**Introductory Note:** This article was originally written for a course that is part of
my physics degree. The original article was handed in as a PDF and typeset using $\LaTeX$.
The text relies heavily on footnotes, which are slightly^{(but only slightly)} easier to read in the original PDF.

# Welcoming Words

Let me start by asking a question: How often have you invoked conservation laws during your career as a
late-night-problem-sheet-specialist^{1}? If the answer is loads, you may be interested in what I have to
tell you^{2}. At the end of this article, you will learn about something called Noether’s Theorem. But
before we get there, a brief detour to something called Calculus of Variations will be necessary [1].
However, I can assure you, my dear reader, that this is a subject of utmost interest and that this
little endeavor will be well worth it.

# Calculus of Variations (a detour)

Imagine a plane. In this plane you find yourself at a point, which is very far removed from the
point where you would like to be^{3}.
Now a natural question arises: What is the best way to get there^{4}?

Here, a keen observer might remark: "That's easy, that's a straight line!". And indeed this
observer would not be wrong^{5}. But can he proof this fact? And more interestingly: What
if the plane is not a plane, but rather a (more) realistic place with mountains and valleys
and other such inconveniences?

This is precisely the kind of question that calculus of variations allows us to answer. The problem
we face is minimizing a functional (a function of functions^{6}). In our case, this is the
length of the path from where we are to where we want to go, as a function of the function that
describes our path.

To recklessly include an equation: Our problem is to find a path $y(x)$ such that

$$ \int_{a}^{b} dx \sqrt{1 + (y^{\prime})^2} = \int_{a}^{b} dx F(x, y, y^{\prime}), $$

which is the distance we would have to walk, is minimized.

This seems hard. Where would we even start^{7}? Well in this case, we start with some wishful
thinking: What if we already knew the solution $y(x)$. This would indeed by very nice, but how does it
help us? Hint: This is where *variations* come in. What if we now *vary* this solution, by
adding on a new, completely arbitrary function $\epsilon \times \eta(x)$^{8}?

With this approach, we find that $y(x)$ has to satisfy the Euler Lagrange Equation^{9} [3]:

$$ \frac{\partial F}{\partial y} - \frac{d}{dx} \frac{\partial F}{\partial y^{\prime}} = 0. $$

Showing this is left to the reader as an exercise^{10}, and can be done
using integration by parts^{11}.

Now solving tricky differential equations like this is not something a lot of people like to do. And while this equation is extremely important, it is not what I want you to take away from reading this article. After all, this is supposed to be a fun read, so I'll take the freedom to leave things like this to the Mathematical Methods lecturer.

Instead, I want to focus on the idea of what a variation is, as something quite similar will turn up for Noether's Theorem. Visually, the idea is to draw the solution (in our very nice case a straight line), and then to make it more fuzzy, by adding extra bumps along the line, but without exactly thinking about what kind of bumps we are adding.

After doing this, we then proceed to squeeze these bumps, until we get back to $y(x)$. If we had actually known the solution, doing this would be quite pointless and tedious. But since we didn't actually know the solution, we managed to learn in using this sneaky method.

# The Hamiltonian Principle (the dry bit)

Chances are, you heard of a guy called Newton^{12}. And since you study physics, you have probably
also heard of his three laws. Now, what if I told you, that instead of fiddling with those, there is
a much nicer^{13} way of doing things?

Enter: The principle of least action [1]. It says that the action of the system is stationary, which a theorist would write as:

$$ \delta \mathcal{A}[y, \dot{y}, t] = \delta \int_{t_0}^{t_1} L(y, \dot{y}, t) \text{ } dt = 0. $$

This is a fancy way of writing, that the functional $\mathcal{A}$ has to obey the Euler Lagrange Equation
from before^{14}.

Now you may well wonder, what this $L$ thing is, and that is indeed a formidable question. For our purposes, we will just define it to be [3]:

$$ L = E_\text{Kin.} - E_\text{Pot.}, $$

the difference between the kinetic and potential energy in our system.
$L$ is called the systems Lagrangian.The full story is, of course, much richer and much more
mathematically engaged than this definition^{15}.

