# The sled and the hole

1. Aug 29, 2004

### Gonzolo

Hi, I'm familiar with most SR, but just introductory GR, so I never got a satisfactory answer to the following problem :

A sled and a hole have similar rest length. Now suppose the sled is travelling near c over the hole. The paradox is that from the hole's reference frame, the sled is shorter, so it should fall in, while in the sled's reference frame, the hole is smaller, so it should go over. Will it fall in it or not?

I understand GR provides the answer (it curves in?), but that it is complicated mathematically. Can someone clear this up? (Mathematically if possible, but not necessarily.) Thanks.

2. Aug 29, 2004

### Nenad

to each other, both objects seem like they have shrunk. But only one pbject actually has, the sled, because only the sled is in actuall motion, and had a distance dilation.

3. Aug 30, 2004

### zefram_c

Eep. You call this a Lorentz invariant treatment? Both objects appear in the other's frame as shrunk. There is no "actual motion" of the sled.
This is another example of a class of paradoxes that arises from the Newtonian assumption of a perfectly rigid body. I was confused by this too, see here: https://www.physicsforums.com/showthread.php?t=34134
Still, in this case, even Newton would agree that the front of the sled falls into the hole, as long as one neglects friction between the sled and the ground. What happens after that is anyone's guess and depends on how you want to model the fall.

4. Aug 30, 2004

### pervect

Staff Emeritus
There is a very interesting part of this problem that probably doesn't matter, and that is what the force of gravity is going to look like in the transformed frame of the sled.

It's probably irrelevant to the actual solution, though.

So I'll try and discuss what I see as the solution first, and then do the interesting but irrelevant part last.

In Newtonian mechanics, one would treat the runners as being rigid bodies.

This is clearly inappropriate in relativity.

So what we have is some sort of distributed spring-mass system, and not a rigid runner.

The propagation velocity of a disturbance will be at the speed of sound in the runners, which will be very low - about 3000 --6000 meters per second, depending on the vibration mode.

If we think of the sled as being feet long, the actual time for anything to happen will be very small.

So let's think of the sled as being very long. If it's a mile long, it will still take the sled only 10 microseconds to transverse a mile long hole at .5c. This is not time for the sled to fall very far.

What will happen, is that basically, the sled runners will deform a little bit in the 10 us that they are unloaded. Exactly how much they will deform isn't clear without a much more detailed analysis of the sled-runner system as a distributed physical system.

But it's pretty clear that the runners will move down some in 10us, though not very far. What happens when the runners move down slightly, then encounter the ground on the other side isn't clear. I'd expect them to disintiegrate, but then I'd expect that real runners wouldn't actually work anyway. But it's pretty clear that the runners will drop a small amount, and that wew could in princple calculate the number.

It's also pretty clear that the physical rigidity of the sled, that was supposed to inspire the whole problem, is an illusion. A sled a mile long is going to sag under it's own weight a lot, what happens at one end will not seriously affect what's happening at the other end.

So much for the solution - now for some of the interesting questions.

How will the gravitational field transform with velocity?

I'm not really sure of the answer here, unfortunately. But to get some idea, one can replace the gravitational field with an electrical one, by imagining that the sled is charged, and is attracted via electrostatic force to some point a long distance away.

In this case we can say that the magnitude of the force will scale up by a factor of gamma in the sled frame.

There will be a magnetic field, but it will be irrelelvant to the problem, as either the sled will be stationary, or the charge will be stationary, so we never have a combination of a magnetic field with a moving charge.

I believe the direction of the force will remain vertical, and not abberate in the sled frame. The hole will be broadside the sled the same time as the point charge we are using to simulate gravity will be. (I think.) And the field will point towards the charge, for the electric field case.

The gravitational situation will hopefully be roughly similar. But while we can judge that the force will increase, I can't say by how much at this point.

5. Aug 30, 2004

### Gonzolo

If so, then it should be quite simple mathematically with say two masses. Relativity doesn't even seem to be needed to solve this. Classical Newton talk would seem to be enough (vertically).

The "paradox" was perhaps simply a demonstration that in relatvity, the concept of a rigid body has to be abolished.

Instead of a point, perhaps a plane? So that the E field is constant in its z? And only defined from a < x < b, 0 elsewhere. Since we only need x and z, two charged dielectric wire should do, neglecting end effects. Moving along different directions, the magnetic field would actually cause them to repel each other.

I don't think so. If there is electrical attraction, and that there is movement, we do have that combination. I don't think the magnetic field would be negligeable.

Yes, I think so too.

Thanks for the input!

Last edited by a moderator: Aug 30, 2004
6. Aug 30, 2004

### Nenad

why is there no motion of the sled when it states that the sled is traveling at very high speed.

7. Aug 30, 2004

### zefram_c

Because in the frame of reference of the sled, the hole is moving towards it. As counterintuitive as it may be, one must accept the idea that (inertial) motion is equivalent to staying put while the world scrolls by you.

8. Aug 30, 2004

### Gonzolo

It is 100% equivalent to say that the sled is at rest and that it is the hole (the surface of the Earth, which can be approximated as being flat compared to the length of the sled and the hole) which is moving.

