# Paradoxes that have come up for special relativity

1. Nov 8, 2004

### Alkatran

I just wanted to see some paradoxes that have come up for special relativity. They're fun to look over and catch where the flaws are. Post your own or any you find! Note that I know the solutions to the one's I'm posting.

There's a bomb attached to the front of a train. When the front of the train enters a tunnel which is the same length as the train, the bomb is armed. When the back of the train enters the tunnel the bomb explodes. Since the train is contracted according to the tunnel, the bomb explodes, but since the tuneel is contracted according to the train the bomb doesn't explode.

A T shaped object is heading towards a U shaped object (open side of U facing the bottom bar of the T, the top bars of the U are the same length as the bottom bar of the T). There is a detonator on the inside of the U at the bottom. The T hits the U. According to the T the U is shorter so the bomb explodes. According to the U the T is shorter so the bomb doesn't explode.
(argh! acceleration + rigids!)

A sled is moving at high speed next to a wall and is then (according to the sled frame) pushed (almost instantly) through a hole in the wall the same length as the sled. According to the wall frame the sled won't fit!

You run towards a barn carrying a pole the same length as the barn. When you are inside the barn the doors at both ends are shut then reopened. According to the barn frame you are contracted and there's plenty of time to shut the doors. According to your frame the barn is contracted and the pole can't fit entirely inside the barn!

2. Nov 8, 2004

### Garth

There have been two recent threads discussing the twin paradox, as to the others I have just one word, "simultaneity" - or rather the lack of it.

Garth

3. Nov 8, 2004

### O Great One

I would like to see the solution to #3, but let's assume that the bottom of the T doesn't quite touch the bottom of the U when they are both at rest relative to each other. How about this paradox:

You are at the bottom of a tall cliff holding a rope which somebody else is also holding onto at the top. You are tugging on the rope at a rate of 1 tug per second according to your frame of reference. The person at the top will feel more tugs than have been generated by you at the bottom. For example, if time is passing twice as fast for the person at the top, then when you have tugged 5 times then the person at the top will have felt 10 tugs.

Last edited: Nov 8, 2004
4. Nov 8, 2004

### Alkatran

Or more generally, the lack of absolute time. It's just hard to wrap your mind around it.

5. Nov 8, 2004

### Alkatran

[hl=#FFFFFF]
Let's say that, according to the rest frame, the T fits perfectly with the U.
From T frame: The bottom of the T touches THEN the massive acceleration makes the U decontract to become the same size and fit perfectly.
From U frame: The T touches the end bars of the U THEN mass acceleration makes the T decontract and become the same size, detonating the bomb.

As for the rope one... I think it's easier to work with light. Now it sounds lik both people aren't moving... so you're working with gravity and general relativity, which I know almost nothing about.

But you could probably have a similar arrangement in SR: a series of things emitting light moving past an observer. Think about it like that.
[/hl]

6. Nov 8, 2004

### Fredrik

Staff Emeritus
I'd like to contribute with two that I think are much more difficult:

6. The rotating disc
Imagine a circular disc rotating around its own axis (e.g. the wheel of a bicycle that's been turned upside down). The distance between any two points on the edge of the disc must be Lorentz contracted! This means that the whole circumference of the disc is shorter when the disc is rotating. But doesn't that mean that the radius must be shorter too?!

(According to GR, a massive disc would cause space to curve around it, but let's assume that we're considering the limit "massâ†’0", so that we can ignore that effect).

7. Blinking lights on a circular path
Suppose that a bunch of light bulbs are moving in a circular path (perhaps attached to the edge of the circular disc in the previous example), and that they were all blinking once every second, all at the same time, before we put them in motion. An observer who's moving at the same speed as the edge of the disc should see that a few light bulbs that are close to each other, and almost stationary in his frame, are blinking at almost exactly the same time. But that must mean that in our system ("the lab frame") those lights are not blinking at the same time. The light bulb that's "in front" of the others (in the direction of motion) must blink after the others. But you can say the same thing about any two adjacent light bulbs...and they're arranged in a circle. :uhh:

Is your head spinning yet? I hope so. It took me about a week to figure this out.

7. Nov 8, 2004

### pervect

Staff Emeritus
I'd like to post an easier version of the rotating disk problem. It's easier because it has more guidance. It's also of some practical interest.

Consider an observer living on a rotating planet. Lets' say the planet has an equatorial radius of 6380 kilometers, and rotates once every 24 hours relative to the fixed stars. (This is not quite Earth, but very close - extra credit for spotting the discrepancy :-)).

