There have been previous questions like this regarding light/photons specifically. My question is what the effect of the amplitude is on, say, a gamma wave? It would change its energy, as $E = h f$. Maybe amplitude doesn't even exist with an EM wave.


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The electric and magnetic field strength ($\bf{E}$ and $\bf{B}$) is a concept of classical electrodynamics. It emerges from quantum electrodynamics as a collective effect only when there are many photons.

When considering only a few photons or even a single photon (like when you have gamma rays) it doesn't make sense to talk about electric and magnetic field or their amplitude.

The energy of a single photon is $E=hf$ (where $f$ is the frequency of the EM wave). Therefore, the amplitude of electric and magnetic field has no effect on the energy of the single photons. Instead, the amplitude effects the number of photons.

The energy density of the EM field (i.e. energy per volume) is $$\rho=\frac{\epsilon_0}{2}\bf{E}^2 + \frac{1}{2\mu_0}\bf{B}^2.$$ Therefore in an EM wave the number $N$ of photons within a volume can be calculated from the volume integral $$\begin{align} N hf &= \int_V \rho\ dV \\ &= \int_V \left( \frac{\epsilon_0}{2}\bf{E}^2 + \frac{1}{2\mu_0}\bf{B}^2 \right)dV \\ &= \frac{1}{2} \left( \frac{\epsilon_0}{2}\hat{\bf{E}}^2 + \frac{1}{2\mu_0}\hat{\bf{B}}^2 \right) V \end{align}$$ where $\hat{\bf{E}}$ and $\hat{\bf{B}}$ are the amplitudes of the EM wave.

  • $\begingroup$ If the amplitude is the number of photons, then I take it it's the RMS value, since I don't think the number of photons would go to 0 and into the negative temporarily. $\endgroup$
    – ettolrach
    Commented Mar 10, 2020 at 11:53
  • $\begingroup$ More precisely: the squared amplitude multiplied by volume. See my updated answer. $\endgroup$ Commented Mar 10, 2020 at 12:18

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