- #1
carllacan
- 274
- 3
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 Aμ like Aμ → A'μ = Aμ + ∂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 = 0
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 ∂i A i = 0 and A0 = 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 = 0
So now we have two conditions on the functions Λ
∂i ∂i Λ = 0
Λ0 = 0
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 Aμ like Aμ → A'μ = Aμ + ∂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 = 0
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 ∂i A i = 0 and A0 = 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 = 0
So now we have two conditions on the functions Λ
∂i ∂i Λ = 0
Λ0 = 0
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?
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