Now let's put a clock on the equator of the planet, and have another clock traverse the planet along the equator at some small velocity 'v' relative to the surface of the planet. Assume a postive sense of the velocity 'v' is when the clock moves in the same direction as the planet's spin (eastward, for the Earth). Ignore any gravitational time dilation effects.

The questions is: Will the two clocks read the same when they are reunited? If not, what is difference of readings of the clocks as a function of velocity 'v'?

8. Nov 8, 2004

### jcsd

Isn't no.6 forbidden in SR as it goes against the assumption that every spatial slice is Euclidean (so the disk will deform in some way to prevent the state of affairs), but allowed in GR where clearly not all spatial slices are Euclidean?

9. Nov 8, 2004

### Alkatran

You're applying absolute time to the ordering. In the lab frame all the lights will blink at the same time because they will all have the same time dilation relative to the lab at any point (not the same spatial contraction, though).

However, according to one of the lights on the apperatus, the light ahead of it blinks before, and the light behind it blinks after. This is because the lights 'now' line is skewed relative to the lab (it is 'higher up' in the direction of the motion, so things ahead of it happen before they would according to the lab)

10. Nov 8, 2004

### Fredrik

Staff Emeritus
Alkatran's solution to number 7 is of course correct. I would just like to add a few words (in white):

I think the best way to see that the light bulbs must blink at the same time in the lab frame is to consider the symmetry of the problem. No direction is any different from any other. If the light bulbs were to blink in any other way, that symmetry would be broken.

The symmetry argument is also valid for a rotating observer, so we can be sure that the light bulbs blink at the same time in both the rotating frame and in the lab frame.

However, they won't blink at the same time in the frame of the inertial observer who's moving at the same speed as the light bulbs. If he's looking at two light bulbs, he will see the one in front of the other blink first. The reason why this is possible is that in his frame, the rotational symmetry is broken by his own motion relative to the rotating disc (or wheel, or whatever).

11. Nov 8, 2004

### RandallB

I've not seen the Train bomb before - but I don't understand why shouldn't the bomb explode in both cases since nothing is disarming the bomb?
Is it that the Bomb is Disarmed when it exits the the Tunnel?
Or is the question does the bomb go off inside or outside the tunnel?

Randall B

12. Nov 8, 2004

Staff Emeritus
It's just the pole and barn case in another guise. Note that the bomb and the trigger (rear of the train) are spacelike related, as are the two ends of the tunnel. Suggest anything?

13. Nov 9, 2004

### Fredrik

Staff Emeritus
I will answer this in white, because i don't want to spoil the fun for everyone else.

The spacelike 3-dimensional hypersurfaces that the rotating observer thinks of as "space" are the same as the spacelike 3-dimensional hypersurfaces that we think of as "space" in the lab frame. They are Euclidean. The solution to this problem is that the distance between different parts of the disc will be stretched by the amount that "cancels" the Lorentz contraction. This has nothing to do with spacetime curvature. It's just a mechanical stretch of the material. When people say that there are no rigid bodies in SR, this is the kind of situation they have in mind.

14. Nov 9, 2004

### somy

I know one with its answer:
we have a solid long cylinder that a long line has drown throuh its side. Then we put a handle at one end and begin to rotate it. And put two observers and two clocks at the two ends.
We know that the cylinder is solid so we can always see the line in two ends at the same time. It seems that we had made a new way to synchronous two clocks at the two ends so it means that we made a mechanism that makes signals with infinite speed.

Ofcourse it has an easy solution!!!

15. Nov 9, 2004

### Alkatran

Assuming the cylinder is long enough and you can see the line even though its very very far away, the line will look curved due to the time it takes the light to reach you. Also, the beginning of the rotation at the opposite end due to rotation at this end is limited by a speed as well (is it the speed of sound through the material?)

16. Nov 9, 2004

### pervect

Staff Emeritus
One important point that should be mentioned, but hasn't been directly mentioned, is that it is impossible to syncrhonize all the lights in a rotating coordinate system use Einsteinian clock synchronization.

Any two points on the circumference of the circle will be moving at slightly different velocities - if we take the lights close enough together, though, we can make this velocity difference small.

When the velocity difference is small, we can then define a co-moving frame, and synchronize the lights in this co-moving frame.

When we carry out this procedure across the whole disk, we will find that we can't synchornize all the lights.

This is related to the fact that if we carry a clock around the rotating disk, it will read a different time when we complete the circumnavigation of the disk.

17. Nov 9, 2004

### Alkatran

You synchronize them before you start the whole thing rotating. The fact is the acceleration shouldn't change the sync at all in the lab frame.

