Time Dilation Equation:

[Vilas Tamhane]

Equation for time dilation is,

 

dT= dT’ *Gamma

 

From a space station M, two spaceships depart for a space voyage and they turn around to return at their mid periods. Gamma for spaceship S1 is 1.5 and it sets for a total time of 2 years as measured on its own clock. Gamma for spaceship S2 is 2 and it sets for a total time of 2 years as per its clock.

 

On their return S1 should find that M-clock shows 3 years and S2 should find that it reads 4 years. How can it be?

 

 

[timo]

It took S2 one year longer to return to the station.

[imatfaal]

Because they won't return at the same time, the pilot of S1 will be back at the spacestation for a year before S2 returns. Take it to extremes - a pilot who travels in S0 at 30mph will travel for 2 years by S0 clock and 2years by M-Clock and will be back for two years (less a miniscule amount) before S2 return and for a year before S1 returns

[Vilas Tamhane]

Because they won't return at the same time, the pilot of S1 will be back at the spacestation for a year before S2 returns. Take it to extremes - a pilot who travels in S0 at 30mph will travel for 2 years by S0 clock and 2years by M-Clock and will be back for two years (less a miniscule amount) before S2 return and for a year before S1 returns

 

 

Interpretation of the time dilation equation is not same as that of length contraction. dT’ is the proper time in the moving frame and dT is not the time duration in moving frame as measured by the stationary observer, but it is the time that is accumulated by the stationary observer. This creates problems.

 

 

Recall twin paradox. If dT’ is 2 years and if Gamma=2, then on arrival, S1 finds that stationary clock has ticked 4 years. Remember that each observer finds clock in the other frame as running slow. So on completing the voyage the twins should not find any time difference. This difference is explained by the turnaround of the twin. If you take a look at the space-time diagram then you will find that 3 years of Dave are 3 years for Bob but when Bob turns around there is a jump in lines of simultaneity. According to traveling Bob, clock of Dave suddenly jumps to higher value of time. Thus bob’s clock doesn’t run slow, it is Dave’s clock which runs fast. And this interpretation creates problem.

 

 

Draw vertical line from O as time axis of a stationary frame and mark an event P on it. At the mid point M, draw horizontal line and mark two points P1 and P2 on it. O-P1-P is the world line of S1 and O-P2-P is the world line of S2. Note that both S1 and S2 meet stationary observer at P and at the same time.

 

 

However if you draw lines of simultaneity for S1 and S2, you will find that there is a small jump for S1 (1 year) and larger jump for S2 (2 years). Thus according to the present convention, age of the both of the twins will advance by 2 years but S1 will find stationary clock advanced by 1 year and S2 will find that it is advanced by 2 years. Though, according to the diagram they meet at a single point P, indicating that the clock of stationary observer has ticked same for both.

 

 

Clearly this cannot be the case and clearly age of S1 and S2 cannot be same. Is there something wrong with the equation? Should it be just like that of length contraction, as,

 

dT=dT’/Gamma ……….2

 

In the equation 2, dT is not the time ticked by the clock in stationary frame. It is the time duration of the moving observer as measured by the stationary observer.

 

 

[Janus]

Here are three space time diagrams of the situation:

 

 

From left to right they are done from the "M" frame, the return leg of S1, and the return leg of S2.

 

The 2yr marks indicate when each ship indicate when the ships turn around by their own clocks (these should read 1 yr, but by the time I realized I labeled it wrong, it would have been more trouble than it was worth to go back and re-do it.)

 

All three show S1 returning to M when M read 3 years and S2 returning when M reads 4 yrs, and each ship returning when it own clock reads 2 yrs.

 

Meaning:

 

According to M, S1 returns and then 1 year later S2 returns each of their clocks reading 2 yrs on return.

According to S1, it returns to the station with its clock reading 2 yrs and then waits(accumulating another year on its clock), for S2 to return with its clock reading 2 yrs on return.

