# EM waves/ photons

For an oscillating charge that produces a spherical wave, the energy of the wave at a point $$r$$ is proportional to its $$A^2$$, where $$A$$ is the amplitude (which we can see from the Poynting vector).

However, the relation $$E=h\nu$$ indicates that it is proportional to frequency. Can anyone explain to me the relationship between a charge with an acceleration $$a$$, the wave's amplitude, and its frequency?

I'm guessing that a higher $$a$$ corresponds to a higher amplitude and a higher frequency.

• Out of courtesy and respect for those who volunteer their time to answer questions, please conform to accepted English (uppercase to start a sentence, and a period at the end) and write equations in MathJax. – garyp Mar 12 at 12:24

Classically the energy of an EM wave is proportional to $$A^2f^2$$, where $$A$$ is the amplitude of the vector potential and $$f$$ is the frequency. Quantum mechanically the energy density is $$Nhf$$. Equating these gives $$N= A^2f/h$$ as the expected number of photons with a standard deviation of $$\sigma=\sqrt N$$. This assumes a monochromatic wave. The relation between charge acceleration and radiated power is given by the Larmor formula.
Pointing vectors is a classical electrodynamics concept, describing the energy flux, which should not be confused with the energy density. $$E=\hbar\omega=hf$$ is a quantum relation between the energy of a photon and its frequency. If we define the operators of the energy density and energy flux in quantum domain (which is rather easy to do), the letter will be expressed as sum over the energies of single photons, see here.
• The direction is of course given by $\vec k$. – my2cts Mar 12 at 10:47