18. Nov 12, 2004

### O Great One

A_______________C_______________B

There is a spaceship at point A and there is a spaceship at point B. A flash of light is emitted from C, which is the mid-point of AB. When the ships see the light they both take off towards each other. Once they pass each other at C, an observer in each spacecraft starts dropping pennies into a bucket at a rate of 1 per second. Once the spacecraft from point B passes point A, the person on board stops dropping pennies. Once the spacecraft from point A passes point B, the person on board stops dropping pennies. So, which spacecraft will have more pennies in its bucket, the ship from A or the ship from B?

I set up this paradox because whenever anyone says that it's impossible to have two clocks each going slower or faster than the other one, the standard response seems to be "We can't compare the clocks because they're separated."

Last edited: Nov 12, 2004
19. Nov 13, 2004

### pervect

Staff Emeritus
The point I wish to make is that the resuting "synchronization" is not Einsteinian. This means that a observer co-moving with a pair of closely spaced clocks on the rotating disk (closely spaced so they have approximately the same velocity) does not think the clocks are isotropically synchronized. (Einsteinian synchronization is the only isotropic synchronization. Anisotropic synchronization means that there is a "preferred direction" in the coordinate system used.)

There is no way to syncrhonize all the clocks on a rotating disk isotropically. This is related to the point I made earlier in the post - if you take a clock very very slowly around the circumference of a rotating disk (or a rotating planet), it will gain or lose a set amount of time depending on how many times around the circumference it has gone (I believe this is called the winding number).

The very slow transport of a clock is equivalent to Einsteinian/isotropic synchronization (consider that in the limt as the velocity goes towards zero, so does the time dilation). Because the clock gains or loses a set amount of time for every trip "around the world", an attempt to syncrhonize all the clocks in an Einsteinian/isotropic manner will always fail.

20. Nov 13, 2004

### Garth

I think you will find the 'winding number' relates to topologically compact spaces - i.e. the number of times your clock might go round a closed universe - just a bit larger than your disk!

Garth

21. Nov 13, 2004

### Janus

Staff Emeritus
Each spaceship will have an equal amount of pennies. But each spaceship will also measure the rate at which pennies are added to the other's bucket as being slower than his own. The answer lies in the fact that each spaceship also determines that he reaches his respective stopping point (A or B) before the other spaceship reaches his.
Thus the ship from point A will, by his own determination, reach point B before the ship from B reaches point A. He will thus determine that the ship from point B conitinues to add pennies to his bucket after he himself has quit adding pennies to his own. By the time that he determines that the other Spaceship has reached his end point and quits adding pennies, the amount of pennies in each bucket will be equal. The reverse is true for the Ship from B, he will determine that he stopped adding pennies before the other ship did. Once again, the apparent paradox appears only when you don't take into account everything Relativity says abput the situation.

22. Nov 13, 2004

### pervect

Staff Emeritus
Well, as I think about it, the formula I gave only works when the clock is constrained to move around the equator only, which means that one is talking about the 1-d perimeter of a 2-d circle. Which probably is compact, topologically.

If the path isn't too complex, one can project the path onto the equatorial plane (a plane passing through the equator of the rotating sphere) and the total delta-t is proportional to the projected area in this plane. This definition may run into trouble when the projected curve is self-intersecting :-(.

23. Nov 14, 2004

### Alkatran

Ah, another simultanety error. As soon as B and A aquire their speeds, things are no longer synchronous according to both of them (we could say they were both moving and intercepted the light from c to make this easier)

A's 'now' line will be titled up so that he sees B further in the 'future' than when A was not moving. So B adds pennies slower than A the whole trip, but gets an initial super-fast penny adding at the beginning.

The same goes for B to A.

24. Nov 15, 2004

### yogi

Originally Posted by O Great One
A_______________C_______________B

There is a spaceship at point A and there is a spaceship at point B. A flash of light is emitted from C, which is the mid-point of AB. When the ships see the light they both take off towards each other. Once they pass each other at C, an observer in each spacecraft starts dropping pennies into a bucket at a rate of 1 per second. Once the spacecraft from point B passes point A, the person on board stops dropping pennies. Once the spacecraft from point A passes point B, the person on board stops dropping pennies. So, which spacecraft will have more pennies in its bucket, the ship from A or the ship from B?

Where is the paradox - if the velocites are equal and the distances are equal and each observer is counting coins according to his own clock - the situation is symmetrical

25. Nov 15, 2004

### O Great One

So, if the ship reaches point A before the other ship reaches point B, then if each ship launches a missle when they reach their respective points then the missles will collide to the right of C. But, if the ship reaches point B before the other ship reaches point A, then if each ship launches a missle when they reach their respective points then the missles will collide to the left of C. So how can both situations be true?

Last edited: Nov 15, 2004