# Time observed in a clock when you start moving relative to it

There's an example given by Brian Green, https://www.youtube.com/watch?v=XFV2feKDK9E&t=17675s To be precise the example start at 04:56:00. In brief it is as follows:

There are two persons Gracie and George at a starting line and they are supposed to calculate how long it takes for Gracie to travel from starting line to finish line. Also there's another team at the finish line which is basically stationary to George (We can call it as George's team) which also calculates total time of travel.

The distance between starting line and finish line is 156 light-minutes. Gracie starts moving with a speed of 12/13c. Now when she reaches the finish line, from George's perspective and also from his team's perspective 169 minutes have passed. Also due to time dilation 65 minutes have passed in Gracie's clock.

from Gracie's perspective, she is stationary but George and his team moving in other direction and hence the clock of George and his team won't be in sync and the clock of George's team will be ahead of George's clock. So according to calculations done Team George's clock will be ahead 144min when Gracie starts moving. And for the remaining journey she will see others clocks moving slower due to time dilation and passing only 25 min. And when we add these two times ( 144 min & 25 min) we get overall 169 minutes. i.e. time passed for George and his team.

Everything's ok till now. My problem is, Suppose there's also a line exactly at the middle of the track and Gracie stop's there. Then the travel time for George and his team will be exactly half as before, 84.5 minutes and for Gracie 12.5 minutes. But now according to Gracie when she starts moving the head start time in team George's clock should also be half (72 minutes) so that they will agree to the clocks.

So How is it that the Team George's clock will be ahead by different amounts depending on where she stops?

Edit: As pointed out by KDP, if there's another team 2 at the middle then their clock will be ahead 72 minutes from Gracie's perspective. But If we consider that team 1's clock is ahead by 144 minutes then when she stops at the middle then 144 minutes (head start time) and 12.5 minutes( time dilated period in team 1's clock by Gracie's perspective) should add to 156.5 minutes in team 1's clock.

So according to her will team 1 and team 2 pass different times when she stops at the middle? How can this be correct ?

Also what does Δt=L×v really denotes? if We add this extra time as per unit of (L) as Gracie moves distance (L) then there wouldn't be a problem here. From Gracie's perspective both team 1 and team 2 clock would read the same times. Then we can avoid the concept of the 'head-start'. Am I correct?

• I added another world line for Gracie in the Geogebra app in my answer. The second red line is her team that is at the finish when she reaches the middle. Their clock reads 32.5 minutes when they arrive next to George's Team 1 who's clock is reading 156.5 minutes.
– KDP
Commented Feb 8 at 6:24
• "Also what does Δt=L×v really denotes?" It is the formula for the "Relativity of Simultaneity". When all the clocks are synchronised in one reference frame and there is a set of spatially separated events that are considered simultaneous in that reference frame, then in another reference frame they are not simultaneous. You can clearly see this in the app as you change v2.
– KDP
Commented Feb 8 at 6:29
• You don't need the concept of a head start or time jumps if you only consider clocks adjacent to the events. Everyone agrees in every reference frame that Gracie's Team 1 clocks were reading 32.5 minutes when they pass George's Team 1 clock when they are reading 156.5 minutes. There is no absolute sense in which clocks are really aging faster when clocks are far apart and moving relative to each other.
– KDP
Commented Feb 8 at 6:38
• Its easy. Her protocol would be : Start moving (add time jump), stop (subtract time jump), start again( (add time jump), stop again (subtract time jump). Jump is always a function of proper distance.
– KDP
Commented Feb 8 at 16:07
• It's not foolish. It is mysterious in nature and I'm glad if my answer and GeoGebra app gave you some insight and progressed your understanding of SR.
– KDP
Commented Feb 8 at 18:43

But now according to Gracie when she starts moving the head start time in team George's clock should also be half (72 minutes) so that they will agree to the clocks.

So How is it that the Team George's clock will be ahead by different amounts depending on where she stops?

Actually, if George has a second team at the half way point, their clocks will be ahead by 72 minutes as you say. The amount trailing clocks of George are ahead by, according to Gracie follows from $$\Delta t = L \times v$$ where L is the rest length of the object or ruler and v is the relative velocity of George. As you can see it depends on L. If L is half the distance then the clocks at that location are only ahead by half the amount.

You might like to play with this Geogebra App that is an interactive spacetime diagram where I have added the times of George's clocks at any given event. v2 is the velocity of an independent observer and can be set to the point of view of either Gracie or George.

For you example here are the results:

Here the blue dots are George's clocks and the blue worldlines belong to George's reference frame. The red line is Gracie's world line. The following chart is the point of view of Gracie:

As you can see George's Team 2 on the half way point (dashed blue world line) have clocks that are always 72 minutes ahead of George at the start line, from Gracie's point of view.