Reciprocal time dilation on the example with two moving trains

[Mike_bb]

Hello!

Let we have two trains: A,B and the clocks are in carriages.

Train A is moving towards train B. Trains are moving parallel to each other.

Measurements of clocks in moving carriages:

Train A carriage 1, Train B carriage 1: 0sec, 0sec (clocks are synchronized)
Train A carriage 1, Train B carriage 2: 2sec, 1.9sec
Train B carriage 1, Train A carriage 2: 2sec, 1.9sec
Train A carriage 2, Train B carriage 2: - ?

I can't understand what are the measurements of clocks in the second carriage of train A/train B?
Could anyone explain?

Thanks.

 

[Ibix]

I don't think there's enough information to answer. This is why the homework forums require a complete statement of the problem, because paraphrases like this often miss out important information.

What is "Train A carriage 1, Train B carriage 2: 2sec, 1.9sec" supposed to mean? Presumably that the readings on some clocks were recorded, but why were they recorded then? Where were they?

 

[Mike_bb]

What is "Train A carriage 1, Train B carriage 2: 2sec, 1.9sec" supposed to mean? Presumably that the readings on some clocks were recorded, but why were they recorded then? Where were they?

This mean that clocks in the carriage 1 of Train A show 2sec while clocks in the carriage 2 of Train B show 1.9sec (time dilation).

 

[Ibix]

This mean that clocks in the carriage 1 of Train A show 2sec while clocks in the carriage 2 of Train B show 1.9sec (time dilation).

As measured by who, using what frame?

 

[robphy]

Draw a position-vs-time graph (aka a spacetime diagram)
which will likely clarify the setup and the problem.

 

[Mike_bb]

What will happen when passenger in carriage 2 of train A and passenger in carriage 2 of train B record measurements of clocks? What measurements will passengers record? Will these measurements be equal to each other? (this question is core)

 

[Ibix]

What will happen when passenger in carriage 2 of train A and passenger in carriage 2 of train B record measurements of clocks?

It depends which frame they choose to use, unless the clocks are at the same place when they measure them.

"At the same time" is a frame dependant concept for things that are not in the same place in relativity, unlike Newtonian physics. If you don't specify the frame, or don't specify that the clocks are passing each other when the times are recorded, then your question has no unique answer.

You also need to specify how the clocks in the different carriages are zeroed (again,because "they were set to zero at the same time" means different things to different frames).

This is what I mean by "there isn't enough information to answer the question".

 

[Dale]

Measurements of clocks in moving carriages:

Train A carriage 1, Train B carriage 1: 0sec, 0sec (clocks are synchronized)
Train A carriage 1, Train B carriage 2: 2sec, 1.9sec
Train B carriage 1, Train A carriage 2: 2sec, 1.9sec
Train A carriage 2, Train B carriage 2: - ?

You need to be much more clear about what these mean. As it is we can only guess.

Right now you are not asking us to explain relativity to you. You are asking us to read your mind about the scenario that you have in mind. We are physicists, not psychics.

1) Identify the reference frame being used
2) Label all important objects (e.g. clocks)
3) Specify the velocity of each object with respect to the frame being used
4) Specify the time of each clock at some event
5) Ask your question

 

[Mike_bb]

Ibix,

It's easier than you think.

The train A are moving towards train B (parallel motion). Passenger in carriage 1 of train A records measurement of his clocks when he see passenger in carriage 2 of train B. Passenger in carriage 2 of train B also records measurement of his clocks. These measurements represent as follows:

"Train A carriage 1, Train B carriage 2: 2sec, 1.9sec"

I want to understand what measurements will passengers record (when passenger in carriage 2 of train A will see passenger in carriage 2 of train B).

Is this explanation clear to you?

 

[Dale]

What do you mean by "see passenger in carriage"? Are the clocks on each train synchronized using the Einstein convention in the respective train's frame? How fast are the trains moving? How far apart are the carriages? How did you get 2 and 1.9?

 

[Mike_bb]

What do you mean by "see passenger in carriage"? Are the clocks synchronized using the Einstein convention in their reference frame?

Passenger see another passenger in the window of carriage. The clocks are synchronized as I mentioned above in my post #1.

 

[Mike_bb]

How did you get 2 and 1.9?

These values are random. I just show that one of these measurement less than another due to time dilation effect.

 

[Matterwave]

It usually helps immensely in describing a scenario by drawing it. Even a very basic one. :)

 

[robphy]

How did you get 2 and 1.9?

These values are random. I just show that one of these measurement less than another due to time dilation effect.

