# Different Clock Rates Throughout Accelerating Spaceship

by 1977ub
Tags: acceleration, clocks, relativity, time dilation
 P: 293 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?
P: 1,399
 Quote by 1977ub As far as B is concerned, a ping is released once per second lets say. You don't reason that A & C will perceive it that way, I take it.
No, as measured by A's clock (assuming it's a normal windup clock, or else an electronic clock), the pings come more often than once per second, and as measured by C's clock, they come less often than once per second.

 Ok let's scale this back to a single ping/boost. While in rest frame, all agree that distances between ships will take light 1 second to reach from B to A or C, let's say. After the first brief round of boosting, Will the ships no longer find their own clocks to be synchronized? Will they no longer find their ships to be the same distance from one another as they did in the rest frame?
Okay, let's suppose that initially A, B, and C are all at rest. Then at t=0 (according to the "launch" frame, they all accelerate suddenly, to get to a new speed of v relative to the launch frame. Call the launch frame F, and the new frame F'.

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:

$t_1 = 0, x_1 = -L$

$t_2 = 0, x_2 = 0$

$t_3 = 0, x_3 = +L$

The coordinates of these events in frame F' are:

$t_1' = \gamma (t_1 - \dfrac{v}{c^2} x_1) = \gamma \dfrac{vL}{c^2}$
$x_1' = \gamma (x_1 - v t_1) = -\gamma L$
$t_2' = \gamma (t_2 - \dfrac{v}{c^2} x_2) = 0$
$x_1' = \gamma (x_2 - v t_2) = 0$
$t_3' = \gamma (t_3 - \dfrac{v}{c^2} x_3) = - \gamma \dfrac{vL}{c^2}$
$x_3' = \gamma (x_3 - v t_3) = +\gamma L$

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.
P: 1,399
 Quote by 1977ub 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.
You're talking about the case where the two rockets accelerate and decelerate identically, as measured in their initial rest frame? In that case, from the point of view of the "launch" frame, the clocks will remain synchronized at all times. From the point of view of those aboard the rockets, the front clock will get ahead of the rear one during acceleration, and then the rear rocket will catch up during deceleration.

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.
Sci Advisor
P: 2,358
 Quote by 1977ub 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? .... If this is true, and I have missed it, is there a clear description somewhere?
Look for "Bell's spaceship paradox". It's described in terms of length contraction instead of time dilation, but it's still a pretty good starting point for understanding just how tricky "synchronized" acceleration can be.
 P: 293 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 relativity-of-simultaneity, 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 trapped-photon clocks that are used to illustrate the non-accelerating 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?
 P: 293 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?
Sci Advisor
Emeritus
P: 7,204
 Quote by 1977ub 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?
That sounds right as there isn't any way to define "at the same time" for spatially separated clocks.

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]
P: 367
 Quote by 1977ub 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?
Let's look at this from the launch frame.

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.
P: 1,399
 Quote by 1977ub 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?
It depends on the acceleration profiles of A&C. You're making the acceleration discrete, so that every time there is a "ping", A&C suddenly change velocity to a new velocity. If they follow identical acceleration profiles, then the distance between A&C will gradually increase, as viewed by the people on board the rockets.

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).
 P: 293 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.
 P: 293 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.
Sci Advisor
Emeritus
P: 7,204
 Quote by 1977ub pervect, 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.
Yes, that's a good way of looking at it. You still do need a definition of simultaneity to use while "boosting. The most common definition of simultaneity used fits the above description of what happens. This definition is to define events in the accelerated "frame" to be simultaneious when they are simultaneous in the co-moving inertial reference frame.
 P: 293 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.
 P: 293 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 "stick-figure" way to illustrate this without performing the integration of the changing velocity.
P: 1,399
 Quote by 1977ub 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.
Well, you can actually calculate it. For simplicity, let F = the rest frame before hurling the boulder. F' = the rest frame afterward. Let $v$ = the relative velocity between the frames. Let $T$ = the time required to hurl the boulder, as measured in frame F. Let $D$ = the distance traveled while hurling the boulder, as measured in frame F. Let $L$ = the distance between people, as measured in frame F.

