0
$\begingroup$

Imagine, 2 persons ('A' & 'B') are 6 light years apart in space, stationary to each other and with no gravitation acting on anybody. Suppose 'B' starts his clock which also shows years, months and days. also lets say he flashes a signal at the same time. When this signal reaches at 'A' he starts his own clock which also shows year, month and time and also flashes a signal. But now another 6 year passes for B to see the clock of 'A' which is just started. now they can see each others clock and the ticks of the both clocks are in sync even though they show different timings.

from 'A's perspective his own clock shows same time as that of the clock of 'B'. and from 'B's perspective 'B' clock running 12 years ahead of 'A's clock. And They just remain in this setting for a brief time.

Now 'B' starts his journey towards 'A' with speed of 0.6c. lets say 'B' achieve his speed (0.6c) in 1 second (for the sake of simplicity). Lets say from 'B's perspective his clock showing 24 years and 'A's clock showing 12 years just before starting the journey.

from 'B' perspective the distance next to him will be contracted to 4.8 light years due to length contraction and hence he'll reach at 'A' after 8 years. so his final clock will show 32 years when he reach at 'A'.

from 'A' perspective the journey of 'B' will start when he see time 24 years in both the clocks but if he remove signal delay from his clock as the journey of 'B' starts, he can say that his actual clock was at 18 years when the journey started and also it'll take actual 10 years for 'B' to reach at him. So when 'B' reaches at the position of 'A' their respective clocks will show 32 years('B's clock) and 28 years ('A's clock).

first of all, IS MY MATH CORRECT ABOUT THE TIMINGS OF CLOCK 'A' AND 'B' WHEN 'B' REACHES AT THE POSITION OF 'A'?

Go further only If I am correct about the former question.

In reality, without the consideration of signal delays, both will perceive each others clocks as ticking faster.

From 'A's perspective the journey actually started when he see 24 years in each clock. And 4 more years pass for him When 'B' reaches at the position of 'A'.
So Does 'A' actually perceive 'B' as moving and the respective relativistic doppler shift only for 4 years? And what can he say about the velocity of the 'B'? As he moved 6 light years in just 4 years.

Also from 'B's perspective the clock of 'A' showing 12 years when he starts his journey. but similar as before if signal delay is removed, he can say that actual time is 18 years in the clock of 'A' just before departure. but then as he takes 8 years to reach at the position of 'A', 'A's clock should be time dilated and should pass only 6.4 years but that is not the case.

is it because of the acceleration of 'B' at the start of the journey there are 3.6 more years in 'A's clock?

$\endgroup$
13
  • $\begingroup$ Doppler shift starts immediately for the accelerating observer. It is delayed by distance/c for the non-accelerating observer. None of this has anything to do with length contraction of the Lorentz transformation in general. The correct transformation to use is the Poincare transformation which takes distance and time delay into account. A paradox is a teaching tool that tries to motivate the student to think about why his reasoning (using Lorentz transformations) was flawed. In this case it's because of the time delay. $\endgroup$ Jan 29 at 6:53
  • 1
    $\begingroup$ Where did you get "from B's perspective B's clock running 12 years ahead of A's clock" and so on? $\endgroup$ Jan 29 at 7:00
  • $\begingroup$ @FlatterMann Thank You! But I am not a physics student as in advanced degree, I know basics. My knowledge is from various videos and articles. I suspect there's no simpler explanation to these phenomenon. $\endgroup$
    – Ankur
    Jan 29 at 7:12
  • $\begingroup$ @Ankur The explanations you can find online vary greatly in quality. It's another one of those "It must be true because I saw it on the internet-not!" kinds of phenomena. There is, unfortunately, no quality control in the realm of self-publishing. $\endgroup$ Jan 29 at 7:15
  • $\begingroup$ @naturallyinconsisten in a sense that 'B' can see 'A's clock and the clocks also show years, month etc. you can suppose that the from 'B' perspective clock ticks were in sync for 12 years before 'B' departures. $\endgroup$
    – Ankur
    Jan 29 at 7:17

1 Answer 1

0
$\begingroup$

first of all, IS MY MATH CORRECT ABOUT THE TIMINGS OF CLOCK 'A' AND 'B' WHEN 'B' REACHES AT THE POSITION OF 'A'?

Yes, up to this point your calculations appear to be correct. When the meet, B's clock will show 32 years and A's clock will show 28 years.

But does 'A' actually perceive 'B' as moving and the respective relativistic doppler shift only for 4 years?

Yes, A only sees a doppler shift for 4 years of his time, but he knows that due to light travel time, B must have started moving 6 years prior to that making a total travel time of 10 years in A's reference frame.

And what can he say about the velocity of the 'B'? As he moved 6 light years in just 4 years.

