Introductory physics education tends to take the following path:

1. A student learns about forces
2. They spend the next few months looking at blocks on ramps
3. Eventually, they learn about using conservation of energy to calculate velocities, etc., and everything becomes infinitely easier

I remember the first time I learned abot conservation of energy -- e.g. the idea that potential energy here equals kinetic energy there -- I was shocked. Why even learn ordinary forces anyways if energy was so much easier to use? And further, why are we even allowed to use these random laws of conservation in the first place? Where did they come from?

In this post I’d like to explore what I think is the most beautiful idea in physics. It’s the idea that “conservation laws”, like the conservation of momentum, are not random, but rather come about due to a symmetry in the physical system we’re looking at.

What are symmetries?

Something exhibits symmetry when we can move, flip, rotate, or do something to it without changing its appearance. This is one of those things that’s much more intuitive to see than to say in words, so take a look at the square below. You can click and drag the square to rotate it. Under what rotations does it appear unchanged? You’ll see the square turn green if so.

Now you probably knew the answer before even rotating the square – if we rotate it by a multiple of $90$ degrees, the square appears unchanged. In this sense, we could write something like $$\text{Square }(0 \text{ rotation}) = \text{Square }(90 \text{ degree rotation})$$

In general, this means a :symmetry is when we can change something (e.g. the angle of rotation) without changing the appearance of the shape. In other words, $$\frac{\Delta \text{ appearance of shape}}{\Delta \text{ angle}} = 0$$

What are conservations?

A quantity $I$ is conserved if it does not change over time – in other words, if $$\frac{dI}{dt}=0$$

The goal is as such: $$\boxed{\begin{aligned} \ &\text{given some physical situation (billiard balls hitting each other, a disk spinning} \\ & \text{ in space, etc.), can we find a quantity } I \text{ that does not change over time?}\end{aligned}}$$

Proof

We’re doing physics, so what equation do we usually start with? Usually it’s Newton’s law, $$\mathbf{F} = ma$$ We’re going to write this equation in a more illuminating way: namely, using energy.

Step 1: The Lagrangian

We will define a quantity called the Lagrangian, $$L = \frac{1}{2}mv^2 - U(x)$$ where $\frac{1}{2}mv^2$ is the total kinetic energy of the particle and $U$ is the total potential energy of the particle (that could be, say, $mgh$ for gravitational potential energy, or whatever potential energy function we want). $L$, then, tells us about the total energy overall.

Now, what happens if we take the derivative of the Lagrangian with respect to $x$ (the position of the particle)? We get $$\boxed{\frac{dL}{dx} = -\frac{dU(x)}{dx}}$$ because the first term is not dependent on $x$.

Similarly, we could take the derivative of the Lagrangian with respect to $v$, the velocity of the particle: $$\boxed{\frac{dL}{dv} = mv = \textbf{p}}$$ which you may notice is just the momentum of the particle.

The crucial insight here is that force is related to both of these quantities. Namely, force is the rate of change of the momentum with respect to time: $$\textbf{F} = \frac{d\textbf{p}}{dt}$$

and in addition, force is also equal to the rate of change of potential energy with respect to $x$: $$\textbf{F}=-\frac{dU(x)}{dx}$$ In summary, we can combine the two boxed equations into one equality (where I will now use partial derivatives since our functions are dependent on both $x$ and $t$): $$\boxed{\frac{\partial}{\partial t}\frac{dL}{dv} = \frac{dL}{dx}}$$ This equation is called the Euler-Lagrange Equation and is already a remarkable result – we can calculate an equation of motion for a particle formulaically by taking derivatives of its energy. For our purposes, though, the conclusion is just that this equation is true; namely, that $$\boxed{\textbf{The Euler-Lagrange equation is just } F=ma.}$$

Step 2: Finding the conservation law

Now that we’ve convinced ourselves that the Euler-Lagrange equation is just a way of rewriting $F=ma$, we are going to prove that given any symmetry $x\rightarrow x(\lambda)$, we get a conserved quantity of the form $$\frac{\partial L}{\partial v}\frac{\partial x}{\partial \lambda} \ \ \ \ \ \text{(this quantity is conserved!!)}$$ Remember that for something to be conserved, :it means its time derivative is zero. Therefore, $$\frac{\partial}{\partial t}\Big(\frac{\partial L}{\partial v}\frac{\partial x}{\partial \lambda}\Big)= 0$$

Let’s expand out the left side of this equation using the product rule:
$$=\Big(\frac{\partial}{\partial t}\frac{\partial L}{\partial v}\Big)\frac{\partial x}{\partial \lambda}+\frac{\partial L}{\partial v}\Big(\frac{\partial}{\partial t}\frac{\partial x}{\partial \lambda}\Big)$$

