How does deceleration affect relative time

[vin300]

Suppose there are two planets A and B, each stationary relative to the other at distance x. Now an observer travels from planet A to B. He sees that the distance between them has shortened. If the clocks on the two planets are synchronised, then with the help of a Minkowski diagram, it could be understood that during the start of the journey, the observer observes the clock on planet B to lag behind the clock on planet A by vl/c², where l is the contracted length.Now the observer reaches planet B, decelerates instantly to a complete stop and observes that planet A jumps back to distance x. Just before deceleration, he sees that the clock on planet A shows time t/γ, where t is the time on his clock. What happens after the deceleration? Does the time on clock A also jump to t or does it remain t/γ? And does the time on clock B, on the planet where he has landed remain (t-vx/c²)/γ or does it jump forward to t?

 

[HallsofIvy]

The rate at which the two clocks are running will now be the same (the traveler will see a "jump" in the rate of the clock on planet B) but the time shown will not change. The traveler will observe that the clock that remained on planet B shows less time has elapsed than the clock he brought with him.

 

[vin300]

That makes sense, but during discussions of the twin paradox, we say that the twin that accelerates from velocity v to -v in no time, sees that the clock on the stay-on-earth twin jumps by some amount, and this looks correct with the aid of a spacetime diagram.

 

[Hurkyl]

If the clocks on the two planets are synchronised, then with the help of a Minkowski diagram, it could be understood that during the start of the journey, the observer observes the clock on planet B to lag behind the clock on planet A by vl/c²,

This looks wrong -- the clock on planet B should look to be ahead of A, not behind.

Don't forget to ensure that the traveler's line of simultaneity should be Minkowski-orthogonal to his trajectory, not Euclidean-orthogonal. i.e. if you draw a diagram with a time-axis that points North and a space-axis that points East, and the trajectories of A and B are both vertical (with B on the East side of A), then the trajectory of our traveller will point North-Northeast, whereas his line of simultaneity will point East-Northeast.

When the traveler lands on B, he sees that the clocks on both planets are once again synchronized, and that his watch is behind. This is easy to verify in the coordinates I describe above.

 

[vin300]

whereas his line of simultaneity will point East-Northeast.

Yes, so that means that the observer measures an amount of time greater than 0 in the rest frame of planet B to be 0 in his frame.So the clock at distance +x lagged behind.

 

[yuiop]

Suppose there are two planets A and B, each stationary relative to the other at distance x. Now an observer travels from planet A to B. He sees that the distance between them has shortened. If the clocks on the two planets are synchronised, then with the help of a Minkowski diagram, it could be understood that during the start of the journey, the observer observes the clock on planet B to lag behind the clock on planet A by vl/c², where l is the contracted length.

l is the proper length, not the contracted length (and the clock on planet B jumps ahead as Hurkyl said).

Now the observer reaches planet B, decelerates instantly to a complete stop and observes that planet A jumps back to distance x. Just before deceleration, he sees that the clock on planet A shows time t/γ, where t is the time on his clock.

Yes.

What happens after the deceleration? Does the time on clock A also jump to t or does it remain t/γ?

The clock on A jumps to t*γ.

And does the time on clock B, on the planet where he has landed remain (t-vx/c²)/γ or does it jump forward to t?

The clock on B remains at t*y (same as shown on clock A), because he is now at rest with planets A and B and the clocks are synchronised in the rest frame of planets A and B.

 

[Hurkyl]

Yes, so that means that the observer measures an amount of time greater than 0 in the rest frame of planet B to be 0 in his frame.So the clock at distance +x lagged behind.

I can't make sense of this.

The simple argument is that the line of simultaneity pierces B's worldline in a larger t coordinate than where it pierces A's worldline. Since the clocks on A and B agree with the t coordinate, then at that instant in the observer's rest frame, B's clocks read a larger value than A's clocks.

 

[vin300]

I can't make sense of this.

OK, I'll explain. Let's ignore length contraction and time dilation for simplicity. The trajectory of the observer points north-northeast.This means that for a constant distance measured in the positive x direction in the frame of planet A, the distance reduces in the frame of the observer with increasing time in the frame of planet A. The distance axis of the observer points east-northeast. For a constant t measured in the frame of planet A, the time reduces with increasing distance in the frame of planet A.

 

[Hurkyl]

OK, I'll explain. Let's ignore length contraction and time dilation for simplicity. The trajectory of the observer points north-northeast.This means that for a constant distance measured in the positive x direction in the frame of planet A, the distance reduces in the frame of the observer with increasing time in the frame of planet A.

This is very confusing.

* constant distance measured in the positive x direction

You mean, in the coordinates I have mentioned, draw a vertical worldline x units East of A, correct?

* "the distance reduces in the frame of the observer with increasing time"

This makes sense interpreted as "in the frame of the observer, the space coordinate of this worldline is decreasing with respect to its time coordinate"

* in the frame of planet A.

I can't make any sense of this.

* For a constant t measured in the frame of planet A

This sentence makes sense; we are looking at a horizontal line in our coordinate chart.

