Maxwell's Derivation of EM Wave Momentum: E/c

[vin300]

How did maxwell derive the momentum of electromagnetic wave to be E/c ?

 

[cepheid]

I don't know how Maxwell derived it, or even whether he derived it. But there are two things I can say about where the result comes from.

First, you can show as a result of the Special Theory of Relativity that the general expression for the energy, E, of a particle is given by:

E² = p²c² + m²c⁴​

where p is the momentum of the particle, m is its mass, and c is the speed of light. So, for a massless (m = 0) particle like a photon, this reduces to

E = pc​

Second, this result is not inconsistent with the results of quantum mechanics. For example, quantum mechanics says that for particle states in which there is a definite momentum, that momentum is given by

$$p = \hbar k$$​

where $\hbar$ is h/2π, h is Planck's constant, and k = 2π/λ is the wavenumber of the particle. However, you may also be familiar with the result from quantum mechanics that the photon energy is given by:

E = hν​

where ν is the frequency of the photon. Now, remember that the relationship between frequency and angular frequency (ω) is ν = ω/2π. So we can write the photon energy as:

E = (hω)/2π = $\hbar \omega$​

I apologize for the hybrid between LaTeX and non-LaTeX equations. I couldn't be sure that hbar character would show up for all users, so I resorted to LaTeX to generate it. Anyway, so we have the results that:
$p = \hbar k$ and $E = \hbar \omega$. How are ω and k related? Well, recall that the speed of a wave is the product of its wavelength and its frequency:

c = λν = (λω)/2π = ω/k​

Therefore:

ω = kc​

and:

$E = \hbar \omega$ = $\hbar kc$ = pc​

Same result.