Why is B's clock in advance instead of A's?

[Benoit]

Are you good with Lorentz' tranformations ?

I tought I was, until I tried to do this exercice (it is really classical):
Two planets, A and B, are at rest with respect to each other, a distance L apart, with synchronized clocks. A spaceship flies at speed v past planet A toward planet B and synchronizes its clock with A’s right when it passes A (they both set their clocks to zero). The spaceship eventually flies past planet B and compares its clock with B’s. We know, from working in the planets’ frame, that when the spaceship reaches B, B’s clock reads L/v. And the spaceship’s clock reads L/γv, because it runs slow by a factor of γ when viewed in the planets’ frame.

How would someone on the spaceship quantitatively explain to you why B’s clock reads L/v (which is more than its own L/γv), considering that the spaceship sees B’s clock running slow?

The answer is :
The person on the spaceship says is: “My clock advances by L/γ v during the whole trip. I see B’s clock running slow by a factor γ , so I see B’s clock advance by only (L/γv)/γ = L/γ^2v. However, B’s clock started not at zero but at Lv/c2. Therefore, the final reading on B’s clock when I get there is Lv/c2 + L/γ^2v = L/v( v^2/c^2 + 1/γ^2) = L/v( v^2/c^2 + (1 − v^2/c^2)) = L/v

My question is why is B's clock in advance, shouldn't it suppose to be A (rear clock ahead) ?

 

[Dale]

In your frame B is the rear clock. I always find it helpful to sketch the spacetime diagram.

 

[Benoit]

In your frame B is the rear clock. I always find it helpful to sketch the spacetime diagram.

It is obvious now, thanks !

 

[Jeronimus]

It helps to know that when two events which are _space separated_ in one inertial frame (F) of reference and happen simultaneously in that frame, they will never happen simultaneously for any other observer which is moving at any given vrel relative to an observer who is in (F).

To make your thought experiment a bit simpler without removing any of its points to make, we can assume that just when the spaceship passes A, A and B clocks happen to be synced at zero, as measured by observers at rest relative to planet A and B.

But this means, that since any observer on the spaceship is NOT at rest relative to A&B, that an observer on the spaceship could not possibly measure clocks on A and B to be at sync just when he passes by.
Doing the lorentz transformations, you get that an observer on the spaceship will measure the clock on planet B to be ahead in time. He could even communicate it quickly to an observer on planet A, that he measures the clock on B to be ahead of time while A measures it to be in sync with his own clock and the clock on the spaceship.

You could even add another spaceship stationed on planet A to the thought experiment, to make it clearer. This spaceship would accelerate (instantaneous to keep it simple) to the same speed of the spaceship passing by A, exactly the moment it arrives at A.
At rest relative to A, before the acceleration, an observer inside the second spaceship would of course also measure the clock at planet B to be in sync. AFTER the instantaneous acceleration however, he is now in the same inertial frame of reference as the passing by spaceship.
He will have to do the same lorentz transformations and arrive at the same conclusion, that just after his acceleration, the clock at planet B is measured to be ahead of time.

And while it is also true that he will measure the clock at B to be ticking slower all his way towards B, at the same rate an observer at rest relative to B measures the incoming spaceships' clocks to tick slower, when they arrive, their clocks will show a lower counter because the slower ticking of B cannot make up for the advancement in time the second spaceship measured on the clock on planet B caused by the acceleration.

So basically, when you accelerate relative to a clock, which was at rest formerly to you and at a distance in front of you, then you will measure this clock to tick faster as long as you keep accelerating(at high enough accelerations and long enough distances). If the clock is at a distance behind you, it will not tick slower but actually rewind back as measured. At least that is what you get when you do the lorentz transformations.

Personally, i see philosophical issues arising with this. If instead of clocks, we consider humans, and the incoming spaceship measures the human on B to be an older instance of let's say Alice, while an observer at rest to A measures a younger instance of Alice then how could someone ask himself "what is Alice thinking/experiencing at this very moment?".
It gets more clearer when we consider the second spaceship. Before the acceleration, an observer inside that spaceship would measure the clock to be at 0 seconds. So when he would ask himself "what does Alice experience at this moment" he would assume the moment at 0 seconds one would think. But when he accelerates, the instance of Alice which is measured to be simultaneous to him, isn't 0 seconds anymore but maybe 2 seconds depending on the distance and vrel.
But he could decide to accelerate back into his former frame again, at rest to planet A, and now he would again be looking at Alice BEFORE 2 seconds, maybe at 0.5 or 1 seconds.
This makes sense only mathematically, but does not really seem to make sense "humanly?".

Can anyone else see this or am i missing something here?

 

[Dale]

Can anyone else see this or am i missing something here?

The reference frame you are describing is not a valid reference frame. Non inertial reference frames are not as easy and natural to define as inertial. If you really want to investigate non inertial frames then I would recommend starting with Dolby and Gulls radar coordinates.

http://arxiv.org/abs/gr-qc/0104077

 

[sydneybself]

Benoit #1 said: "
"I see B’s clock running slow by a factor γ"
Wrong. B's clock has not been affected by Time Dilation, since it has not been moving through space. It's clock has advanced by L/V, (as has A's although it can't be perceived either from B or S). Assuming there is no finite distance between B and S, and thus no time delay when information is transmitted between them, when the observer at S observes at B's clock it will read L/V and when the observer at B looks at S's clock it will read L/γv.

 

[PeterDonis]

B's clock has not been affected by Time Dilation, since it has not been moving through space

With respect to the common rest frame of A and B, yes. But not with respect to the spaceship frame; in that frame, the ship is at rest and A and B are moving.

 

[Jeronimus]

The reference frame you are describing is not a valid reference frame. Non inertial reference frames are not as easy and natural to define as inertial. If you really want to investigate non inertial frames then I would recommend starting with Dolby and Gulls radar coordinates.

http://arxiv.org/abs/gr-qc/0104077

part of this paper reads

"if Barbara’s hypersurfaces of simultaneity at a certain time depend so sensitively on her instantaneous velocity as these diagrams suggest, then she would be forced to conclude that the distant planets swept backwards and forwards in time whenever she went dancing!"

which is pretty much what it boils down to. Understanding the full paper is a whole other matter however. Maybe in my next lifetime.

 

[Dale]

Understanding the full paper is a whole other matter however. Maybe in my next lifetime.

My honest recommendation would be to table the paper and focus on understanding non inertial objects from inertial reference frames only. There really is no good reason to worry much about non inertial frames. Keep the paper in the back of your mind more as a cautionary tale about how difficult defining non inertial frames can be.