# Travelling back down a train - relatively speaking.

Discussion in 'Physics & Math' started by Confused2, Jul 4, 2020.

1. ### Confused2Registered Senior Member

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607
Let us imagine we have two stations 1 km apart.
The train runs between the two stations at velocity v.
The train driver measures the time interval between the first station passing his window and the second station passing his window.
The train driver compares the time he measures with the interval shown on the station clocks which are synchronised in the station frame.
The train driver finds that his elapsed time is less than the interval indicated by the station clocks by an amount we have calculated a few times already.
We later learn the the train itself was 1km long and as the driver passed the first station he sent a stopwatch backwards down the train at velocity -v. In the train frame the stop watch has velocity |v| and using clocks synchronised in the train frame we can show that the stopwatch will run slow by the amount we have already calculated a few times - so relative to the station clocks the stopwatch runs even slower than the drivers own clock.

Meanwhile, seen from the station frame the stopwatch travelling back down the train has velocity +v-v = 0 so runs at the same rate as the station clocks.

The stopwatch is slower than the driver's clock.
The stopwatch is faster than the driver's clock.
I don't understand the question.
Hello Mum.

3. ### HalcRegistered Senior Member

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227
So far so good.

This is wrong. Frame reference is omitted.
The stopwatch is slower than the driver clock in the train frame. It is of course not true in the track frame.
Again, a meaningless statement without frame reference. It is faster than the driver clock in the track frame.

The two statements, with frame references, are in no way contradictory. Without the frame references, the statements are ambiguous.

5. ### Janus58Valued Senior Member

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2,155
But not synchronized in his frame[/quote]
The train driver finds that his elapsed time is less than the interval indicated by the station clocks by an amount we have calculated a few times already.
[/QUOTE]
If he notes the time on the first station's clock as he passed it, and then compares this to the time reading of second station's clock as he passes it, and then takes the difference between these readings, he will say that this difference is more than the time interval he measured on his own clock. But if compares how much time passed on either clock in the interval between passing stations, he will note that both clocks accumulated less time than his clock.

7. ### Confused2Registered Senior Member

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607
I must admit I (still) think the implied frames are sufficient.
To clarify.
Alice stays on the station.
Bob drives the train at velocity v relative to Alice (he might be a muon in disguise) see muon experiment.
Colin runs back down the train at velocity -v relative to Bob (he might also be a muon in disguise) see muon experiment.
Bringing them all back together at a later stage (or simply set v=-v when Bob reaches the second station)
Bob has aged less than Alice. (Bob moved distance x at velocity v in Alice's frame)
Colin has aged less than Bob. (Colin moved distance x at velocity -v in Bob's frame)
Colin is the same age as Alice. (Colin has not moved relative to Alice)

8. ### Confused2Registered Senior Member

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607
Following on from the last post.
We gave Alice,Bob and Colin the same number of muons in a bag at the start.
When they meet up later and count their muons a simple analysis suggests that Bob has more muons than Alice and Colin has more muons than Bob. And (clearly) Alice and Colin have the same number of muons.
Not sure where you are heading with that.

9. ### Neddy BateValued Senior Member

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2,112
Which station? I assume the first one passed by Bob.

This sounds like Colin stays with Alice the whole time, unless Alice was not at the first station as I assumed.

Alice and Colin remain at the first station, (I assume), but Bob arrives at the second station. So they are not "all back together" in the same place.

10. ### Confused2Registered Senior Member

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607
Alice is at the first station passed by Bob.
Colin runs towards the rear (at velocity -v) of the train driven by Bob at velocity v. This has the effect that he doesn't move wrt Alice.

They can be brought back together either by setting v=-v or Bob stops the train at the second station and walks back to meet Alice and Colin.

11. ### Neddy BateValued Senior Member

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2,112
Okay, thanks for the clarification. So, there is no difference between Alice's and Colin's clocks. So why add Colin? We already knew that Bob would have said that Alice's clock ran at a slower rate, by the same factor that he says that Colin's clock runs at a slower rate.

When Bob stops the train at the second station, he should know that both Alice's and Colin's clocks are in synch with the second station's clock which he is co-located with. So he should know that both Alice's and Colin's clocks have accumulated more ticks than Bob's. So everyone stops their clocks, and Bob walks back to show them his clock, and verify everything.

Yes this means that Bob must come to terms with the idea that, according to him, Alice's & Colin's clocks must tick fast during Bob's deceleration phase, in order to become in synch with the second station clock.

Questions?

Last edited: Jul 4, 2020
12. ### Confused2Registered Senior Member

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607
I see what you're saying - actually not but I'll work on it. Thanks. No questions at present.

13. ### Neddy BateValued Senior Member

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2,112
Sounds good.

It's basically this: Since we know the two station clocks are synchronised in the station frame, then we know that when Bob comes to a complete stop at the second station, since Bob is then at rest in the station frame, then whatever time is displaying on the second station clock must also be the time currently being displayed on the first station clock. Whatever time Bob might have calculated (using SR) as the time on the first station clock (i.e. Alice's/Colin's clock time) before Bob started to decelerate the train was a valid calculation, but only in the train frame, so it no longer holds after the deceleration when he is no longer at rest in the train frame. I prefer to think of it as Bob jumping off the train, so that the train frame goes on with its own calculations, unaffected by Bob's change of frame.

14. ### Confused2Registered Senior Member

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607
This works for me in a way that deceleration didn't.