# Thread: A simple special relativity calculation (I guess)

1. Of my own devising it goes like this.I hope it is "realistic",easy and revealing (at least to me)

A "train" with a hole in the middle of its rear carriage speeds through the station at a speed of c/2.
Its length is 1 light second.

There is a stationary observer at the station

Inside the train a beam of light goes from the back of the train to the front and is reflected back to the rear (so it travels for 2 seconds in the train according to the reference frame of the train -and always in the direction of or against the direction of the motion of the train)
Then the beam passes through the hole at the rear and completes its journey to the observer at the station-it exits the train.

How much time has elapsed on the observer's watch (to include the time the light beam traveled in the carriage as well as the return journey. ?

I think I can see that the answer should be a little under 4 seconds and I also understand that t(prime) =t/sq.rt.(1-1/4) =1.155t but I am not sure as to how to put it all together.

Could the answer be 2 times the reciprocal of 1.155 plus 2 seconds? ie 3.732 seconds?

Or is the 2 nd half of the journey (outside the train) going to "simply" take another 1.732 seconds instead of 2 seconds -giving a total of about 3.46 seconds ?

Or maybe something different again?

Sorry I have no graphic skills to draw the schema but I hope it was clear enough (.

2.

3. If you are considering the time it takes for the light travels after it leaves the train unitl the it reaches the observer, You have to state where the station observer is in relation to the train when the light starts its trip.

If the observer is even with the rear of the train when the light starts, then the total round trip back to the observer is 3.465 sec. It takes 2 sec in the train for the light to make the round trip. At 0.5c this works out to ~2.31 sec for the station observer. In this time, it takes the rear of the train has moved 0.5c X 2.31s = 1.155 light sec, so the light will take that much longer to reach the observer.

The other way to work it out is to consider that according to the station observer, the train is length contracted to 0.866 light seconds in length. Thus to find half the time for the round trip of the light, I take 0.866ls/(1c-.5c)= 1.732 sec ( The length of the train divided by the difference between the speed of light and the speed of the train). I double this to get the total round trip time of 3.464 sec (The slight difference in answers is due to rounding.)

4. thanks .Yes the (stationary) observer is directly behind the train at the station when it "departs".

I suppose I could have simply said that a particle of light is aimed at a mirror which is 1 light second distant away and also moving away at c/2.(removing all the chassis of the train and only leaving the mirror at the front)

The observer records the time for the beam of light to be reflected back to him.

Would that give the same answer ?

5. Originally Posted by geordief
thanks .Yes the (stationary) observer is directly behind the train at the station when it "departs".

I suppose I could have simply said that a particle of light is aimed at a mirror which is 1 light second distant away and also moving away at c/2.(removing all the chassis of the train and only leaving the mirror at the front)

The observer records the time for the beam of light to be reflected back to him.
Would that give the same answer ?
In your first scenario, you said that the train was 1 light sec long in the reference frame of the train. Now, since the train is moving at 0.5c relative to the station, this means that it is 0.866 light sec long in the frame of the station. So for the station observer, the answer is the same as if the mirror started 0.866 light sec away, not 1 light sec.

If the mirror started 1 light sec away as measured by the station frame, then the round trip would take

2(1 ls/(1c-0.5c)) = 4 sec.

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