# In a moving light clock, does the velocity of the clock add to the velocity of the light?

Currently going through the class Space, Time and Einstein from worldscienceu. On module Time in Motion an example is given of 2 light clocks, one moving and one stationary.

The point is made that as seen in the above image, the light of the moving clock has to travel a greater distance thus making the moving clock tick slower. Is the velocity of the moving clock added to the vertical velocity of the light to obtain that oblique trajectory?

• Downvoted why ? – silkAdmin Jul 13 '16 at 2:26
• Logically speaking, almost every popular science explanation including the one you're talking about is logically flawed, because they rely on so many assumptions beyond the postulates of relativity that they can't be used as valid justification for the effect. For instance, why is it the same vertical distance in both cases? And why can we ignore the time taken for absorption and re-emission? And how do we measure time? If you make a box such that light takes 1μs to travel from top to bottom, this doesn't guarantee the same emission and absorption delay. – user21820 Jul 13 '16 at 10:02
• Even the derivations given at en.wikipedia.org/wiki/… each involve extra assumptions. Also, the moving clock in our reference frame isn't made by anything to tick faster or slower just like that. Rather, from our viewpoint it ticks slower than the stationary one. From its viewpoint our clock is ticking slower than itself. (By the way I didn't downvote, and I can't really tell why you got downvoted.) – user21820 Jul 13 '16 at 10:44
• I've removed your second question since we prefer to have one question per post. You can post it separately, but I think you might have in mind the situation shown in the image in this other answer of mine. – David Z Jul 13 '16 at 15:53

The simple answer is yes. The velocity of moving clock does get added to the velocity of light but the velocities don't add up in a Galilean (linear) fashion. So it would be incorrect to use the phrase "add" here. Instead we can say the velocity of moving clock changes velocity of light pulse. Note that the speed of light pulse still remains $$c$$, but its velocity changes (i.e., only the direction changes).

Hopefully you understood the qualitative reasoning - According to the stationary observer, the light pulse has to travel at an oblique angle in order to catch up to the top half of the light clock.

Here's a quantitative explanation which will show you how to calculate the exact angle at which the light pulse will have to travel as seen by stationary observer.

$$\mathbf u_\parallel = \frac{\mathbf u_\parallel' + \mathbf v}{1 + \frac{\mathbf v \cdot \mathbf u_\parallel'}{c^2}}, \quad \mathbf u_\perp = \frac{\sqrt{1-\frac{v^2}{c^2}}\mathbf u_\perp'}{1 + \frac{\mathbf v\cdot \mathbf u_\parallel'}{c^2}}$$

Here, the primed frame of reference is that of the moving light clock and the unprimed frame of reference is that of stationary observer. So, if the moving light clock is moving at a velocity $$v$$, then for the light pulse we have $$u_\parallel'=0, u_\perp'=c$$. Plugging in these values in above two formulas we get,

$$\mathbf u_\parallel = v,\quad \mathbf u_\perp = \sqrt{1-\frac{v^2}{c^2}}\mathbf c$$

These are the components of velocity of the light pulse as seen by the stationary observer. You can calculate the oblique angle by using $$\theta=\arctan{\frac{u_\perp}{u_\parallel}}$$. Also, the magnitude of this velocity i.e., $$\lVert \mathbf u\rVert = \sqrt{\mathbf u_\parallel ^2 + \mathbf u_\perp ^2} = c$$. So you can see that the speed of light is consistent. Hope this helps!