How can the Pole&Barn Paradox be solved?

[ghwellsjr]

By using the same math, you can easily prove that:


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

So, the trains line up perfectly as measured from T1. You can repeat the same exact reasoning and you'll get the same result from T2.

Only in the scenario that you decided the OP had in mind where the two trains are traveling at the same speed in opposite directions in the track frame where "as soon as" is defined but for which there is no clue that the OP actually had that in mind.

 

[Doc Al]

I challenged you first, solve the problem using length contraction ONLY. Please show your math, words don't count.

It's trivial!

At the same moment--according to the T1 frame--that the front of T1 passes the rear of T2, the gun at the rear of T1 fires. In the T1 frame the distance (at that instant) between the front and rear of T2 is L/γ, yet the distance from the gun and the rear of T2 is L. The bullets miss the train. Done!

 

[GAsahi]

It's trivial!

At the same moment--according to the T1 frame--that the front of T1 passes the rear of T2, the gun at the rear of T1 fires. In the T1 frame the distance (at that instant) between the front and rear of T2 is L/γ, yet the distance from the gun and the rear of T2 is L. The bullets miss the train. Done!

It might be trivial but it is wrong. The two trains match up perfectly in all frames of reference.
The bullet has a component along the direction of the trains. EVEN IF the perpendicular distance between the trains goes to 0 in the limit, the bullet will still hit the other train. ALWAYS. This is what naive application of length contraction does to you.

 

[GAsahi]

Only in the scenario that you decided the OP had in mind where the two trains are traveling at the same speed

Incorrect. Doesn't have to be the same speed, the path I posted works for different speeds just the same.

in opposite directions in the track frame where "as soon as" is defined but for which there is no clue that the OP actually had that in mind.

He was quite precise. The bullet is fired when the tail of one train coincides with the front of the other. The OP did not specify the frame, I specified it to be the frame of the track. Now, can you please post your solution?

 

[Doc Al]

It might be trivial but it is wrong. The two trains match up perfectly in all frames of reference.

Please define clearly the scenario you are analyzing.

The bullet has a component along the direction of the trains. EVEN IF the perpendicular distance between the trains goes to 0 in the limit, the bullet will still hit the other train. ALWAYS. This is what naive application of length contraction does to you.

Applying the principle of charity, I will assume you are discussing a different scenario that I am. Please define what you are talking about. (That bit about the bullet's having a component along the direction of the trains is irrelevant.)

Certainly there's nothing 'naive' about applying length contraction to the scenario that I am discussing.

 

[Doc Al]

He was quite precise. The bullet is fired when the tail of one train coincides with the front of the other. Now, can you please post your solution?

When according to whom?

The frame was not defined in the OP. I took (as an example) the frame of T1, which is common in these kinds of problem. What frame are you taking?

 

[GAsahi]

Please define clearly the scenario you are analyzing.

Applying the principle of charity, I will assume you are discussing a different scenario that I am. Please define what you are talking about.

See the post above yours. The bullet is fired when the tail of T1 coincides with the front of T2 as measured in the frame of reference of the track. Post 22 gives you the precise math, please go back and read it.

(That bit about the bullet's having a component along the direction of the trains is irrelevant.)

Actually, it isn't. Neglecting it combined with your naive application of length contraction is what led you to your incorrect resolution.

Certainly there's nothing 'naive' about applying length contraction to the scenario that I am discussing.

You got the wrong answer. I suggest that you go over the math and you think some more. I can see that you are a "PF Mentor", so you certainly must know about solving problems correctly.

 

[Doc Al]

See the post above yours. The bullet is fired when the tail of T1 coincides with the front of T2 as measured in the frame of reference of the track. Post 22 gives you the precise math, please go back and read it.

OK, now you've defined your scenario. And this, like the other, is perfectly analyzable using only length contraction. In this case, of course, the trains contract by the same amount, thus both ends line up at the same time. Do you seriously need "the full Lorentz transform" to analyze this trivial situation?