No one measures B to travel 6 lightyears in 4 years. A measures B to travel 6 lights in 10 years (v=0.6c) and B measures A to travel 4.8 lightyears in 8 years (v=0.6c).

Also from 'B's perspective the clock of 'A' showing 12 years when he starts his journey. but similar as before if removes the signal delay he can say that actual time is 18 years in the clock of 'A' just before departure. but then as he takes 8 years to reach at the position of 'A', 'A's clock should be time dilated and should pass only 6.4 years but that is not the case.

When B accelerates, he has to re-synchronise his clocks and when he does that he will see that A's clock is ahead by a factor of 3.6 years due to the relativity of simultaneity. Instead of reading 18 years at the time of acceleration, A's clock now reads 21.6 years in B's reference frame and A's clock advances by 6.4 years in the 8 years measured by B for A to travel to B, giving the expected 0.8 time dilation factor.

If an object of length l is moving relative to us and there are clocks at either end of it, then the relativity of simultaneity formula $$ \Delta t = v l $$ tells us how much the clock at the rear is ahead of the clock at the front. In our case A and B are 6 ly apart and the relative velocity is 0.6c so the result is $v l = 0.6 * 6 = 3.6 years$. Add this to 18 and you get 21.6 years. That is the missing 3.6 years you are looking for. A's clock advances 28-21.6 = 6.4 years in B's reference frame and this is equal to 8 years time dilated by a factor of 0.8 as expected.

is it because of the acceleration of 'B' at the start of the journey there are 3.6 more years in 'A's clock?

The acceleration per se does not affect the time dilation. (See my 5 counter examples to the claim that acceleration affects time dilation.)

It is the fact that B changes to a different frame of reference that artificially creates a forward jump in A's time from B's point of view.

B makes odd measurements because B is not a valid inertial reference frame but accelerates in the middle. The best way to analyse the problem is to have a third observer "C" that is always moving at 0.6c relative to A. C is a valid inertial reference frame and things make sense if you draw the time space diagram from C's point of view.

Also, if you want to make things simpler for yourself to understand and make it easier for people to understand you, you really should use the standard form of time synchronisation. That means when B receives a timing signal from A of 0 years, B should set his own clock to +6 years, to allow for the light travel time. Now the clocks are properly synchronised in B's refence frame. When both clocks are stationary, A sees signals from B as 6 years behind and B sees signals from A as 6 years behind and things are symmetrical. Both consider each others clocks to be running at the same rate and reading the same time.

In relativity, an inertial reference frame is not just one clock, but an array of rulers and clocks that potentially stretches to infinity.

When both are stationary with respect to each other, A sets up his clocks and can lay them out all the way to B's location and synchronises all his clocks to each other. B does the same and sets clocks out at regular intervals all the way to A's location and synchronises all his clocks. Now for every clock belonging to A there a clock belonging to B right next to it and there really is no good reason to set all of B's clocks to read 6 years more than all of A's clocks when they are right next to each other and stationary with respect to each other. It just adds a layer of confusion.

If you want to use the standard Lorentz transformations it makes sense to use the standard clock synchronization method as that is assumed in the equations.

$\endgroup$
12
  • $\begingroup$ Thank you for the reply and suggestions. Just to confirm, suppose 'B' decelerates to speed 0 in 1 second when reaching at the position of 'A' and that won't add any extra time in the clock of 'A', right? Because they are stationary to each other after that. $\endgroup$
    – Ankur
    Jan 29 at 13:07
  • $\begingroup$ If B decelerates in 1 second of A's time, then A's clock advances one second and B's clock advances less than 1 second. If B stops instantly there will be no difference in times. the deceleration can be avoided by A and B taking photos of each other's clocks as they pass. To be precise, as B decelerates, his velocity is varying and you would have to take that into account by integrating, but the result can be purely accounted for in terms of the velocity at any instant. In practise what happens in that one second is negligible for an experiment over many years. $\endgroup$
    – KDP
    Jan 29 at 13:21
  • $\begingroup$ Sorry I meant 1 second in 'B's time. $\endgroup$
    – Ankur
    Jan 29 at 13:27
  • $\begingroup$ In 1 second of B,s time, the required constant proper acceleration to go from 0.6c to zero is a = atanh (0.6) $\approx 0.693$. The time that A's clock advances during the acceleration is $1/a*sinh(a*1) \approx 1.08202$ seconds. If you want to get into the intricacies of calculations in the accelerating context, then really that should be a separate question. $\endgroup$
    – KDP
    Jan 29 at 14:00
  • $\begingroup$ Nope. This is enough. Thanks again! $\endgroup$
    – Ankur
    Jan 29 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.