General solution to the wave equation of electromagnetic field

[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 $$d^3\mathbf{x}$$

 

[DrClaude]

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

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

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?

 

[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?

 

[dextercioby]

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

 

[sardar]

instead?

I can provide an explanation for the general solution to the wave equation of the electromagnetic field. The solution provided is a standard form of the solution, which is known as the plane wave solution. In this solution, we consider the electromagnetic field as a superposition of many plane waves with different frequencies and wave vectors. The integral in the solution represents the summation of all these plane waves.

Now, coming to the question about the differentials, it is important to note that the integral represents the sum over all possible values of the wave vector $\mathbf{k}$. Therefore, the differential used in the integral should also be in terms of the wave vector $\mathbf{k}$, which is represented by $d^3\mathbf{k}$. This is because the wave vector $\mathbf{k}$ is a three-dimensional vector and the integral is over all possible values of $\mathbf{k}$, hence the $d^3\mathbf{k}$. This is also known as the Fourier transform in the wave vector space.

On the other hand, the differential $d^3\mathbf{x}$ represents the volume element in real space, and it is not applicable here because the integral is not over all possible values of position in space, but over all possible values of the wave vector $\mathbf{k}$.

In summary, the use of $d^3\mathbf{k}$ in the integral is correct and necessary to represent the summation over all possible values of the wave vector $\mathbf{k}$. It is important to understand the physical meaning and context of the differential used in the integral.