I Can the A and φ potentials be fixed in a region of space?

[cuallito]

Can the magnetic and electric potentials (A and φ) be fixed to zero, or at least some constant value in a region of space? Naively, I'd think something like this might work (a hollow conducting sphere connected to a voltage source connected to ground, would the potentials inside the sphere be constant?):

 

[Orodruin]

The potentials are only physical up to gauge equivalence. Hence, they cannot be uniquely determined by physics alone without also adhering to some gauge fixing conditions. This also means that the question in itself is ill defined.

 

[cuallito]

I take your point, but wouldn't the potentials equaling one constant value in one gauge equal another constant value in another gauge?

 

[Orodruin]

I take your point, but wouldn't the potentials equaling one constant value in one gauge equal another constant value in another gauge?

No.

 

[cuallito]

Can you please explain why?

 

[Ibix]

I take your point, but wouldn't the potentials equaling one constant value in one gauge equal another constant value in another gauge?

Consider $A'^\mu=A^\mu+v^\mu$ where $v^x=y$ and all other components are zero. The Lorenz condition is that $\partial_\mu A^\mu=0$, which is satisfied by $A'^\mu$ if it is satisfied by $A^\mu$, since $\partial_\mu v^\mu=\partial y/\partial x = 0$, but $A'$ is manifestly not constant if $A$ is. And this is in the same gauge.

 

[Orodruin]

Can you please explain why?

Suppose that $\vec A$ is a vector potential of $\vec V$, i.e., $\vec V = \nabla\times \vec A$. Then, if we let $\vec A' = \vec A + \nabla \phi$ for any scalar field $\phi$, then $\nabla\times \vec A' = \nabla\times(\vec A + \nabla\phi) = \nabla\times\vec A = \vec V$ so $\vec A'$ is also a vector potential of $\vec V$. As long as you pick $\phi$ such that $\nabla\phi$ is not constant both potentials cannot be constant.

Consider $A'^\mu=A^\mu+v^\mu$ where $v^x=y$ and all other components are zero. The Lorenz condition is that $\partial_\mu A^\mu=0$, which is satisfied by $A'^\mu$ if it is satisfied by $A^\mu$, since $\partial_\mu v^\mu=\partial y/\partial x = 0$, but $A'$ is manifestly not constant if $A$ is. And this is in the same gauge.

That would be because the Lorenz gauge is only a partial gauge fixing and leaves a residual gauge freedom.

 

[Orodruin]

Also, in the language of differential forms, where the 4-potential $A$ is a 1-form on spacetime containing both the scalar and vector potentials of electromagnetism, we have the field tensor $F = dA$. Adding the differential of any spacetime function $\phi$ to $A$, i.e., $A' = A + d\phi$ leaves the same field tensor $F = dA' = dA + d^2 \phi = dA$ as $d^2 = 0$.

 

[Ibix]

That would be because the Lorenz gauge is only a partial gauge fixing and leaves a residual gauge freedom.

So I am wrong to say "the same gauge" because the Lorenz gauge condition specifies a family of possible gauge fixings, of which I picked two?

 

[Orodruin]

So I am wrong to say "the same gauge" because the Lorenz gauge condition specifies a family of possible gauge fixings, of which I picked two?

If by ”the same gauge” you intend a gauge that completely fixes the gauge freedom then yes. The Lorenz gauge still leaves the freedom to add the gradient of any harmonic function.

 

[cuallito]

Suppose that $\vec A$ is a vector potential of $\vec V$, i.e., $\vec V = \nabla\times \vec A$. Then, if we let $\vec A' = \vec A + \nabla \phi$ for any scalar field $\phi$, then $\nabla\times \vec A' = \nabla\times(\vec A + \nabla\phi) = \nabla\times\vec A = \vec V$ so $\vec A'$ is also a vector potential of $\vec V$. As long as you pick $\phi$ such that $\nabla\phi$ is not constant both potentials cannot be constant.

Thanks, it's a good thing we have people like you on here. Is $\phi$ here the same thing as the electric potential, or is it any arbitrary scalar potential you can add to $\vec A$ ?

 

[Orodruin]

Thanks, it's a good thing we have people like you on here. Is $\phi$ here the same thing as the electric potential, or is it any arbitrary scalar potential you can add to $\vec A$ ?

It is an arbitrary scalar function. I was only discussing the case of a vector field with a vector potential. The case of electromagnetism is analogous and what I discussed in the follow-up post.

 

[vanhees71]

The electromagnetic potentials $\Phi$ and $\vec{A}$ are not uniquely defined by the physical situation and thus are not observable fields in Maxwell's theory. The electromagnetic field $(\vec{E},\vec{B})$ are observables.

The potentials are introduced as "auxiliary" mathematical quantities to fulfill the homogeneous Maxwell equations identically, i.e., you set
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{E}=-\partial_t \vec{A}/c -\vec{\nabla} \Phi.$$
Then indeed the two homogeneous Maxwell equations,
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\partial_t \vec{B}/c=0$$
are identically fulfilled (check this!).

But for given $\vec{E}$ and $\vec{B}$ also
$$\Phi'=\Phi+\partial_t \chi/c, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
lead to the same $\vec{E}$ and $\vec{B}$ as $\Phi$ and $\vec{A}$ for any arbitrary field $\chi$ (check this!). That's the famous gauge invariance of Maxwell's electromagnetism.

 

[Orodruin]

Note to OP. This

The electromagnetic potentials $\Phi$ and $\vec{A}$ are not uniquely defined by the physical situation and thus are not observable fields in Maxwell's theory. The electromagnetic field $(\vec{E},\vec{B})$ are observables.

The potentials are introduced as "auxiliary" mathematical quantities to fulfill the homogeneous Maxwell equations identically, i.e., you set
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{E}=-\partial_t \vec{A}/c -\vec{\nabla} \Phi.$$
Then indeed the two homogeneous Maxwell equations,
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E}+\partial_t \vec{B}/c=0$$
are identically fulfilled (check this!).

But for given $\vec{E}$ and $\vec{B}$ also
$$\Phi'=\Phi+\partial_t \chi/c, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
lead to the same $\vec{E}$ and $\vec{B}$ as $\Phi$ and $\vec{A}$ for any arbitrary field $\chi$ (check this!). That's the famous gauge invariance of Maxwell's electromagnetism.

is the same as this

Also, in the language of differential forms, where the 4-potential $A$ is a 1-form on spacetime containing both the scalar and vector potentials of electromagnetism, we have the field tensor $F = dA$. Adding the differential of any spacetime function $\phi$ to $A$, i.e., $A' = A + d\phi$ leaves the same field tensor $F = dA' = dA + d^2 \phi = dA$ as $d^2 = 0$.

in different notation.

 

[vanhees71]

You can formulate Maxwell's electrodynamics entirely in the Cartan calculus. It's very elegant and modern but not so common. An interesting textbook, providing an axiomatic approach to classical electrodynamics, based on Cartan calculus is

F. W. Hehl, Y. N. Obukhov, Foundations of classical electrodynamics, Springer (2003)