Here’s a brief and high-level derivation of Noether’s theorem!
This is one of the most important results in modern physics, as it describes a very deep connection between the symmetries of a physical system, and those quantities in it that are conserved over time. Here, we use symmetries to mean transformations that leave the system’s physics unchanged.
Say we have a set of fields $\phi_a(x)$, with spacetime coordinates $x^\mu$ and a metric signature $(+,-,-,-)$.
The action is then defined as
$$S(\phi) = \int{Ld^4x}= \int{L(\phi_a, \partial_\mu(\phi_a), x^\mu)d^4x}$$
Making the action stationary $(\delta S=0)$ under arbitrary changes in the field $\delta\phi_a$ gives us the Euler-Lagrange equations, using the standard derivation:
$$\frac{\partial L}{\partial\phi_a} - \partial_\mu \left(\frac{\partial L}{\partial(\partial_\mu\phi_a)}\right)=0$$
Cool! Now, let’s say we have some continuous symmetry with parameter $\epsilon$, such that
$$\begin{aligned} x^\mu&\to x’^\mu =x^\mu+\epsilon\delta x^\mu\ \phi_a(x)&\to\phi’_a(x’)=\phi_a(x)+\epsilon\delta \phi_a \end{aligned}$$
Note that $\delta x^\mu, \delta\phi_a$ are both functions of $x$!
The full change in the Lagrangian $L$ (to first order) is $\Delta L=L’(x)-L(x)$. Let’s define this to be equal to some function $W^\mu$ of both the fields and coordinates:
$$\Delta L=\epsilon \partial_\mu W^\mu(x)$$
Therefore, the variation in the Lagrangian density is
$$\delta L=\frac{\Delta L}{\epsilon}=\partial_\mu W^\mu(x)$$
Taking the differential for the equation for the action and substituting gives
$$\delta S=\int{\partial_\mu W^\mu d^4x}$$
This is an integral of a divergence!
Now we actually need to flesh out some of these terms. Firstly, given that $L=L(\phi_a, \partial_\mu(\phi_a), x^\mu)$, taking the derivative gives
$$\delta L=\frac{\partial L}{\partial\phi_a}\delta\phi_a+\frac{\partial L}{\partial(\partial_\mu\phi_a)}\delta(\partial_\mu\phi_a)+\partial_\mu L \delta x^\mu$$
Now, let’s look at the middle term on the right. Given that a field can vary even if we don’t change $x$, we have
$$\partial_\mu\phi_a\to\partial_\mu\phi_a+\partial_\mu(\delta \phi_a)$$
If we do move the coordinates and Taylor expand to first order, we get
$$\phi_a(x+\delta x)\approx \phi_a(x)+\delta x^\nu \partial_\nu \phi_a(x)$$
Taking the derivative of the second term gives
$$\partial_\mu(\delta x^\nu \partial_\nu \phi_a(x))=(\partial_\mu \delta x^\nu)\partial_\nu\phi_a+\delta x^\nu\partial_\mu\partial_\nu\phi_a$$
However, in the middle term at the top of this section, we have $\delta(\partial_\mu\phi_a)$, which is a change at the same coordinate $x$. As a result, we need to subtract off the effect of the coordinate shift!
$$\delta(\partial_\mu\phi)=\partial_\mu(\delta\phi_a)-(\partial_\mu\delta x^\nu)\partial_\nu\phi_a$$
Now we can substitute this into the $\delta L$ equation and gather terms to get
$$\delta L=\left[\frac{\partial L}{\partial \phi_a}-\partial_\mu \frac{\partial L}{\partial(\partial_\mu\phi_a)}\right]+\partial_\mu\left[\frac{\partial L}{\partial(\partial_\mu\phi_a)}\delta\phi_a+L\delta x^\mu\right]-\frac{\partial L}{\partial(\partial_\mu \phi_a)}\partial_\nu\phi_a\partial_\mu(\delta x^\nu)$$
However, we can substitute in the Euler-Lagrange equation at the top to get the first bracket to vanish!
Now, recall that we defined $\delta L=\partial_\mu W^\mu$, leaving us with
$$\partial_\mu \left[\frac{\partial L}{\partial(\partial_\mu\phi_a)}\delta\phi_a+L\delta x^\mu-W^\mu\right]=\frac{\partial L}{\partial(\partial_\mu\phi_a)}\partial_\nu\phi_a\partial_\mu(\delta x^\nu)$$
For all rigid symmetries, which must include all internal symmetries and constant translations, we know $\partial_\mu\delta x^\nu=0$, leaving the RHS zero:
$$\partial_\mu \left[\frac{\partial L}{\partial(\partial_\mu\phi_a)}\delta\phi_a+L\delta x^\mu-W^\mu\right]=0$$
We define the Noether current $j^\mu=\frac{\partial L}{\partial(\partial_\mu\phi_a)}\delta\phi_a+L\delta x^\mu-W^\mu$, leaving us with the local conservation law
$$\partial_\mu j^\mu=0$$
This conservation law directly leads to a conserved quantity: if we integrate over a spatial surface (i.e. at constant time), we get
$$Q:=\int{j^0(x)d^3x}\implies \dot{Q}=0$$
And so $Q$ is the conserved quantity!
In summary, if we have a physical system and we know:
Then we can calculate the Noether current $j^\mu$, and calculate the conserved quantity $Q$ by integrating!
Let’s look at some examples!
The classic one is time translation symmetry leading to energy conservation.
The Lagrangian of a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-V(\phi)$$
Our symmetry is $t\to t+\epsilon$, leaving the others unchanged, giving:
$$\begin{aligned} \delta x^\mu&=(1,0,0,0)\ \delta \phi&=0 \end{aligned}$$
Therefore, the Noether current is
$$j^\mu=L\delta x^\mu_0$$
I’ll be honest here, I’m not sure why $W^\mu$ is zero.
Now,
$$j^0=\frac{1}{2}\dot{\phi}^2+\frac{1}{2}(\nabla\phi)^2+V(\phi)$$
This is the standard form of the Hamiltonian density, and therefore integrating over a surface tells us that the total energy in a system is constant!
We can also look at how translation invariance is linked to conservation of momentum. Again, the Lagrangian of a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-V(\phi)$$
Say we have a translation in some direction $k$, such that
$$\begin{aligned} \delta x^\mu&=(0,…1,…,0)\ \delta \phi&=0 \end{aligned}$$
As before, the Noether current is
$$j^\mu=L\delta x^\mu_k$$
The conserved charge is then the integral:
$$P_k=\int{ L\delta _k^0d^3x}=-\int{\frac{\partial L}{\partial(\partial_0\phi)}\partial_k\phi d^3x}=\int{\dot{\phi}\partial_k\phi d^3x}$$
And this is linear momentum!