I’ve read that the Poynting vector points into an incandescent light bulb at all instants of time, whether there is direct/constant or alternating current through the wire. This means the bulb is obtaining energy from the electromagnetic field.

But, the bulb is getting hot, and it is emitting light, so shouldn’t there be Poynting vectors pointing outside the bulb? To indicate the bulb is releasing energy, in form of heat and light.

Furthermore, shouldn’t the net Poynting vector be zero at the bulb? To indicate that all of the energy the bulb receives is again released (as heat and as light). I mean, if the Poynting vector wasn’t zero, it’d mean energy is being stored in the bulb, but that can’t happen in an incandescent bulb.

More generally, since (ideal) resistors can’t store energy, I think the net Poynting vector at the resistor should also be zero.

What are your thoughts? Is my analysis wrong? How?

  • $\begingroup$ en.m.wikipedia.org/wiki/Joule_heating have you look at this article explaining resistive heating? $\endgroup$
    – Triatticus
    Commented Nov 25, 2021 at 18:47
  • $\begingroup$ @Triatticus What does that add to the discussion? The article says that energy is dissipated in resistive elements, the same thing I said. $\endgroup$
    – alejnavab
    Commented Nov 25, 2021 at 18:49
  • 4
    $\begingroup$ We just ignore the Poynting vector of the thermal radiation in this analysis (and we can't give an exact form for it anyway – just a time averaged one) and at least part of the energy won't be transported away by radiation, but by convection and heat conduction – so the time averaged Poynting vector, will still point towards the filament even when including the thermal radiation. $\endgroup$ Commented Nov 25, 2021 at 18:51
  • $\begingroup$ @AlejandroNava but did you read through the entire article? It gives the reason for joule heating down to the microscopic level it's also general to any element since all elements have some level of resistance to them. $\endgroup$
    – Triatticus
    Commented Nov 25, 2021 at 20:00
  • $\begingroup$ @Triatticus I suppose you mean the Microscopic description section. That section explains how colliding electrons produce heat, which I knew. As Sebastian Riese and The Photon pointed out, my mistake was that I thought that the Poynting vector includes all types energy, when it actually only includes electromagnetic energy (and thus also light), not heat (thermal energy). $\endgroup$
    – alejnavab
    Commented Nov 25, 2021 at 21:59

1 Answer 1


The Poynting vector only accounts for electromagnetic energy transfer. It doesn't account for kinetic energy, which is what the heat produced by a resistor is.

In the case of an incandescent bulb, the light emitted is only a small fraction of the power input. Maybe a few per cent. We could calculate the contribution of the light to the Poynting vector if we want to, but often we're actually interested in the power delivered to the bulb at low frequencies, so it would be counterproductive to do so.


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