Consider a cylindrical conductor made of a single material (say, copper) with conductivity $\sigma$. Assume there is a current through the conductor and, to avoid complications, assume that the current is constant.

Question: Given that the vast majority of the EM (electromagnetic) energy flows near a conductor surface, why does the spatial temperature distribution ("heat-map") of the conductor (due to Joule heating) decrease radially outwards?

P.S. Note that I may simply be mistaken that the "heat-map" decreases radially outwards.

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    $\begingroup$ Joule heating depends on the $\vec J \cdot \vec E$ product, not the energy density or the Poynting vector. Note that the latter two are non-zero outside the conductor. $\endgroup$
    – Puk
    May 8, 2022 at 16:13
  • $\begingroup$ @Puk Ok, but what about the spatial distribution inside the conductor? Is it radially decreasing (more heat near the center, if you take a cross-section) or just uniform in the conductor, or ...? $\endgroup$ May 8, 2022 at 16:41
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    $\begingroup$ For DC, $\vec J\cdot \vec E$ is more or less uniform in the conductor (disregarding temperature-dependent conductivity and the usually negligible Hall effect), so the conductor is heated uniformly. However, the conductor cools from its surface, so the temperature is higher near its axis. $\endgroup$
    – Puk
    May 8, 2022 at 16:50
  • $\begingroup$ @Puk You should post that as an answer. Yet, there is indeed some counterintuitive point in this: the fact that the conductor is heated more or less uniformly (and then becomes radially decreasing due to cooling) does seem a bit against the location of spatial EM energy near the surface. One would expect quite the opposite to happen: since most of the EM energy is near the surface, why isn't the surface incredibly hot? Conductors can get crazy hot, which requires lots of energy, so... where is this thermal energy coming from, if not from the EM field? $\endgroup$ May 8, 2022 at 16:57
  • $\begingroup$ @Puk Also, for DC, isn't it valid to say that $\vec{J } = \sigma \vec{E}$ ? But shouldn't $\vec{E}$ vanish (or tend to vanish) near the axis of the conductor? In an ideal conductor, after all, $\vec{E} = \vec{0}$ anywhere inside the conductor. $\endgroup$ May 8, 2022 at 17:07

2 Answers 2


For DC current, the current density $\vec J$ and the electric field $\vec E$ are both more or less uniform in the conductor. This means the Joule heating power density $\vec J \cdot \vec E$ is also approximately uniform in the conductor. However, the conductor cools from its surface, so the temperature is higher near the conductor's axis.

There are some second-order effects that cause $\vec J$ to fail to be exactly uniform. Most conductors have a positive temperature coefficient of resistance, which means the conductivity is lower around the conductor's axis (where the temperature is high), so the current density is lower here. Another effect is the usually minute Hall effect caused by the conductor's own magnetic field, resulting in some radial carrier redistribution and a corresponding radial variation in conductivity. But normally the temperature will still be higher near the center.

The energy density doesn't directly influence where the heat is generated. Note that if you regard the energy as being stored in the fields, the energy density is non-zero even outside the conductor, but this doesn't mean heat is generated here. You would see the effect of Joule heating on the Poynting vector in the form of a radially inward component as in the figure below, representing the flow of electromagnetic energy into the conductor where it is converted to heat.

enter image description here

I. Galili and E. Goihbarg, "Energy transfer in electrical circuits: A qualitative account", Am. J. Phys. 73, 141 (2005)

As for your question about why $\vec E \ne 0$ in the conductor, in general $\vec E = 0$ in a conductor is only valid in electrostatics where there is no current flow. A non-zero $\vec E$ is needed to push charges around and create current (except in superconductors).

  • $\begingroup$ So, as crazy as it sounds, the wire is heated from the outside-in by the EM field. But since $\vec{S}$ is always directed inward the wire, energy transfer is forced near the axis. Correct? $\endgroup$ May 8, 2022 at 18:30
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    $\begingroup$ If we regard the energy as being stored in the fields, yes, the inevitable conclusion is that there is an inward energy flow into the wire from outside, consistent with Poynting's theorem. Note however that $S$ is not necessarily strictly radially inward: it will normally have a component along the wire as well, corresponding to energy transfer along the wire. This component is determined by external circuit the wire is connected to. $\endgroup$
    – Puk
    May 8, 2022 at 18:41
  • $\begingroup$ Well, I would say it is counterintuitive. From your last statement, I would assume it is possible to design an external circuit such that $\vec{S}$ is almost perfectly parallel to the wire, and this should minimize the wire heating. On the other hand, with a circuit where $\vec{S}$ is almost perfectly radially inward, the same wire should heat much more. $\endgroup$ May 8, 2022 at 18:53
  • $\begingroup$ Can we regard the energy as being stored elsewhere (not in the fields), and still be consistent? $\endgroup$ May 8, 2022 at 18:54
  • $\begingroup$ I'm not claiming that all this is intuitive, just that it works out mathematically. Where exactly the energy is stored and which way it is flowing outside wires are rarely of practical concern and are not useful considerations when designing circuits, at least for DC. Ultimately the wire heats up because of inelastic collisions of charge carriers in the wire. This is quantified by $\vec J$ and $\vec E$, which we have a more direct control over than $\vec S$. Reducing the inward component of $\vec S$ just means reducing the current, or equivalently, the voltage across the wire. $\endgroup$
    – Puk
    May 8, 2022 at 19:06

