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Suppose I have a photon with energy $E$ (in its rest frame) moving along the $z$ direction with momentum $\mathbf{p} = p_z\mathbf{\hat{z}}$:

The Lorentz transform in the laboratory frame should be: $$ E'=\gamma(E-p_zv) $$

so using $p_z = E / c$ and $v=c$ for light, I get that $E'=0$.

How is this possible? Is there a physical meaning?

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    $\begingroup$ There is no photon rest frame, it violates the principle postulate of SR: [\begin paraphrase] the speed of light is 'c' in all frames of reference [\end paraphrase]. $\endgroup$ – kηives Apr 10 '14 at 15:31
  • $\begingroup$ $\gamma = \left(1-{v \over c}^2\right)^{-1/2}$ is indeterminate when $v=c$, so the equation says $E'={0 \over 0}$. $\endgroup$ – George G Apr 10 '14 at 15:35
  • $\begingroup$ Exactly I was expecting that something had to blow up to infinity. Thanks $\endgroup$ – SuperCiocia Apr 10 '14 at 18:36
  • $\begingroup$ @kηives (@kives ?): "There is no photon rest frame" -- Correct; compare with this question about phtotons and mutual rest. "the princip[al] postulate of SR: [\begin paraphrase] the speed of light is 'c' in all frames of reference [\end paraphrase]." -- Emphasis on inertial frames of reference; i.e. any system whose members are mutually at rest. There is a system "comoving with the photon"; it's just not an inertial system. $\endgroup$ – user12262 Apr 11 '14 at 5:20
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It is incorrect to state that there is a photon with energy $E$ in its rest frame, but it would be correct to talk about a photon with energy $E$ in some frame. In this case, the equation that you wrote would be valid to express the energy of that photon observed from a different frame (that you choose to characterize with a prime) that moves away with normalized velocity $\beta=\frac{v}{c}$,

$\displaystyle E^\prime = E \, \gamma \, ( 1 - \beta ) = E \, \sqrt{\frac{1 - \beta}{1 + \beta}}$.

This expression follows from exactly what you wrote. This result is the well known expression of the Doppler effect, and is valid for any $\beta < 1$. In the case of $\beta = 1$, you have found that $E^\prime = 0$. You cannot have an observer moving at the speed of light with respect to any other observer, so no one will ever measure $E^\prime = 0$ anyway. The somehow paradoxical idea that some observer would measure a zero energy is protected by the fact that no observer can ever reach the necessary speed for this to happen. However, you can think of a reference frame in the mathematical sense, moving at that speed. In this case your conclusion would be the same that it would be for a massive particle: in the rest frame where that particle is at rest, the energy is the rest energy, i.e. its mass (times $c^2$). In the (non-physical) frame with the photon being at (non-physical) rest, the energy of that photon would be its mass, i.e. $0$.

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What you are calculating is the invariant mass, for a photon the invariant mass is 0. This transformation is only 1 component of the Energy-Momentum 4-vector. Use the full form and you will get the result you want. The full form is given right at the end of this page

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