Different Clock Rates Throughout Accelerating Spaceshipby 1977ub Tags: acceleration, clocks, relativity, time dilation 

#37
Feb213, 09:01 PM

P: 297

Another related question:
So two ships are side by side. They accelerate identically for a period of time, then decelerate identically for a period of time, finally coming to a stop. Since they are side by side, they both agree that their clocks are perfectly synchronized the whole time. However, If instead of being side by side, one of the ships is spacially separated from the other *along the axis of travel*, we can no longer say this? Instead we must say that during the acceleration, the ship at the rear will determine that the ship at the front has a faster ticking clock, and during the deceleration phase, presumably, the other way around, since all will be agree that clocks are synchronized after both have stopped. If this is true, and I have missed it, is there a clear description somewhere? 



#38
Feb213, 09:08 PM

P: 1,657

Assume that there is some distance L between A and B and between B and C. So let e_1 = the event at which A suddenly changes velocity, e_2 = the event at which B suddenly changes velocity, e_3 = the event at which C suddenly changes velocity. The coordinates of these events in frame F are: [itex]t_1 = 0, x_1 = L[/itex] [itex]t_2 = 0, x_2 = 0[/itex] [itex]t_3 = 0, x_3 = +L[/itex] The coordinates of these events in frame F' are: [itex]t_1' = \gamma (t_1  \dfrac{v}{c^2} x_1) = \gamma \dfrac{vL}{c^2}[/itex] [itex]x_1' = \gamma (x_1  v t_1) = \gamma L[/itex] [itex]t_2' = \gamma (t_2  \dfrac{v}{c^2} x_2) = 0[/itex] [itex]x_1' = \gamma (x_2  v t_2) = 0[/itex] [itex]t_3' = \gamma (t_3  \dfrac{v}{c^2} x_3) =  \gamma \dfrac{vL}{c^2}[/itex] [itex]x_3' = \gamma (x_3  v t_3) = +\gamma L[/itex] So from the point of view of the crew, immediately after the acceleration, it appears that: C accelerated first, then B, and finally A. So from their point of view, the distance between C and B increased, and the distance between B and A increased. So they conclude that if they want to keep the distance between the rockets constant in future jumps, they will have to tell C to accelerate a little softer or a little later than B, and A should accelerate a little harder or a little earlier than B. 



#39
Feb213, 09:16 PM

P: 1,657

I'm not sure what's a good reference book for this stuff. If you google "rocket, acceleration, relativity", you get plenty of hits, but I'm not sure what articles are the most definitive. 



#40
Feb213, 09:33 PM

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#41
Feb213, 10:12 PM

P: 297

Nugatory,
Yes, I have read about Bell's Spaceship Paradox. I actually found it a bit surprising that people didn't get it at first. Acceleration leads to velocity leads to length contraction measured in the rest frame. If on the other hand you try to coordinate acceleration such that in the rest frame the two ships are seen as being at a constant distance, you are stretching the line! The whole issue I'm dealing with here I am somehow not prepared for, however. I understand that length contraction is closely related to relativityofsimultaneity, and that as a moving rod gets shorter in the RF, the clocks at either end of the rod get out of synch in RF with the tail clock reading later than the front clock, and this can be seen the in resolution to the ladder paradox. I understand that clocks in the moving ship that seem out of synch in the reference frame seem perfectly in synch on board the ship and vice versa. I guess i'd like to use something like the trappedphoton clocks that are used to illustrate the nonaccelerating phenomena of SR. You can continue to look at them from the resting frame, but do they actually show you how a clock ticks in an accelerating ship? After all, on the ship, the photon's path now curves. There no longer would appear to be any "regular" tick that one can find on board the accelerating frame to compare with one's own frame, right? You have to use the equivalence principle to cut up the acceleration into individual SR slices? 



#42
Feb213, 10:33 PM

P: 297

stevendaryl,
"So they conclude that if they want to keep the distance between the rockets constant in future jumps, they will have to tell C to accelerate a little softer or a little later than B, and A should accelerate a little harder or a little earlier than B." I'm defining everything as automatic. There are 3 ships with observers. A & C are set to fire the booster when they get a ping. B is set to send a ping and then wait the set time which light takes to get from B to A in the frame, and then fire booster. B is to ping once per second. Start program. Now, you agree with me that from the rest frame, the distance between A to C contracts as the whole parade accelerates. Are you telling me that from the POV of observers on the ships, that the AC distance *increases* as the ships accelerate? 



