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From this article,

Let’s say Alice is holding a light clock, and Bob is watching her run by, while holding it, with speed V. Alice is standing still (according to Alice), and the time,$ \tau$, between ticks is easy to figure out: it’s just $\tau = \frac{d}{C}$. From Bob’s perspective the photon in the clock doesn’t just travel up and down, it must also travel sideways, to keep up with Alice. The additional sideways motion means that the photon has to cover a greater distance, and since it travels at a fixed speed (EEP y’all!) it must take more time

My question is, why does photon need to travel sideways to hit the Alice light clock?

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  • $\begingroup$ See my comment under Michael Seifert's answer. $\endgroup$ – bright magus Oct 3 '15 at 7:37
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Because in Bob's reference frame, the light clock has moved sideways a certain distance during the time of flight of the photon. So to hit the "top" mirror, the photon must have travelled some horizontal distance (depending on Alice's speed) as well as the vertical distance; the total displacement is a diagonal.

BTW, the diagrams on the page you linked to are (in my opinion) not all that elucidating. There are some better ones here.

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  • $\begingroup$ As an aside: it is remarkably hard to find a good light-clock illustration that isn't on a website owned by someone who's trying to disprove relativity or peddle some pet theory. $\endgroup$ – Michael Seifert Sep 29 '15 at 13:46
  • $\begingroup$ So it's a distance in 2d space? $\endgroup$ – PaulD Sep 29 '15 at 13:59
  • $\begingroup$ And how does this changes time for both mirrors, if photon still has to travel both mirrors? $\endgroup$ – PaulD Sep 29 '15 at 14:09
  • $\begingroup$ To answer your first question: yes, the photon appears to travel a "diagonal" path relative to the motion of the light-clock, so its path is no longer one-dimensional. I'm not sure what your second comment means. $\endgroup$ – Michael Seifert Sep 29 '15 at 15:57
  • $\begingroup$ I meant, if the distance is equal for both clocks, how does moving of one clock changes time for both of them? Photon still have to travel same distance for both of them (no matter if it changes). Since then, time passed from one tick to another is same for two mirrors. $\endgroup$ – PaulD Sep 29 '15 at 16:04

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