Actually, it isn't. Neglecting it combined with your naive application of length contraction is what led you to your incorrect resolution.




You got the wrong answer. I suggest that you go over the math and you think some more. I can see that you are a "PF Mentor", so you certainly must know about solving problems correctly.

Next time you might like to read the thread that you are posting in. I clearly defined my scenario. And, given that scenario, my answer is correct. Nothing naive about it!

 

[ghwellsjr]

What you said is that you could solve the "paradox" using length contraction ONLY. Can you please post the math that supports your point?

Where did I ever say that? I never said that. I said there was no paradox stated. I said the OP described two different scenarios using length contraction correctly in both of them. If he chooses to define "as soon as" in the frame of T1, his description is the "solution". Nothing more needs to be solved. If, on the other hand, he chooses to define "as soon as" in the frame of T2, then that different definition of the different scenario produces a different "solution" which again leaves nothing more to be solved. But you defined "as soon as" in the track frame with the added and unwarranted assumption that the two trains were traveling at the same speed in opposite directions (as speed that you could have numerically calculated precisely, by the way).

...which is precisely what I did by solving the problem from the perspective of the track. Math included.

You invented your own scenario that was not defined by the OP and you declared the "solution" without using any math, just like the OP did for his two scenarios.

Not at all. I solved the problem in the frame of the track and AFTER that, I transformed it in the frame of T1. You can also transform into the frame of T2 just the same.

That's what I said. Once you define a scenario in one frame, nothing further needs to be solved, the "solution" does not need a second frame or the Lorentz transform. But if you want, just for the fun of it, you can transform the coordinates from the frame of definition into any other frame to show that the "solution" is the same.

 

[ghwellsjr]

See the post above yours. The bullet is fired when the tail of T1 coincides with the front of T2 as measured in the frame of reference of the track. Post 22 gives you the precise math, please go back and read it.

OK, now you've defined your scenario. And this, like the other, is perfectly analyzable using only length contraction. In this case, of course, the trains contract by the same amount, thus both ends line up at the same time. Do you seriously need "the full Lorentz transform" to analyze this trivial situation?

Saying that he measures the time in the track frame is not enough, he also assumed that the two trains were traveling in opposite directions at the same speed in that frame, something that was not stated by the OP.

And then he did exactly what he says we shouldn't do: he used the fact that the two trains were length contracted by the same amount and declared that the other ends of the two trains coincided as his solution, without using any math or the LT. Then he used the Lorentz transform to show that the same solution applied in one of the train's rest frame.

 

[Doc Al]

Saying that he measures the time in the track frame is not enough, he also assumed that the two trains were traveling in opposite directions at the same speed in that frame, something that was not stated by the OP.

You are correct, of course. Again, I was giving what I thought was the most charitable interpretation of what he was doing.

And then he did exactly what he says we shouldn't do: he used the fact that the two trains were length contracted by the same amount and declared that the other ends of the two trains coincided as his solution, without using any math or the LT. Then he used the Lorentz transform to show that the same solution applied in one of the train's rest frame.

Yep.

 

[ghwellsjr]

GAsahi, do you not agree that we could have done exactly what you did using train T1's rest frame to define "as soon as" and used the fact that only T2 would have been contracted and the bullet would not hit T2 (just like the OP said) and then transformed that scenario into the rest frame of T2 to show again that the bullet wouldn't hit T2?

And then do you not agree that we could have done the same thing again using train T2's rest frame to define "as soon as" and used the fact that only T1 would have been contracted and the bullet would hit T2 (just like the OP said) and then transformed that scenario into the rest frame of T1 to show again that the bullet would hit T2?

 

[GAsahi]

OK, now you've defined your scenario. And this, like the other, is perfectly analyzable using only length contraction.

I very much doubt it. You will get another incomplete result and another incorrect conclusion.

In this case, of course, the trains contract by the same amount, thus both ends line up at the same time. Do you seriously need "the full Lorentz transform" to analyze this trivial situation?