Now I state, without proof^{16}, the following: Any motion of a system $y(t)$ that obeys the principle
of least action, obeys Newtons laws^{17} [1, 3].

# Noether's Theorem (finally, the grand reveal)

We have now covered the setup necessary for introducing the setup of Noether's Theorem: Symmetry
variations^{18}.

Symmetry variations are like the variations in the calculus of variations, but with two important differences:

- They are not required to vanish at the boundary
^{19}. - The function we called $\eta$ before now has a special form.

We define, for the continuous coordinate transformation $y(t) \rightarrow y^\prime(t) = f(y, \dot{y})$, the symmetry variation:

$$ \delta_S y(t) = y^\prime(t) - y(t), $$ as this difference tends to $0$ [1].

Armed with this nifty definition^{20}, let me show you something magical: Consider the case, when
the action is symmetric under a continuous transformation of variables (like translation), then
we have:

$$ \delta_S \mathcal{A} = 0. $$

Now we can state Noether's Theorem: If the action is symmetric under a continuous transformation, we get
a constant of motion^{21}. And on top of knowing that there *is* a conserved quantity, we can
also directly write down *what* this quantity is.

Let me take you through what this means mathematically^{22}:

$$ \delta_S \mathcal{A} = \delta_S \int_{t_0}^{t_1} L(y, \dot{y}, t) \text{ } dt $$ $$ = \int_{t_0}^{t_1} \frac{\partial L}{\partial y} \delta_S(y) + \frac{\partial L}{\partial \dot{y}} \delta_S(\dot{y}) \text{ } dt $$ $$ = \left[ \frac{\partial L}{\partial \dot{y}} \delta_S(y) \right]_{t_0}^{t_1} + \int_{t_0}^{t_1} \left(\frac{\partial L}{\partial y} - \frac{d}{dt}\frac{\partial L}{\partial \dot{y}}\right) \delta_S(y) \text{ } dt $$ $$ = 0. $$

We now recognize the Euler Lagrange Equation under the integral sign^{23}, and remember^{24}, that the start
and end times $t_0$ and $t_1$ are arbitrary to find:

$$
\frac{\partial L}{\partial \dot{y}} \delta_S(y) = \text{Constant},
$$
which is Noether's Equation^{25}!

# Conservation of Momentum (a much-needed example)

This probably looks very abstract, so we will consider the following example: Take a system having constant potential $E_\text{Pot.}(y) = V$. Then, consider continuously translating the system with the transformation $y(t) \rightarrow y^\prime(t) = y(t) + \epsilon$, where $\epsilon \in \mathbb{R}$ is an arbitrary constant, tending to $0$.

For the Lagrangian, we have $L = \frac{1}{2}m\dot{y}^2 - V$, where all the symbols have their usual meaning. Since $L$ does not depend on $y$, the symmetry variation is $\delta_S L = 0$, and thus we find for the action $\delta_S \mathcal{A} = 0$. Finally, invoking Noether's Theorem, we know:

$$ \quad\text{ } \frac{\partial L}{\partial \dot{y}} \times \delta_S(y) = \text{Constant} $$ $$ \Leftrightarrow \frac{1}{2}m2\dot{y} \times \epsilon = \text{Constant} $$ $$ \Leftrightarrow m \dot{y} = p = \text{Constant}, $$ which is simply the statement that linear momentum is conserved.

This simple example demonstrates the power and elegance of Noether's Theorem in particular and applying mathematics
in general. Using only very few assumptions,
namely the principle of least action, it is possible to capture the rich structure of the world
we inhabit^{26}.

