# U(1) invariance of classical electromagnetism

• A
• Featured
bhobba
Mentor
what does it mean to say about the "classical electromagnetism" to be U(1) invariant?

Very nice thread - loved it.

Personally though I prefer the reverse - getting EM from local U(1) invariance of QM rather than asking what does it mean to be U(1) invariant.

It's even found in popular books eg Victor Stenger - The Comprehensible Cosmos - page 251 under Electromagnetism:.
https://www.amazon.com/dp/1591024242/?tag=pfamazon01-20

He covers the other stuff too (symmetry breaking and all that), again at the populist level using just a little bit of math and calculus.

This is an advanced thread, but I mention it for those not advanced but would like a gentle introduction - you just need a bit of calculus.

Thanks
Bill

vanhees71
Gold Member
I also think that QT clarifies many troubles of classical physics, not only phenomenologically but also from a mathematical point of view. While gauge invariance is pretty straight forward also in classical electrodynamics, the true trouble is the idea of "point charges", which just doesn't fit into the classical field-theoretical framework. The mechanics should be rather treated in the sense of continuum mechanics, which is the natural setup for classical field theory.

Another example is statistical mechanics. It starts already with the definition of the relation between phase-space volume and "number of microstates", for which you obviously need a "natural" measure for phase-space volume, which is immediately delivered by QT in terms of Planck's constant ##h=2 \pi \hbar##.

In many respects QT is easier than classical physics, but of course from a somewhat advanced point of view ;-)).

Ben Niehoff
Gold Member
Classically, it is true that there is no reason for the gauge group to be ##U(1)## rather than ##\mathbb{R}## (which is simply the universal cover of ##U(1)##).

For non-Abelian groups, the kinetic term ##F^2## has its gauge indices contracted via the group's Killing form, and then questions of compactness translate into questions of whether the kinetic term has the correct sign (to prevent runaway solutions, or, quantum-mechincally, in order to have a minimum energy).

For Abelian groups, however, the Killing form is flat. So to determine whether such groups are compact or noncompact requires further input (in this case, the quantization of angular momentum).

Note that classical Yang-Mills theory (with non-Abelian gauge group) exists just fine. It just doesn't describe reality very well.

dextercioby and vanhees71
thierrykauf
Gold Member
$$\bar{\psi}\gamma^\mu \partial_\mu \psi$$ becomes $$\bar{\psi}e^{-i\phi(x)}\gamma^\mu[\partial_\mu + e\frac{\partial\phi(x)}{\partial x}]e^{i\phi(x)}\psi$$ under an internal rotation with one angle, hence the (1) of U(1). When the electron is placed in an external electromagnetic field, the coupling shows up as $$j^\mu A_\mu$$ that is $$\bar{\psi}\gamma^\mu ieA_\mu\psi$$. This has the same form as the U(1) transformed Dirac derivative term. So if you started with an electron already coupled to an E&M field, the Lagrangian would automatically be U(1) invariant. The word gauge is a misnomer. Weyl introduced it because he had conformal invariance in mind. But the name stuck. Even when it means invariance by rotation. Here gauge refers to invariance of the second kind, that is when the phase is space dependent. There are more thorough answers on this thread.