Then identify a number of events:
1. $e_1$: the rear person starts to hurl the boulder.
2. $e_2$: the rear person finishes.
3. $e_3$: the front person starts to hurl the boulder.
4. $e_4$: the front person finishes.

Our assumptions are that the corresponding actions are synchronized in frame F. So we have the coordinates for these events in frame F:
1. $x_1 = 0,\ \ t_1 = 0$
2. $x_2 = D,\ \ t_2 = T$
3. $x_3 = L, \ \ t_3 = 0$
4. $x_4 = L+D, \ \ t_4 = T$

Now use the Lorentz transforms to see the coordinates in frame F':
1. $x_1' = 0, \ \ t_1' = 0$
2. $x_2' = \gamma (D - vT), \ \ t_2' = \gamma (T - \dfrac{vD}{c^2})$
3. $x_3' = \gamma L,\ \ t_3' = -\gamma \dfrac{vL}{c^2}$
4. $x_4' = \gamma (L+D - vT), \ \ t_4' = \gamma (T - \dfrac{v(L+D)}{c^2})$

While in frame F, $t_1 = t_3$ and $t_2 = t_4$, in frame F', the order of events is: $t_3'$, then $t_1'$ or $t_4'$ and then finally $t_2'$. (The order of $t_1'$ and $t_4'$ depends on the size of $L, D,$ and $T$.)

So what things look like in frame F' is this:
1. Initially, both people are traveling at speed v in the -x direction.
2. $t' = t_3'$: the front person throws a boulder. His speed in the -x direction starts slowing down. But the rear person continues to travel at speed v in the -x direction.
3. $t' = t_1'$: the rear person starts to throw a boulder, as well.
4. $t' = t_4'$: the front person comes to rest.
5. $t' = t_2'$: the rear person comes to rest.

So between times $t_3'$ and $t_2'$, 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.
P: 293
 Quote by stevendaryl So we have the coordinates for these events in frame F: $x_1 = 0,\ \ t_1 = 0$ $x_2 = D,\ \ t_2 = T$ $x_3 = L, \ \ t_3 = 0$ $x_4 = L+D, \ \ t_4 = T$
But this is what I'm not sure I can believe anymore. Specifically that t2 and t4 are both equal to the same amount of time in the RF. Once the hurlers start pushing the boulders, the vehicles star moving, and their clocks no longer appear to be synchronized with one another from RF. Frontward clock now appears to be set later than Rearward clock. Thus RF must conclude that boulders are finished being hurled at different RF times.
P: 1,399
 Quote by 1977ub But this is what I'm not sure I can believe anymore. Specifically that t2 and t4 are both equal to the same amount of time in the RF. Once the hurlers start pushing the boulders, the vehicles star moving, and their clocks no longer appear to be synchronized with one another from RF.
You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).

In scenario (1):
• The clock in the front runs faster than the clock in the rear, according to the original rest frame, and also according to those aboard the rockets.
• The distance between the front and rear contracts, according to the original rest frame.
• The distance between the front and rear remains constant, according to those aboard the rockets.
• The acceleration felt by those in the front is less than the acceleration felt by those in the rear.

In scenario (2):
• The clocks in the front and rear run at the same rate, according to the original rest frame.
• The front clock runs faster than the rear clock, according to the people in the rockets.
• The distance between the front and rear remains constant, according to the original rest frame.
• The distance between the front and rear expands, according to those aboard the rockets.
• The acceleration felt by those in the front is the same as the acceleration felt by those in the rear.
P: 293
 Quote by stevendaryl You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).
In my later posts, I settled on a 3rd scenario I found more fundamental, in which I do not presume to know how things will turn out, but instead start with identical programmed equipment. We've got identical springs compressed holding identical boulders. Both parties in their identical but separate vehicles activate the release on their own copy of the equipment at t0 which both parties in their vehicles and an observer in the rest frame agree is simultaneous, since at t0 all are at rest wrt one another. So how does that pan out? Presumably to the rest observer, the boulders *complete* their release at different times, since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

 Related Discussions Special & General Relativity 3 Special & General Relativity 17 Special & General Relativity 44 Special & General Relativity 7 Special & General Relativity 7