’m not only talking about multiplication (it’s just that we usually omit the symbol that signals the operation). The operation could be addition, multiplication, or more exotic things, like turning a number into a complex number, or associating an element with an ordered pair. There are many possibilities (but not all of them are isomorphisms!).

On the right hand side, we have $\phi(a)\phi(b)$. Once again, this means that we are performing an operation between $\phi(a)$ and $\phi(b)$. The crucial part here is that this operation doesn’t *need* to be the same as the one in $X$. Remember, we are in $Y$ now, so the operation could be different!

Put simply, the idea of $\phi(ab) = \phi(a) \phi(b)$ is that any time we combine two elements in $X$ and then use the isomorphism to go to $Y$, it’s the same thing as going to $Y$ first in each individual element, and *then* putting them together.

With that out of the way, let’s start looking at some isomorphisms.

## Flipping Switches

We’re in no rush, so let’s look at an example that is fairly straightforward. Consider your classic light switch. It can be in two positions. Either the switch is “ON”, or the switch is “OFF”. However, the more important thing is that we have some additional structure on the light switch. There are two things we can do with a light switch. We can “flip” it from one setting to the other, or we can leave it be. Let’s give these operations some names. We will call a flip $\uparrow$, and doing nothing will be called $n$ (for neutral, or “nothing”). Then, we can associate certain operations on the light switches. I’ll use the symbol $*$ to denote doing two operations, one after the other.

First, we can do nothing twice, which results in nothing happening. Therefore, we can say that $n*n = n$. Similarly, we know that if we flip a switch and then do nothing, or do nothing and then flip a switch, the result is that the switch gets flipped once. Therefore, we can encode that mathematically by saying $n \, \,* \uparrow = \, \uparrow* \, \,n = \, \uparrow$. Finally, the last thing we can do is to flip the switch twice, which amounts to the same thing as doing nothing at all. As such, we have $\uparrow * \uparrow = n$.

From these four combinations of operations, we’ve completely specified the structure of our system. Why? Because any more complicated series of moves we can do can be broken down into a combination of these operations. This is nice, because it prevents us from having to deal with many more elements. Indeed, our set of moves we can do on the light switch has only two elements: flipping a switch, and doing nothing.

If we want to be more formal about this, we can call this set with the operation $*$ a *group*. The way I like to think about a group is that it’s a certain set endowed with an operation that tells us how to “stick” elements together. The structure I was talking about earlier with respect to isomorphisms has to do with this operation. We don’t *just* have some random elements in a set. We have some way of connecting them together. Furthermore, as we saw in the previous paragraph, no matter how we stuck the elements together, we never formed a new element. We always stayed within our set. This is an important property of the operation that our group has.

Now, let’s look at something completely different. Let’s consider the set of integers under modular two arithmetic. If you’re not familiar with modular arithmetic, the quick summary is this. Modular arithmetic considers what the remainder of all integers are when divided by a certain modulus (integer greater than one). For modular two, we only have the elements $0$ and $1$, because every other integer can be reduced to this by dividing by two until the remainder is $0$ or $1$ (this is equivalent to saying all numbers are even or odd). Therefore, we only have these two elements.

Now, let’s consider what happens if we look at this set ${0,1}$ with the operation of addition. Then, we get the following four combinations:

The first three operations should be clear, but the last one might not be. To see it, remember that $1+1 = 2$. But $2$ is divisible by $2$ with no remainder left over, so it is equivalent to zero.

You might be starting to see where this is going. We’ve just defined two sets with different operations that seem to share a lot of structure in common. They both have two elements, and the way the elements combine together is very similar. So can we find an isomorphism between them? You bet!

Since we are dealing with only two elements in each set, we can specify the function when applied to each element (this wouldn’t be the case if we were working with sets that had many or even infinitely many elements in it). We’ll identify the flip with $1$, and doing nothing as $0$. Therefore, we can write out the isomorphism as such:

Since each element in our set for light switches corresponds to exactly one unique value of $0$ or $1$, we can see that our function is both one-to-one and onto, so it is bijective. What we now want to show is our other property, that $\phi(ab) = \phi(a)\phi(b)$. We will only show one, but the others are just as simple to check. If we take the combination $\uparrow * \, \, n$, we get:

As you can see, this works beautifully. Also note that the “operations” for each set aren’t the same thing (one is flipping a switch, and one is doing regular addition), but the structure of the operation gets preserved. Therefore, we’ve got ourselves an isomorphism.

Of course, this is just a simple example (and I’ve left out a few technical details), but I hope it illustrates to you the quite ingenious correspondences we can make between structures.

Let’s go to a bit more complicated example!

## Number Systems and Algebra

In secondary school, you’ve most likely studied equations of the form $y=ax + b$, $y=ax^2 +bx + c$, and maybe even $y=ax^3 + bx^2 + cx + d$. There’s even a chance that you remember the name of these functions: polynomials. You probably won’t be surprised to hear that polynomials form a much larger family of functions than you’ve seen in secondary school. Indeed, a general polynomial $p(x)$ can be written as such:

I know that I zone out when I start seeing too many terms on a line, so I’ll try to avoid it as much as possible. However, my point is that any polynomial has this form. We then know how to add and multiply any two polynomials together. For addition, we add the coefficients of the like terms together, and for multiplication, we slowly and painfully multiply everything out and then collect the terms. It might be painful, but it can be done. The important part is that no matter if you add or multiply polynomials together, the result is *also* a polynomial.

