When does the vector potential A affect the E field?

[DoobleD]

This is probably fairly simple, but I have a hard time to grasp the concept of the vector potential A in electromgnetism. Especially, in the following equation for the electric field :

$$\vec{E} = - \nabla V - \frac{\partial \vec{A}}{\partial t}$$

When does the second term is not 0 ?

 

[Orodruin]

When the vector potential is time-dependent.

 

[DoobleD]

When the vector potential is time-dependent.

Aha, thanks, but that's obvious in the equation. I'm looking to what makes it time-dependent.

EDIT : Well actually, I suppose this is when B is time-dependent ?

 

[Orodruin]

You need charge distributions or currents that change over time. Otherwise everything can be treated as static.

 

[Dale]

Aha, thanks, but that's obvious in the equation. I'm looking to what makes it time-dependent.

EDIT : Well actually, I suppose this is when B is time-dependent ?

See Jefimenko's equations.

 

[vanhees71]

As the most simple example take the free plane-wave modes of the electromagnetic field. In the radiation gauge, i.e., with the potentials fulfilling
$$A^0=V=0, \quad \vec{\nabla} \cdot \vec{A}=0,$$
You have
$$\vec{A}(t,\vec{x})=\vec{A}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}),$$
where $\vec{A}_0 \cdot \vec{k}=0$ and $\omega=c |\vec{k}|$. Then
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}=\mathrm{i} \omega/c \vec{A}_0 \exp(\cdots), \quad \vec{B}=\vec{\nabla} \times \vec{A} = \mathrm{i} \vec{k} \times \vec{A}_0 \exp(\cdots).$$

 

[Metmann]

If the underlying space is simply connected, the time derivative of the vector potential vanishes if the time derivative of the magnetic field vanishes:

$\mathrm{d}_\mathrm{s}B = 0 \Rightarrow B = \mathrm{d}_\mathrm{s} A$, where the subscript s stands for "spatial". Or in other notation $\vec{\nabla} \times \vec{A} = \vec{B}$. You can pull the time derivative into $\mathrm{d}_\mathrm{s}$.

The time derivative of $\vec{B}$ vanishes if and only if $\vec{j} + \frac{\partial \vec{E}}{\partial t} = 0$ (in natural units with $c=1$).

If the underlying space is not simply connected, there does not a priori exist a vector potential, and hence the right side of your equation does not necessarily exist.

 

[DoobleD]

Thanks for the answers and insights ! It does help.

 

[Jano L.]

The time derivative of $\vec{B}$ vanishes if and only if $\vec{j} + \frac{\partial \vec{E}}{\partial t} = 0$ (in natural units with $c=1$).

That's wrong. According to Maxwell's equations, the time derivative of B vanishes if and only if
$$
\nabla \times \mathbf{E} = \mathbf{0}.
$$

 

[Metmann]

That's wrong

Sure, I would edit it, but somehow this isn't possible any longer. But thanks for correcting it.