Vectorpotential as a function of (t- x/c)

[silverwhale]

Homework Statement

I am working through chapter 47 of the Landau Lifschitz. And there is the following argument:

The vector potential is a function of $$t - \frac{x}{c}$$
From the defining equations for the electric and magnetic fields:
$$\vec{E} = - \frac{1}{c} \frac{\partial \vec{A}}{\partial t}, \vec{B} = \nabla \times \vec{A}$$
follows
$$\vec{E} = - \frac{1}{c} \vec{A'}$$
$$\vec{B} = \nabla \times \vec{A} = \nabla (t- \frac{x}{c}) \times \vec{A'} = - \frac{1}{c} \vec{n} \times \vec{A'}$$
$$\vec{B} = \vec{n} \times \vec{E}$$

I can't follow his argument.
Why did the equation for the electric field change from a time derivative of A to a derivative over (t- x/c).
And where does that $$\nabla (t - x/c)$$ come from?
Finally where does that vector n come from?

Any help would be greatly appreciated!

 

[gabbagabbahey]

Homework Statement

I am working through chapter 47 of the Landau Lifschitz. And there is the following argument:

The vector potential is a function of $$t - \frac{x}{c}$$
From the defining equations for the electric and magnetic fields:
$$\vec{E} = - \frac{1}{c} \frac{\partial \vec{A}}{\partial t}, \vec{B} = \nabla \times \vec{A}$$
follows
$$\vec{E} = - \frac{1}{c} \vec{A'}$$
$$\vec{B} = \nabla \times \vec{A} = \nabla (t- \frac{x}{c}) \times \vec{A'} = - \frac{1}{c} \vec{n} \times \vec{A'}$$
$$\vec{B} = \vec{n} \times \vec{E}$$

I can't follow his argument.

Why did the equation for the electric field change from a time derivative of A to a derivative over (t- x/c).

Let's define $t_r\equiv t-\frac{x}{c}$. Then apply the chain rule:

$$\begin{aligned}\frac{\partial}{\partial t}\textbf{A}(\textbf{x},t_r) &= \frac{\partial t_r}{\partial t}\frac{\partial}{\partial t_r}\textbf{A}(\textbf{x},t_r)+ \frac{\partial \textbf{x}}{\partial t}\cdot\mathbf{\nabla}\left(\textbf{A}(\textbf{x},t_r)\right) \\ &= (1)\textbf{A}'(\textbf{x},t_r)+(\mathbf{0})\cdot\mathbf{\nabla}\left(\textbf{A}(\textbf{x},t_r)\right) \\ &= \textbf{A}'(\textbf{x},t_r)\end{aligned}$$

And where does that $$\nabla (t - x/c)$$ come from?Finally where does that vector n come from?

Again, use the chain rule: In index notation w/ Einstein summation convention,

$$\begin{aligned}\mathbf{\nabla}\times\textbf{A}(\textbf{x},t_r) &= \textbf{e}_i\epsilon_{ijk}\partial_j A_k(\textbf{x},t_r) \\ &= \textbf{e}_i\epsilon_{ijk}\left[(\partial_j\textbf{x})\cdot\mathbf{\nabla}A_k(\textbf{x},t)\right]_{t=t_r} + \textbf{e}_i\epsilon_{ijk}\frac{\partial}{\partial t_r} A_k(x_m\textbf{e}_m,t_r) \partial_j t_r \\ &= \textbf{e}_i\epsilon_{ijk}\left[\textbf{e}_j\cdot\mathbf{\nabla}A_k(\textbf{x},t)\right]_{t=t_r}+ (\mathbf{\nabla}t_r)\times\textbf{A}(x_m\textbf{e}_m,t_r) \\ & = \left[\mathbf{\nabla}\times\textbf{A}(x_m\textbf{e}_m,t)\right]_{t=t_r} -\frac{1}{c} (\mathbf{\nabla}x)\times\textbf{A}(x_m\textbf{e}_m,t_r) \\ & =\end{aligned}$$

I don't have the text with me, but it looks like Landau is describing a case where radiation is emitted in the $\textbf{n}\equiv \mathbf{\nabla}x$ direction, and you are only interested in a region where the non-retarded magnetic field is zero.