Vector and scalar potentials for an EM plane wave in a vacuum

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Homework Help Overview

The discussion revolves around the application of vector and scalar potentials in the context of electromagnetic plane waves in a vacuum. Participants are analyzing the implications of various equations, including the Lorentz gauge, Gauss's law, and the Ampere-Maxwell equation, in relation to the behavior of electric and magnetic fields.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the vector potential A from the electric field E and questions the validity of their approach when checking the Ampere-Maxwell equation. They express confusion regarding the imaginary nature of A and the implications of setting J to zero in a vacuum.

Discussion Status

Participants are actively engaging in clarifying misunderstandings related to the vector potential and its treatment in the equations. Some guidance has been provided regarding the necessity of including the displacement current term in the Ampere-Maxwell equation, which has led to a realization for one participant.

Contextual Notes

There is an ongoing exploration of the assumptions made about the vector potential A and its components, particularly in relation to the wave equation and the treatment of vector calculus in Cartesian coordinates.

Natchanon
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Homework Statement
Note: E, B, A, J and x are vectors, and w is omega, the angular frequency.
Determine the vector potential A(x,t) and scalar potential V(x,t) in the Lorentz gauge, for the linearly polarized plane wave described by
E(x,t) = E_0 e^(i(kz-wt)) x_hat
B(x,t) = B_0 e^(i(kz-wt)) y_hat,
with the boundary condition that the potentials must be finite at infinity. (Hint: Let V = 0)
Relevant Equations
1. E and B above
2. Lorentz gauge: ∇⋅A = -μ_0 ε_0 ∂V/∂t
3. Gauss' law in terms of V: -∇^(2)V + μ_0ε_0∂^(2)V/∂t^(2) = ρ/ε_0
4. Ampere-Maxwell law in terms of A: -∇[SUP]2[/SUP]A + μ[SUB]0[/SUB]ε[SUB]0[/SUB]∂[SUP]2[/SUP]A/∂t[SUP]2[/SUP] = -μ[SUB]0[/SUB]J
5. B_0 = E_0 / c
6. c = w/k
Lorentz gauge:A = -μ0ε0∂V/∂t
Gauss's law: -2V + μ0ε02V/∂t2 = ρ/ε0
Ampere-Maxwell equation: -2A + μ0ε02A/∂t2 = μ0J
I started with the hint, E = -V - ∂A/∂t and set V = 0, and ended up with
E0 ei(kz-ωt) x_hat = - ∂A/∂t
mult. both sides by ∂t then integrate to get A = -i(E0/ω)ei(kz-ωt) x_hat
Now this looks good to me at first, as it satisfies B = × A , which gives the B(x,t) equation in the homework statement. And since we let V = 0, it satisfies Gauss' law in vacuum where ρ = 0. But when i checked the Ampere-Maxwell equation, setting J = 0 because we're in vacuum, the first term is fine as it gives 0, but the second term gives μ0ε0iE0ωei(kz-ωt), which isn't equal zero.
Where did I go wrong? Is it my method of finding A or do I misunderstand the Ampere-Maxwell equation? Or something else? Is A even supposed to be imaginary?
 
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Natchanon said:
the first term is fine as it gives 0

It does not. Please show your computation.

Note that the equation is just a wave equation for ##\vec A##.
 
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Orodruin said:
It does not. Please show your computation.

Note that the equation is just a wave equation for ##\vec A##.
Ok, I forgot and treated A as a scalar when calculating ∇^2. So,
2A = (∇⋅A) - × ( × A) = (∂Ax/∂x) - (× B)
First term, we have partial A_x / partial x, but x-component of A doesn't depend x, so it's zero, thus, the first term is zero. The second term is μ0J, which is zero, unless there is something I misunderstand. So even after finding vector laplacian of A, the first term is still 0 for some reason. I don't know what's going on here.
 
it is ##\nabla \times B=\mu_0(J+\epsilon_0\frac{dE}{dt})## so its not just ##\mu_0J##..

(You forgot the displacement current term dE/dt)
 
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Delta2 said:
it is ##\nabla \times B=\mu_0(J+\epsilon_0\frac{dE}{dt})## so its not just ##\mu_0J##..

(You forgot the displacement current term dE/dt)
Thank you. It all makes sense now!
 
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Natchanon said:
Ok, I forgot and treated A as a scalar when calculating ∇^2.
As long as you do things in a Cartesian basis you can factor out the basis vectors and treat the components of ##\nabla^2 \vec A## separately. The issue you were having is solved by #4.
 

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