# train gets struck by lightning

by bfusco
Tags: special relativity
 P: 119 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.
 HW Helper Sci Advisor Thanks P: 7,848 If you treat it classically then you'll obviously get the answer 0. I doubt that's what's wanted.
 P: 119 right, thats where my thoughts keep taking me, but logic dictates that thats not what the teacher wants. lol
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## train gets struck by lightning

 Quote by bfusco 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.
Hi bfusco!

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?
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 Quote by I like Serena Hi bfusco! 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?
That's an interesting point. You are not told the the relative proximities of the observers to the two ends of the train. This means you do not have enough information to answer the question. If you assume the two observers are effectively at the same point (when?) then just considering the speed of light will again say it's simultaneous, and you will need to consider relativity to get a definitive answer.
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 Quote by haruspex That's an interesting point. You are not told the the relative proximities of the observers to the two ends of the train. This means you do not have enough information to answer the question. If you assume the two observers are effectively at the same point (when?) then just considering the speed of light will again say it's simultaneous, and you will need to consider relativity to get a definitive answer.
Let's assume the observer is either in front of the train or behind the train.
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.
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 Quote by I like Serena Let's assume the observer is either in front of the train or behind the train.
I see no justification for that. You'd also have to assume a position for the observer on the train. If the person on the train is up front and the observer on the ground is somewhere in front of the train you're back to simultaneity in the classical view.
 P: 119 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?
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 Quote by bfusco 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?
Let's ignore the speed of the train for now, since it is negligible (by a factor of ##10^7##) relative to the speed of light.
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.
P: 119
 Quote by bfusco 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?
also, upon thinking about it more, to calculate the difference in time the observer on the train records i have to take into account that the light from the right, or in front of him travels a lesser distance. and it still would depend on that distance "l", except now that total distance traveled from the light ahead of the train observer would be say W=L-l
P: 119
 Quote by I like Serena Let's ignore the speed of the train for now, since it is negligible (by a factor of ##10^7##) relative to the speed of light. 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.
im starting to get confused about the location of the people. just to be clear im picturing this event like this: there is a person standing in the center of the length of the train on the roof. so on the roof, 100m to the left is the end of the train, 100m to the right is front of train. the person off to the side of the tracks is also standing in the center of that 200m's when the lightning strikes. is this in agreement with everyone else?
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 Quote by bfusco im starting to get confused about the location of the people. just to be clear im picturing this event like this: there is a person standing in the center of the length of the train on the roof. so on the roof, 100m to the left is the end of the train, 100m to the right is front of train. the person off to the side of the tracks is also standing in the center of that 200m's when the lightning strikes. is this in agreement with everyone else?
If that is the case, then both observers will see the lightning strikes at exactly the same time (ignoring relativity which really has a negligible effect at this speed).

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.
P: 119
 Quote by I like Serena If that is the case, then both observers will see the lightning strikes at exactly the same time (ignoring relativity which really has a negligible effect at this speed). 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.
for the stationary observer, which is the one im assuming is behind the train, there would be a time difference between the two lightning strikes because the strike on the front of the train would have to travel an additional 200m to get to him. if there is an observer behind the train than the initial conditions of the problem need to be corrected, the only way an observer in a non-moving inertial frame can experience 2 lights flashing from different locations at the same time is if he is equal distances away.
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 Quote by bfusco for the stationary observer, which is the one im assuming is behind the train, there would be a time difference between the two lightning strikes because the strike on the front of the train would have to travel an additional 200m to get to him. if there is an observer behind the train than the initial conditions of the problem need to be corrected, the only way an observer in a non-moving inertial frame can experience 2 lights flashing from different locations at the same time is if he is equal distances away.
If an observer behind the train sees the 2 flashes at the same time, that means that the flash at the front of the train occurred first.
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 non-moving inertial frame.
It is impossible to decide what non-moving is, except by defining a specific frame as non-moving.
Did you want to define the earth frame as non-moving?
P: 119
 Quote by I like Serena Btw, I'm afraid there is no such thing as a non-moving inertial frame. It is impossible to decide what non-moving is, except by defining a specific frame as non-moving. Did you want to define the earth frame as non-moving?
well i thought generally the earth frame is considered non-moving, and i felt is was implied by the stating the train is traveling 120km/h. however, since you mentioned it i now see what you meant by the observer behind and in front of the train and now i am more confused. if we are to include that possibility i am lost as to where to begin this problem.
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 Quote by bfusco well i thought generally the earth frame is considered non-moving, and i felt is was implied by the stating the train is traveling 120km/h. however, since you mentioned it i now see what you meant by the observer behind and in front of the train and now i am more confused. if we are to include that possibility i am lost as to where to begin this problem.
Yeah, I guess you'd define the earth-frame as non-moving.

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?
 HW Helper P: 6,164 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.
P: 119
Yeah, I guess you'd define the earth-frame as non-moving.

 Quote by I like Serena 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?
i think so...taking into account the height of the train "h" and the distance between the observer and the train to be "x", i would say the time it takes for the closer lightning strike to reach the observer behind the train would be t1=[sq.rt(x^2+h^2)]/c. the time t2, for the further lightning strike to reach the observer would be ("L" being the length of the train), t2=[sq.rt((x+L)^2+h^2)]/c, so the difference in time for the ground observer behind the train would be Δt=t2-t1.

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=t2-t1

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