9. Aug 30, 2004

### Fredrik

Staff Emeritus
This is one of my favorite relativistic paradoxes. I prefer a slightly different version of it though. I prefer to have some other force push the sled through the hole, instead of gravity. (As I'm sure most of you already know, special relativity describes a universe where there is no gravity, so it seems kind of paradoxial to have the sled fall through the hole). I will talk about this version of the problem instead:

Suppose the sled is very thin (from left to right) and that it's moving very fast (in its "forward" direction), parallell to, and very close to, a very long and very thin wall. There's a hole in the wall. The length of the hole in the wall's rest frame is the same as the length of the sled in the sled's rest frame. When the sled has reached the hole, another object hits the sled from the side very hard, pushing it rapidly towards the hole. The question is, will the sled go through the hole?

First of all I'd like to say that general relativity is completely irrelevant here. You don't need general relativity to explain any of the "paradoxes" of special relativity. They can all be explained using only special relativity.

OK, here's the solution to this problem:

In the sled's rest frame, the hole is shorter than the sled. There's no question about that. In the wall's rest frame, the sled is shorter than the hole. There's no question about that either.

The trick here is to notice that if the front and the rear of the sled begin their acceleration towards the wall at the same time in one of the two frames, they will begin their acceleration towards the wall at different times in the other frame.

Suppose that the whole sled gets a push towards the wall at one point in time in the sled's rest frame. In this case the sled will not fit through the hole. In the sled's rest frame, the explanation is simply that the hole is shorter than the sled. But in the wall's rest frame, the rear begins its acceleration towards the wall before the front of the sled. First the rear slams into the wall before the hole. At a slightly later time, the front begins its acceleration towards the wall, but by now the front of the sled has moved past the hole, so it slams into the wall after the hole.

Suppose instead that the whole sled gets a push towards the wall at a point in time in the wall's rest frame. In this case the sled will go through the hole. In the wall's rest frame, the explanation is simply that the sled is short enough to fit through the hole. But in the sled's rest frame, the front begins its acceleration towards the wall before the rear. The front goes through the hole first, and when the rear goes through, the front will already have passed the entire hole. It will look almost as if the sled (which is longer than the hole) is rotated so it can go through the hole. But it's not being rotated. Both the front and the rear sides of the sled will be perpendicular to the wall the whole time.

10. Aug 30, 2004

### Gonzolo

Thanks, that makes a lot of sense. Except for :

I don't see what you mean by this.

11. Aug 31, 2004

### Fredrik

Staff Emeritus
You're right. That doesn't make much sense. If the front of the sled gets a push to the side before the rear of the sled, it will be rotated, at least if we assume that the internal forces that try to keep the sled shaped like a sled are able to do their job.

I was thinking of the sled as an object consisting of thin slices (perpendicular to the wall) that could slide on each other, and I imagined that each slice would move sideways at different times (in the sled's rest frame). So in my mind the sled was being stretched into another shape, not rotated.

The reason I had a non-physical "sliced sled" in mind is that I thought it was easier to understand what it would look like in both of the scenarios I described.

A physical sled that goes through the hole wouldn't look like a rectangle from above, in the wall's rest frame. The left and right sides would be parallell to the wall, but the front and the rear would not be perpendicular to it.

A physical sled that fails to go through the hole would look just as strange in the wall's rest frame. The left and the right side wouldn't be parallell to the wall, but the front and the rear would be perpendicular to the wall.

One thing I didn't fully understand myself when I wrote my previous posting is that there's a very significant difference between the two alternatives I described. In one of them (sled failing to go through the hole) you only have to do the work that's required to move the sled sideways, but in the other (sled successfully going through the hole), you also have to do work to either rotate the sled or stretch it into another shape.

So if the question of whether the sled goes through the hole or not includes an implicit assumption that we're not allowed to forcefully stretch or rotate the sled, the answer is simply that the sled does not go through the hole.

12. Aug 31, 2004

### Gonzolo

Ok, I see what you mean, I was imagining the same thing, except that I called it "curve" (in the limit of infinitely small slices).

Do you interpret the fact that the sled can go through (as slices) as proof that no rigid bodies exist (that the intermolecular forces are necessarily "springs" perpendicular to the wall?). Or is the deformation completely due to some "curvature of space-time"?

13. Aug 31, 2004

### Fredrik

Staff Emeritus
No. If the sled goes through the hole in that particular way, that only proves that this sled isn't rigid. But the sled doesn't have to go through the hole in that particular way. It can be just as rigid as the rigid bodies of non-relativistic mechanics. If it is, the external force that pushes the sled through the hole will indirectly cause it to rotate (because internal forces will keep the sled shaped like a sled).

It's not at all due to curvature of spacetime. This is just a mechanical deformation of a non-rigid body moving in a flat spacetime.

It's a common misconception that general relativity is needed to resolve some of the "paradoxes" of special relativity. There are no real paradoxes in SR. If there were, the theory would be useless (not just a little useless, but completely useless).

Last edited: Aug 31, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook