- #1

- 274

- 3

## Main Question or Discussion Point

I've looked everywhere and I haven't found an explanation of why is it useful to introduce gauge conditions. I've also searched in this forum, and none of the existing threads I've read answered my question. I apologize if there is and I have failed to find it.

My problem is that, as I see it, gauge fixing fixes nothing. We can still perform transformations that will not affect the field. Let me expose my reasoning:

The electromagnetic field potential is undefined up to the derivative of a function because if we transform A

Now A needs to satisfy this condition, and so does A', so we have

∂

∂

∂

and now we have restricted the kinds of Λ we can use for the gauge transformations, but we still haven't ruled out all of them, so we can impose Coulomb's conditions ∂

Now A and A' need to satisfy these conditions, so we have

∂

∂

∂

So now we have two conditions on the functions Λ

∂

Λ

and we have restricted the transforms we can do, but we haven't ruled out all of them, since we can still transform using a Λ that satisfies those conditions. Shouldn't the gauge fixing rule out every possible function Λ, so there is one and only one potential for every field?

My problem is that, as I see it, gauge fixing fixes nothing. We can still perform transformations that will not affect the field. Let me expose my reasoning:

The electromagnetic field potential is undefined up to the derivative of a function because if we transform A

^{μ}like A^{μ}→ A'^{μ}= A^{μ}+ ∂^{μ}Λ the field is the same. In order to fix this we impose ∂_{μ}A^{μ}= 0.Now A needs to satisfy this condition, and so does A', so we have

∂

_{μ}A'^{μ}= 0∂

_{μ}A^{μ}+ ∂_{μ}∂^{μ}Λ = 0∂

_{μ}∂^{μ}Λ = 0and now we have restricted the kinds of Λ we can use for the gauge transformations, but we still haven't ruled out all of them, so we can impose Coulomb's conditions ∂

^{i}A_{i}= 0 and A_{0}= 0.Now A and A' need to satisfy these conditions, so we have

∂

_{i}A'^{i}= 0∂

_{i}A^{i}+ ∂_{i}∂^{i}Λ = 0∂

_{i}∂^{i}Λ = 0So now we have two conditions on the functions Λ

∂

_{i}∂^{i}Λ = 0Λ

_{0}= 0and we have restricted the transforms we can do, but we haven't ruled out all of them, since we can still transform using a Λ that satisfies those conditions. Shouldn't the gauge fixing rule out every possible function Λ, so there is one and only one potential for every field?

Last edited: