1. Jun 2, 2012

### Sagar_C

I have studied in some details pole&barn paradox and I thought I could solve the following paradox, but sadly I couldn't! Please help me with it so that I can understand special relativity better. Thanks in advance for any help.

The paradox: There are two trains T1 and T2 of equal proper length "L" (say) running on two parallel tracks in opposite direction with a relative velocity V such that due to length contraction one appears of length L/2 w.r.t. the other. Train T1 has a gun right at the "back end" which can shoot perpendicularly right towards the track of train T2. Suppose, it has been arranged for the the gun to shoot "as soon as" (has to be defined properly, I guess) the front end of T1 coincides with the back end of train T2. Now one can see that in the frame of T1, T2 will appear contracted (to L/2) so that T2's front wouldn't have crossed the back-end of T1 when gun shoots, and thus gunshot will not hit T2. But in the frame of reference of T2, T1 will appear contracted (to L/2) and thus, the gunshot will hit T2. So is T2 hit or not hit?

2. Jun 2, 2012

### ghwellsjr

You guessed right, except that there is no "proper" way to define "as soon as". Which ever frame you choose to define "as soon as" in determines whether the gunshot hits T2, as you have already pointed out.

3. Jun 2, 2012

### GAsahi

1. You can't have different outcomes for the same event, all observers need to agree that the projectile hits (or doesn't hit) the other train. In this case, they will agree that it hits (reasons are explained below)

2. It is a very bad idea to use length contraction in solving relativity problems. Length contraction is a consequence of the full-fledged Lorentz transforms and it is applicable ONLY when one marks BOTH ends of an object SIMULTANEOUSLY as viewed from a SINGLE frame of reference. This is not the case in your example.

3. The projectile is shot perpendicular to the frame attached to T1, i.e. it has zero speed component along the train T1. This is no longer true when judged from the frame of the track (or from the frame of train T2), the projectile has a component along train T2. In other words, the projectile is "angled" in the direction of motion of train T2. This is true in Galilean relativity just the same as it is true in SR. Since the projectile is angled in the direction of the incoming train (T2), it WILL hit it.

4. Jun 2, 2012

### Sagar_C

Thanks. Can I get rid of issue of angle by saying that the trains are almost touching each other?

5. Jun 2, 2012

### Warp

As far as I can see, this is basically the exact same problem as the pole-in-barn one. The solution to that one is, as I assume you know, that from the barn's perspective its doors close and open at the same time, but from the pole's perspective the back door closes and opens first, and then after the pole has advanced enough, the front door will do the same.

In your case the problem is that the back end of T1 disagrees about the moment when the ends meet, compared to the front end of T1. (In other words, the front end of T1 sees the event to happen at a different time than the back end of T1 does.)

Or if we state it a bit differently, if you were to command T1 to shoot at the right moment from an external point of view (where it looks like the trains are moving at the same speed), from the back end of T1 it would look like you are commanding it "too late".

6. Jun 2, 2012

### GAsahi

It will not help you. You still can't use length contraction to solve the problem. Here is a simple reason why you cannot: from the point of view of an observer on the track, both trains appear length contracted (to the same extent), yet both ends coincide when the trains pass each other. One more time : length contraction is a BAD idea when trying to solve relativity problems. My advice for you is to forget about it.

7. Jun 2, 2012

### Sagar_C

While I agree with you regarding the "bad" idea, have a look at the article "Beyond the pole-barn paradox: How the pole is caught" (just google). There length contraction has been put to very good use to address the paradox. Is there any flaw there? Thanks again.

8. Jun 2, 2012

### ghwellsjr

Concerning 1: Yes, all observers (and by that I think you mean all reference frames because all observers are in all reference frames) will agree on the outcome. However, you cannot say, "in this case" because no case was stated as I pointed out in my previous post.

Concerning 2: When the length of an object is viewed from a frame in which the both ends are simultaneous, you have just specified the rest frame of the object in which there is no length contraction, it has what is called its Proper Length. Length contraction is a perfectly valid explanation but it must always be stated what frame of reference is being used if it is a different one from the one in which its length is defined.

Concerning 3: In problems involving parallel trains, it is always assumed that we disregard the perpendicular travel times of bullets or light, especially when the distance between the train tracks is not specified. But even if we do consider this extra detail, the problem has still not been defined adequately to provide an answer.

9. Jun 2, 2012

### ghwellsjr

No, you don't want to forget about length contraction, you want to understand it.

You can get rid of the angle issue simply by ignoring it (just like we ignore gravity issues in thought problems and a host of other issues that would make our thought problems too cumbersome to analyze), we all know what you meant and if someone brings it up, just say you are ignoring it.

10. Jun 2, 2012

### Staff: Mentor

You can. That's a good simplification of the thought experiment, takes the angle complications away without losing the important aspects of the thought experiment.

(If you really wanted to be mathematically rigorous about this, you'd say that the tracks are close enough that you can ignore the small corrections from the angle effects. This procedure is valid because you can make the angle effect as small as you want by moving the tracks close enough together, so no matter how small your tolerance for error, you can find a separation that brings the angle effect in under that tolerance).

11. Jun 2, 2012

### Sagar_C

Please allow me to ask an intermediate step. Is it valid to assume that as T1's tip touches T2's tail, T1's tip sends a (fastest possible) signal as light pulse towards its tail to command the gun to shoot? I am under the impression that this mode of signal (being the speed of light) will have same speed in both the train's frame. However, in T1's frame this signal travels L distance to reach the gun, but in T2's frame this signal travels only L/2 to reach the gun. Is this concept right?

