# Is the “How to break the speed of light” minute physics video wrong?

I am referring to this video, on YouTube, by minutephysics, which has quite a lot of views.

In the video it states that if you flick your wrist while pointing a laser that reaches the moon, that the spot of light on the moon will travel 20 times the speed of light. Now don't get me wrong, I do like their videos, just this one seemed a bit fishy to me. At first I thought it all practically made sense, then I realised something...

In my mind, I would think that light particles (photons) travel from the laser to the moon and bounces off the moon and back to your eye (it doesn't just stay there, in place, so you can move it around). Now, what he is stating is that if you flick your wrist these photons that have travelled to the moon will move along with your wrist. Wouldn't these photons be bouncing off of other objects or still travelling to the moon by the time you flick your wrist? i.e. dissipating, therefore new photons will be travelling to the moon (from the laser directly).

For example: let's say you point the laser at the moon, and once it reaches the moon, you wait a couple of seconds and then flick your wrist. The laser that you have flicked will emit photons in every direction that your wrist was in, correct? i.e. The photons would shoot out in a straight line (unless disrupted) continuously, with your wrist taking no affect on the speed of the photons.

So back to the question, is this video wrong?

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 – Qmechanic♦ Jan 5 at 1:05

The photons move at the speed of light in a straight line from the laser to the moon and back. The spot on the moon can move faster than light. There is no law against that. The spot is not a physical object, just an image. When you turn your wrist nothing happens to the photons which are already on the way to the moon - they continue on the same trajectory. But new photons are emitted in the new direction of your laser. It's like waving a garden hose back and forth.

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So basically, the spot isn't moving at all; it's rather being re-created in a different position continuously. – miguel.martin Jan 4 at 10:13
Yes, exactly right. – Michael Brown Jan 4 at 10:15
+1 Nice answer. I always try to filter these faster than light confusions by asking "is it just a locus?", because loci can do whatever they like! – twistor59 Jan 4 at 12:03
Which from memory is part of the point being made in the video. – Daniel Blay Jan 10 at 13:38

I have a question though. I tried doing a back-of-the-envelope calculation.

Suppose the wall you are shining a laser on is a distance d away from your hand. Assuming instantaneous reflection, the reflect light reaches you at $t_1 = 2d/c$.

Now suppose you move your hand vertically upwards by a distance $\mathrm{d}s$ with a velocity $v$. So a time $\mathrm{d}t = \mathrm{d}s/v$ elapses while you move your hand up. Suppose you do this at $t_0 = f\times t_1$, where $f$ is a number between 0 and 1. Then the time it takes the reflected light from the new position to reach you is $$t_2 = f\times t_1 + \mathrm{d}t + 2d/c$$

Then the velocity of the spot on the wall is: $$\frac{\mathrm{d}s}{t_2-t_1} = \frac{\mathrm{d}s}{f\times t_1 + d\times t}$$ But $d$ appears in the denominator for this, suggesting that the speed of the spot on the wall decreases as $d$ increases.

But I know from actual life that the speed of the spot increases as you increase the distance from the wall. What am I doing wrong?

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 I've latex'ed your equations. It should still accurately represent your answer, unless I made a mistake. Please edit it in that case. – Michael Brown Jan 10 at 11:03 Could you explain the movement of your hand a little more? Because if you move your hand up, you're creating a diagonal line of photons and the spot moves at the same speed as your hand, no matter what the distance $d$ is. ¿Do you mean you rotate you hand up, just like the video presuposes? – MyUserIsThis Jan 10 at 11:23 @Ferfer93 You're correct. If the hand is always horizontal and the vertical position of the hand is $z(t)$ then the apparent position of the spot is always $z(t-2d/c)$. Then $v_\text{spot}(t)=\dot{z}(t-2d/c)=v_\text{hand}(t-2d/c)$: the velocity of the spot is the same as that of the hand only retarded by the light travel time. Only rotating the hand will do anything non-trivial. – Michael Brown Jan 10 at 11:52 This is probably a stupid question, but I don't understand how rotation changes the situation. – user34801 Jan 11 at 12:07