In another thread, I came across the question of how the photon's energy and momentum appears in the moving frame. The question is best explained with the standard light clock. Lets have 2 horizontal mirrors, that is, one above the other. In the rest frame, the photon is set up to be linearly polarized such that its electric field goes in and out of the screen/page. The points of maximal and minimal intensity are shown as circles in the left part of the picture. Photon is moving up. (the green curve is simply a visual cue) Now there are 2 lines of reasoning that contradict each other: 1. Middle picture When viewed from a moving frame, the maxima and minima of the electric intensity should be at the same points, that is, at the same height. It means that the color of the photon should be red-shifted, as its wavelength is stretched by the factor ##\gamma##. 2. Right picture The 4-momentum of the photon is, in the rest frame, (E, 0, E, 0). Boosting by ##v## and ##\gamma## in the x-direction, we get $$E'=\gamma(E-v\cdot 0)=\gamma E$$ $$p_x'=\gamma(0-v\cdot E)=-\gamma v E$$ $$p_y'=p_y=E$$ Since the wavelength is ##\lambda'=1/E'=1/(\gamma E)=\lambda/\gamma<\lambda##, the photon appears blue-shifted. From PeterDonis' comments in another thread, it seems that the second way is correct, but can please someone pinpoint the error in reasoning (1)?