Time Dilation in Non-Stationary Reference Frame: A, B, C

[joshuadeepak]

TL;DR

Simply put, my question is to understand the time dilation in one moving frame of reference with respect to another moving frame of reference. But applying the relative velocity as given by special theory of relativity isn't justifying it. I want to know the concept clearly and where I went wrong.

Let's consider three observers A, B and C. The experiment starts at t = 0.

A is 'absolutely' stationary.
B immediately (please imagine it) starts moving at speed v1 with respect to A.
C immediately starts moving at speed v2 with respect to A in the same direction as B.

Let's say A measures a time of t0.
In that 'duration' B would measure a time,
t1 = t0 * sqrt(1 - (v1^2 / c^2))

And C would measure the time as:
t2 = t0 * sqrt(1 - (v2^2 / c^2))

Now, the speed of observer C as measured by observer B would be:
v21 = (v2 - v1)/(1 - (v2 * v1 / c^2))

- Checkpoint 1

------------------------------------------------
Now, remove observer A out of existence (although, retain t0, v1 and v2) and we are now in observer B's reference frame, so B is stationary for us now.

Can I write, the time measured by observer C as:

t2 = t1 *sqrt(1 - (v21^2 / c^2))

If so, by plugging in the expressions of t1 and v21 in terms of t0, v1 and v2 in the right hand side of the equation, I don't get back the expression for t2 (in checkpoint 1)

- Checkpoint 2

Please tell me where I've gone wrong in understanding the concept.
Thank you!

 

[malawi_glenn]

You need to incorporate the positions of B and C and consider the full Lorentz-transformation.

 

[PeroK]

TL;DR Summary: Simply put, my question is to understand the time dilation in one moving frame of reference with respect to another moving frame of reference. But applying the relative velocity as given by special theory of relativity isn't justifying it. I want to know the concept clearly and where I went wrong.

Let's consider three observers A, B and C. The experiment starts at t = 0.

A is 'absolutely' stationary.
B immediately (please imagine it) starts moving at speed v1 with respect to A.
C immediately starts moving at speed v2 with respect to A in the same direction as B.

Let's say A measures a time of t0.
In that 'duration' B would measure a time,
t1 = t0 * sqrt(1 - (v1^2 / c^2))

This would be the time on B's clock, as measured by A.

Note that there is a difference between an isolated clock and the "coordinated" time in a frame of reference. An isolated clock measures the coordinate time at only one spatial point (in a reference frame in where it is at rest). You need to consider other clocks at different points to measure the time at those points.

If we have such clocks at rest at different points reference frame B (and synchronized in reference frame B), then these clocks are not synchronized in reference frame A.

And C would measure the time as:
t2 = t0 * sqrt(1 - (v2^2 / c^2))

As above.

Now, the speed of observer C as measured by observer B would be:
v21 = (v2 - v1)/(1 - (v2 * v1 / c^2))

- Checkpoint 1

That's correct.

------------------------------------------------
Now, remove observer A out of existence (although, retain t0, v1 and v2) and we are now in observer B's reference frame, so B is stationary for us now.

Can I write, the time measured by observer C as:

t2 = t1 *sqrt(1 - (v21^2 / c^2))

This is the time on C's clock as measured by B when $t_1$ has elapsed on B'c clock.

If so, by plugging in the expressions of t1 and v21 in terms of t0, v1 and v2 in the right hand side of the equation, I don't get back the expression for t2 (in checkpoint 1)

I assume you are trying to compare measurements of C's clock in the reference frames of A and B here. These are different measurents relative to different reference frames.

- Checkpoint 2

Please tell me where I've gone wrong in understanding the concept.
Thank you!

You can generate a similar "problem" by considering B and C moving at the same speed relative to A in opposite directions. The B and C clocks would be synchonised in reference frame A, but there would be relative, mutual time dilation of the B and C clocks as measured by each other.

 

[joshuadeepak]

You need to incorporate the positions of B and C and consider the full Lorentz-transformation.

Thanks for your valuable input.
Didn't know that there is more to it. Will explore in depth.

 

[malawi_glenn]

Thanks for your valuable input.
Didn't know that there is more to it. Will explore in depth.

Get Morins book "relativity for the enthusiastic beginner"

 

[joshuadeepak]

This would be the time on B's clock, as measured by A.

Note that there is a difference between an isolated clock and the "coordinated" time in a frame of reference. An isolated clock measures the coordinate time at only one spatial point (in a reference frame in where it is at rest). You need to consider other clocks at different points to measure the time at those points.

If we have such clocks at rest at different points reference frame B (and synchronized in reference frame B), then these clocks are not synchronized in reference frame A.

As above.

That's correct.

This is the time on C's clock as measured by B when $t_1$ has elapsed on B'c clock.

I assume you are trying to compare measurements of C's clock in the reference frames of A and B here. These are different measurents relative to different reference frames.

You can generate a similar "problem" by considering B and C moving at the same speed relative to A in opposite directions. The B and C clocks would be synchonised in reference frame A, but there would be relative, mutual time dilation of the B and C clocks as measured by each other.

