Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question already has an answer here:

If I make a huge laser with a figure for shadow in front of the laser, and I shine it on to the moon, will I see the light from the laser AND the shadow moving the same speed? (I read somewhere the shadow could move with faster-than-light speed)

If I move the laser in such a way, that the light should overcome the proportional distances on the moon, will I see it, or will I see a delay?

share|improve this question
add comment

marked as duplicate by Qmechanic May 6 '13 at 6:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

One way to think of a "moving shadow" is by following the last photon that was allowed through. In that case, the speed of a shadow is exactly the speed of light.

On the other hand, you could also define the speed of a shadow as the speed of the boundary between dark and light. In that case there is no thing that's actually moving, so there's no bound on the speed of the shadow -- it can go as fast as you want it to. (Well, it's limited by your device and the distance to the moon, which serves to multiply the speed because you're presumably using the distance to multiply the angle.)

But it's really important to remember that this isn't defying the law that nothing can go faster than light. Again, there is no thing that's moving any faster than light. The shadow in this second sense is just a pattern that you define, and it was created by actual particles that moved at the speed of light.

share|improve this answer
3  
As an other example, group velocity for a wave can exceed the speed of light. –  Mostafa May 5 '13 at 23:09
    
That's a really important example, too. [But I think it might be a bit harder to follow. :) ] –  Mike May 5 '13 at 23:11
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.