# Are all solutions of Maxwell's equation related by a gauge transformation?

Consider Maxwell's equation (without source):

$$\partial_\mu F^{\mu \nu} = 0 \implies \partial_\mu \partial^\mu A^\nu = \partial_\mu \partial^\nu A^\mu.$$

Can we find a pair of classical field configurations $$A^\mu(x),A'^\mu(x)$$ such that they both satisfy the equation above (assuming similar boundary conditions) but are not related to each other by a gauge transformation of the type:

$$A'^\mu(x) = A^\mu(x)+\partial^\mu \varphi(x) \quad ?$$

If it's impossible, how could we argue/show this?

Answer: Thanks to my2cts' answer, I've found 2 solutions $$A$$ and $$A'$$ not related by a gauge transformation : $$A^\mu=(0,e^{-i(t-y)},0,0)$$ and $$A'^\mu=(0,0,e^{-i(t-x)},0)$$. It makes sense since they both give rise to different EM fields, which are invariant under gauge transformations.

• I don't think one can answer this question until the boundary conditions are defined. Feb 14 '19 at 5:52
• I think that reading the comments and answer to my question posed in the following link answer this question. physics.stackexchange.com/q/452308 Feb 14 '19 at 7:20
• @akhmeteli I completely agree and rarely if one ever sees them mentioned in the context of the vector potential but why? Feb 14 '19 at 12:54
• Just for the record: there an even more trivial example. $A=0$ is a solution, the set of all solutions that differ from it by a gauge transformation are $A=\mathrm{d}\phi$. All such solutions have $F=0$. Clearly there exist solutions with non zero $F$. Feb 14 '19 at 18:09

• @my2cts If I understand your point, the idea is that we can find 2 linearly independent solutions $E,B$ and $E',B'$ that solve Maxwell's equation, and since the EM fields are unaffected by a gauge transformation we cannot relate them to each other? Feb 14 '19 at 15:24