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## Main Question or Discussion Point

Suppose we're given the action

[itex]

S=-mc\int ds + \frac{q}{c}\int A_{\mu}(x) dx^{\mu}-\frac{1}{4c}\int d^{D}xF_{\mu \nu}F^{\mu \nu}

[/itex]

The first two integrals are over the particle's worldline while the last is over spacetime. So I'm able to successfully vary the action with respect to the gauge potential to achieve Maxwell's equations but I'm confused as to what the gauge potential is in this case.

If the last term weren't there, this action would describe the physics of a charged particle in some external field [tex] A_{\mu}(x) [/tex]. However, when we vary the action to yield Maxwell equation, we conclude that the current [tex] j^{\mu}(x) [/tex] *of the particle itself* is the source of the EM field. This is very confusing to me.

In the above action is the field the field caused by the particles motion or is it an external field? Or am I asking a silly question?

[itex]

S=-mc\int ds + \frac{q}{c}\int A_{\mu}(x) dx^{\mu}-\frac{1}{4c}\int d^{D}xF_{\mu \nu}F^{\mu \nu}

[/itex]

The first two integrals are over the particle's worldline while the last is over spacetime. So I'm able to successfully vary the action with respect to the gauge potential to achieve Maxwell's equations but I'm confused as to what the gauge potential is in this case.

If the last term weren't there, this action would describe the physics of a charged particle in some external field [tex] A_{\mu}(x) [/tex]. However, when we vary the action to yield Maxwell equation, we conclude that the current [tex] j^{\mu}(x) [/tex] *of the particle itself* is the source of the EM field. This is very confusing to me.

In the above action is the field the field caused by the particles motion or is it an external field? Or am I asking a silly question?