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I just finished reading Feynman's Lectures on Physics vol.I, §34-9: "The momentum of light". The author explains that there is a relation between the wave 4-vector $k^{\mu}$ and the energy-momentum 4-vector $p^{\mu}$ of an EM wave, namely

$$p^{\mu}=\hbar k^{\mu}, $$ or equivalently $$\tag{deB}W=\hbar \omega, \mathbf{p}=\hbar \mathbf{k},$$

and those equations are called de Broglie relations.

However, as I learned in my classical electromagnetism course, flux of energy in such a wave is quantified by Poynting's vector, yielding formulas such as the following:

$$\tag{1} I=\frac{1}{2 \mu_0 c} E_0^2, $$

where $I$ stands for "average intensity" of the wave and $E_0$ for "maximum amplitude of electric field".

Question Where is $\omega$? It does not appear in formula (1) nor in any other formula based on Poynting's vector. But as of equations (deB) it should do so. Am I wrong?

Thank you.

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2 Answers 2

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The section you are referring to clearly states that those equations do not apply to the wave, but to the "particles" of light, the photons. The resolution is that two waves of the same amplitude but different frequencies contain different numbers of photons. This has interesting consequences, for instance it means it is possible to communicate via radio waves carrying miniscule amounts of power while an optical signal of similar intensity would be drowned out by shot noise.

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  • $\begingroup$ I re-read more carefully and noticed what you mention. Indeed, the author clearly distinguishes the classical framework from the quantum one. Thank you very much! $\endgroup$
    – Giuseppe Negro
    Feb 3, 2012 at 1:59
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zephyr's answer is correct, but here's a slightly different way of explaining the same thing.

One of the predictions of quantum electrodynamics is that the amount of energy in an EM wave is quantized. This means that the wave can only carry certain discrete amounts of energy. The allowed energies of an EM wave are multiples of a single base energy, which is $\hbar\omega$. So the formula that Feynman gives, $E = \hbar\omega$ (I notice you used $W$ but it's much more common to write $E$), actually tells you the fundamental "unit" of energy carried by the wave, not the total amount of energy it carries. The formula for the total energy carried by the wave is more properly written $E_n = n\hbar\omega$ where $n$ is some integer. It's this energy $E_n$ that is related to the intensity $I$. This does imply that the electric field of an EM wave is also quantized.

It's worth remembering that any results that come from classical electrodynamics, including the formula $I = E_0^2/2\mu_0 c$, "know" nothing about the quantization of energy. As far as classical EM theory is concerned, $I$ can be anything at all. The contribution of the quantum theory is to say that $I$ can only have certain values.

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    $\begingroup$ ($W$ is there for coeherence with the book's notations). Your point is clear, thank you. So a quantized formula for intensity might look like $$(\text{no. of quanta per unit time per unit surface})\cdot(\hbar \omega)?$$ (I'm afraid I'm on slippery ground now...) $\endgroup$
    – Giuseppe Negro
    Feb 3, 2012 at 1:57
  • $\begingroup$ Yep, that's correct. $\endgroup$
    – David Z
    Feb 3, 2012 at 7:01

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