1.5.2 The Energy and Momentum of a Photon (Where m = 0) (Special Relativity)
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This article is from the Relativity and FTL Travel FAQ, by Jason W. Hinson jason@physicsguy.com with numerous contributions by others.
1.5.2 The Energy and Momentum of a Photon (Where m = 0) (Special Relativity)
We should quickly note the case where the rest mass of an object is
zero (such is the case for a photon--a particle of light). Given the
equation for the energy in the form of Equation 1:8 (E = gamma*m*c^2), one
might at first glance think that the energy was zero when m = 0. However,
note that massless particles like the photon travel at the speed of light.
Since gamma goes to infinity as the velocity of an object goes to c, the
equation E = gamma*m*c^2 involves one part which goes to zero (m) and one
part which goes to infinity (gamma). Thus, it is not obvious what the energy
would be. However, if we use the energy equation in the form of Equation 1:7
(E^2 = p^2*c^2 + m^2*c^4), then we can see that when m = 0 then the energy
is given by E = p*c).
Now, a photon has a momentum (it can "slam" into particles and change
their motion, for example) which is determined by its wavelength (lambda) in
the equation p = h/lambda (where h = 6.626E(-34) Joules is called Planck's
constant). A photon of wavelength lambda thus has an energy given by E = p*c
= h*c/lambda, even though it has no rest mass.
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