- #1
mahblah
- 21
- 2
Homework Statement
The wave equation is
[itex]\nabla^2 \mathbf{A}(\mathbf{r},t) = \frac{1}{c^2} \frac{\partial^2 \mathbf{A}(\mathbf{r},t)}{\partial t^2}[/itex]
I want to get a solution for the vector potential A.
Homework Equations
we can use the Fourier transformation
[itex] \mathbf{A}(\mathbf{k},\omega) = \int{d\mathbf{r}}\int{dt \mathbf{A}(\mathbf{r},t) \exp{[-i(\mathbf{k \cdot r} - \omega t)]}} [/itex]
[itex] \mathbf{A}(\mathbf{r},t) = \int{\frac{d\mathbf{k}}{(2\pi)^3}} \int{\frac{d\omega}{2 \pi} \mathbf{A}(\mathbf{k},\omega) \exp{[-i(\mathbf{k \cdot r} - \omega t)]}} [/itex]
to get
[itex] \left( \mathbf{k}^2 - \frac{\omega^2}{c^2}\right) \mathbf{A}(\mathbf{k},\omega) =0 [/itex]
(so [itex] \omega = c k [/itex] )
The Attempt at a Solution
Now the the solution is (formally)
[itex] \mathbf{A}(\mathbf{r},t) = \sum_{\lambda =1,2} \int{\frac{d\mathbf{k}}{(2\pi)^3}} \int{\frac{d\omega}{2 \pi} A_\lambda(\mathbf{k},\omega) \hat{\epsilon}_\lambda(k) \cos{(\mathbf{k \cdot r} - \omega t + \varphi_\omega)} \delta(\omega -ck)} [/itex]
but i don't understand well why we have 2 polarization vector and form where they come...
thanks all,
MahBlah