Wrong again, the trains do not have the same speed, my solution works just as well.

Next time you might like to read the thread that you are posting in. I clearly defined my scenario. And, given that scenario, my answer is correct. Nothing naive about it!

Your scenario may be correct, it is your conclusion that is wrong. You are missing some very important components in your naive application of length contraction. It is precisely this insistence on applying length contraction that produces the "paradoxes".

 

[GAsahi]

It's trivial!

At the same moment--according to the T1 frame--that the front of T1 passes the rear of T2, the gun at the rear of T1 fires. In the T1 frame the distance (at that instant) between the front and rear of T2 is L/γ, yet the distance from the gun and the rear of T2 is L. The bullets miss the train. Done!

You are missing a couple of components in your solution, this is why you arrive to the wrong conclusion.
For you to apply length contraction in order to claim that "the length of T2 is L/γ", you need to mark both ends of T2 simultaneously in the frame of T1. But the order to "fire" needs to travel from the front of T1 to its rear and you well know that such a signal takes some time. So, in frame 1, you may be marking both ends of T2 simultaneously but you fire AT SOME LATER TIME. While the signal to fire travels from the front of T1 to its rear, T2 CONTINUES MOVING IN THE SAME DIRECTION. If you add the missing pieces to your solution, you will find out that you were wrong all along, contrary to your claims, the bullet HITS T2.
Look at it another way, in the frame of the track, the bullet hits T2, so you need to arrive (via proper computation) to the SAME conclusion in ALL frames. Otherwise, your attempt to apply length contraction only is plain wrong.

 

[Doc Al]

Your scenario may be correct, it is your conclusion that is wrong. You are missing some very important components in your naive application of length contraction. It is precisely this insistence on applying length contraction that produces the "paradoxes".

In my scenario, length contraction allows you to instantly get 'the answer' by viewing things from one frame. (The frame in which the defining events are simultaneous.) The apparent paradox is generally obtained from naively applying length contraction in a different frame without also taking into account the relativity of simultaneity (as I pointed out in post #19).

 

[Sagar_C]

Okay, so in view of the helpful comments let me rewrite the paradox and the way I was trying to solve it.

The paradox: There are two trains T1 and T2 of equal rest 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 at the same time when the front of T1 passed the rear of T2 according to T1 clocks. 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?

My solution:

Let us define there different events:

Event E0: Tip of T1 meets tip of T2.
Event E1: Tip of T1 meets tail of T2.
Event E2: Tail of T1 meets tip of T2.
Event E3: Tail of T1 meets tail of T2.
Event E4: Gun at tail of T1 shoots at T2.

We choose E0 as the point where origins of both the frames of references of T1 and T2 coincide.

Now I argue when E4 occurs between E2 and E3, the bullet from the gun in the tail of T1 will shoot T2.

In the frame of reference of T1: T2 has length L/2 (velocity is, say, v=√3/2 in the units of c). Time when E1 occurs is t1=L/2v and time when E2 occurs is t2=L/v. The time when E3 occurs is t3=(L+L/2)/v=3L/2v. In this frame, E4 by design occurs when E1 occurs i.e. at t4=L/2v. Since t4 doesn't lie between t2 and t4. Actually, t4<t2,t3. Hence, Bullets doesn't hit T2.

In the frame of reference of T2: T1 has length L/2. Time when E1 occurs is t1'=L/v and time when E2 occurs is t2'=L/2v. E3 occurs at t3'=3L/2v. Finding t4' corresponding to E4 is what remains now. E1 and E4 are simultaneous in T1's reference frame but not so in T2's frame. It can be argued (by simple thought experiment) that E4 actually occurs before E1 in T2's frame. Using either thought experiment and Lorentz transformations, I end up finding t4'=-L/2v: t4'<t2',t3'. Bullets doesn't hit T2. (PLEASE ALSO SEE POST 54).

Thanks for cooperating with me.