#43
Feb313, 01:27 AM

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Did you ever read about the relativity of simultaneity? Einstein's original explanation can be found at http://www.bartleby.com/173/9.html, there are many others out there. This is a key issue in understanding relativity, and it doesn't involve acceleration at all. Instead we must say that during the acceleration, the ship at the rear will determine that the ship at the front has a faster ticking clock, and during the deceleration phase, presumably, the other way around, since all will be agree that clocks are synchronized after both have stopped. If this is true, and I have missed it, is there a clear description somewhere?[/QUOTE] 



#44
Feb313, 04:27 AM

P: 440

When the ships are accelerating, the two clocks are ticking at the same rate. The amount of light, that is on trip, trying to catch the fleeing front ship, is increasing. This explains why rear clock seems slow as seen from the front. When the ships are decelerating, the two clocks are ticking at the same rate. The amount of light, that is on trip, trying to catch the fleeing front ship, is decreasing. This explains why rear clock seems fast as seen from the front. 



#45
Feb313, 07:51 AM

P: 1,657

If instead, you try to keep the distance between A&C constant, as viewed by the people aboard the rockets, then you have to accelerate A (the rear rocket) slightly more than C (the front rocket). 



#46
Feb313, 08:48 AM

P: 297

How's this.
RF (rest frame) : events further along x are simultaneous. observer moving along x axis: events further along x that RF views as simultaneous seem to be later as a linear function of x distance, so that clocks RF views to be moving at same rate are still seen as moving at the same rate, but set later. observer accelerating along X: events further along x that RF views as simultaneous seem to be *increasingly* later, such that clocks that RF sees as both synchronized and ticking at the same rate appear to be ticking faster the further up x axis. 



#47
Feb313, 09:07 AM

P: 297

pervect,
[ That sounds right as there isn't any way to define "at the same time" for spatially separated clocks. ] Yes, good point. I was thinking of programmed maneuvers which are intended to be simultaneous and are deemed so in the rest frame. I definitely understand the relativity of simultaneity. I've become very accustomed to remembering that on our moving frame, things that seem simultaneous are not seen as so on the rest frame and vice versa. This very habit had made it harder for me to think of things getting out of synch on the accelerating ship, for observers on the ship. But of course, this is the difference between moving and accelerating. In the boosting example, during periods of "coasting" people along the ship will find their clocks to tick at the same rate. However during periods of "boosting" the clock up ahead will appear to all aboard to be ticking faster. 



#48
Feb313, 09:19 AM

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#49
Feb313, 11:24 AM

P: 297

Two people are in ships sitting at different points along x. They find that their clocks are synchronized. They agree that one second into the future, they will each hurl an identical boulder backward along the x axis with the same force. After they do so, they find that their clocks are no longer synchronized. I have found this puzzling, but certainly:
1) They can't be synchronized in both the rest and the moving frame. 2) They absolutely *have* to be synchronized in the rest frame due to the symmetry of the actions. 3) Therefore they can't be any longer synchronized in the moving frame. This I find very simple and persuasive. Furthermore, as the vehicle accelerates, this effect would multiply, and the clocks would become ever more desynchronized to those on the ship, suggesting differing clock rates along the ship. Looks like I'll need to delve further into the equations of GR to be convinced that the differing clock rates at different altitudes are a similar phenomenon. 



#50
Feb513, 09:18 AM

P: 297

2) Actually I take it back. Since the hurling process takes time, and they are both moving wrt rest frame by the end of the boulder hurling process, I expect the forward person to finish hurling later that the back individual as seen from the rest frame.
3) So this simple logic doesn't convince me that they must be out of synch in the moving frame. Perhaps there's no "stickfigure" way to illustrate this without performing the integration of the changing velocity. 



#51
Feb513, 09:55 AM

P: 1,657

Then identify a number of events:
Our assumptions are that the corresponding actions are synchronized in frame F. So we have the coordinates for these events in frame F:
Now use the Lorentz transforms to see the coordinates in frame F':
While in frame F, [itex]t_1 = t_3[/itex] and [itex]t_2 = t_4[/itex], in frame F', the order of events is: [itex]t_3'[/itex], then [itex]t_1'[/itex] or [itex]t_4'[/itex] and then finally [itex]t_2'[/itex]. (The order of [itex]t_1'[/itex] and [itex]t_4'[/itex] depends on the size of [itex]L, D, [/itex] and [itex]T[/itex].) So what things look like in frame F' is this:
So between times [itex]t_3'[/itex] and [itex]t_2'[/itex], the front person is traveling slower than the rear person, so experiences less time dilation. So when they come to rest, the front clock will have gained more time than the rear clock, and also, the distance between the rear and the front will have increased. 



#52
Feb513, 01:53 PM

P: 297





#53
Feb513, 02:32 PM

P: 1,657

In scenario (1):
In scenario (2):




#54
Feb513, 04:53 PM

P: 297




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