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This is probably a silly question, and likely reflects a poor understanding, in general, of electromagnetic radiation, but here goes anyway:

I have encountered two equations that I am struggling to unite into a unified concept, in the context electromagnetic radiation.

The first equation is the Planck-Einstein Relationship:

$$E=h\nu,$$ where $E$ is energy, $h$ is Planck's constant, and $\nu$ is the frequency of the photon.

The second equation is the average intensity of an electromagnetic wave $I_{\text{avg}}$:

$$I_{\text{avg}}=\frac{E_0 B_0}{2\mu_0}$$ where $E_0$ is the maximum electric field strength, $B_0$ is the maximum magnetic field strength, and $\mu_0$ is the permeability of free space.

Intensity, $I$, has units of energy per time per unit area (or, joules per second per meter$^2$).

Now, from what I understand, the first equation describes how much energy an incoming photon has at a particular frequency. I believe electromagnetic radiation propagates through space as a photon. Therefore, somewhere embedded within the $I_{\text{avg}}$ equation should be the energy of a photon, right?

So, here is my confusion. I work with MRIs and generate radiofrequency pulses in the $250\text{ MHz}$ range in order to carry out my experiments.

I was under the impression that the "$250\text{ MHz}$" referred to the frequency at which the electric field and magnetic fields oscillate. How does this number relate to the $\nu$ in the Planck-Einstein Relationship? Does $\nu = 250\text{ MHz}$ ? i.e. is the photon's frequency the same thing as the electric and magnetic fields' oscillations?

Further, it seems like there must be two components that contribute to the energy of an electromagnetic wave. Firstly, the frequency of the oscillation. But secondly, the amplitude of the electric field and magnetic field. In the $I_{\text{avg}}$ equation, these two amplitude terms are being considered in the form of $E_0$ and $B_0$. However, I do not see where the frequency of the fields' oscillations is coming into play.

Any clarification would be greatly appreciated! Cheers~

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When you have the equation $E=h\nu$, that’s for one photon. Now an EM wave is made of gazillions of photons so the energy is given by $E=Nh\nu$. This isn’t a helpful formula so we look at energy per second per unit area $S(t,\nu)$. This will be given by $$S(t,\nu)=n(t)h\nu$$ where $n(t)$ is the number of photons passing through a unit area per second. And this is what the amplitude is related to. So you can see how the energy is related both to the amplitude and the frequency.

I was under the impression that the "250 MHz" referred to the frequency at which the electric field and magnetic fields oscillate. How does this number relate to the $\nu$ in the Planck-Einstein Relationship? Does $\nu=250$ MHz? i.e. is the photon's frequency the same thing as the electric and magnetic fields' oscillations?

Yes! The frequency of the macroscopic field oscillation is exactly the frequency of the photon, or more precisely, energy/h of the photon. You can read up more about why this is so here, although you’d need to know a good amount of QM.

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  • $\begingroup$ Great! Exactly what I was looking for. So does that mean that there is some proportionality constant (let's call it $k$) such that $n(t)h\nu=k*I_{\text{avg}}$?...or is it as straightforward as $n(t)h\nu = I_{\text{avg}}$ $\endgroup$
    – S.C.
    Commented Feb 28, 2020 at 5:29
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    $\begingroup$ $n(t)h\nu$ is the intensity of light. And the time average intensity and intensity is related by a factor of $1\over 2$ and this has no time dependence. $\endgroup$
    – Superfast Jellyfish
    Commented Feb 28, 2020 at 5:39
  • $\begingroup$ hmmm. Is this also correct? $\frac{\int_0^t n(t)}{t}*h\nu = I_{\text{avg}}$ $\endgroup$
    – S.C.
    Commented Feb 28, 2020 at 5:47
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    $\begingroup$ Yes. But because of the periodic nature of $n(t)$ with usually a high frequency ($250$Mhz in your case), it is sufficient to average over one time period. And the time dependence goes as $cos^2(2\pi\nu t)$ whose average over a cycle is 1/2. $\endgroup$
    – Superfast Jellyfish
    Commented Feb 28, 2020 at 6:08
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    $\begingroup$ Awesome. Thank you so much. Cheers~ $\endgroup$
    – S.C.
    Commented Feb 28, 2020 at 6:08

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