Frequently used way to derive Lorentz gamma seems flawed

Question

One of the most frequently used methods to explain the deduction of Lorentz' gamma is the idea to watch a photon traverse a distance of $c\cdot t$ in a moving object (e.g. a train) from outside the object. In the outside observer's frame the photon travels a distance of $c\cdot t'$, such that we assume $(c\cdot t)^2 + (v\cdot t')^2 = (c\cdot t')^2 \Rightarrow \dfrac{t'}{t} = \gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

So far, so good. But why must the photon traverse a path perpendicular to the object's motion? And what if we tilt the ``photon-clock'' a little bit, say by an angle of $\alpha$:

Then we gat the relationship: $(c\cdot t')^2 = (c\cdot t)^2 + (v\cdot t')^2 - 2(c\cdot t)(v\cdot t')\cdot \cos(\pi/2+\alpha)$ which can be solved for gamma as: $\gamma = \dfrac{2\sin(\alpha)\dfrac{v}{c}+\sqrt{4\sin^2(\alpha)\dfrac{v^2}{c^2}+4(1-\dfrac{v^2}{c^2})}}{2(1-\dfrac{v^2}{c^2})}$ or simpler $\gamma = \dfrac{\sin(\alpha)\beta+\sqrt{\sin^2(\alpha)\beta^2+(1-\beta^2)}}{(1-\beta^2)}$

Obviously, for $\alpha==0$ we get $\gamma = \dfrac{\sqrt{(1-\beta^2)}}{(1-\beta^2)} = \dfrac{1}{\sqrt{(1-\dfrac{v^2}{c^2})}}$ as expected. But for other angles we get other values of gamma:

Of course, I don't doubt that $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, but I wonder where the mistake in my calculations is, or if that standard approach to calculating gamma by watching a moving train with a travelling photon in it, only works in the case the photon's motion is perpendicular to the train's motion.

Answer 1

Let's look at the red side of the triangle in your diagram where $\alpha \neq 0$. In the frame where the clock is moving (the "lab frame"), this is the distance between the emitter and the receiver at the time the light is received—with all measurements made in the lab frame. By labeling it as "$ct$", you have made the implicit assumption that this red distance is equal to the distance between the emitter and the receiver as measured in a frame moving with the light-clock (the "clock frame").

The problem is that this is not necessarily the case. In the case where $\alpha = 0$, the distance between the emitter and the receiver is perpendicular to the relative motion, and it can be argued (via symmetry considerations) that distances measured perpendicular to the frames' relative motion must be the same in both frames.^* But if $\alpha \neq 0$, there's a component of this distance that is parallel to the relative motion, and we are not guaranteed that the red distance in your diagram—which is measured in the lab frame—is in fact the same as the distance measured in the clock frame.

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^*Any source that doesn't make an argument concerning the invariance of perpendicular displacements before introducing the light-clock argument is taking questionable shortcuts at best.