Examining these two terms closely, what do we notice? We’ve actually seen the first term already. Remember the Euler-Lagrange equation? That’s just a rewriting of Newton’s Law, so we know it must be true, and that told us that $$\frac{\partial}{\partial t}\frac{dL}{dv} = \frac{dL}{dx}$$ so we can actually just replace the first parantheses with $\frac{\partial L}{\partial x}$:
$$=\frac{\partial L}{\partial x}\frac{\partial x}{\partial \lambda}+\frac{\partial L}{\partial v}\Big(\frac{\partial}{\partial t}\frac{\partial x}{\partial \lambda}\Big)$$

The second parantheses can also be simplified – let’s rewrite it in a more illuminating way:
$$=\frac{\partial L}{\partial x}\frac{\partial x}{\partial \lambda}+\frac{\partial L}{\partial v}\Big(\frac{\partial}{\partial t}\frac{\partial}{\partial \lambda}\Big)x$$ Notice that $\frac{\partial}{\partial t}x$ is just velocity $v$, so we get
$$=\frac{\partial L}{\partial x}\frac{\partial x}{\partial \lambda}+\frac{\partial L}{\partial v}\frac{\partial v}{\partial \lambda}$$

This is just the derivative of the Lagrangian with respect to $\lambda$ after doing the chain rule! In other words, it’s just $$\frac{d}{d\lambda} L(x, v) \stackrel{?}{=} 0$$

---

What were we trying to prove again? We wanted to show that this derivative was 0, because that means that the initial quantity $$\frac{\partial L}{\partial v}\frac{\partial x}{\partial \lambda}$$ is conserved. The work above showed that
$$\textbf{if} \ \ \ \ \frac{d}{d\lambda} L(x, v) = 0$$
$$\textbf{then} \ \ \ \ \ \frac{\partial L}{\partial v}\frac{\partial x}{\partial \lambda} \ \ \ \ \text{ is conserved}$$

In other words, if the Lagrangian is symmetric under a transformation of $\lambda$, then we get a conserved quantity. This beautiful idea is called Noether’s Theorem, and is can be summarized in the following sentence: $$\boxed{\text{Symmetries give rise to conservation laws.}}$$

This is a deep mathematical idea and the proof presented here is extremely simplified – however, I’d like to note that this theorem is universally true, much beyond the scope of this post. For our purposes, let’s see the theorem in action.

Putting the theorem to action

We’re going to start with the simplest symmetry: translational symmetry. A translational symmetry means if we shift x by some amount, the amount of energy (the Lagrangian) stays the same. In other words, $$L(x) = L(x+\lambda)$$
Therefore, there is a conserved quantity associated with the shift $x\rightarrow x+\lambda$ that can be calculated by substituting $x+\lambda$ in for $x$:
$$\begin{aligned}\frac{\partial L}{\partial v}\frac{\partial x}{\partial \lambda} = & \ \frac{\partial L}{\partial v}\frac{\partial}{\partial \lambda}(x+\lambda) \\ \\ = & \ \frac{\partial L}{\partial v} \\ \\ = & \ \frac{\partial}{\partial v}\Big(\frac{1}{2}mv^2 - U(x) \Big) \\ \\ = & \ \boxed{mv} \end{aligned}$$
so we have found that under translational symmetry, the conserved quantity is momentum!

$$\boxed{\text{Translational symmetry yields conservation of momentum.}}$$

Interconnectedness

This is a result that comes about out of just math and a bit of logic, yet uncovers a profound relationship between seemly disconnected ideas. My favorite part of physics is seeing one idea float into the lap of another – perhaps by some kind of divine luck, or perhaps hinting that there exists a deeper connection that silently and rhythmically breathes life into everything we experience. At its core, physics is about uncovering that deep connection. I hope you learned something in this article!

On clear days, I can see that I’m a small and simple part of a big and complex ancient structure. In some mysterious way we are connected to each other. And we grow from that connection.
– June Huh

:x caveat

The theorem below actually only applies to “continuous” symmetries. The example of a square rotating is not a continuous symmetry because it only happens for 90, 180, 270, and 360 degree rotations (this makes it a :discrete symmetry). An example of a “continuous” symmetry would be the rotation of a circle; clearly, any angle you rotate a circle at would yield an identical looking circle, whether it be 1 or 1.1 or 1.003437 or 29482742 degrees.

:x discrete

individually separate / distinct; a square has 4 separate rotational symmetries.

:x conservationDef

Maybe it’s useful to think about an example. What does “energy conservation” mean? It means that at any random time t, the energy is the same; in other words, the energy is not changing over time, which we write as saying its time derivative is zero: dE/dt=0.