- #1
victorvmotti
- 155
- 5
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 $$d^3\mathbf{x}$$
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 $$d^3\mathbf{x}$$