 

[GAsahi]

GAsahi, do you not agree that we could have done exactly what you did using train T1's rest frame to define "as soon as"

Sure. This is what the Lorentz transforms are for.

and used the fact that only T2 would have been contracted and the bullet would not hit T2 (just like the OP said) and then transformed that scenario into the rest frame of T2 to show again that the bullet wouldn't hit T2?

Problem is that you, DocAl and the OP all arrive to the wrong conclusion ("...bullet doesn't hit T2"), when in reality, a correct application of special relativity, proves the opposite.

And then do you not agree that we could have done the same thing again using train T2's rest frame to define "as soon as" and used the fact that only T1 would have been contracted and the bullet would hit T2 (just like the OP said) and then transformed that scenario into the rest frame of T1 to show again that the bullet would hit T2?

See? This is precisely what I keep explaining: naive application of length contraction ONLY leads to contradictions.

 

[Doc Al]

For you to apply length contraction in order to claim that "the length of T2 is L/γ", you need to mark both ends of T2 simultaneously in the frame of T1.

Of course.

But the order to "fire" needs to travel from the front of T1 to its rear and you well know that such a signal takes some time.

Nope. In my scenario, the 'firing' is simultaneous with the front of T1 passing the rear of T2. (As viewed from T1.)

You are analyzing a different scenario.

 

[Sagar_C]

Just to add to post number 51. I think in the following way I can also show that bullet misses hitting T2 even in T2's frame:

In the frame of T2: (x,t)=(L,L/v) is the event when tail of T2 touches tip of T1. Calling the right-moving (T1) frame a primed we can write corresponding (x',t') (primes used here are not be confused with the primes used in post 51) as

x'=γ(x-vt)=0
t'=γ(t-vx)=L/(γv) ------ Just to keep in mind for future that γ=2 if v=√3/2, c=1.

So in the primed (T1-frame) gunshot is at (0-L,L/(γv)) — I could write so because the distance of gun from T1's tip in rest frame is L and gunshot is simultaneous event in T1's frame. So let us see how this event looks like in unprimed or T2's frame:

x=γ(x'+vt')=γ(-L+v.L/(γv))=L-γL<0 always for v<c. Hence, when the gun shoots the location of the gun is to the left of the origin (tips of both T1 and T2 meeting) and hence even in T2's frame, T2 survives the hit. Also,

t=γ(t'+vx')=γL(1/γv-v) which is -L/2v when v=√3/2.

 

[Doc Al]

What you want to determine, from each frame's viewpoint, is where the gun is when it goes off. If the gun goes off while it is between the tip and tail of T2, then the bullets will hit. Otherwise, not.

 

[Sagar_C]

What you want to determine, from each frame's viewpoint, is where the gun is when it goes off. If the gun goes off while it is between the tip and tail of T2, then the bullets will hit. Otherwise, not.

I think that is what I should have done. I was always trying to find "when"! Could you clarify my following confusion, sir?

Assuming v=√3/4 in the units of c so that L contacts to L/2 in the moving frame. In the frame T1, I think I understand what is happening. I am confused a bit what happens in the T2's frame. When (and where) both their tips meet (regarded as the origins in both the frames), T2 sees T1 is rushing right with the velocity v towards T2. T1's length is L/2 in T2's frame. I found that when T1's tip meets the tail of T2, the gun shoots and at that moment gun is to the left of T2's tip. Hence, T2 survives. But time when the gun shoots is 5L/8v in T2's frame. So the tail of T1 should have moved 5L/8 by that time. Which means, 5L/8 being greater than L/2 (length of T1 in T2's frame), tail of T1 has crossed tip of T2. How can the gun and the tail where the gun is kept have two different locations?

Edited to add: Everything is fine! I was taking v=√3/4 when γ=2, whereas it should have been v=√3/2

Thanks a bunch everyone.