There’s another assumption that barely crosses our minds, but let’s make it explicit. The coefficients for our polynomials will be anything in the reals. This is to distinguish it from *other* kinds of objects that have coefficients only in the integers, for example. Since our coefficients are real, we will call this set $\mathbb{R}[x]$.

But $\mathbb{R}[x]$ isn’t only a set. Because we established certain kinds of operations on it (namely, adding and multiplying) that result in new polynomials in our set, $\mathbb{R}[x]$ is actually called a *ring* (and in particular, an integral domain). The name isn’t terribly important, but I think it’s good to be aware of the terms. In essence, being a ring means that the set $\mathbb{R}[x]$ is closed under our two operations, and has a few more technical features.

This is all well and good, but you may be wondering where in the world am I going with this. Where is the isomorphism?

To set this up, remember how we worked in modular two arithmetic, which essentially compressed all numbers into either $0$ or $1$. We’re going to do the same thing here, but with a slightly different flavour. What we’ll do is consider all of the polynomials of the form $p(x) = (1+x^2)f(x)$. In other words, all of the polynomials which have $(1+x^2)$ as a factor. Then, we will that *any* polynomial with this factor is identified to be zero. Since $f(x)$ can be any polynomial, this means $(1+x^2)=0$, or equivalently, $x^2=-1$. This is really nice, because anytime we see $x^2$ in our polynomials, we can “swap” it for $-1$.

So how does this change $\mathbb{R}[x]$? Remember that all of our polynomials have the form:

We know that the first and second terms won’t be affected, since there are no $x^2$ terms in them. However, every *other* term in $p(x)$ can be simplified. In fact, since every power of $x$ is either even or odd, any power that is larger than one will be reduced to a power that is less than or equal to one. As such, when we replace all of the $x^2$ in the polynomials with $-1$, we will get a new set of polynomials in the form:

Do you know any other sets with this kind of description? It’s the complex numbers $\mathbb{C}$, of course! The condition that $x^2=-1$ might have given it away, but I think this is just a lovely correspondence that just pops out without you really knowing about it until the end.

The isomorphism now would be of the form $\phi(a+bx) = a+bi$, but this can also be established using the first isomorphism theorem of rings, which we won’t get into here. What I want you to get from this is that we can establish a correspondence between doing some sort of pseudo-modular arithmetic (with a strange modulus) on the set of polynomials, and end up with the complex numbers.

## Coda

There are many more isomorphisms, but I think we can call it a day. I *love* it when we can find correspondences between objects that somehow capture a notion of being the “same”, yet still seem different. It’s when we view them from a different perspective that we see that there differences aren’t really fundamental.

Lastly, the solution to the problem I posed at the beginning of the essay. The isomorphism between the graphs is given by the following. Send vertex $a$ to $5$, $b$ to $2$, $c$ to $6$, $d$ to $1$, $e$ to $7$, $f$ to $4$, and $g$ to $3$. The way I look at the problem is to imagine trying to find a way to twist and bend one graph into the other. However, this only works when you know that there is an isomorphism beforehand!

### Scrabble Misfortunes

My family and I play a lot of Scrabble, which is a word game where you have to use tiles with letters on them to make words on the board. The board containes multipliers that also give you more points, as well as the tiles themselves, which each have a certain number of points associated with them.

In Scrabble, you usually want to start first, because the first person to play gets to double the points they make. For example, if they played the word FISH, which gives a total of $4+1+1+4=10$ points, their score would be twenty points. As such, being first is a nice boost to get you started in the game.

Recently though, one of my family members had two games where they were the first to start, and during both of these games, they couldn’t actually play a word! This is because there were no vowels in the initial seven tiles they picked. Consequently, they couldn’t play a word, since we don’t recognize any word that doesn’t contain a vowel. This meant that they had to skip their turn on both occasions, which obviously frustrated them.

This got me thinking: how likely was this to happen? In other words, was this event something that happened oftem? I couldn’t remember having this happen before this brief spell of bad luck, so I thought this would be a good application of conditional probability.

There are 100 tiles in total in the game of Scrabble. Out of those 100 tiles, 54 of them are consonants (I’m counting the two blank tiles as vowels, since they can be used as vowels if needed). Then, if we let $A$ be the event that I don’t pick any vowels in seven letters, then this is equal to saying that I want to pick *only* consonants. If we apply some basic counting principles from discrete mathematics, then we know that the total number of ways to get seven consonants from fifty-four tiles is $54 \choose 7$. Similarly, the number of ways to pick seven letters from the total bag of tiles is $100 \choose 7$. Finally, since it is probably safe to assume that there’s an equally likely chance of picking any given tile, the probability of picking only consonants is then:
In other words, there’s about a one-in-one-hundred chance of being unlucky and not being able to start the game. I’d say that this probably matches fairly well with my experience, though I do seem to have noticed this happen more as of recently. Perhaps I’m simply seeing events now that I’ve focused on them.