I am so frustrated over this problem and am losing my sleep over it. :-(

12. Jun 2, 2012

### Staff: Mentor

There's nothing wrong with setting up such a situation. It's perfectly unambiguous. (But it's different than your original set up.)
Sure. Both frames will see the light pulse move at speed c.
Almost. Don't forget that in T2's frame, the rear of T1 is moving towards the oncoming light pulse, so the light pulse has even less distance to cover to reach the rear of T1.

13. Jun 2, 2012

### Sagar_C

Thanks for the reply. I couldn't grasp how this is different form the original set up?

14. Jun 2, 2012

### Staff: Mentor

Your original set up had the gun at the rear of T1 fire 'as soon as' the front of T1 passed the rear of T2. That needed a bit more definition, which could have been done by simply adding that the gun would fire at the same time that the front of T1 passed the rear of T2 according to T1 clocks. (You'd have to arrange for that in advance, of course, but it's certainly doable.)

But in your new scenario there will be a delay for the signal to pass from front to rear of T1 no matter whose frame you view things from.

15. Jun 2, 2012

### Sagar_C

Oh! I never thought that was doable! I thought at best there would be time difference needed for light to travel the length. I think I have a wrong concept in my head. Would appreciate a little more clarification.

16. Jun 2, 2012

### Staff: Mentor

All you'd need to do is have synchronized clocks at the front and rear of T1. Arrange things so that the front of T1 passed the rear of T2 when the T1 front clock showed 1pm (for example). Easily done (in a thought experiment, at least). Then just have the gun at the rear of T1 fire when the T1 rear clock also shows 1pm.

17. Jun 2, 2012

### Staff: Mentor

What isn't doable is to send information from point A to point B faster than light. But you can do something at A and something at B at some times which are too close together for light to go from A to B as long as information doesn't have to go between them. E.g. you could get information to come from some 3rd point C at light speed to both A and B.

18. Jun 2, 2012

### Sagar_C

This was helpful. Is there anywhere this paradox has been solved? I found it as a problem in a (youtube) lecture on Relativity by Prof. R. Shankar of Yale. I have almost given up! I think my basics are too weak to address this.

19. Jun 2, 2012

### Staff: Mentor

Where exactly is the paradox? First you have to agree on a setup. Then you can unambiguously figure out if the bullets will hit train T2 or not. Then the trick is to see if you can understand that result from both frames. (Both frames must agree that the bullets either hit or missed T2.)

The stumbling block for most is that they forget about the relativity of simultaneity. If two things are simultaneous according to T1, then they will not necessarily be simultaneous for T2.

20. Jun 2, 2012

### Staff: Mentor

In keeping with Doc Al's comment, any time you are ever presented with a paradox in special relativity, chances are better than two to one that the key is the relativity of simultaneity. That is almost always not correctly specified in the problem setup (as was done here) or it was neglected in the analysis.

The relativity of simultaneity is the single most difficult concept of SR.

21. Jun 2, 2012

### GAsahi

Sure there was, it is the case expressed by the OP.

This is outright incorrect, there is an infinity of frames , all DIFFERENT from the proper frame of the object, where both ends of the object can be marked simultaneously without any difficulty.

No one claimed that length contraction is invalid, all I told the OP is that attempting to use length contraction instead of using the full-fledged Lorentz transforms often leads to incorrect solution, as it happened in the case of his (false) "paradox".

Says who?

Last edited: Jun 2, 2012
22. Jun 2, 2012

### GAsahi

OK,

Words rarely convey the precise meaning, so we'll resort to the actual language employed by physics, math.

In the frame of the track, at time $\tau_1$ the rear of train $T1$ coincides with the front of train $T2$:

$$x_{T1,rear}=x_{T2,front}$$

In the frame attached to $T1$ the positions of the rear of train $T1$ and the front of train $T2$ are:

$$X_{T1,rear}=\gamma(v_1)(x_{T1,rear}-v_1 \tau_1)$$

and

$$X_{T2,front}=\gamma(v_1)(x_{T2,front}-v_1 \tau_1)$$

$$X_{T1,rear}=X_{T2,front}$$

You need to use the full-fledged Lorentz transforms in order to solve the problem correctly.

Last edited: Jun 2, 2012
23. Jun 3, 2012

### ghwellsjr

The OP never said that both trains were traveling at the same speed in the track frame, only that they were going in opposite directions at a specific relative speed. He then mentioned frames for both trains but he didn't say which one defines "as soon as". As a matter of fact, from all the details he gave, the trains could be traveling at almost any speed as long as their relative speed was 0.866c.
Now you're talking about an infinity of frames whereas before you said there was a single frame:
Why did you say "a SINGLE frame"?
If he had defined the frame in which "as soon as" was applicable, then no other frame would have been necessary to solve the problem, he could have simply used the knowledge of length contraction to know whether or not the other train was hit. The only reason to use the Lorentz transform is to see how the same solution "looks" in other frames.
Einstein set the precedent in his 1920 book involving observers on trains and observers on embankments, never once concerned with the issue that they were not in the same vertical plane.

24. Jun 3, 2012

### Staff: Mentor

I agree, and the same with the time dilation formula. I think that students should always be recommended to use the full Lorentz transform. In any case where the simplified length contraction or time dilation formulas are appropriate then they will automatically fall out Lorentz transform, but using the full transform will avoid situations where the simplified formulas are used inappropriately.

25. Jun 3, 2012

### GAsahi

You are splitting hairs while you are perpetrating outright mistakes.

No, he couldn't, I just disproved this misconception using detailed math. Why do you persist in pushing fringe ideas?

Last edited: Jun 3, 2012