
#1
Feb513, 07:34 AM

P: 9

Hello
1 km long train traveling on a straight track at a constant speed V of 90 percent of the speed of light. At some point in the ground near the track there is a flashlight, when the last wagon passes the flashlight, a beam is sent to the engine. on the engine there is a facility that once it receives the beam its stops the train. Relatively to a passenger sitting at the last car, the train is in rest, so when the train will stop, the distance from the last wagon to the flashlight will last 0.9 km Relative to an observer standing on the ground next to the flashlight, if we take into account the Lorentz contraction the train length is 0.43 km, so the time it takes the light to reach the engine will be 0.43/0.1C second. During this time the train traveling at 0.9C, so the train begins to stop after 3.87 km. My question is when the train really start breaking down? Does the answer changes if the flashlight is located not on the ground but on the train? Apologized for my English 



#2
Feb513, 07:40 AM

P: 718

I don't completely follow the scenario you're describing, however if it's a wellposed relativity question then the answer will be: they are both right, according to their own reference frames. Questions like, "What time does X really happen at?" don't make sense in relativity, because time is relative.




#3
Feb513, 07:46 AM

P: 9

Yes,
But we're the train will be eventually after the stop? 



#4
Feb513, 08:04 AM

P: 87

Who is right? 



#5
Feb513, 08:41 AM

P: 9

I can't fit the numbers.
The Lorentz factor is 0.43 so the ground expand is 2.32 giving a distance of 2.08. 



#6
Feb513, 09:13 AM

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P: 4,518

Knowing that, you could recast your scenario into the following: a light is turned on at the rear of a 1 km long train. When the light reaches the front of the train, it suddenly starts accelerating backwards. Now ask your question. 



#7
Feb513, 09:21 AM

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Call the length of the train in its rest frame [itex]L[/itex], and the train's velocity relative to the ground [itex]v[/itex]. Set the zero of time in both frames (train and ground) to the event where the back end of the train passes the flashlight and the flashlight beam is emitted. The time in the train's rest frame when the flashlight beam reaches the front end of the train (where we assume the engine is located) is simply [tex]t' = \frac{L}{c}[/tex] where [itex]c[/itex] is the speed of light. The time in the ground frame when the flashlight beam reaches the front end of the train is more complicated, because the train is moving in that frame as well as the light beam. The light beam will catch up to the train at a rate [itex]c  v[/itex], and the distance it has to catch up is the length of the train in the ground frame, which is [itex]L / \gamma[/itex], where [itex]\gamma = 1 / \sqrt{1  v^2 / c^2}[/itex]. So the time in the ground frame will be [tex]t = \frac{L}{\gamma \left( c  v \right)} = \frac{L}{c} \sqrt{\frac{1 + v / c}{1  v / c}}[/tex] If you have specific numbers, you can plug them into these formulas, but the formulas make it clear that the two times are different (and the time in the ground frame is longer). 



#8
Feb513, 09:56 AM

P: 9

My problem is that according to the formulas we have 2 different answers regarding the distance between the flashlight and the train when the train begin to brake. We can't say that both are true, because eventually the train stops in specific place. 



#9
Feb513, 10:18 AM

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Could you please restate your question with a scenario that does not include any extraneous or irrelevant or inconsistent information? 



#10
Feb513, 10:36 AM

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First, you've allowed for length contraction but not time dilation. Second, you have to be very careful about the relativity of simultaneity: you on the ground and the train passenger will not agree about the time at which the "train started braking" event happens, nor do both ends of the train start slowing simultaneously. Be careful with both of these effects, and the contradiction will disappear. As an aside, you're attacking this problem the hard way. You're much more likely to get the whole picture right if you try one or both of: 1) Draw a spacetime diagram showing the path of the two ends of the train and the light flash through spacetime. 2) Set up the entire problem either using the frame in which the platform is at rest or the frame in which the train is at rest (doesn't matter which, as long as you're consistent). After you have the times and positions along the track of all the relevant events (flash emitted and flash received at head of train) use the Lorentz transforms to see what their values are in the other frame. 



#11
Feb513, 11:00 AM

P: 9

But I didn't want to get relative answers like "it depend on the observer" , so I added to the story that the train brakes which will bring to one distinguish answer. (but I've got those kind of answers anyway). 



#12
Feb513, 11:13 AM

P: 9

Everyone here have good suggestions, but if someone can came out with a number, this will be grate. 



#13
Feb513, 11:21 AM

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Working in the rest frame of the ground. Put the flashlight at x = 0 and suppose it flashes at t = 0 just as the rear end of the train passes. The history of the light pulse is x = ct, while the worldline of the front of the train is x = 0.43 + 0.9 ct. They intersect at x = 0.43 + 0.9 ct = ct, or x = 4.3 km, which is the place on the track where you'll see the brake marks.




#14
Feb513, 11:53 AM

P: 9

Now, how can we get the same result with the passenger point of view? 



#15
Feb513, 11:58 AM

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In the train frame, that distance is simply [tex]d' = v t' = \frac{L v}{c}[/tex] which gives 0.9 km, as above, if we plug in your numbers. In the ground frame, the back end of the train moves at v and the flashlight is stationary (note that, as ghwellsjr points out, this *does* change if the flashlight is on the train, but that version seems much less interesting since the distance from flashlight to back end of train is zero always), so we have [tex]d = v t = \frac{L v}{c} \sqrt{\frac{1 + v / c}{1  v / c}}[/tex] So the ratio of distances is the same as the ratio of timesthe distance is *longer* in the ground frame than it is in the train frame. We can't go further in the analysis without making some assumption about how the train decelerates. Let's assume that its deceleration is constant, and that it takes a time T to decelerate from v to zero, as seen in the ground frame. That means it will cover a distance [tex]D = \frac{1}{2} v T[/tex] in the ground frame while braking. In the "train frame", the train is no longer at rest during this process; this may be a key point that you are missing. The train is not moving inertially during the deceleration, so it isn't at rest in any inertial frame during that process; but we can analyze things in the frame in which the train was at rest until it started decelerating. When we do that, we find that the train covers a distance [tex]D' = \frac{D}{\gamma} = \frac{1}{2} \frac{v T}{\gamma}[/tex] during the deceleration. So the total distance between the flashlight and the back end of the train in each frame will be: Ground frame: [tex]d + D = \frac{L v}{c} \sqrt{\frac{1 + v / c}{1  v / c}} + \frac{1}{2} v T[/tex] "Train" frame: [tex]d' + D' = \frac{L v}{c} + \frac{1}{2} \frac{v T}{\gamma}[/tex] Each term in the ground frame is strictly larger than the corresponding term in the "train" frame, so the distance in the ground frame will be larger. But as noted above, the train does not remain at rest in the "train" frame, so the shorter distance in that frame is *not* "the distance according to the train". 



#16
Feb513, 12:23 PM

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#17
Feb513, 01:01 PM

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While the train is moving, train guy and ground guy will report different distances between the flashlight and the mark (because of length contraction) and also different times between the flash of light leaving the flashlight and the brakes being applied to make the mark (because of time dilation). However, they will agree that the light signal traveled at the speed of light; distance covered divided by time in transit comes out to be c for both of them. 



#18
Feb613, 02:43 PM

P: 9

OK, I figure it out myself.
We can't assume that the "back end" of the train stops immediately when the engine stops. But if we look at the engine distance from the flashlight according to the passenger point of view, the distance is 1.9 km, but like mpv_plate mention there is a "ground expanding" factor which is 1/0.435889=2.294162 if we multiply this factor on our 1.9 we get 2.29416*1.9=4.35889 the same distance the observer on the ground see. 


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