# Joule effect in Poynting theorem

I came across two different statements of the Poynting theorem. In one of them the power dissipated through Joule effect is made explicit.

The one without explicit Joule effect is:

$$\frac{dW}{dt} = \frac{d}{dt}\int_V u_{EB} d\tau + \Phi[\overrightarrow{S}]$$

The second one is:

$$\frac{dW_0}{dt} = \int_V \sigma |\overrightarrow{E}|^2 d\tau + \frac{d}{dt}\int_V u_{EB} d\tau + \Phi[\overrightarrow{S}]$$

where the first integral in the RHS is the power dissipated through Joule effect. These two statements of the theorem are from Griffiths' "Introduction to Electrodynamics" and from "Electromagnetic Energy Transmission and Radiation", Adler, Chu, Fano, respectively.

The difference arises from the definition of the power delivered to the system per unit time: Griffiths defines it as:

$$\frac{dW}{dt} = -\int_V \overrightarrow{E} \cdot \overrightarrow{J} d\tau$$

while Adler, Chu, Fano define it as:

$$\frac{dW_0}{dt} = -\int_V \overrightarrow{E} \cdot \overrightarrow{J_0} d\tau$$

with $$\overrightarrow{J} = \overrightarrow{J_0} + \sigma \overrightarrow{E}$$.

By replacing $$\overrightarrow{J}$$ with $$\overrightarrow{J_0} + \sigma \overrightarrow{E}$$ in the first statement of the Poynting theorem, one can reconstruct the second one.

My doubt is: shouldn't the power dissipated through Joule effect already be included in the flux of the Poynting vector (it is electromagnetic radiation in the IR afterall...)?

And, if that is the case, let $$\overrightarrow{S} = \overrightarrow{S_J} + \overrightarrow{S_r}$$ with subscript J idicating the part of the Poynting vector due to the Joule effect's radiation and subscript r indicating the reminder. Then, by linearity of the flux: $$\Phi[\overrightarrow{S}] = \Phi[\overrightarrow{S_J}] + \Phi[\overrightarrow{S_r}]$$. But $$\Phi[\overrightarrow{S_J}]$$ also is the power dissipated through Joule effect (let's call it $$W_J$$), so we have:

$$W_J = \Phi[\overrightarrow{S_J}] = \int_V \sigma |\overrightarrow{E}|^2 d\tau$$. The second statement of the Poynting theorem becomes:

$$\frac{dW_0}{dt} = W_J + \frac{d}{dt}\int_V u_{EB} d\tau + W_J + \Phi[\overrightarrow{S_r}]$$

and the Joule effect is counted twice. Where did I mess up?

## 1 Answer

The phrase “power dissipated through the Joule effect” does not, to me, evoke the notion of radiation. To me, this means heat. That is, electromagnetic fields drive currents which dissipate heat into the material. That’s where the conductivity comes in: it’s a statement of the extent to which radiation is or is not re-emitted after driving a current. That is to say, the radiation from the $$\sigma E$$ current is not the same as the power dissipated by the Joule effect. In fact, since energy is conserved, they are complementary. The radiation goes into $$S$$, the heat into $$W_J$$.

• Thank you! I get your point. Unfortunately I am still a bit confused: let's say we deliver power to a resistor which heats up due to Joule effect. We know that the power emitted as thermal radiation increases with temperature, so the EM radiation thorugh a closed surface that contains the resistor increases (and accordingly the flux of $\overrightarrow{S}$ should increase). If the power dissipaded thorugh Joule effect is already counted in the integral $\int_V \sigma E^2$, the increasing flux of the Poynting vector seems redundant to me. Sep 22, 2021 at 7:35
• @LucioPhys oh, you’re talking about blackbody radiation?? Now I see. Blackbody radiation does indeed carry energy away from a hot device. However, the extent to which this occurs depends on factors not included explicitly in your equations: is the resistor emissivity high or low? To include this, we would need an additional term in the Poynting theorem’s energy balance. $\sigma E$ does not include blackbody physics. Sep 22, 2021 at 11:19