# Does a conducting wire give off measurable radiation?

In the Drude model (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum

$$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.

Accelerating charges give off Larmor Radiation. So does a DC carrying wire give off radiation one can measure?

I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:

$$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$

Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?

• Every dc current carrying wire heats up, so yes, it does give off radiation, and much more than what you have been calculating. Check your heating appliances for their electrical power ratings! – CuriousOne Sep 15 '14 at 20:49
• @CuriousOne, I was going to preempt this response in my OP but decided against it. What you're talking about is Joule heating, which is collisions with the lattice ions, and then (I assume) subsequent Blackbody radiation. That will definitely give off radiation, but isn't the mechanism I described. – YungHummmma Sep 15 '14 at 21:30
• How are you going to separate the two? Moreover, how is the Larmor component that you estimate going to escape the bulk? – CuriousOne Sep 15 '14 at 21:31
• @CuriousOne, well my question didn't really address separating them, it was just asking if the concept is sound. Do you mean separating the two concepts, or separating the radiation that would be emitted from both of them? The Larmor radiation will obviously escape if it comes from near the surface, and if the material is translucent to the wavelength(s) emitted, the whole thing. – YungHummmma Sep 16 '14 at 2:11
• The mean free path of electrons in metals for classical particle diffusion models would be very short (1nm). The potential drop in a metal wire over that distance is on the order of maybe nV... which is negligible compared to the thermal energy of the electrons at 300K. I think you need to think about a metal as the excitation of a many-particle electron sea, which delivers the correct results for many phenomena, in which case the Larmor model may not work. It's a very different deal in insulators. – CuriousOne Sep 16 '14 at 3:05

If you carry your example forward, then the total power emitted by a 1 m long copper wire that is carrying 1 A is just the sum of the power emitted by each electron. How many charge carriers are there in a 1 m copper wire with a resistance of 100 Ohm (since you had a field of 100 V/m)?

Resistivity of copper is $1.7\cdot 10^{-8}~ \Omega\cdot m$, so a 1 meter long $100 ~\Omega$ wire has an area of $\frac{1.7 \cdot 10^{-8}\cdot 1}{100} = 1.7\cdot 10^{-10}~m^2$ and a volume of $1.7\cdot 10^{-10}~m^3$; with one conduction electron per atom, and $8.5\cdot 10^{28}$ atoms of copper per cubic meter, that means there are about $1.5\cdot 10^{19}$ electrons participating in the conduction.

Multiplying the power of one electron by the number of electrons gives an estimate of the power of the emitted EM radiation due to a 1 A current as 1.5 nW. That is the estimated power emitted due to the motion of the electrons.

That is small, but not nearly as small as the number you first estimated. I imagine you would have to do a lock-in type experiment, where you turn the current on and off many times over a long time, and average the signal measured during "current on" and subtract from it the signal measured during "current off". Obviously you have to correct for the heating of the wire, which will itself result in significant power emission (black body radiation). In particular, for a symmetrical waveform (current on / current off) you would get a "sawtooth" thermal profile, except that there would be a small amount of curvature on the sawtooth; and unless you measured and corrected for this, you would have a hard time extracting those nano watts from the experiment. It probably means you need a setup that is a little more complex than the typical lock-in.

At room temperature, the emission of the copper wire mentioned (cross section $A$, length $\ell$) would be

$$P = 2 \pi r \ell \sigma T^4 = \sqrt{4\pi A} \ell \sigma T^4 = 21~ \rm{mW}$$

Clearly you would have to do this experiment at very low temperatures in order not to be overwhelmed by black body radiation.