# General solution to the wave equation of electromagnetic field

1. Nov 10, 2014

### victorvmotti

Suppose that we have the four-vector potential of the electromagnetic field, [texA^i[/tex]

The wave equation is given by $$(\frac {1}{c^2} \frac {\partial^2}{\partial t^2}-\nabla^2) A^i=0$$

Now the solution, for a purely spatial potential vector, is given by

$$\mathbf{A}(t, \mathbf{x})=\mathbf{a_k} \exp{i(\pm \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}}); \mathbf{k}.\mathbf{a}=0$$

To find the general solution we write the superposition as

$$\mathbf{A}(t, \mathbf{x})=\int (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}})) \frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

My question is that where this $$\frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

comes from? Shouldn't it be $$d^3\mathbf{x}$$

2. Nov 10, 2014

### Staff: Mentor

It comes from the density of states in $\mathbf{k}$-space. If you were to integrate over $d^3\mathbf{x}$, you wouldn't get the vector potential as a function of $\mathbf{x}$, would you?

3. Nov 10, 2014

### victorvmotti

Why we do not say that $$\mathbf{A}(t, \mathbf{x})=\sum_{\mathbf{f(k)}\mathbf{g(k)}} (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}}))$$

Isn't superposition simply given by the above sum? Why and how we transform to the integral?

4. Nov 10, 2014

### dextercioby

Isnt' k a continuous variable? How could you sum after its continuous values?