it sure does!

Same with two rays of laser light - if you point them at an angle to eachother and then increase the angle until it is over 180 deg.... the crossing point gets faster and faster and exceeds the speed of light.

This is no problem, of course, as no mass is moved around.

I think I'd like to disagree.

Since the time the lasers or the sun were turned on the emitted light has only travelled a finite distance. However, infinite speed is reached only at an infinite distance from the source.

Another question is whether the shadow can travel faster than c if the sun has been "on" for a sufficiently long time.

There are phenomena with superluminal shadows in gas clouds around young stars. For your shadow, the chances are low to become superluminal. During the process of sunset the visible segment of the sun's disk becomes smaller (and thus less luminous) and your shadow fades away. The more distant parts of the shadow fade away by a further reason: place an observer at the distant place near the end of your shadow. He will see you in front of the sinking sun and only obstructing a small part of the light comming from there. Therefore, the longer the shadow becomes, the lower will be its contrast to the non-shadow region arround. But if the sun would be a laser beem and there would be no atmosphere and we would place photodetectors on the ground looking to the sun's light spot we would observe shadow speeds finally in access of c.

Shadows are dark areas surrounded by light. Therefore the shadow moves exactly with the same speed of the light. In other words, even if from pure geometrical considerations the point in which the shadow appears should move with infinite speed, the shadow is really formed only when the light outside its border will arrive.

Francesco,

shadow formation has nothing to do with light fronts or with light moving from outside the shadow into it and thus changing its position. Light was there all the time and is simply reshaped by obstruction. For finding out how exactly this works, the 'purely geometric considerations' are sufficient, wheras your way of taking into account the 'motion of light' leads you into the woods.

Yes dear Ulrich, i think so.

Ans: No, limited by velocity of light.

Why: A shadow appears to the observer when no photon reaches his/her eyes. An area that appears illuminated, when photons reach that location and the reflected light is recorded by your eyes. So, if you suddenly stop the photons that goes to a very far away location (say 1000s km away), the shadow can only appear after a time decided by the time taken for the last photon to reach your eye from the remote location. In short, the speed of shadow is limited by the velocity of light, "C".

I do not know, whether the discussion is still open. Anyway:

(and I assume that there is no atmosphere because otherwise there is absorption and fading, etc.)

Since Earth is a rotating (almost) sphere there are two shadows involved:

1. The shadow behind the terminator (day-night border) that moves with (almost) constant velocity depending on the date and the latitude.

2. Your shadow.

Your shadow cannot be longer than the distance between your feet and the terminator. That means that in the beginning the tip of your shadow moves away from you but its length is limited because of the Earths curvature and finally, when the tip of your shadow hits the terminator the shadow gets shorter again until the terminator moves past you when your shadow is finally gone.

All these movements are _much_ slower than the speed of light. E.g., the maximal velocity of the terminator on Earth is about 462 m/s.

Wonderfull,

I was waiting since months for an interesting answer.

Go on with discussion.

I'd like to modify the question a bit. Assume an ideally flat, infinitely large disk world with a hole in it. Assume also a point-source of light that has been 10 m above this world for an infinitely long time and a diskworldian of 1 m height standing 10 m away from the hole. Now the light source suddenly decides to dissappear into the hole with finite velocity. As the light source approaches the height of 1 m above the ground, does  the shadow of the diskworldian move with infinite velocity along the surface of the disk world as pure geometry would predict or does the speed of the shadow's edge never exceed the speed of light? We'll also assume perfect vakuum on this particular world.

Well, everybody knows that the strong magical field of the disk world slows light down to about the velocity of sound, and that causes all sorts of strange phenomena. ;)

To give my answer to Philip's question:

Since there is (according to the description) only one diskworldian (lets call him Rincewind) on the disk world he will be the only one to judge the length of his shadow at any specific time. He is a very magical creature, so the velocity of light on the disc world will be very small, lets say 10ms-1. For convenience (his and ours), Rincewind has marked the disk world in distances of 1 light seconds (aka 10m) radially, so that he can easily measure the length of his shadow at any time. Let us assume that the light source (aka sun) sinks with a velocity of 1ms-1.

At t=0s the sun is 10m above the disc world, and Rincewind's shadow is 10/9 m long. That is when we start. Now the sun starts to sink.

At t=8s the light source is 2m above ground. Geometry tells us that the length of the shadow should be 10m. Let us calculate, when the last photon emitted from the sun hits the point that is at the tip of the shadow. Apparently, this photon left the sun at t=8s. It travels sqrt(404) about 20.1m to hit the tip of the shadow roughly at t=10.01s. From there it scatters from the surface of the disc world getting reflected just into Rincewind's eye after travelling roughly another 10m at about t=11s. After that for Rincewind the shadow will have reached a length of 10m. Before that time photons having been emitted from the sun in the past will still reach Rincewind's eyes from a distance of 10m.

At t=8.5s the light source is 1.5m above ground. Geometry tells us that the shadow should be 20m. A similar calculation as before tells us that the last photon emitted from the sun reaches the tip of the shadow after travelling just over 30m at t=11.5s. After reflection it reaches Rincewind after another roughly 20m at t=13.5s. Then the shadow as seen by Rincewind will have a length of 20m.

At t=8.75s the light source is 1.25m above ground. The shadow should be 40m long, but the last photon from the sun reaches the tip of the shadow at around t=13.75s, and the reflection is back roughly at t=17.75s. This is the time when the shadow is seen to be 40m long.

The sun reaches 1m at t=9s.

In general, H(t)=10-t is the height of the sun.

l(t)=10/(9-t) is the geometric length of the shadow.

The time when the last photon from point l(t) arrives at the eyes (feet) of Rincewind is T(t)=t+1/(9-t)+(10-t)/(10(9-t)) sqrt(t^2-18t+82)

So Rincewind sees the tip of the shadow in position l(t) at time T(t).

For computing the velocity of the percieved shadow it is necessary to compute the visible tip of the shadow at time t to be x(t)=l(T-1(t)) which is a rather complicated formula. But this function has the following properties:

1) x(9s) = 4.95m (the apparent shadow length when the sun reaches 1m)

2) x'(9s) = 9999/5980 ms-1 = 1.67207 ms-1

3) x(10s) = 7.028m (the apparent shadow length when the sun sets (0m))

4) x'(10s) = 2.50884 ms-1

5) x(Infty) = Infty

6) x'(Infty) = 5ms-1

So the maximal velocity of the shadow tip is half the speed of light which is reached only at t=Infty, when finally the (percieved) shadow gets infinitely long.

For the "real shadow" seen by an observer moving along with the shadow the following holds:

U(t)=t+(10-t)/(10(9-t)) sqrt(t^2-18t+82) is the time when the last photon reaches l(t). For computing the velocity of the shadow it is necessary to compute the visible tip of the shadow at time t to be y(t)=l(U-1(t)) which is another rather complicated formula. But this function has the following properties:

1) y(9s) = 6.12m

2) y'(9s) = 2.77 ms-1

3) y(10s) = 9.95m

4) y'(10s) = 5.01 ms-1

5) y(Infty) = Infty

6) y'(Infty) = 10 ms-1

So the tip of the shadow (even when the observer moves with its tip) never goes faster than the speed of light.

Of course, the limits c/2 and c, respectively, do not change, no matter how tall Rincewind is, how far he stands away from the center, how high the sun is initially, how fast the sun sinks (as long as it sinks slower than the speed of light), or how large the speed of light is - only the formulas get more complicated.