What is the correct condition for the winning pulse in special relativity?

[PineApple2]

Hello. I have a question about a problem from Morin's book, Introduction to classical Mechanics.
Problem 11.6. The problem and it's proposed solution are attached.

The condition for the (light-speed traveling) pulse to "win the race" with the bomb is, by the book,
<br /> L/c &lt; L(1-1/\gamma)/v<br />
meaning that the travel time of the pulse to the front of the tunnel would be less than the time of the bomb reaching the front.

But I thought the condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's. that is,
<br /> \frac{L/\gamma}{c-v}&lt; L(1-1/\gamma)/v<br />
It appears to be leading to the same expression, but still, the condition is different.
Why is the proposed condition correct?
(it seems to me that it is enough for the pulse to get to the front of the train, even before the front of the train makes it to the end of the tunnel).

 

[bcrowell]

Hi, PineApple2,

Welcome to PF!

[...]meaning that the travel time of the pulse to the front of the tunnel would be less than the time of the bomb reaching the front.

But I thought the condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's.

He's writing down the condition under which the light pulse and the deactivation signal arrive at the far end of the tunnel simultaneously. Therefore the two phrases you wrote in bold are synonymous.

There are two mistakes in your expression L/\gamma/(c-v).

One is that the signal doesn't just have to travel L/\gamma. Since the train is moving while the signal travels, the signal has to travel more than the the length of the train. It has to travel the full length of the tunnel, so this should be L, not L/\gamma.

The other mistake is the use of c-v for the speed of the pulse. Velocities don't add linearly the way they do in Newtonian mechanics. In particular, the speed of light is the same in all frames of reference, so this should be c, not c-v.

An easier way to verify the consistency of the answer in the two frames is the following. Let event A be the emission of the pulse, and let event B be the arrival of the front of the train at the far end of the tunnel. In the tunnel's frame, A precedes B. In the train's, B precedes A. Observers in different frames can only disagree on the time-order of two events if the events are spacelike in relation to each other. So A and B are spacelike, which means no signal can get from A to B.

 

[PineApple2]

Hi Ben,
Thanks for replying.

You wrote:
"He's writing down the condition under which the light pulse and the deactivation signal arrive at the far end of the tunnel simultaneously. Therefore the two phrases you wrote in bold are synonymous."

but what I don't understand is why they should be simultaneous. a "stronger" condition would be that the signal arrives at the front of the train, before the front arrives to the end of the tunnel, in order to deactivate the bomb.

so theoretically there can be a situation that the light signal arrive at the front of the train, and the train has yet to arrive the end of the tunnel.
The expression I wrote \frac{L/\gamma}{c-v} was not a mistake: I know that the train keeps going and the pulse is traveling at speed c, but I've written the relative speed the pulse needs to have in order to close the gap: the pulse needs to close the initial gap of L/\gamma in a relative speed of c-v, so this is the time it takes him.
an alternative derivation without relative speed is:
L/\gamma+vt=ct
which leads to the same result

Thanks,
PineApple.

 

[Bill_K]

I agree with you, PineApple2, your answer is correct, Morin's is not.

 

[PineApple2]

Ok. So now I think that logically I'm right, but technically the condition that the pulse hits the front of the train can never occur before the front of the train arrives at the far end of the tunnel, because
<br /> t=\frac{L/\gamma}{c-v}=\frac{L}{C}\sqrt{\frac{1+v/c}{1-v/c}}&gt;\frac{L}{c}<br />
independent of the train length L. (I don't understand why this is but I just showed it). So it is enough to use the condition L/c.

If anyone has a better suggestion I'd be happy to hear that.

PineApple.

 

[harrylin]

Hello. I have a question about a problem from Morin's book, Introduction to classical Mechanics.
Problem 11.6. The problem and it's proposed solution are attached.

The condition for the (light-speed traveling) pulse to "win the race" with the bomb is, by the book,
<br /> L/c &lt; L(1-1/\gamma)/v<br />
meaning that the travel time of the pulse to the front of the tunnel would be less than the time of the bomb reaching the front.

But I thought the condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's. that is,
<br /> \frac{L/\gamma}{c-v}&lt; L(1-1/\gamma)/v<br />

The bomb doesn't travel to the front of the train, the bomb already is there. The race is between the bomb and the pulse to the end of the tunnel.

It appears to be leading to the same expression, but still, the condition is different.
Why is the proposed condition correct?
(it seems to me that it is enough for the pulse to get to the front of the train, even before the front of the train makes it to the end of the tunnel).

Yes, the two conditions boil down to the same. The explosion event is only conditioned by the question: at the moment that the bomb reaches the end of the tunnel, has the signal already reached that same point or not. That is also explained in the answer.
If the pulse reaches the end of the train before the train reaches the end of the tunnel, then it also reaches the end of the tunnel before the train does - and the other way round.Harald

 

[PineApple2]

The bomb doesn't travel to the front of the train, the bomb already is there. The race is between the bomb and the pulse to the end of the tunnel.

You clearly did not read what I wrote. I didn't say the bomb was traveling to the front of the train, I said the race was between the bomb at the front of the train and the signal.

You are saying nothing new

 

[ghwellsjr]

I'd like to chime in on this but I cannot open the pdf. Could you please post what it says.

 

[BruceW]

Train in a tunnel **
A train and a tunnel both have proper lengths L. The train moves toward the tunnel
at speed v. A bomb is located at the front of the train. The bomb is designed to explode when the front of the train passes the far end of the tunnel. A deactivation
sensor is located at the back of the train. When the back of the train passes the near
end of the tunnel, the sensor tells the bomb to disarm itself. Does the bomb explode?

 

[harrylin]

You clearly did not read what I wrote. I didn't say the bomb was traveling to the front of the train, I said the race was between the bomb at the front of the train and the signal.

No, you clearly did not read what you wrote. In standard English,
"I thought the condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's."
is short for:
"I thought the condition should be that the travel time of the pulse to the front of the train would be shorter than the bomb's travel time to the front of the train." I now understand that you meant "to the front of the tunnel"; however that doesn't change my answer to your question.

You are saying nothing new

Apparently you did not read that I wrote:

"the two conditions boil down to the same. The explosion event is only conditioned by the question: at the moment that the bomb reaches the end of the tunnel, has the signal already reached that same point or not. That is also explained in the answer.
If the pulse reaches the end of the train before the train reaches the end of the tunnel, then it also reaches the end of the tunnel before the train does - and the other way round."

Good luck,
Harald