(Electric) Scalar and vector potential

[lailola]

Homework Statement

In the problem, the electric scalar and vector potentials are,
$\phi=0, \vec{A}=A_0 e^{i(k_1 x-2k_2y-wt)}\vec{u_y}$

I have to find E, B and S.

Then, I have to calculate $\phi '$ that satisfies $div\vec{A}+\frac{\partial \phi '}{\partial t}=0$ Then calculate E and B.

Is it possible to find $\vec{A}'$ and $\phi'$ that satisfy the previous equation and produce the same E and B as $\vec{A}$ and $\phi$?

Homework Equations

$\vec{E}=-grad\phi-\frac{\partial \vec{A}}{c\partial t}=0$
$\vec{B}=rot(\vec{A})$

The Attempt at a Solution

Using the equations I find:

$\vec{E}=A_0wi/c e^{i(k_1x-2k_2y-wt)}\vec{u_y}$

$\vec{B}=A_0ik_1 e^{i(k_1x-2k_2y-wt)}\vec{u_k}$

$\vec{S}=\frac{c}{4\pi} \vec{E}x\vec{B}=1/(4\pi) A_0^2wk_1sin^2(k_1x-2k_2y-wt)\vec{u_x}$

For the next part I find,

$\phi ' = - 2K_2 c A_0/(w) e^{i(k_1x-2k_2y-wt)}+ constant(x,y)$

Then, I calculate E and B as before.

I don't know how to answer the last part. Any idea?

Thank you.

 

[vela]

Homework Statement

In the problem, the electric scalar and vector potentials are,
$\phi=0, \vec{A}=A_0 e^{i(k_1 x-2k_2y-wt)}$

Your expression for the vector potential isn't a vector.

I have to find E, B and S.

Then, I have to calculate $\phi '$ that satisfies $div\vec{A}+\frac{\partial \phi '}{\partial t}=0$ Then calculate E and B.

Is it possible to find $\vec{A}$ and $\phi'$ that satisfy the previous equation and produce the same E and B as $\vec{A}$ and $\phi$?

Homework Equations

$\vec{E}=-grad\phi-\frac{\partial \vec{A}}{c\partial t}=0$
$\vec{B}=rot(\vec{A})$

The Attempt at a Solution

Using the equations I find:

$\vec{E}=A_0w/c e^{i(k_1x-2k_2y-wt)}\vec{u_y}$

$\vec{B}=A_0ik_1 e^{i(k_1x-2k_2y-wt)}\vec{u_k}$

$\vec{S}=\frac{c}{4\pi} \vec{E}x\vec{B}=1/(4\pi) A_0^2wk_1sin^2(k_1x-2k_2y-wt)\vec{u_x}$

For the next part I find,

$\phi ' = - 2K_2 c A_0/(w) e^{i(k_1x-2k_2y-wt)}+ constant(x,y)$

Then, I calculate E and B as before.

I don't know how to answer the last part. Any idea?

Thank you.

 

[Chopin]

Your expression for the vector potential isn't a vector.

Presumably $A_0$ is itself a vector, which makes that definition perfectly valid.

I don't know how to answer the last part. Any idea?

Thank you.

Imagine taking your current definitions for $A$ and $\phi$ (I'm renaming your $\phi'$ to $\phi$ for simplicity, and so that I can reuse the symbol $\phi'$ below), and adding new quantities to them. So something like:
$$A \rightarrow A + A'\\ \phi \rightarrow \phi + \phi'$$

Now plug those definitions into your previous equations, and see if you can use them to come up with some constraints on the forms that $A'$ and $\phi'$ would have to take, in order to cause $E$ and $B$ to still come out the same.

 

[vela]

Presumably $A_0$ is itself a vector, which makes that definition perfectly valid.

I don't think so. A₀ appears in the expressions for the electric and magnetic fields along with unit vectors. It's pretty clear that the OP meant for A₀ to denote a scalar.

 

[Chopin]

Hmm, you're right. Guess I didn't look close enough at those E/B solutions.

lailola, can you clarify what the original problem is, and how you came up with your definitions for E and B?

 

[lailola]

I don't think so. A₀ appears in the expressions for the electric and magnetic fields along with unit vectors. It's pretty clear that the OP meant for A₀ to denote a scalar.

I forgot to write the vector in the expression of the vector potential. I've written it.

 

[lailola]

Presumably $A_0$ is itself a vector, which makes that definition perfectly valid.



Imagine taking your current definitions for $A$ and $\phi$ (I'm renaming your $\phi'$ to $\phi$ for simplicity, and so that I can reuse the symbol $\phi'$ below), and adding new quantities to them. So something like:
$$A \rightarrow A + A'\\ \phi \rightarrow \phi + \phi'$$

Now plug those definitions into your previous equations, and see if you can use them to come up with some constraints on the forms that $A'$ and $\phi'$ would have to take, in order to cause $E$ and $B$ to still come out the same.

Ok, thank you!