Energy flux for an Em wave is defined by the poynting vector $(1/\mu_0) E \times B$
The energy is also proportional to $E^2 $
If you take for example the simplest form of generating light, the hertzian dipole. E and B are both proportional to $\omega$ Thus the poynting vector is proportional to $\omega^2 $or $(2\pi f)^2 $and the energy in the EM wave is also proportional to $ \omega^2$
How does This relate to the equation E=hf
Why isn't it $f^2$?
Is my conclusion that the energy in an EM wave proportional to $\omega^2$ true? as this is the conclusion from the hertzian dipole, is it true in general?
Why is this different from the energy of a single photon
Imagine a source produces radio waves with a very low frequency, Now let's imagine this source produces lots of these waves such that they are inphase and constructively interfere with eachother. the new wave has the same low frequency, yet HIGH AMPLITUDE. classical em says the energy has increased and so has the intensity, without a frequency change
However E=hf seems to say that the energy and therefore intensity is invariant. what am I not understanding?
Best Answer
You are confusing the electromagnetic field energy in a volume with the energy of single photon.
Light is made up of photons. The energy of a single photon with frequency $f$ is given by $E=hf$, which is very tiny. For example: a red photon has energy $3\cdot 10^{-19}$ J, and a violet photon has energy $6\cdot 10^{-19}$ J.
On the other hand, the energy of the electromagnetic field (with electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$) in a volume $V$ is $$E=\left(\frac{\epsilon_0}{2}\mathbf{E}^2 + \frac{1}{2\mu_0}\mathbf{B}^2\right)V.$$ This field contains zillions of photons. Increasing the fields $\mathbf{E}$ and $\mathbf{B}$ (while keeping the frequency constant) will increase the number of photons, but will not change the energy per photon. So there is no contradiction.