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EM waves/ photons

Physics Asked by jensen paull on April 10, 2021

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 $E=hnu$ 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.

2 Answers

Pointing vectors is a classical electrodynamics concept, describing the energy flux, which should not be confused with the energy density. $E=hbaromega=hf$ 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.

Remark: and oscillating charge has a direction of oscillations (which means that it is effectively an oscillating dipole) - therefore the electromagnetic field emitted by such a charge never has a spherical symmetry, but rather cylindrical one.

Answered by Vadim on April 10, 2021

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.

Answered by my2cts on April 10, 2021

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