How Do Gaussian Profiles Affect Solutions to Maxwell's Equations in a Vacuum?

[JustinLevy]

Electromagnetic waves

Homework Statement

Find the solution of Maxwell's equations in vacuum for a continuous beam of light of frequency $\omega$ traveling in the z direction with a gaussian profile in the x and y directions.

Homework Equations

Maxwell's equations in vaccuum.
$$\nabla \cdot \vec{E} = 0$$
$$\nabla \cdot \vec{B} = 0$$
$$\nabla \times \vec{E} = - \frac{\partial}{\partial t} \vec{B}$$
$$\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \vec{E}$$

These of course can be combined to give the wave equation:
$$\nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E} = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{E}$$

The Attempt at a Solution

I already know what the plane wave solutions look like in the z direction:
$$\vec{E}(x,y,z) = \vec{E}_0 \cos(kz - \omega t)$$
where $$k = \omega/c$$

No polarization was specified, so let's choose linear polarization in the x direction. Our solution should then be similar for polarization in the y direction, and we can get the general solution of any polarization by adding these with arbitrary amplitudes and phases.

So, for linear polarization in the x direction and generalizing the above with dependence in the x-y direction it would be something like:
$$\vec{E}(x,y,z) = \hat{x} \ E_0 \ g(x,y) \ \cos(kz - \omega t)$$

Now seeing the constraint on g(x,y) we find:
$$0 = \nabla \cdot \vec{E} = E_{0} \ \cos(kz - \omega t) \frac{\partial}{\partial x}g(x,y)$$

Which seems to say there can't be any dependence on x! What!?

What am I doing wrong?
I've tried searching around the net and haven't found any good hints on this problem either. Please help.

 

[anathema]

Firstly, your attempt at a solution looks very good and you are on the right track. However, there are a few things that need to be clarified and some steps that need to be taken to arrive at the correct solution.

Firstly, the form of the solution for the electric field in vacuum for a continuous beam of light can be written as:

\vec{E}(\vec{r},t) = \hat{x} \ E_0 \ f(x,y,z) \ \cos(kz - \omega t)

where \vec{r} is the position vector, t is time, and f(x,y,z) is a function that determines the spatial profile of the beam. This function f(x,y,z) can be chosen to be a Gaussian function in the x and y directions, as specified in the problem, but it is not necessary to have a dependence on z as well.

Next, we need to apply Maxwell's equations to this solution to determine the correct form for f(x,y,z). Starting with the first equation \nabla \cdot \vec{E} = 0, we can write:
\frac{\partial}{\partial x} (E_0 \ f(x,y,z) \ \cos(kz - \omega t)) = 0

This implies that the function f(x,y,z) must be independent of x, which is what you found in your attempt at a solution. However, this does not mean that there can't be any dependence on x at all. It simply means that f(x,y,z) can be a function of y and z, but not x.

Similarly, applying the second Maxwell's equation \nabla \cdot \vec{B} = 0, we find that f(x,y,z) must also be independent of y.

Next, we can apply the third and fourth Maxwell's equations, \nabla \times \vec{E} = - \frac{\partial}{\partial t} \vec{B} and \nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial}{\partial t} \vec{E} respectively, to determine the form of f(x,y,z). These equations will yield a differential equation for f(x,y,z) that can be solved to find the correct form of the Gaussian profile in the x and y directions.

I hope this helps to

 

[ogb p]

You are on the right track with your solution for the electromagnetic wave in vacuum with a gaussian profile in the x and y directions. However, there are a few things to consider:

1. The constraint on g(x,y) does not necessarily mean there can't be any dependence on x. It means that the derivative of g(x,y) with respect to x must equal zero, which can only be true if g(x,y) is a constant or a function of y only. Therefore, g(x,y) can still have dependence on y and be a valid solution.

2. Another important factor to consider is that the wave equation you have written assumes a monochromatic wave with a single frequency. A gaussian profile in the x and y directions implies a superposition of waves with different frequencies, so the wave equation needs to be modified to take this into account.

3. One way to approach this problem is to use Fourier analysis to decompose the gaussian profile into a sum of plane waves with different frequencies and amplitudes. Then, you can use the plane wave solutions you already know to construct the complete solution for the gaussian profile.

I hope this helps guide you in the right direction. Keep in mind that finding the complete solution for a continuous beam of light with a gaussian profile can be a complex problem, and there may be multiple valid solutions depending on the specific conditions and assumptions made. Good luck with your research!