Then make the arithmetic easier by using (say) 5.0 s and 4.0 s.

 

[Mike_bb]

It usually helps immensely in describing a scenario by drawing it. Even a very basic one. :)

case 1 : Train A carriage 1, Train B carriage 1: 0sec, 0sec (clocks are synchronized)
case 2 : Train A carriage 1, Train B carriage 2: 2sec, 1.9sec
Train B carriage 1, Train A carriage 2: 2sec, 1.9sec
case 3: Train A carriage 2, Train B carriage 2: - ?

 

[Dale]

Passenger see another passenger in the window of carriage. The clocks are synchronized as I mentioned above in my post #1.

Presumably at all times any passenger can look out the window and see all of the carriages on the other train. So every passenger will record all possible times of all the carriages on the other train.

Which clocks are synchronized using what convention? From what you wrote in the above posts it is not possible to tell if you only mean that clock A1 is synchronized with clock B1 as they pass, or if you mean that all of the A clocks are synchronized with each other and all of the B clocks are synchronized with each other.

 

[FactChecker]

Train A carriage 2, Train B carriage 2: - ?

At what time, by whose set of synchronized clocks, at which carriage location?
When A carriage 1 reads 2 sec, A carriage 2 also reads 2 sec.
When B carriage 1 reads 2 sec, B carriage 2 also reads 2 sec.

 

[Ibix]

View attachment 373472

case 1 : Train A carriage 1, Train B carriage 1: 0sec, 0sec (clocks are synchronized)
case 2 : Train A carriage 1, Train B carriage 2: 2sec, 1.9sec
Train B carriage 1, Train A carriage 2: 2sec, 1.9sec
case 3: Train A carriage 2, Train B carriage 2: - ?

So you mean that the clock times are recorded when the clocks in pairs of carriages pass each other.

Fine - that has an answer. The only remaining question is how the clocks are zeroed. Are they all zeroed at the same time according to observers using the rest frame of one train, or using the frame where the trains are moving at equal and opposite velocities? Or are the clocks zeroed simultaneously using the rest frame of the train they are in?

 

[robphy]

Here's a spacetime diagram drawn on rotated graph paper.
(All light-clock-diamonds have the same area
and all light-clock-diamonds have lightlike edges, by Lorentz invariance.)

Note that, in addition to time-dilation,
there is length-contraction and relativity-of-simultaneity
(which are not as easily represented on your spatial boxcar drawings).

I chose nice numbers to make the arithmetic simple.
Let Bob travel at (3/5)c to the right on Alice's diagram,
with boxcars spaced (for convenience) 3 of Bob's sticks apart.
Alice is at rest, with boxcars spaced identically with 3 of Alice's sticks.

When the trains first meet, the lead clocks read 0.
In Alice's frame, when her clock reads 0, all of Alice's boxcar-clocks also read 0.
In Bob's frame, when his clock reads 0, all of Bob's boxcar-clocks also read 0.

I think you should be able to read off all of the clock readings when the various boxcars meet.

Notice a bunch of Minkowski-right-triangles similar to
a 3-5-4 Minkowski-right-triangle [which is characteristic of relative-velocity (3/5)c].

Notice that, for v=(3/5)c, the Doppler factor is k=2.
The Doppler factor is an eigenvalue of the Lorentz boost.
So, to get Bob's diamond from Alice's diamond,
stretch by k=2 along the forward future-lightlike direction, and
shrink by k=2 (to preserve area) along the backward future-lightlike direction.

---

And here's the (4/5)c case, with boxcar spacing 4 sticks,
4-5-3 Minkowski-right-triangles, and Doppler k=3.

These spacetime-diagrams on rotated graph paper are essentially
"spacetime-diagrams decorated by the light-signals in a ticking light clock",
which provide a construction for doing graphical calculations in special relativity.

Essentially, once you know what the light-clock-diamonds are,
use them like coordinate-bricks to lay out
so many ticks-of-time and so many sticks-of-space.

If you choose nice-numbers (velocities with rational Doppler-k factors),
and convenient choices based on the associated pythagorean triple,
you can lay out diagrams to count off your answers.
Once you've developed your geometric and physical understanding and intuition,
you can better appreciate the associated formulas (often involving right-triangle leg-ratios as hyperbolic-trigonometric functions) as applied to whatever values you are given in the problem.

Upon looking for certain triangles,
these diagrams will summarize and support any explanation you may wish to use:
time-dilation, length-contraction, doppler, lorentz-transformation,
some combination, etc.