Twin paradox and jumping time

[vilas]

Bob sets on a voyage with a velocity 0.8c for total 6 years, leaving back Dave on the space station. On return Dave finds him younger by 4 years. Reciprocal results in time measurements in relativity make the theory redundant but here we see application of time dilation equation. This is explained by jumping of Bob’s line of simultaneity at a point where Bob turns around. This is however a basic and correct interpretation of relativity and it has nothing to do with changing frame.

Assume for example, Bob starts at a point P, far away from O. clocks of both are synchronized. Just before starting reading of both the clocks are same and both are on the space axis ox. However when Bob sets in motion instantly, x-axis of Dave is not a line of simultaneity for Bob. It is inclined upwards from the point P. Therefore if zero is the starting time for Bob, it is some positive value for Dave as measured by Bob.

However there is no reason why we should draw lines of simultaneity. Difference in time coordinates occur due to the distance dependent term in the equation. This term is eliminated when we count time duration and not instantaneous value of time. In twin paradox, total duration is counted and so it should not be distance dependent.

So only way we can draw space-time diagram is the usual triangle for Bob’s world line. Now, using scale, we mark 10 dots for 10 years on time axis of Dave and total 6 dots on the world line of Bob. Thus when Bob spends 6 years in his spaceship, Dave has spent 10 years.
In the Bob’s frame, this triangle is the mirror image and now Bob’s 10 years are Dave’s 6 years. No doubt, between the two, they know who is accelerated, but in any of the equations of SR, changes do not depend on history of acceleration.

 

[Mentz114]

In the Bob’s frame, this triangle is the mirror image

No it is not. The situation is not symmetrical unless both travellers have identical worldlines ( apart from opposite directions).

Be aware that the time shown on the clocks is the proper time summed between the start and end events along each worldline.

$$\tau = \int_{start}^{finish}\ \sqrt{1-\left(\frac{dx}{dt}\right)^2}\ dt$$

 

[vilas]

No it is not. The situation is not symmetrical unless both travellers have identical worldlines ( apart from opposite directions).

Be aware that the time shown on the clocks is the proper time summed between the start and end events along each worldline.

$$\tau = \int_{start}^{finish}\ \sqrt{1-\left(\frac{dx}{dt}\right)^2}\ dt$$

The equation merely counts (integrates) time of Dave and Bob, along their world lines in the frame of Dave. In the frame of Bob situation is exactly reciprocal.

 

[JesseM]

However there is no reason why we should draw lines of simultaneity. Difference in time coordinates occur due to the distance dependent term in the equation. This term is eliminated when we count time duration and not instantaneous value of time. In twin paradox, total duration is counted and so it should not be distance dependent.

I have no idea what you're talking about here, can you give a more detailed numerical example, showing exactly how you propose to "count time dilation", and what equation you are referring to that has a "distance dependent term"?

 

[JesseM]

The equation merely counts (integrates) time of Dave and Bob, along their world lines in the frame of Dave. In the frame of Bob situation is exactly reciprocal.

But in the frame of Bob, what are the "start" times for Dave's worldline and Bob's? If you want to integrate from events along both worldlines which occur at the same starting times, then simultaneity is relevant.

 

[vilas]

I have no idea what you're talking about here, can you give a more detailed numerical example, showing exactly how you propose to "count time dilation", and what equation you are referring to that has a "distance dependent term"?

t=(t'+vx'/c^2)*Gamma. This is a time coordinate transformation equation. When Dave counts time of Bob, we simply get t=t'*Gamma. This is because clock is placed at same place x'.

 

[JesseM]

t=(t'+vx'/c^2)*Gamma. This is a time coordinate transformation equation. When Dave counts time of Bob, we simply get t=t'*Gamma. This is because clock is placed at same place x'.

OK, your earlier post would have been clearer if you had referred to specific equations like "Lorentz transformation equation" and "time dilation equation". But I still don't understand why you think simultaneity is irrelevant, how can either of them apply the time dilation equation to the other to find the time experienced by each during Bob's return voyage, without first figuring out the event on Dave's worldline that's simultaneous with Bob turning around?

 

[vilas]

OK, your earlier post would have been clearer if you had referred to specific equations like "Lorentz transformation equation" and "time dilation equation". But I still don't understand why you think simultaneity is irrelevant, how can either of them apply the time dilation equation to the other to find the time experienced by each during Bob's return voyage, without first figuring out the event on Dave's worldline that's simultaneous with Bob turning around?

I get what you say but some might not. It is therefore necessary to elaborate the point. Yes we cannot avoid simultaneity because the equation t=t’*Gamma is applicable when there is no change in velocity and the term vx’/c^2 gets eliminated during counting of time duration. This term is not eliminated when v changes sign at the point of turn about. Thanks.