Why does the electromagnetic field adjust when a charged field changes phase?

[thehangedman]

I have a question regarding gauge invariance. When a charged field changes phase:

y -> e^it * y

The electromagnetic field adjusts to make the equations work:

A_m -> A_m - idt / dx_m

What I don't understand is why, purely from a physics standpoint, this would happen? That is, is there some other reason beyond just making the equations work, the electromagnetic field would adjust? Is this due to the two fields being coupled? If so, what would happen if there are two particles? What if one particle phase shifts by "t" and the other by "k" (t != k)?

 

[Bob_for_short]

...The electromagnetic field adjusts to make the equations work:

A_m -> A_m - idt / dx_m

No, they add this to A_m in order to compensate the equation change, to keep it form-invariant. Of course, one can safely let the former A_m. The physics does not depend on the phase factor.

On the other hand, it shows the "gauge" liberty in choosing A_m. Usually it is namely A_m which is chosen (=fixing a gauge) rather than the phase factor.

 

[thehangedman]

I'm sorry for misspeaking. I know the equations "work" without it. What you are suggesting is that in essence, the shift done to the electromagnetic field is done to keep the equations easy to work with. You could leave them there, but then your equations get messy. Having the extra phase term is a pain, so adjusting A to remove it makes life easier. I was under the impression there was more than just a mathematics reason for keeping the equations phase invariant. Is that not correct?

 

[Bob_for_short]

There is no other reasons.

I will tell you that I think of all this. You may choose any variable changes for solving your problem. They are not obliged at all to preserve the original equation form. The only requirement is to be non-singular (reversible).