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Train gets struck by lightningby bfusco
Tags: special relativity 
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#1
Feb1413, 04:34 PM

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1. The problem statement, all variables and given/known data
A train with proper length of 200m is traveling at the speed of 120km/hr. An observer on the ground sees two lightening striking the two ends of the train at exactly the same time. What the time interval of the lightening strikes observed by the observer on the train? 3. The attempt at a solution in this class we are currently going through special relativity, however for this question the speed of the train is so slow compared to the speed of light i feel as though i can approach it classically. this idea of special relativity is sorta new for me so if someone understands what the teacher is trying to get at with this question and wouldnt mind pushing my thought process in the correct direction i would greatly appreciate it. 


#2
Feb1413, 05:17 PM

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If you treat it classically then you'll obviously get the answer 0. I doubt that's what's wanted.



#3
Feb1413, 05:20 PM

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right, thats where my thoughts keep taking me, but logic dictates that thats not what the teacher wants. lol



#4
Feb1413, 05:54 PM

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Train gets struck by lightning
How about taking into account that the speed of light is finite? No need to worry about time dilations and length contractions, but how about plain old classical speed? 


#5
Feb1413, 06:02 PM

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#6
Feb1413, 06:08 PM

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At the very least it'll give an upper bound for the time interval an observer in the train will observe. It will only be simultaneous for both observers (without relativity) if the observers are in the same position relative to the direction of the train. 


#7
Feb1413, 06:12 PM

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#8
Feb1413, 06:16 PM

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well the way i am thinking is that since the train is moving with a velocity of 120km/h towards say the right, the light from the bolt behind needs to travel that extra distance covered to reach the train observer. i just have a problem representing that mathematically.
i want to say that the light from behind needs to travel the proper length of the train "L" plus that traveled distance "l" to get a total distance "S", S=L+l (1). however "l" is equal to the velocity of the train times the time it took for the light to cover a distance "l", l=vt (2), plugging the "l" into the first equation you get S=L+vt (3). the next part would be to solve for that "t". this is where im getting a little messed up. since l=120(km/h)*t, and the time for light to travel that distance is t'=l/c,...would i have to use the Lorentz transformation here? 


#9
Feb1413, 06:25 PM

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Any effect by the theory of relativity will be astronomically less than all "normal" effects. The difference in time between the lightning strikes for an observer behind the train is the length of the train divided by the speed of light. 


#10
Feb1413, 06:27 PM

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#11
Feb1413, 06:30 PM

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#12
Feb1413, 06:36 PM

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I'd go for an observer behind the train and an observer in the middle of the train. But we might as well also consider the case of an observer behind the train and an observer at the front end of the train. That would represent the scenario where the difference in time would be greatest. 


#13
Feb1413, 06:42 PM

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#14
Feb1413, 06:50 PM

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You should be able to calculate how much later the second flash at the end of the train came... Btw, I'm afraid there is no such thing as a nonmoving inertial frame. It is impossible to decide what nonmoving is, except by defining a specific frame as nonmoving. Did you want to define the earth frame as nonmoving? 


#15
Feb1413, 06:56 PM

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#16
Feb1413, 07:01 PM

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So can you answer the question how much time passes between the first and second strike if an observer behind the train observes them at the same time? 


#17
Feb1413, 07:09 PM

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In retrospect I'm starting to wonder.
You are so persistent in putting both observers in the same place. And you are so eager to apply the Lorentz transformation. It suggests that you are supposed to use it and to make that assumption. 


#18
Feb1413, 07:12 PM

P: 128

Yeah, I guess you'd define the earthframe as nonmoving.
if we ignore the height of the train and assume all this happens at say eye level, t1=x/c and t2=(x+L)/c, and Δt=t2t1 


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