# How does gauge invariance determine the nature of electromagnetism?

Gold Member
In his book, "The greatest story ever told", Lawrence Krauss states: "Gauge invariance ... completely determines the nature of electromagnetism."

My question is simple: How?

I have gone back thru the math. Gauge invariance allows us to use the Lorenz gauge with the vector and scalar potentials (I find these equations easier to visualize than the relativistic ones with tensors), which when used with Maxwell's equations (the ones with sources) to discover that the potentials obey wave equations with sources. So the wave equations + the Lorenz gauge are equivalent to Maxwell's equations. It seems to me this is a long way from Krauss's statement.

Or maybe I am expecting too much.

(Btw, is there a page somewhere showing how to code formulas on this forum?)

Mentor
You are correct, that is a long way from Krauss’ statement. In fact, you can derive Maxwell’s equations from the gauge invariance, see: https://aapt.scitation.org/doi/abs/10.1119/1.17279?journalCode=ajp as well as the original paper that this simplified version is based on.

One key thing to recognize is the importance of Noether’s theorem. Any invariance of the Lagrangian leads to a conserved quantity. In the case of the EM gauge invariance the corresponding conserved quantity is the charge. The conservation of charge itself can be used to derive half of Maxwell’s equations.

• • anorlunda and vanhees71
Gold Member
2022 Award
Krauss's statement can be better understood within quantum mechanics than within classical physics. The full story is revealed in relativistic QFT only, but the principle can also be made clear from non-relativistic physics.

$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi(t,\vec{x}).$$
This equation can be derived from Hamilton's principle, using the Lagrangian
$$\mathcal{L}=\mathrm{i} \hbar \psi^* \partial_t \psi -\frac{\hbar^2}{2m} (\vec{\nabla} \psi^*) \cdot (\vec{\nabla} \psi).$$
Here ##\psi^*## and ##\psi## can be varied independently, and it's easy to see that variation with respect to ##\psi^*## leads to the free-particle Schrödinger equation.

Now the Lagrangian (and thus the action) is invariant under global transformations,
$$\psi(t,\vec{x}) \rightarrow \psi'(t,\vec{x})=\exp(-\mathrm{i} q \alpha) \psi(t,\vec{x}), \quad \psi^*(t,\vec{x}) \rightarrow \psi^{\prime *}(t,\vec{x})=\exp(\mathrm{i} q \alpha) \psi(t,\vec{x}).$$
Then Noether theorem tells us that there's a conservation law,
$$\partial_t \rho+\nabla \cdot \vec{j}=0$$
with
$$\rho=q|\psi|^2, \quad \vec{j} = -q \frac{\mathrm{i} \hbar}{2m} (\psi^* \vec{\nabla} \psi-\psi \vec{\nabla} \psi^*).$$
Now the general idea of "local gauge symmetries" is to make such a symmetry "local". By this we ask, how we can extend the above symmetry to get a symmetry for space-time dependent parameters, ##\alpha \rightarrow \alpha(t,\vec{x})##. As it turns out, to this end we simply have to introduce a scalar and a vector field ##\Phi## and ##\vec{A}## and make everywhere, where there's a partial derivative wrt. ##t## and ##\vec{x}##
$$\partial_t \rightarrow \partial_t + \mathrm{i} q \Phi, \quad \vec{\nabla} \rightarrow \vec{\nabla}+\mathrm{i} q \vec{A}.$$
Then, with space-time dependent ##\alpha##,
$$(\partial_t + \mathrm{i} q \Phi) \psi'(t,\vec{x})=\exp(-\mathrm{i} q \alpha) [\partial_t\psi + (\mathrm{i} \Phi - \mathrm{i} q \partial_t \alpha) \psi],$$
and this stays invariant when we also transform the scalar potential by
$$\Phi \rightarrow \Phi'=\Phi+\partial_t \alpha.$$
In the same way we can show that we must transform the vector potential by
$$\vec{A} \rightarrow \vec{A}+\vec{\nabla} \alpha.$$
The new Lagrangian then gives the free-Schrödinger-field Lagrangian augmented by the interaction with the electromagnetic field.

To also get Maxwell's equations you have to add the corresponding free-field Lagrangian, which in the end gives a coupled system of equations with the electromagnetic charge and current densities expressed in terms of the Schrödinger field ##\psi##.

All this can of course easily made relativistic by starting from the Klein-Gordon or the Dirac equation and following exactly the same steps.

The formulas are best typed in LaTeX code. It's easy in the forum:

https://www.physicsforums.com/help/latexhelp/

Last edited:
• Dale