mitochan said:
Mass does not apply for a single photon, so let us prepare N photon gas of same wavelength and no total momentum in a theorotetical massless sphere vessel.
Now assume instead a light-clock at rest in frame ##S## containing 2 photons, each with an energy of ##L##, moving always in opposite direction between the mirrors at the top and at the bottom. Then the energy-content of this light-clock is ##E_0 = 2L##.
In this reference frame ##S##, the two mirrors are removed at the same time. One photon escapes now from the light clock in y-direction, the other in (-y)-direction.
Assume a frame ##S'##, moving constantly with ##v## in positive x-direction. Two observers are at the same x'-coordinate, but at y'-coordinates with opposite signs.
Described in ##S##, those oberserves in ##S'## (with time-dilated clocks) must receive the photons each blue-shifted with energy
##L' = L \gamma (1 - \frac{v}{c} * \cos{90°}) = \gamma L##, according to
Einstein's Doppler-formula.
(In frame ##S'##, the blue-shift comes from aberration.)
So in frame ##S'##, the energy content of the light clock must have been ##E = 2L' = \gamma E_0## before loosing the photons. That means:
$$E_{kin} = mc^2(\gamma -1)= E -E_0 = E_0(\gamma -1)$$
Plausibility-check to classical mechanics (using for x << 1: ##\frac{1}{\sqrt{1-x}}\approx 1+\frac{1}{2}x##):
##mc^2(\gamma -1) \approx \frac {1}{2}mv^2##
This is a similar, but simpified argumentation as in Einstein 1905:
https://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf