Photon momentum

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

[tex] p = \hbar k [/tex]​

where [itex] \hbar [/itex] 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 = [itex] \hbar \omega [/itex]​

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:
[itex] p = \hbar k [/itex] and [itex]E = \hbar \omega [/itex]. 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:

[itex]E = \hbar \omega [/itex] = [itex] \hbar kc [/itex] = pc​

Same result.
 

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