Let’s choose a cylindrical coordinate system $\rho,\theta,z$ along the axis of the wire. We’ll assume that the wire is the the positive conductor in some electric circuit, and therefore has a small nonzero net charge; see e.g. the illustration at this other answer.

Model the electric field as

\begin{align} \vec E_\text{inside} &= \frac1\sigma \vec J = E_\text{in}\hat z \\ \vec E_\text{outside} &= E_\text{out} \frac R\rho \hat \rho \end{align}

where $\vec J$ is the current density, $\sigma$ is the conductivity of the wire, $R$ is the wire’s radius, and the field magnitudes $E_\text{in}$ and $E_\text{out}$ are the solutions to freshman homework problems.

Actually, we can see that this approximation is not strictly allowed. A surface charge, such as on the edge of our wire, introduces a discontinuity in the field normal the surface, the $\hat \rho$ direction. However the field components parallel to the surface, in the $\hat\theta$ and $\hat z$ directions, must be continuous. So our outside field must actually be

$$ \vec E_\text{outside} = E_\text{out}\frac R\rho \hat\rho + E_\text{in} \hat z $$

Our assumption that the field outside the wire is radial is therefore equivalent to an assumption that the conductivity $\sigma$ of the wire is large.

The magnetic field at some distance $\rho$ from the axis is

$$ \vec B = \frac{\mu_0}{2\pi} \frac{I_\text{enclosed}}{\rho} \hat \theta $$

Outside of the wire, $\rho > R$, the enclosed current is just the entire current, $I = \pi R^2 J$. Inside the wire, only the inner part of the current contributes, and the field becomes

$$ \vec B_\text{inside} = \frac{\mu_0}{2\pi}\frac{\pi\rho^2 J}{\rho}\hat\theta \propto \rho\hat\theta $$

We can find the direction of the Poynting vector $\vec S \propto \vec E \times \vec B$ everywhere. Outside of the wire there is a component proportional to

$$\vec S_\text{downstream} \propto E_\text{out} \hat\rho \times \hat\theta = E_\text{out} \hat z$$

That is to say, the power transport outside of the wire is parallel to the direction of the current in the “supply” wire of the circuit. The current-return wire will have a small net negative charge, $\vec E{}_\text{outside}^\text{return} \propto -\hat\rho$, and energy will flow antiparallel to the current — also towards the load.

There is also a radial power flow proportional to

$$ \vec S_\text{radial} \propto E_\text{in} |\vec B| (-\hat\rho) $$

which is responsible for heating the wire. To find how much energy is deposited per unit volume, let’s consider a cylindrical shell with radius $r$, thickness $d\rho$, length $L$, and volume $2\pi rL\,d\rho$. The power deposited in this shell is given the difference between the power entering from its outside and the power exiting towards its inside:

\begin{align} P &= \int_\text{outer} \vec S \cdot d\vec A - \int_\text{inner} \vec S \cdot d\vec A \\ &= S_\text{outer} A_\text{outer} - S_\text{inner} A_\text{inner} \end{align}

Outside of the wire, we have $S \propto B \propto r^{-1}$, and area $A \propto r$, so no energy is deposited in the current-free space outside of the wire. However, inside the wire we have

$$ \vec S \cdot d\vec A \propto BA \propto r^2 $$

The power in a shell with thickness $d\rho$ therefore goes like

\begin{align} B_\text{outer} A_\text{outer} - B_\text{inner} A_\text{inner} & \propto (r + d\rho)^2 - r^2 \approx 2r\, d\rho \end{align}

Since the power per unit volume and the volume itself are both proportional to $r\,d\rho$, we have uniform power deposition in the metal.

You ask why the wire is hotter at the center than at the edges. This is a heat-transport problem: the heat can only leave the wire through its surface, so the surface must cool before the interior.

  • $\begingroup$ Unfortunately, I can only accept one answer. But I will carefully check out the math you posted in your very precise answer. $\endgroup$ May 8, 2022 at 19:23

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