* the time reduces with increasing distance in the frame of planet A

The time of what? Distance of what?
______________

Finally, what does this have to do with clocks on B?
I've attached the space-time diagram, with A, B, and O(bserver) labelled, along with O's line of simultaneity. Observe that, in O's frame, the two large red dots both lie on the same line of simultaneity -- they occur at the same time. One dot is "the clock at A reads t=0" and the other is "the clock at B reads t=1".

 

[yuiop]

So the clock at distance +x lagged behind.

I think you are missing the fact that if the observer is traveling with a velocity of v as measured in the rest frame of planets A and B, then the planets are traveling with a velocity of -v in the rest frame of the traveling observer. If there was a rod connecting planets A and B, then in the rest frame of the traveling observer, planet B is at the trailing edge of the moving rod, and a clock on planet B is advanced by a factor of +vl/c^2 relative to a clock on planet A (at the leading edge of the rod).

 

[vin300]

I've attached the space-time diagram, with A, B, and O(bserver) labelled, along with O's line of simultaneity. Observe that, in O's frame, the two large red dots both lie on the same line of simultaneity -- they occur at the same time. One dot is "the clock at A reads t=0" and the other is "the clock at B reads t=1".

Your spacetime diagram is of course correct. The two red dots lie on the same line of simultaneity in O's frame. What does this mean? It means that in the rest frame of A and B, the event that occurs at time t=1 at B is ascribed time t'=0 by the observer, since it is simultaneous with the event at the origin. As seen from the observer's frame, the time at B is lesser than the time at B in the synchronized rest frame of A-B. Let's also draw a line of simultaneity that passes through t=0 at B. It could be seen that this line passes through the t' axis lower than t'=0. That is to say that the event at B at t=0 is ascribed a time t' that is lesser than 0 by the observer. So the observer sees that along the x-axis of frame A, the time shown by the clocks are lesser by vx/c^2 than the time shown by the clock at A. This time has to be multiplied by γ, because the clocks in frame A run at a rate lesser by a factor γ.

 

[yuiop]

Your spacetime diagram is of course correct. The two red dots lie on the same line of simultaneity in O's frame. What does this mean? It means that in the rest frame of A and B, the event that occurs at time t=1 at B is ascribed time t'=0 by the observer, since it is simultaneous with the event at the origin. As seen from the observer's frame, the time at B is lesser than the time at B in the synchronized rest frame of A-B. Let's also draw a line of simultaneity that passes through t=0 at B. It could be seen that this line passes through the t' axis lower than t'=0. That is to say that the event at B at t=0 is ascribed a time t' that is lesser than 0 by the observer. So the observer sees that along the x-axis of frame A, the time shown by the clocks are lesser by vx/c^2 than the time shown by the clock at A.

You are getting yourself tied up in knots here.You can use the Lorentz transforms to work out what the answer should be. Let us say planet A and B are at rest in reference frame S and are separated by a distance Δx as measured in S. A is it the origin of S and B is at coordinates (x,t) = (x,0). The observer O, travels in the positive x direction with velocity v = 0.8c from A to B as measured in the rest frame of A and B (frame S).

O is at rest in frame S'. In this reference frame A is initially at (x',t') = (0,0 and B is at (x',t') = (x/γ,0). We want to work out the time (t) on clock B when clock A shows zero and the clocks at rest in frame S' are showing time t' = 0, so we use the reverse Lorentz transform:

$$t = \gamma (t'+vx'/c^2)$$

(Note that the reverse transform is simply the normal transform with the primed and unprimed symbols swapped and the sign of the velocity reversed.)

In this case t'=0, x' = x/γ, so we get:

$$t = \gamma (0+vx/(\gamma*c^2)) = \gamma * v*x/(\gamma*c^2))= v*x/c^2$$

Note that this result is positive and the B clock is ahead of the A clock by vx/c^2 according to observer O.
Also note that x and v are both measured in the rest frame of planets A and B. (x is the proper distance between those planets and not the length contracted distance measured by O.)

This time has to be multiplied by γ, because the clocks in frame A run at a rate lesser by a factor γ.

This is not correct. The Δt = vΔx/c^2 formula calculates the time on the clocks in frame A and no further transformation is required.

Now consider a worked example for a slightly different scenario. A and B are 1 light year apart as measured in S. We send a light signal from A to B and if the clocks are correctly synchronised in S, we would expect the light signal to arrive at B one year after it left A. Agree?

Now if O is traveling at 0.8c from A to B as measured in S, then from O's point of view, B is traveling at -0.8c and in the opposite direction to the light signal sent from A. Call O's rest frame S'. The closing speed is 1.8c and the distance is 0.6 light years, so the signal arrives at B in 1/3 of a year, all as measured in S'. In this time we would expect the B clock to advance by (1/3)/γ = (1/3)*0.6 = 0.2 years. Since clock B was initially ahead of clock A by vx/c^2 = +0.8 years in S', clock B will show 1 year when the light arrives. If clock B was initially behind clock A by 0.8 years in S', then clock B would show -0.6 years when the light signal arrived, which causes all sorts of problems.

 

[vin300]

You're right.