gauge symmetries? wtf is that

How symmetrical the big disks you put in your ears are

quite symmetrical desu

>>9081205
some spooky thing you shouldn't know

I wish I knew more about them OP, they seem to point at some deep physics.

I have only encountered gauge symmetries in classical electromagnetism when formulating fields in terms of potentials. Assuming there are no magnetic charges or currents (easy since these haven't been experimentally found) and linear isotropic media defined by epsilon and mu, maxwell's equations look something like

[math]\nabla \times \vec{E} = - \mu \frac{\partial \vec{H}}{\partial t} [/math]
[math]\nabla \times \vec{H} = \vec{J} + \epsilon \frac{\partial \vec{E}}{\partial t} [/math]
[math]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon}[/math]
[math]\nabla \cdot \vec{B} = 0[/math]

The divergence of B is zero, so it can be rewritten as the curl of some vector function A, because the divergence of a curl is always 0.

[math]\vec{B} = \nabla \times \vec{A} \Rightarrow \vec{H} = \frac{1}{\mu}\nabla \times \vec{A}[/math]

Plugging this into the curl equation for E

[math]\nabla \times \vec{E} = - \mu \frac{\partial}{\partial t}\left(\frac{1}{\mu}\nabla \times \vec{A}\right)[/math]
[math]\nabla \times (\vec{E} + \frac{\partial \vec{A}}{\partial t}) = 0[/math]

The term in parenthesis can be rewritten as the gradient of some scalar function V, because the curl of a gradient is always 0.
[math]\vec{E} + \frac{\partial \vec{A}}{\partial t}) = - \nabla V[/math]
[math]\vec{E} = -\nabla V \frac{\partial \vec{A}}{\partial t})[/math]

Now, the curl of A determines B, but there are no constraints on what the divergence of A is. Since the curl of a gradient is always zero, we can freely add the gradient of any function to A to yield A' and B remains the same.

[math]\vec{A}' = \vec{A} + \nabla \lambda[/math]

But since we derived V using A, this means V changes too. I'm about out of post, but the punch line is:

[math]V' = V - \frac{\partial \lambda}{\partial t} [/math]

This lambda represents what we call a gauge symmetry. A and V turn out to be members of a family of functions with continuous symmetry.

>>9081293
wow, thanks for the answer! My brain is still processing it