 

[GAsahi]

Just to add to post number 51. I think in the following way I can also show that bullet misses hitting T2 even in T2's frame:

In the frame of T2: (x,t)=(L,L/v) is the event when tail of T2 touches tip of T1. Calling the right-moving (T1) frame a primed we can write corresponding (x',t') (primes used here are not be confused with the primes used in post 51) as

x'=γ(x-vt)=0
t'=γ(t-vx)=L/(γv) ------ Just to keep in mind for future that γ=2 if v=√3/4, c=1.

So in the primed (T1-frame) gunshot is at (0-L,L/(γv)) — I could write so because the distance of gun from T1's tip in rest frame is L and gunshot is simultaneous event in T1's frame. So let us see how this event looks like in unprimed or T2's frame:

x=γ(x'+vt')=γ(-L+v.L/(γv))=L-γL<0 always for v<c. Hence, when the gun shoots the location of the gun is to the left of the origin (tips of both T1 and T2 meeting) and hence even in T2's frame, T2 survives the hit. However,

t=γ(t'+vx')=γL(1/γv-v) which is 5L/8v when v=√3/4 which takes me back to the post 51 and makes me suspect that the condition I stated there: "Now I argue when E4 occurs between E2 and E3, the bullet from the gun in the tail of T1 will shoot T2." is fishy! I can not figure out how. :(

Trying to solve the problem in frames 1 and 2 is messy. On the other hand, solving the problem in the track frame is very clean. Since you persist in trying to solve the problem in the frames 1,2, you need to consider the following part that you keep missing:

-if the gun is fired when the front of T1 passes the tail of T2 as measured from T1, then , in frame T2, the gun is fired TOO EARLY by the amount of time:

$$dt_2=\gamma(V)(dt_1-Vdx/c^2)=\gamma(V)(0-VL/c^2)=-\gamma(V)VL/c^2$$

The "minus" sign shows exactly that, in T2, the gun was fired BEFORE the front end of T1 aligned with the tail end of T2 by an amount of $\gamma(V)VL/c^2$. If you factor that in, the conclusions in frames 1 and 2 agree.
The lesson from this is:

1. Choose frames carefully, the calculations can be more difficult in certain frames compared to others.

2. Whatever frame you choose, the OUTCOME needs to be the same. If it isn't, it means you did something wrong.

3. It is NEVER a good idea (no matter what the others told you) to use ONLY length contraction or ONLY time dilation since you will most likely miss the THIRD relativistic effect (relativity of simultaneity). On the other hand, using the full - fledged Lorentz transforms guarantees that you don't miss anything.

 

[Doc Al]

I think that is what I should have done. I was always trying to find "when"! Could you clarify my following confusion, sir?

Assuming v=√3/4 in the units of c so that L contacts to L/2 in the moving frame. In the frame T1, I think I understand what is happening. I am confused a bit what happens in the T2's frame. Then both their tips meet, T1 is rushing right with the velocity v towards T2. T1's length is L/2 in T2's frame. I found that when T1's tip meets the tail of T2, the gun shoots and at that moment gun is to the left of T2's tip. Hence, T2 survives. But time when gun shoots is 5L/8v. So the tail of T1 should have moved 5L/8(>L/2) by that time. Which means, 5L/8 being greater than L/2 (length of T1 in T2's frame), tail of T1 has crossed tip of T2. How can the gun and the tail where the gun is kept have two different location?

Things get tricky when you view them from T2's frame. You have to take into account all three effects: length contraction, time dilation, and the relativity of simultaneity. It's a good exercise, but the easy way to do it is to use the Lorentz transformation which takes care of everything at once.

Here's how I'd do it. I will assume that in T1's frame, T2 is moving along the +x axis at speed v. Let the common origin of the frames be at the instant when the tip of T1 passes the tail of T2. The coordinates of that event are (0,0) in both frames.

The key event is the firing of the gun. The x,t coordinates of that event in the T1 frame are (L,0). Use the LT to find the coordinates of that same event in the T2 frame. You'll get: (γL, -γvL/c^2). Now the question is: Where does T2 say the gun is when it fires? Obviously at x' = γL. And since the tip of train T2 is at x' = L, the bullets miss hitting the train.

