What could be and how to come up with a mathematical formula for the local electric field inside a blackbody cavity? I asked a similar question here more than two years ago. I did not get an answer to my complete satisfaction. I would like to reiterate the problem again.
The local electric field of a monochromatic radiation is nonzero and varies sinusoidally in a predictable fashion. For example, the electric field of an ideal monochromatic radiation is described by $${\bf E}({\bf r},t)={E}_0\hat{{ \varepsilon}}\cos({\bf k}\cdot{\bf r}-\omega t),$$ at any location ${\bf r}$, is nonzero at any time $t$ and varies with time in a predictable manner. Here, $E_0$ is a fixed number and so is $\omega$ (the frequency of the radiation), and $\hat{\varepsilon}$ represents the constant polarization vector.
In contrast, assuming that the electric field at any location of a Blackbody cavity is due to an incoherent superposition of electric fields of all frequencies, polarizations (and all amplitudes?), can we rigorously come up with a mathematical expression for the local electric field at any time $t$ for the blackbody radiation? I am interested in getting a mathematical formula that properly represents the local electric field of incoherent thermal radiation.
 A: Using Cartesian coordinates you could write something like:
\begin{eqnarray} 
\vec{E}(\vec{r},t) &  =  & \int E_{0}(\omega) \left(\cos (\omega[t - x/c]) \hat{j} + \cos(\omega[t -x/c] +\phi_{x,\omega}) \hat{k}\right)\ d\omega \nonumber \\
           & + & \int E_0(\omega)\left( \cos (\omega[t - y/c]) \hat{i} +  \cos(\omega[t -y/c] +\phi_{y,\omega}) \hat{k}\right)\ d\omega \nonumber \\
           & + & \int E_0(\omega)\left( \cos (\omega[t - z/c]) \hat{i} +  \cos(\omega[t -z/c] +\phi_{z, \omega}) \hat{j}\right)\ d\omega . \nonumber 
\end{eqnarray}
This represents the sum of three unpolarised beams of light travelling in three directions, each with equal power, summed over all frequencies. In each term, the $\phi_{i,\omega}$ term represents a random phase between $0$ and $2\pi$.
The time averaged value of $\vec{E}\cdot \vec{E}$ is
$$\left< \vec{E}\cdot \vec{E}\right> = 3 \int E_0^2(\omega)\ d\omega    $$
and so the time-averaged energy density of the electromagnetic fields would be
$$ 6\epsilon_0 \int E_0^2(\omega)\ d\omega = \int \frac{\hbar \omega^3}{\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right)\ d\omega\ , $$
where the term on the right is the total energy density of a blackbody radiation field and the extra factor of 2 on the left hand side accounts for equal energy density in the magnetic field.
Thus
$$E_0(\omega) = \left[ \frac{\hbar \omega^3}{6\epsilon_0\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right) \right]^{1/2} $$
Of course, the time-average of the electric field itself is zero.
