Now this might be a silly question but it's actually bugging me, this one might be easier to understand if you have kids that watch (or used to watch) Peppa Pig. In one of the episodes, about shadows, the kids try to run away from their shadows, they try to move faster and faster and of course failing. Then comes in Mr Elephant saying something along the lines

It doesn't matter how fast you go, you can't run away from your shadow

At this point I would just like to say that I'm aware that this is a kids cartoon and all but in theory: is it possible?

For example if you would be traveling at a higher than speed of light (without anything in your pathway and with a constant source of light) would the position of the shadow be offset or moving away from ones current position? Or do elephants actually know a thing or two about physics?

• Things like us don't move faster than light. And they can't. – ACuriousMind Aug 14 '14 at 13:33
• Also, it's not possible for us to run faster than the speed of light - or escape our shadow. However, one interesting thing to note (also perhaps for @ACuriousMind) is that shadows can move faster than the speed of light. – Danu Aug 14 '14 at 13:35
• @Danu: Shadows don't move, they aren't objects ;) – ACuriousMind Aug 14 '14 at 13:38
• Does it help if I tell you that you are a few nanoseconds or so faster than your shadow? :) – PhotonBoom Aug 14 '14 at 14:18
• Does an Event Horizon count as a shadow? en.wikipedia.org/wiki/… – Aron Aug 14 '14 at 17:01

Note: This is basically item 3 in jkel's answer.

If you move at an appreciable fraction of the speed of light, then your shadow can appear to be "trailing" you, although it will always be "attached" to your feet if you're on flat ground.

Suppose a person is moving in the direction shown below and that there are plane waves coming in from an angle.

The left image is a top view, and the person is moving toward the right of the screen. The right image, where the person is moving into the screen, shows better the angle that the light is making relative to the horizontal.

Now think about the light rays that hit the person's head, body, and feet all at the same time, as shown in the diagram below.

The dashed lines show the path light would travel if the person weren't there. But the light does get blocked. Since light has a finite speed, it will take a longer amount of time for the shadow of the person's head to appear on the ground compared to the shadow from the person's feet.

Now all of this is for light that strikes the person at the same time. So far, so good.

Okay, by similar reasoning, the shadows on the ground at any given time are created by light striking the person at different times. That is, light that is missing and creating the shadow of a person's head must have hit the head earlier than light that is missing and creating the shadow of the person's feet. In other words, at any given instant in time, the shadow of the person's head is behind where you might expect, since that missing light struck the person's head at an earlier time.

If you put it all together, you get something like this at any instant in time:

The shadow will come out from the feet, but the head will be "behind" where you might expect it to be if the person was stationary.

• Liked this answer best! – hyp Aug 15 '14 at 8:01
• The shadow is a region where photons are blocked from reflecting from the ground. If I move into a position it actually takes a split second for the shadow to form as there are photons behind me that are still travelling and will still illuminate the ground. Only when all of these photons have been reflected and no new photons arrive will the shadow form. Because of this your shadow will indeed become detached from you if you move fast enough. – Brice M. Dempsey Aug 15 '14 at 8:46
• If the light runs parallel to the ground, the shadow will be detached even if the person is not moving at all. Then you can easily move away from the screen on which the shadow is projected, thus raising the distance. – Holger Aug 15 '14 at 14:00
• Do you have an idea of what would happen if the runner always was accelerating? I have in mind the fact that one can stay ahead of a beam of light if one is constantly accelerating (i.e. a hyperbolic trajectory) and has a sufficient head start. – Semiclassical Aug 16 '14 at 12:24

Here is how you can "run faster than" (or at least, get away from) your shadow: you jump at sunset (I just realized 15 minutes after posting that this is the point that @jkej's answer made as possibility #2)

Your shadow will detach from your feet, and it will "run away" from you. In the frame of reference of the shadow, you are running away from it.