## References

**[1]** V. I. Arnold. *Mathematical Methods of Classical Mechanics (2nd Edition).* Springer-Verlag, 1989.

**[2]** S. V. Fomin I. M. Gelfand. *Calculus of Variations.* Prentice-Hall International, Inc., 1963.

**[3]** Mary L. Boas. *Mathematical Methods in the Physical Sciences (2nd Edition).* John Wiley & Sons, Inc., 1983.

## Footnotes

**1:** Asking for a friend. **[back]**

**2:** If the answer is not loads, you should probably postpone reading this until you have gone back to your problem sheets and reconsidered your life choices. Especially those, that lead to you reading this article. At least those are the ones I would focus on. **[back]**

**3:** Yes, that happens to me all the time too. **[back]**

**4:** If the first question that comes to your mind is *"How did I get here?!?!?"*, I recommend less alcohol. **[back]**

**5:** However, he would be a horrible person for stealing my punchline. **[back]**

**6:** For the notation lovers: $ F:\mathcal{D}^n \rightarrow \mathbb{R} $, where $ \mathcal{D}^n $ is the set of continuous functions $ f_\mathcal{D} : [a,b] \rightarrow \mathbb{R} $ with norm $ \sum_{i=0}^n \max_x |f^{(i)}(x)| $ [2]. (Note: If you like equations like this, you should probably read a proper book [2] instead of this article.) **[back]**

**7:** Other that the obvious: Did Euler happen to solve this problem for me? **[back]**

**8:** $\eta(x)$ has to go to $0$ at the ends of our interval, such that we still end up with a path from A to B. It should also be nice, i.e. continuous, defined on the interval, etc. So not quite arbitrary, but who has time for so many details? **[back]**

**9:** Now that we included one equation already, a few more won't do any harm, will they? **[back]**

**10:** Ah, everyones favorite! **[back]**

**11:** You will also need to recognize, that at the solution $y(x)$ we have: $\epsilon = 0$ and $\frac{d}{d\epsilon}F[y(x) + \epsilon \times \eta(x), y^\prime(x) + \epsilon \times \eta^\prime(x)] = 0$. **[back]**

**12:** I hear, he's a big deal. Discovered gravity, or something like that [1]. **[back]**

**13:** Nicer^{TM}, when you don't have to draw funny force diagrams, but instead get to solve partial differential equations. **[back]**

**14:** You will now begin to appreciate how nice this formulation of classical mechanics truly is. Especially, after you had a chance to meet some partial differential equations in the differential equations course. Even more so, if they are not separable. Those, are the nicest^{13}! **[back]**

**15:** If interested, read a proper book [1] and stop wasting your time here. **[back]**

**16:** And without any shame. **[back]**

**17:** Furthermore, anyone who uses the principle of least action, instead of Newtons laws, has a happier life. Or not. Actual results may vary based on age, health, and personal preference. **[back]**

**18:** Just when you thought we were done. Ha, got you! More random things that look like they are pulled from thin air, right until the end. Although, personally, this is what I enjoy about mathematics: The moment of revelation, when everything suddenly makes sense and has a precise purpose. Mathematicians like writing in this way: Do all the setup first, then pull the pieces together. Why exactly they like this style of writing will forever be a mystery. (However, my money is on this: It makes them look smarter.) **[back]**

**19:** This was mentioned before, but will only have been read by those select few, who suffered through my attempt at humor in the footnotes. **[back]**

**20:** Rigor not included. **[back]**

**21:** Okay, I admit it. I lied. Instead of dragging you through all this setup, I could have just said something like: "If there is a symmetry in the laws of nature *(read: System Lagrangian)*, then we get conserved quantities *(also called Noether Charges)*". But that would not have made for a very long article, would it? **[back]**

**22:** The long equation that follows, makes it seem like the solution is fully spelled out. However, there will be a sneaky jump right at the end, where I brush over the fact that $f(t_a) - f(t_b) = 0 \quad \forall \quad t_a, t_b \in \mathbb{R}$ implies $f(t)$ is a constant in $t$. **[back]**

**23:** This is trivial^{TM}. (Proof by intimidation, a classic.) **[back]**

**24:** Thanks to the many footnotes, I may have successfully drained your attention enough for you to not remember. If that is the case, please refer to the Internet for help. **[back]**

**25:** You may now be at a point where you think: "Wow, so what?". And you would not be entirely wrong to think that. However I feel strongly that I did not succeed in writing this article if you are not feeling inspired now. So please take a moment to feel inspired. **[back]**

**26:** To $\pm 7.5\text{mm}$. Never forget that physics is an experimental science. (Even though I personally prefer to not be the one conducting them.) **[back]**