Special relativity - Gauge invariance

[Aleolomorfo]

Homework Statement

In an inertial reference frame $S$ is given the four-potential:
$$A^\mu=(e^{-kz}, e^{-ky},0,0)$$
with $k$ a real constant.

1. $A^\mu$ fullfills the Lorentz gauge? And the Coulomb gauge?
2. Which is the four-potential $A'^\mu$ in a reference frame $S'$ which is moving relatively to $S$ with velocity $v$ along the z-axis?
3. $A'^\mu$ fullfills the Lorentz gauge? And the Coulomb gauge?

Homework Equations

Lorentz gauge: $\partial_\mu A^\mu=0$
Coulomb gauge: $\vec{\nabla}\cdot\vec{A}=0$

The Attempt at a Solution

First of all I have some doubts about the form of the four-potential. Usually a four-potential is given in this way: scalar potential $\phi$ entry, vectorial potential $\vec{A}$ entries (x, y and z components). In this situation we have $\phi = e^{-kz}$. However, there is a function of only $y$ as $A_x$. So I have thought that the four-vector is given back to front $(A_z,A_y,A_x,\phi)$. The result of the first question depend on this choice. I have other doubts about this exercise but first I would like to solve this problem first and then writing the others.

 

[Orodruin]

So I have thought that the four-vector is given back to front $(A_z,A_y,A_x,\phi)$. The result of the first question depend on this choice. I have other doubts about this exercise but first I would like to solve this problem first and then writing the others.

I would say it is more common to place the time-component first. However, this is a convention and you must check how it is done in whatever text you are taking the problem from.

 

[George Jones]

I think that the (standard) order is correct, so that the is mixing (that gives the desired result) between the t and x components upon a change of frame.

I hope that I am not giving too much away, but I think that the point of the exercise is that the Lorenz (not Lorentz) condition equation is covariant, while the Coulomb gauge equation is not. Consequently, if the the Lorenz condition is satisfied in one inertial frame, i is satisfied in all inertial frame. The Coulomb gauge, however, can be satisfied in one frame and not satisfied in another.

 

[Aleolomorfo]

Lorenz (not Lorentz)

Yes, sorry. I always forget that there is not the "t".

I think that the (standard) order is correct, so that the is mixing (that gives the desired result) between the t and x components upon a change of frame.

So if the order is correct, in $S$ both the Lorenz gauge and the Coulomb gauge are satisfied. Since the Lorenz gauge is satisfied in $S$ I can say that it is satisfied in all inertial references, and so also in $S'$. Instead for the Coulomb gauge I have to write $A'$:
$$A'^\mu=(\gamma e^{-kz},e^{-ky},0,-\gamma v e^{-kz})$$
And so in $S'$ the coulomb gauge is not satisfied since $\vec{\nabla'}\cdot A'\neq 0$. However also the Lorenz gauge is not satrisfied in this frame, where is the mistake?

 

[Orodruin]

Please show your computations.

 

[Aleolomorfo]

Reference $S$ ($c=1$)
$$A^\mu=(e^{-kz}, e^{-ky},0,0)$$
Lorenz gauge: $\partial_\mu A^\mu=\frac{\partial\phi}{\partial t}-\frac{\partial A_x}{\partial x}-\frac{\partial A_y}{\partial y}-\frac{\partial A_z}{\partial z}=\frac{\partial }{\partial t}e^{-kz}-\frac{\partial}{\partial x}e^{-ky}=0$. Lorenz gauge is satisfied.
Coulomb gauge: $\vec{\nabla}\cdot\vec{A}=0$ (calculus is the same as for lorenz gauge)

To write $A'^\mu$ I use Lorentz transformations along z-axis for four-vector: $A'^\mu=(\gamma(\phi-vA_z),A_x,A_y,\gamma(A_z-v\phi))$. The result is:
$$A'^\mu=(\gamma e^{-kz},e^{-ky},0,-\gamma v e^{-kz})$$
I repeat the same calculus done in $S$ and I obtain what I stated in the previous message:

And so in S′S′S' the coulomb gauge is not satisfied since →∇′⋅A′≠0∇′→⋅A′≠0\vec{\nabla'}\cdot A'\neq 0. However also the Lorenz gauge is not satrisfied in this frame, where is the mistake?

 

[George Jones]

I repeat the same calculus done in $S$ and I obtain what I stated in the previous message:

With respect to what did you differentiate the components of A'?

 

[Aleolomorfo]

With respect to what did you differentiate the components of A'?

I differentiated with respect to $t,x,y,z$ and now I see this is not correct. I have to differentiate with respect to $t',x',y',z'$. So:
Lorenz condition in $S'$: $\frac{\partial\phi'}{\partial t'}-\frac{\partial A'_x}{\partial x'}-\frac{\partial A'_y}{\partial y'}-\frac{\partial A'_z}{\partial z'}$. But my potential is given in terms of the old coordinates so I have to change the derivatives.
Since $t=\gamma(t'+vz')$ and $z=\gamma(z'+vt')$:
$$\frac{\partial}{\partial t'}=\frac{\partial}{\partial t}\frac{\partial t}{\partial t'}+\frac{\partial}{\partial x}\frac{\partial x}{\partial t'}+\frac{\partial}{\partial y}\frac{\partial y}{\partial t'}+\frac{\partial}{\partial z}\frac{\partial z}{\partial t'}=\gamma\frac{\partial}{\partial t}+\gamma v\frac{\partial}{\partial z}$$
Mutatis mutandis:
$$\frac{\partial}{\partial z'}=\gamma v\frac{\partial}{\partial t}+\gamma\frac{\partial}{\partial z}$$
But If I see if the lorenz conditon is satisfied I will find:
$$\partial'_\mu A'^\mu=\gamma\frac{\partial\phi'}{\partial t}+\gamma v\frac{\partial \phi'}{\partial z}-\gamma v\frac{\partial A'_z}{\partial t}-\gamma\frac{\partial A'_z}{\partial z}$$
I do not write the derivates respect $x$ and $y$ since they are obviously zero.
$$\partial'_\mu A'^\mu=\gamma^2 v (-k)e^{-kz}+\gamma^2 v (-k)e^{-kz}$$
But this is not zero. Maybe I have done a calculus mistake but I have checked twice and I did not find it; or maybe there is a conceptual mistake.

 

[George Jones]

$\frac{\partial\phi'}{\partial t'}-\frac{\partial A'_x}{\partial x'}-\frac{\partial A'_y}{\partial y'}-\frac{\partial A'_z}{\partial z'}$.

Do you really want to use this combination of signs and index placement? If so, you have to change the upstairs indices in

$$A'^\mu=(\gamma e^{-kz},e^{-ky},0,-\gamma v e^{-kz})$$

to downstairs indices.

 

[Aleolomorfo]

Ok, I've found the solution. Thank you very much indeed to George Jones for his help!