Unfortunately, it won't last... the elephant, in the end, is still right. Unless, of course, you jump just as the sun sets: your shadow would disappear before you land again. Timing may be tricky, but with a good pogo stick you might just do it.

update for @PlasmaHH:

Source of the elephant picture: http://wallpho.com/173361-cartoon-elephant-id-91474.htm

• I had given a +1 had you used elephants in your illustration ;) – PlasmaHH Aug 14 '14 at 18:56
• @PlasmaHH - you have your wish... – Floris Aug 14 '14 at 19:08

Depending on exactly what you mean by away from your shadow, I can think of a number of methods:

1. Position yourself in the shadow of some object larger than you. This would result in your shadow disappearing altogether; the ideal solution in my opinion, but perhaps considered cheating by the likes of Mr Elephant.
2. Position yourself in such a way that your shadow falls very far from you, for instance on top of a tall building or a mountain. This works even better when the sun is low. This would not necessarily involve any running, but there is nothing stopping you from reaching this position by means of a light jog. There is really no limit to how far you can get from your shadow using this method, but at distances on the order of 100 m or more your shadow will be blurred beyond recognition
3. In case of interpreting "running away from your shadow" very strictly as "running so fast that your shadow is far away from where it would be if you had been standing still", this would be limited by how fast you would "run" and how far your shadow is from you through this formula: $$d=\frac{v}{c}L$$ where $d$ is the displacement of your shadow as defined above, $v$ is your running speed (assuming you are running orthogonally to the sun light), $c$ is the speed of light and $L$ is the distance between you and your shadow. No relativistic effects have been considered here. As you see, for this method it is also helpful to position yourself as in method 2, since this would give you a large $L$. As a quick, unrealistic example, let's assume you were "running" with a speed of 1% of the speed of light on top of the Burj Khalifa with a 60$^\circ$ solar zenith angle. Then your extremely vague and blurry shadow would be "lagging" by approximately 16 m on the ground 1600 m away from you.
• I just realized that your point 2 is a lot like my answer... I was creating the diagrams as you posted this. So you get my upvote. – Floris Aug 14 '14 at 16:07
• @Floris Yes, I think several of us were writing at the same time and we're all touching on the same solutions in principle. I think it's good with several styles of explaining the same thing. Your answer has the advantage of a figure for instance. I was too lazy for that. :) – jkej Aug 14 '14 at 16:17
• You could also wear a coat made of a metamaterial that bends light around you... ;) – moonshadow Aug 14 '14 at 16:25
• 4. Running [in the direction] away from your shadow (but never gaining any distance) – Izkata Aug 14 '14 at 17:22
• MIght be easier to discuss in terms of ultrashort light pulses or in terms of a high-speed shutter disk (toothed wheel). In both those standard laboratory situations, there's light (or dark) on the target while there's the opposite condition at the test plane. – Carl Witthoft Aug 14 '14 at 17:30

Your shadow, surely, is not just the apparent darkening of a 2D region of a diffuse surface you happen to be standing in front of; it is the entire volume of space that your presence is preventing light from reaching, the extrusion of your silhouette from the light source out to infinity (for a point light source, or possibly to a point a finite distance away in the case of an area light source larger than you and sufficiently close).

Moving sufficiently fast with respect to the light source will cause this volume to deform in all kinds of peculiar ways, but can never detach it from you.

This is borderline philosophical stuff, but...

If you want to put some distance between you and your shadow, you have to fly. Actually, you have to put some distance between and the opaque surface right below you, so swimming in a tank would do the trick as well. This will disconnect you from your shadow.

If you have a jetpack, or if you are in an ultralight/paraglider/etc., and the only light source is the sun, then the higher you go, the farther you will be from your shadow. You don't need to be particularly fast this way - even if you are close to stalling, you will still be able to distance yourself from your shadow.

You could also hide behind an opaque object so that you get no direct light from the light source. That way you won't have a shadow of your own.

Alternatively you could just become transparent, but that may not be very healthy.

There is a philosophical question of what happens to your shadow, after it would travel faster than the speed of light. This is easily realized with the thought experiment of an object is moving around a point source of light.

Given that the radial velocity of a rotating body is $\omega = vr$, then the shadow projected at a distance $r$ can then be rotating/traveling faster than the speed of light. This is the extrapolation of the equation: assume the body at $1m$ is traveling at a tangential velocity of $v=.5c$, then assume there is a second concentric circle where the shadow is being projected at $r=2m$. The projected shadow along the second orbit is moving at $v=c$. Moving the second concentric circle where the shadow is projected any further out would mean that it would be traveling faster than $c$.

So, assume that in the above experiment, where the projected shadow is moving faster than $c$, there would be an event horizon between you and your projected shadow, meaning you could not observe it and therefore you would have perceived to outrun it.

• Why would there be an event horizon between you and the shadow moving faster than $c$? – NeutronStar Aug 17 '14 at 16:19
• Indeed. And why $r = 2m$ ? Are you trying to bring black holes into this? And why would the rotating body cause the shadow to move? A perfect rotating sphere would have a stationary shadow. – HDE 226868 Aug 17 '14 at 17:00