According to S2 it returns with its own clock reading 2 yrs to find that S1 is already there waiting for it, reading 3 yrs on its clock.

 

There is no paradox or contradiction to any of this.

Edited July 6, 2011 by Janus

[Vilas Tamhane]

Here are three space time diagrams of the situation:

 

 

 

From left to right they are done from the "M" frame, the return leg of S1, and the return leg of S2.

 

The 2yr marks indicate when each ship indicate when the ships turn around by their own clocks (these should read 1 yr, but by the time I realized I labeled it wrong, it would have been more trouble than it was worth to go back and re-do it.)

 

All three show S1 returning to M when M read 3 years and S2 returning when M reads 4 yrs, and each ship returning when it own clock reads 2 yrs.

 

Meaning:

 

According to M, S1 returns and then 1 year later S2 returns each of their clocks reading 2 yrs on return.

According to S1, it returns to the station with its clock reading 2 yrs and then waits(accumulating another year on its clock), for S2 to return with its clock reading 2 yrs on return.

According to S2 it returns with its own clock reading 2 yrs to find that S1 is already there waiting for it, reading 3 yrs on its clock.

 

There is no paradox or contradiction to any of this.

 

 

Thanks for your efforts, but my diagram suggested was different. I don’t know yet how to transfer images.

 

S1 and S2 travel in the same direction. Turnaround points P1 and P2 are on the same horizontal line and so their return world lines meet at the same point P on the time axis of M. I have drawn the picture but I don’t know how to transfer it.

 

 

[Janus]

Thanks for your efforts, but my diagram suggested was different. I don’t know yet how to transfer images.

 

S1 and S2 travel in the same direction.

Okay, I changed my diagram to reflect that they travel in the same direction. It does not change the fact the conclusion of when S1 and S2 return to the space station or the respective clock readings of S1, S2, or M when these events occur.

 

 

Turnaround points P1 and P2 are on the same horizontal line and so their return world lines meet at the same point P on the time axis of M. I have drawn the picture but I don’t know how to transfer it.

 

Then you did it wrong.

 

You can choose a frame where the turn around points occur simultaneously, and draw a ST diagram from that frame, but it would look like this:

 

 

And you would still arrive at the same conclusions as in my other diagrams. There is no frame in which S1 and S2 return to the space station simultaneously.

Edited July 7, 2011 by Janus

[REXITIVITY]

I am a timeologist, and wonder if you have the basic concept of why time changes and how much. The only thing that determines your time, is light; and your speed in relationship to it. Light is the invarable, time is the variable. Thus if you are going very very fast, your local time would be much different than earth time...1 day would equal 1000 years on earth at 1/2 mile below the speed of light. The planet of the apes premise of astronauts returning centuries later is possible. The astronaut could experience a couple of days by his watch, and the earth 2000. Thus your question about relative time all hinges around relative speed. The one post that imagined what time would do if you could exceed the speed of light was accurate. Time would stop for the universe...because it is all relative to light. Your space would become huge and the universe would appear to shrink, like the view from a plane as it takes off. You could cross it and enter back with no time off the clock...is that too fantastic to believe? The big question is: can we exceed the speed of light? There is evidence we can...and that Mass would not increase as einstein said. I suspect I will get someone's guffaw with that...rightly so...but what if it was mathmatically provable?

[ajb]

I am a timeologist...

 

 

See the Urban Dictionary if anyone is unsure what that means, follow this link. It is not a term I am familiar with myself.

[Vilas Tamhane]

Okay, I changed my diagram to reflect that they travel in the same direction. It does not change the fact the conclusion of when S1 and S2 return to the space station or the respective clock readings of S1, S2, or M when these events occur.

 

 

Then you did it wrong.

 

And you would still arrive at the same conclusions as in my other diagrams. There is no frame in which S1 and S2 return to the space station simultaneously.

 

 

Yes, I can see that I goofed up. If I draw the diagrams which I have suggested then, S1 and S2 will arrive simultaneously but would age differently and so the time dilation equation holds good.