Solution to EM Wave: Questions & Answers

[KFC]

I know that in free space, the general solution of the wave equation about electric field is of sine and cosine form. One can also write it in complex form as

$$E = E_0 \exp(i\vec{k}\cdot\vec{r} - i\omega t)$$

I have two queations about this solution

1) If consider the polarization, how should I rewrite the solution?

2) In the presence of polarization, will the angular frequency $$\omega$$ depend on polarization?

 

[KFC]

I know that in free space, the general solution of the wave equation about electric field is of sine and cosine form. One can also write it in complex form as

$$E = E_0 \exp(i\vec{k}\cdot\vec{r} - i\omega t)$$

I have two queations about this solution

1) If consider the polarization, how should I rewrite the solution?

2) In the presence of polarization, will the angular frequency $$\omega$$ depend on polarization?

I just find the solution involves polarization in another textbook, it reads

$$\vec{E} = \tilde{E} \exp(i\vec{k}\cdot\vec{r} - i\omega t)$$

where $$\tilde{E}$$ is complex amplitude represent the amplitude and the polarization. The book read: for finding the general multi-mode solution, we have to add up all possible solutions of above form

$$\sum_{k}\sum_p\tilde{E}_{k,p} \exp(i\vec{k}\cdot\vec{r} - i\omega_{k, p} t)$$

Now the complex amplitude depends on wavenumber k and p (polarization I guess). Well, I don't understand why a 'p' there? If we need to sum over p, so what values of p is allowed? In the book, it said p can be 1 or 2, but why is that? The last question is about the frequency, why it also depends on p?

 

[Thaakisfox]

That is because the EM wave is a transverse wave. The direction of the wave propagation is "k". So the E vector is in the plane perpendicular to the "k" vector. So E can be given in a basis of two vectors in the plane, representing polarization.
Explicitly:
Let $$\textbf{n}$$ be a unit vector parallel to the direction of wave propagation, that is parallel to $$\textbf{k}$$. This means that we can write $$\textbf{k}=k\textbf{n}$$ where k is the absolute value of the k vector.

Now let's take a unit vector in the plane perpendicular to the k vector, that is let:

$$\textbf{k}\cdot \textbf{e}^1(\textbf{n})=0$$

We want an orthonormal basis in this plane. But we know that we can generate a unit vector orthogonal to e^1 via the cross product, so let:

$$\textbf{e}^2(\textbf{n})=\textbf{n}\times\textbf{e}^1(\textbf{n})$$

Now since the E vector is in this plane, it can be given as a linear combination of the above two unit vectors. So we have for E:

$$\textbf{E}(\textbf{r},t)=A_1(\textbf{k})\textbf{e}^1(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}+A_2(\textbf{k})\textbf{e}^2(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}$$

Using summation notation:

$$\textbf{E}(\textbf{r},t)=\sum_{p=1}^2A_p(\textbf{k})\textbf{e}^p(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}$$

So the polarization state can be represented by the vector (A_1,A_2). This is called the Jones vector. hence the p index.

 

[KFC]

That is because the EM wave is a transverse wave. The direction of the wave propagation is "k". So the E vector is in the plane perpendicular to the "k" vector. So E can be given in a basis of two vectors in the plane, representing polarization.
Explicitly:
Let $$\textbf{n}$$ be a unit vector parallel to the direction of wave propagation, that is parallel to $$\textbf{k}$$. This means that we can write $$\textbf{k}=k\textbf{n}$$ where k is the absolute value of the k vector.

Now let's take a unit vector in the plane perpendicular to the k vector, that is let:

$$\textbf{k}\cdot \textbf{e}^1(\textbf{n})=0$$

We want an orthonormal basis in this plane. But we know that we can generate a unit vector orthogonal to e^1 via the cross product, so let:

$$\textbf{e}^2(\textbf{n})=\textbf{n}\times\textbf{e}^1(\textbf{n})$$

Now since the E vector is in this plane, it can be given as a linear combination of the above two unit vectors. So we have for E:

$$\textbf{E}(\textbf{r},t)=A_1(\textbf{k})\textbf{e}^1(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}+A_2(\textbf{k})\textbf{e}^2(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}$$

Using summation notation:

$$\textbf{E}(\textbf{r},t)=\sum_{p=1}^2A_p(\textbf{k})\textbf{e}^p(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}$$

So the polarization state can be represented by the vector (A_1,A_2). This is called the Jones vector. hence the p index.

Thanks a lot. Very clear :)

But in your explanation, it seems that the angular frequency only depends on k but not polarization (so no p index in $$\omega$$)?

By the way, if it is not plane wave, do we still have something like polarization?