August 3, 2020

Noether current $j$ is a conserved property. (It is constant with respect to time.)

$$ \frac d{dt}j=0 $$As with Langrangian action, the math works out. But what, conceptually, *is* Noether current? Why is it conserved?

In the previous post, I showed that the Lagrangian $L$ is a way of calculating temporal velocity as an outside observer. So $\frac{\partial L}{\partial\dot q}$ is the derivative of a particle's temporal velocity $L$ with respect to its spatial velocity $\dot q$.

What is $\delta q$? A small change in spatial position. A particle moves in the direction of maximizing temporal velocity $L$. Therefore $\delta q$ is oriented along the direction of the particle's path.

Putting this together, Noether current $j\equiv\frac{\partial L}{\partial\dot q}\delta q$ equals proper temporal velocity. Of course it is conserved! The conservation of Noether current is just a mathy way of saying that every clock counts forward at one second per second in its own reference frame.

The equation for Noether current is **useful** because it is expressed in an observer's spatial reference frame instead of proper time. We thus derive Noether's theorem.

So in summary, Noether’s Theorem merely says that whenever there is a continuous symmetry in the action, there is a corresponding conserved quantity.

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A Simple Introduction to Particle PhysicsarXiv:0810.3328

There is no such thing as a point particle. Everything is nonlocal waves. Therefore it is imprecise to describe temporal velocity $L$ of a point particle. Instead of temporal velocity $L$, we should think in terms of the conceptually equivalent temporal velocity density $\mathcal L$. $L$ is the integral of $\mathcal L$ over space 空.

$$ L=\int_空\mathcal L $$We can plug $\mathcal L$ into the Euler–Lagrange equations.

$$ \frac{\partial\mathcal L}{\partial\phi}=\partial_\mu\left(\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\right) $$We can abbreviate the partial derivatives with respect to Minkowski spacetime with the d'Alembert operator $\square=\frac{\partial^2}{\partial t^2}-\nabla^2$.

$$ \begin{eqnarray*} \frac{\partial\mathcal L}{\partial\phi}&=&-\square\mathcal L \\ \square\mathcal L+\frac{\partial\mathcal L}{\partial\phi}&=&0\\ \end{eqnarray*} $$When we use temporal velocity density $\mathcal L$ instead of temporal velocity $L$, Noether current $j$ becomes Noether current density $j\equiv\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi$. Since $j$ is spread out, it is the total $Q=\int_{空}j$ that is globally conserved instead of $j$. Examining this locally, any decrease in $j$ flows to adjacent space. Any decrease to $j$ must flow into adjacent space and vice-versa. The absolute value of $j$ is called "charge density" and the gradient of charge density is called "current density".

Everything that follows works out numerically. What follows is an intuitive explanation.

Remember from before that $\mathcal L$ is the temporal velocity density of a field, roughly $\alpha$. By analogy, $\dot\phi$ is the spatial velocity density of the same field. In special relativity, a particle's momentum is proportional to $\frac v\alpha$. So it follows that momentum density $\Pi^\mu\equiv\frac{\partial\mathcal L}{\partial\dot\phi_\mu}$.

The Hamiltonian density is a little more complicated. In classical mechanics the Hamiltonian is just the sum of kinetic and potential energies. But, as we showed last time, the sum turns into a difference when we examine things relativistically. The Hamiltonian thus has two parts. On the left side, we multiply momentum density $\Pi^\mu$ by $\dot\phi_\mu$ to get kinetic energy density. On the right side, the Lagrangian $\mathcal L$ serves as its own field. Thus $\mathcal H\equiv\frac{\partial\mathcal L}{\partial\dot\phi_\mu}\dot\phi_\mu-\mathcal L$.

Hamiltonian field theory is important to quantum field theory.

We have now defined the time evolution of fields entirely in terms of wave divergence equations, albeit subject to some constraints. There are two parts to wave divergence: that dependent on the field's absolute value and that dependent on changes to the field. In a static field there is no magnetism so all that matters is the absolute value of the electric field $-\bar\nabla\phi$.

Suppose we live in 4-dimensional spacetime with 3 spatial dimensions. For a spherically-symmetric charge, the divergence of $\phi$ is the total charge density contained by the sphere. This is Gauss' law $\bar\nabla\cdot\bar E=\rho$. Coulomb's law $F=\frac{q_1q_2}{4\pi r^2}$ comes from applying Gauss' law to a spherically symmetrical field and dividing by $4\pi r^2$, the surface area of a sphere. Maxwell's equations follow from applying special relativity to Coulomb's law.