Okay, interesting point that I did not think of before. I understand what you're trying to say from your alternative example:

You can generate a similar "problem" by considering B and C moving at the same speed relative to A in opposite directions. The B and C clocks would be synchonised in reference frame A, but there would be relative, mutual time dilation of the B and C clocks as measured by each other.

In short, I can summarise your answer as:
Time dilation is very "relative/ specific" to the frame of reference.
It sounds very non intuitive and I must ask a
follow up question in your alternative example, partly based on the 'time travel' excerpt, if A, B and C are of same age in the beginning.
So, B and C would age similarly and lower with respect to A.
But to B, C would look younger and vice versa?
How is that possible?

Thank you for your reply.
Looking forward to your response again.

 

[Sagittarius A-Star]

Can I write, the time measured by observer C as:
t2 = t1 *sqrt(1 - (v21^2 / c^2))

No. The usual time-dilation formula is only valid for a time-interval between two events, that happen in the "moving" frame at the same x'-coordinate.

Consider two events with a temporal interval of $\Delta t'$ and a spatial distance $\Delta x'$, as expressed in the coordinates of the primed frame $S'$. If these events are i.e. two ticks of a clock at rest in $S'$, then $\Delta x'=0$.

From the inverse Lorentz transformation for time follow with $\Delta x':=0$ the time-dilation formula:
$\Delta t = \gamma (\Delta t' + v \Delta x' / c^2) = \gamma \Delta t'$.

 

[PeroK]

Okay, interesting point that I did not think of before. I understand what you're trying to say from your alternative example:

In short, I can summarise your answer as:
Time dilation is very "relative/ specific" to the frame of reference.
It sounds very non intuitive and I must ask a
follow up question in your alternative example, partly based on the 'time travel' excerpt, if A, B and C are of same age in the beginning.
So, B and C would age similarly and lower with respect to A.
But to B, C would look younger and vice versa?
How is that possible?

Velocity-based time dilation is symmetric. If A and B are moving at relative to each other, then each measures the other's clock to be running slow. That's fundamental. This applies to inertial motion: that is to say, motion with no proper acceleration. We are also talking here about Special Relativity: that is to say, flat spacetime with no gravity (curvature of spacetime).

How is it possible? There are no contradictions in this model of spacetime. The key point is that time and space are actually a four-dimensional spacetime, where different reference frames with relative motion have a different view of what is time and what is space. This leads to the basic concepts of time dilation, length contraction and the relativity of simultaneity. Everything is held together by the Lorentz Transformation. Or, alternatively, if you prefer, by using Minkowski Geometry to describe spacetime.

That's one of the starting points for learning Special Relativity.

The first chapter of Morin's book, which covers all this in detail, is free online here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[Sagittarius A-Star]

The first chapter of Morin's book, which covers all this in detail, is free online here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

Unfortunately the free chapter 1, although it explains i.e. good the "relativity of simultaneity", does not contain Lorentz transformation nor Minkowski geometry. That comes in chapter 2.

 

[Renato Iraldi]

TL;DR Summary: Simply put, my question is to understand the time dilation in one moving frame of reference with respect to another moving frame of reference. But applying the relative velocity as given by special theory of relativity isn't justifying it. I want to know the concept clearly and where I went wrong.

Let's consider three observers A, B and C. The experiment starts at t = 0.

A is 'absolutely' stationary.
B immediately (please imagine it) starts moving at speed v1 with respect to A.
C immediately starts moving at speed v2 with respect to A in the same direction as B.

Let's say A measures a time of t0.
In that 'duration' B would measure a time,
t1 = t0 * sqrt(1 - (v1^2 / c^2))

And C would measure the time as:
t2 = t0 * sqrt(1 - (v2^2 / c^2))

Now, the speed of observer C as measured by observer B would be:
v21 = (v2 - v1)/(1 - (v2 * v1 / c^2))

- Checkpoint 1

------------------------------------------------
Now, remove observer A out of existence (although, retain t0, v1 and v2) and we are now in observer B's reference frame, so B is stationary for us now.

Can I write, the time measured by observer C as:

t2 = t1 *sqrt(1 - (v21^2 / c^2))

If so, by plugging in the expressions of t1 and v21 in terms of t0, v1 and v2 in the right hand side of the equation, I don't get back the expression for t2 (in checkpoint 1)

- Checkpoint 2

Please tell me where I've gone wrong in understanding the concept.
Thank you!

Your error consists of the interpretation of the formula for the sum of relativistic velocities.
v= (v1 + v2)/(1+v1v2)
Here v1 is the speed of B with respect to A and v2 is the speed of C with respect to B.
It is the speed with respect to the ground of a ball thrown at speed v2 inside the train going at v1 with respect to the ground.
You mistakenly add two velocities with respect to the same reference system.

 

[Renato Iraldi]

I want to clarify because I was a little confused.
The sum of velocities must be done in the way v13=v12+v23
in the article it is made v23 =v12+v13
In this case the formula is not applicable.