What does it mean by independent (in gauge fixing of EM field)

[kof9595995]

What does it mean by "independent"(in gauge fixing of EM field)

In my textbook, it gives the Coulomb gauge [tex]\phi = 0,\nabla A = 0[/tex] and says they will kill two degrees of freedom of the four potential and leave two independent components. I understand [tex]\phi = 0[/tex] will kill one degree of freedom, but I'm not so sure about [tex] \nabla A = 0[/tex]. Usually the scenario is,let say we have 3 unknowns, and they are linked by a equation, such that given any two, the third unknown will be uniquely determined, that's what we called "kills a degree of freedom". However in [tex] \nabla A = 0[/tex], given [tex]{A_x},{A_y}[/tex], we will only know [tex]\frac{{\partial {A_z}}}{{\partial z}}[/tex], so A_z is not uniquely determined by A_x and A_y. So how should I interpret this?

 

[kof9595995]

Well, perhaps it's more appropriate to put the post in classical physics. I met this question when reading some QFT, that's why I put it here.

 

[dextercioby]

No, you should view the [itex] \nabla\cdot\vec{A} = 0 [/itex] as a constraint equation, in the sense that the 3 spatial components of the vector potential are not entirely independent, but are related throogh their first derivatives.

 

[kof9595995]

I know, but in this case what's the precise definition of "independent" or "dependent"? If the third unknown can't be uniquely determined by the the first two, in what sense have we killed a degree of freedom?

 

[dextercioby]

That's seen in the momentum representation, in which the gauss' law becomes [itex] \sum_{i=1}^{3} k_i A_i = 0 [/itex], where the new A's are the Fourier transformed of the initial ones. Now you can see that all 3 components are not really independent.

 

[kof9595995]

I see, that seems clearer, thank you

 

[kof9595995]

Emm, wait, that doesn't really resolve the problem, in momentum representation,specifying A_x, A_y can only determine k_z*A_z, not A_z, so it's no better than position representation.

 

[dextercioby]

Yes, but [itex] k^{\mu}k_{\mu} = 0 [/itex] from which you can express [itex] k_z [/itex] in terms of [itex] \omega, k_x, k_y [/itex].

 

[kof9595995]

Emm, I was wrong, because k_z is just a independent variable, so if we know k_z*A_z, we know what A_z is.

 

[kof9595995]

Interestingly, in k space we uniquely determine A_z, but in position representation we don't. This is because when doing the Fourier transform, we assume A_z goes to 0 at z=infinity, realizing this, we can just as well uniquely determine A_z in position representation