According to T2, the gun fires long before the tip of T1 passes the tail of T2.

Edit: I just saw that you found your error. Good!

 

[Dale]

Sure it is, it shows how the application of the full-fledged Lorentz transform proves that , from the frame of reference of $T1$, the projectile hits $T2$

It doesn't show any such thing. It shows merely that the intersection of two worldlines (the rear of T1 and the front of T2) is a frame-independent occurence. I.e. if they are at the same position at the same time in one frame then they are at the same position at the same time in all frames.

The scenario had to do not only with what was happening at the rear of T1 and the front of T2, but also what was happening "simultaneously" at the front of T1 and the rear of T2. Your math did not address that at all. And as ghwellsjr correctly pointed out "simultaneously" was not specified sufficiently to determine the outcome of the scenario.

 

[Sagar_C]

Things get tricky when you view them from T2's frame. You have to take into account all three effects: length contraction, time dilation, and the relativity of simultaneity. It's a good exercise, but the easy way to do it is to use the Lorentz transformation which takes care of everything at once.

Here's how I'd do it. I will assume that in T1's frame, T2 is moving along the +x axis at speed v. Let the common origin of the frames be at the instant when the tip of T1 passes the tail of T2. The coordinates of that event are (0,0) in both frames.

The key event is the firing of the gun. The x,t coordinates of that event in the T1 frame are (L,0). Use the LT to find the coordinates of that same event in the T2 frame. You'll get: (γL, -γvL/c^2). Now the question is: Where does T2 say the gun is when it fires? Obviously at x' = γL. And since the tip of train T2 is at x' = L, the bullets miss hitting the train.

According to T2, the gun fires long before the tip of T1 passes the tail of T2.

Thank you. I have solved the problem now. Really thanks for all the inputs. I have modified the posts above including the correcting the mistake that I was taking wrong v for γ=2. Sorry for bothering you with such a trivial mistake.

 

[Dale]

By using the same math, you can easily prove that:


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

So, the trains line up perfectly as measured from T1. You can repeat the same exact reasoning and you'll get the same result from T2.

Which also doesn't provide enough information to answer the question. Since we need to know if the gun on the T1 rear goes off before or after the intersection events that you have described, and that depends on an unspecified notion of simultaneity with the T1 front event.

 

[Dale]

The two trains match up perfectly in all frames of reference.

What do you mean by this?

 

[GAsahi]

It doesn't show any such thing. It shows merely that the intersection of two worldlines (the rear of T1 and the front of T2) is a frame-independent occurence. I.e. if they are at the same position at the same time in one frame then they are at the same position at the same time in all frames.

Good, you got this correct.

The scenario had to do not only with what was happening at the rear of T1 and the front of T2, but also what was happening "simultaneously" at the front of T1 and the rear of T2. Your math did not address that at all. And as ghwellsjr correctly pointed out "simultaneously" was not specified sufficiently to determine the outcome of the scenario.

Sure it does, u need to read post 22 from the beginning. It says clearly that the front of T1 lines with the rear of T2 at the SAME location at time $\tau_1$ as measured from the track frame. Please go back and re-read the post.

 

[Dale]

Sure it does, u need to read post 22 from the beginning. It says clearly that the front of T1 lines with the rear of T2 at the SAME location at time $\tau_1$ as measured from the track frame. Please go back and re-read the post.

Where does it say that? I have been through post 22 a half dozen times and cannot find one single mention in either text or math of the front of T1, only the rear of T1. If it indeed says it at all then it certainly doesn't say it clearly.

 

[Dale]

There are two trains T1 and T2 of equal rest 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 at the same time when the front of T1 passed the rear of T2 according to T1 clocks. 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

This is a completely specified scenario and a correct analysis. All other frames will necessarily also agree that the shot will not hit.

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?

T2 is not hit. Although T1 is contracted in T2's frame, the shot and the passing of the ends are not simultaneous. The gun fires too late, thereby missing T2.

I can't check your math right now, but it looks like you got the right conclusion.