Given a plane wave characterized by Ex and By

[zhillyz]

Homework Statement

Given a plane wave characterized by E_x , B_y , propagating in the positive z-direction,

E = E₀sin[\frac{2\pi}{\lambda}(z-ct)]\widehat{x}

show that it is possible to take scalar potential ϕ = 0 . Find a possible
vector potential A for which the Lorentz gauge is satisfied.

Homework Equations

i)E= -∇\varphi

ii)∇E = \frac{\rho}{ε}

iii)B = ∇×A

iv)∇A = \frac{-1}{c^2}\frac{\delta\varphi}{\delta t}

The Attempt at a Solution

So basically I am thinking combining the above first two equations which results in the laplace operator but I am not sure how this ties into phi equating to 0. Then there is equation 4 the Lorenz gauge and I assume values of A and phi are plugged into see if it balances. But its the first step i am not sure on. How to show the scalar potential can be zero..

-∇^2\varphi = \frac{-\rho}{\epsilon}

 

[gabbagabbahey]

Homework Statement

Given a plane wave characterized by E_x , B_y , propagating in the positive z-direction,

E = E₀sin[\frac{2\pi}{\lambda}(z-ct)]\widehat{x}

show that it is possible to take scalar potential ϕ = 0 . Find a possible
vector potential A for which the Lorentz gauge is satisfied.

Homework Equations

i)E= -∇\varphi

ii)∇E = \frac{\rho}{ε}

iii)B = ∇×A

iv)∇A = \frac{-1}{c^2}\frac{\delta\varphi}{\delta t}

The Attempt at a Solution

So basically I am thinking combining the above first two equations which results in the laplace operator but I am not sure how this ties into phi equating to 0. Then there is equation 4 the Lorenz gauge and I assume values of A and phi are plugged into see if it balances. But its the first step i am not sure on. How to show the scalar potential can be zero..

-∇^2\varphi = \frac{-\rho}{\epsilon}

For any vector potential and scalar potential combination to be valid, the resulting fields must simultaneously satisfy all of Maxwell's equations, so I would say that the other 2 Maxwell's equations would also be "relevant equations" for this problem.

Also, the divergence of a field \mathbf{E} is usually written as \mathbf{\nabla} \cdot \mathbf{E}, not \mathbf{\nabla}\mathbf{E} (which is how one writes the gradient of a vector field, which is a 2nd rank tensor).

Now, in general you have \mathbf{E}= -\mathbf{ \nabla } \varPhi -\frac{ \partial \mathbf{A} }{ \partial t} (this should also have been listed as a relevant equation ). What do you get for \mathbf{A} when you choose \varPhi = 0? Does it satisfy the Lorentz Gauge? Does the resulting \mathbf{B}-field, together with the given \mathbf{E}-field satisfy all of Maxwell's equations?

 

[zhillyz]

Yeah I knew the correct way to write divergence but couldn't see the middle dot symbol anywhere haha :) thanks.

Okay that makes sense, there are a list of four maxwell equations that can be manipulated and I need to make sure all are satisfied when I pick values of A and phi. The question states that I have to do this with phi equating to zero.

Sooo plug chug and see what happens? Finishing with my equation 'iv)'?

======

E = E₀sin[\frac{2π}{λ}(z−ct)]xˆ

E = -∇\Phi -\frac{\delta A}{\delta t} NB. Set '\Phi' = 0 so;

E = -\frac{\delta A}{\delta t} and then

-∫E.\delta t = A and then

B = ∇×A, ∇.B = 0 = ∇.(∇×A) which will test if B is correct and finally both values into Lorenz gauge to equal 0?

 

[zhillyz]

If my understanding is correct then after I have took the negative integral of E and then found the curl of it I get a cosine function in the y-axis as I would expect but the following divergence of this function would give me a scalar function instead of the zero I am looking for?

EDIT***
Never mind no it doesn't haha. I think I have solved it. Thank you.

 

[gabbagabbahey]

Yeah I knew the correct way to write divergence but couldn't see the middle dot symbol anywhere haha :) thanks.

No problem, the \LaTeX code for it is just "\cdot". And for the partial derivative symbol, it is "\partial".

E = -\frac{\delta A}{\delta t} and then

-∫E.\delta t = A and then

B = ∇×A, ∇.B = 0 = ∇.(∇×A) which will test if B is correct and finally both values into Lorenz gauge to equal 0?

Not quite sure what -∫E.\delta t = A means (how does one integrate with respect to a partial differential?), don't you mean \mathbf{A} = -\int \mathbf{E} dt?

Other than that it looks like you've probably done everything correctly (you didn't actually post what you got for \mathbf{A} or \mathbf{B}, so I can't be 100% sure). Just 2 things that you should keep in mind:

(1) Make sure you check all 4 of Maxwell's equations - you made no mention of checking the curl and divergence of \mathbf{E}

(2) Is your choice of \mathbf{A} unique? What about the "constant" of integration, are there restrictions placed on it by Maxwell's equations or the Lorentz gauge condition?

 

[zhillyz]

Thank you for the correct Latex code and I did indeed mean A = - ∫Edt.

For A I got \frac{\lambda}{2\pi}E₀cos[\frac{2\pi}{\lambda}(z-ct)]\widehat{x}

For B I got -E₀sin[\frac{2\pi}{\lambda}(z-ct)]\widehat{y}

And thank you for all your help I shall check all of Maxwell's equations and integration constants.

Kind Regards Hilly

 

[gabbagabbahey]

Thank you for the correct Latex code and I did indeed mean A = - ∫Edt.

For A I got \frac{\lambda}{2\pi}E₀cos[\frac{2\pi}{\lambda}(z-ct)]\widehat{x}

For B I got -E₀sin[\frac{2\pi}{\lambda}(z-ct)]\widehat{y}

You are missing a factor of \frac{1}{c}, and you'll want to recheck the sign of B.

And thank you for all your help I shall check all of Maxwell's equations and integration constants.

You're welcome! And remember the integration "constant" is not just a scalar in this case, and can depend on position (just not time).

 

[zhillyz]

I am in desperate need of a warning before posting :), the factor of \frac{1}{c} I do have in my answer, I merely forgot it in this post :)whoops!

And double checking the sign of B I came to the same negative sine hmm.. Am I going wrong somewhere?

The operator ∇×A ends up as equating to \frac{\partial{Ax}}{\partial{z}}..

1)everything before the cosine has no variable z so is treated as constant
2)the function inside cosine is \frac{2z\pi}{\lambda} - \frac{2ct\pi}{\lambda}
3)so second part has no z and first part has derivative \frac{2\pi}{\lambda}
4)cosine changes to negative sine?

 

[gabbagabbahey]

And double checking the sign of B I came to the same negative sine hmm.. Am I going wrong somewhere?

Sorry, I meant A (and subsequently B)

 

[zhillyz]

Right you are, :) thank you again.