# Thermal Electric Field

Say I have a blackbody isotropic radiator. I can define the spectral power radiated from the body as $$P=k_B T df$$ where $df$ is the bandwidth of interest. Thus the total energy radiated is just $k_B T$. I can also define the energy of an electric field across volume V as: $$U=\epsilon_0 \int{E^2dV}$$ Which led me to wondering, is there any meaning behind the expression: $$E=\sqrt{\frac{k_B T}{\epsilon_0 dV}}$$ I would interpret the above expression as the peak electric field that could exist across a volume due to thermal energy. Does that make sense? If there is a background thermal energy, does it make sense that there should also be a thermal electric field? Or is there a better way to define it?

Say I have a blackbody isotropic radiator. I can define the spectral power radiated from the body as $$P=k_B T df$$ where $df$ is the bandwidth of interest. Thus the total energy radiated is just $k_B T$.

With this definition of power per frequency range, you will have energy $k_B T$ for each frequency interval whose width is one Hertz. There is a big problem with this idea. Since there is infinity of disjoint intervals of such kind, the total energy comes out infinite. That is unrealistic result, energy is always finite in real situations.

In the theory of thermal radiation, this problem is avoided by considering only such distributions of energy on the frequency spectrum which lead to finite total energy. This implies, mathematically, that $P$ cannot be the same for all frequencies $f$, but ultimately it has to decrease as frequency increases. The blackbody radiates according to Planck's formula and from certain point on the energy of radiation in a box per unit frequency interval decreases as frequency increases.

I can also define the energy of an electric field across volume V as: $$U=\epsilon_0 \int{E^2dV}$$ Which led me to wondering, is there any meaning behind the expression: $$E=\sqrt{\frac{k_B T}{\epsilon_0 dV}}$$

If I understand this right, you are considering an electric field that has energy $k_B T$ for some temperature $T$ per some volume $dV$.

I would interpret the above expression as the peak electric field that could exist across a volume due to thermal energy. Does that make sense?

There is no basis for such interpretation in your description, as far as I can see. The maximum magnitude of electric field in thermal radiation isn't really limited by anything, as long as the space where this maximum is attained is kept small enough so total energy is preserved.

For the situation as I understood it, the formula $$E=\sqrt{\frac{k_B T}{\epsilon_0 dV}}$$ gives a kind of average of electric field - the precise term is root mean square value (rms). That is because it can be obtained as square root of average (integral) of square of $E$ over the volume $dV$.

If there is a background thermal energy, does it make sense that there should also be a thermal electric field? Or is there a better way to define it?

Yes, the heat radiation in all space is considered to be electromagnetic radiation, which consists of electric and magnetic field that vary in time and space in a random manner.

• I do like your interpretation of rms field. One more thing that bothered me was that if I put a dipole in this electric field, the potential energy would be U = pE. It seems odd to me the potential energy of a dipole in a "thermal electric field" could exceed thermal energy in the first place. – Kthaxt Feb 8 '18 at 23:48
• Well, if you put a dipole somewhere, the electric field will change in its vicinity and is not completely random thermal radiation any more. By putting it there, the field and its energy changes. – Ján Lalinský Feb 9 '18 at 0:06
• Well sure, but I don't think that changes the question. If there exists a background random field, and a I place a dipole in it, the new field is just a superposition of those fields. The thermal noise field doesn't go away. So it seems reasonable to define the potential energy of the dipole in that thermal field. But the result, in which the potential energy can be larger than kBT, seems odd. Maybe there's some time averaging I need to account for. – Kthaxt Feb 10 '18 at 20:29
• Indeed, if you're thinking that energy corresponding to a single degree of freedom should be comparable to $k_BT$, this is true in classical statistical physics, but only in the sense that expected average of kinetic energy of single degree of freedom is of order $k_BT$. The value of kinetic energy fluctuates in time and there is no hard limit to its magnitude. – Ján Lalinský Feb 10 '18 at 20:39