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by lenkite 2507 days ago
The speed of shadow can be faster than light though.

https://physics.stackexchange.com/questions/335537/can-a-sha...

2 comments

I think those answers are actually wrong. When the object casting the shadow moves, the shadow remains in the same place for an observer in it until the light from the source reaches the observer inside the shadow.

I know that's a weird explanation, so consider:

  t0: S~~~>O     U (shadow exists)

  t1: S~~~~~~>   U (shadow exists)
      
  t2: S~~~~~~~~~>U (shadow !exist)
Where "S" is a light source, "O" is an opaque object, "~" are photons traveling to the right, "U" is the observer, and "t0", "t1", and "t2" are times (increasing).

At time "t0", "U" thinks he's in the shadow.

At time "t1", "U" thinks he's in the shadow.

At time "t0", "U" thinks he isn't in the shadow, since the photons are now hitting him.

A similar calculation/thought-experiment can be done for shadows with "angular momentum", in case you think the tangential velocity of the shadow will exceed the speed of light.

Thanks for that nice diagram! What happens when the light source moves and the shadow is far bigger than the object casting it ? Wouldn't the speed of the shadow on the surface be faster than the speed of light ?
For example, when a light source is super close to an object and the shadow gets super big really far away, and the light source moves?

  t0: SO          U1 (U1 in shadow)
                  
  t1: S~~~~~~~~~~>U1 (shadow leaves top first)
       O~~~~~~>   .
         ~>      .
                .
              .
             '
            U2

Dotted line is the path of a fixed point on the shadow during the time S moves.

I didn't do the precise math, but I'm pretty sure the tangential velocity of the shadow along the dotted line won't be greater than the speed of light. The curvature of the "wave front" formed by the tips of the arrows ">" above will be lesser than the dotted line curvature, so the photons near the top of the diagram hit the dotted edge before the ones towards the bottom. This is because the source, S, takes time to move away from O.

Note that the wavefront formed by the photons moves radially outward from S, but ascii art is limiting.

No information can be conveyed by the wavefront and so nothing is actually moving than the speed of light. What you diagrammed is called the Lighthouse Paradox:

* https://en.wikipedia.org/wiki/Lighthouse_paradox

There are similar things that appear to exceed the speed of light:

* https://en.wikipedia.org/wiki/Faster-than-light

See "group" and "phase" velocities for similar things to the lighthouse.

A shadow isn’t actually a thing. It’s an image, like a mouse pointer. If I had a sufficiently large screen, and I made the mouse pointer jump (by setting its position) to the other end of the screen, I could calculate its speed and make that number higher than the speed of light. But has anything actually moved? No it hasn’t, because a mouse pointer, like a shadow, is only an illusionary image of a thing, not an actual thing.
An absence of something can only be defined relative to that something and never taken just by itself. So a shadow only exists as a function of light (no light) or a consequence of the absence of light. So it would not travel faster than light in a way that can carry additional information.

Quantum entanglement works faster than light but cannot carry any information. As such the speed of causality (and implicitly of light) is still the real limit.

Thanks for this clarification. I have been wondering about this for awhile and your explanation helped a lot.
from the point of view of someone in the shadow, and subsequently not in the shadow, the speed of light is still the limit. If the light source is 1LY away, it will take 1LY for me to notice that I am no longer in shadow, regardless of how fast the shadow moves. There's no way of measuring the shadow movement that isn't limited by the speed of light.