Different Clock Rates Throughout Accelerating Spaceship

[Nugatory]

I think it's because lower clock really is slower than upper clock.

Try writing down a rigorous definition of "really is slower", one that I could use to construct an experiment that will allow all observers to agree about which one is "really slower". It can be done, but it's harder than it sounds, and when you succeed you'll have a better sense of why the question posed in #13 of this thread isn't all that well-formed.

 

[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?

 

[stevendaryl]

As far as B is concerned, a ping is released once per second let's 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.

 

[stevendaryl]

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.

 

[Nugatory]

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.

 

[1977ub]

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?

 

[1977ub]

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?

 

[pervect]

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]

 

[jartsa]

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.

 

[stevendaryl]

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).

 

[1977ub]

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.

 

[1977ub]

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.

 

[pervect]

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.

 

[1977ub]

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.

 

[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.

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.

 

[stevendaryl]

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.

 

[1977ub]

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. $x_4 = L+D, \ \ t_4 = T$

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.

 

[stevendaryl]

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.

 

[1977ub]

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.

 

[PAllen]

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.

If I understand it, your 3d scenario is identical to stevendaryl's scenario (2). In which case it doesn't correspond in a meaningful way to an idealization of real rocket or to a tall building sitting on the surface of large planet (these are scenario (1) as stevendaryl labels them).

You've got boulders expelled by identical springs at t0 per starting rest frame. This means they must be the same distance apart and moving at the same speed per this starting rest frame. If they each expel another boulder at t0+1 per their own watches (which are still in synch per the rest frame (but slow), but not per each other), again their speed and distance and clocks are in synch per the starting rest frame. Per each other, their clocks are out of synch and they have moved further apart.

 

[1977ub]

If I understand it, your 3d scenario is identical to stevendaryl's scenario (2).

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

3) Therefore, the final moment of boulder release from the spring will be different between the vehicles as seen by the rest frame.

I am describing the case where the apparatus, the program, the intent of the two vehicles is the same, but I think that they end up out of synch at then end of the acceleration period, as seen in the rest frame, therefore we can't in a general way say that they have "accelerated in exactly the same way in the rest frame." The intended so, they began so, but because the acceleration takes time, and they are moving during that time, they have not ended up so. They *began* accelerating in exactly the same way in the rest frame. They entire acceleration process did not take place in exactly the same way in the rest frame.

 

[PAllen]

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

No. They would be out of synch with starting frame if they followed a synchronization procedure while they were in motion. Since they started synchronized in the initial frame, and follow identical physical process, they remain in synch in the initial frame, and out of synch with each other.

3) Therefore, the final moment of boulder release from the spring will be different between the vehicles as seen by the rest frame.

Nope. It will be in synch from the initial frame; it will appear out of synch to each vehicle, if they were to apply a synchronization procedure.

 

[stevendaryl]

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?

That's not a 3rd scenario, that's scenario 2. If the front and rear follow exactly identical actions that are (initially) simultaneous in the initial rest frame, then they will always be synchronized in the initial rest frame, and the distance between them will remain constant, according to the initial rest frame.

Presumably to the rest observer, the boulders *complete* their release at different times

No. If they are doing the same actions, starting at the same time, they will finish at the same time.

...since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

The person in the front and the rear will not view their clocks as synchronized, but they will continue to be synchronized according to the initial rest frame.

 

[stevendaryl]

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

Why do you think that? If the front and rear are accelerating identically, then they will always be traveling at the same velocity, according to the initial rest frame.

 

[1977ub]

ok. thanks.

 

[Austin0]

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.

How do they determine these effects within the frame?
Measure relative clock rates and distance??

 

[1977ub]

ok hope this is ok for this same thread. I'm moving in a more basic direction for understanding acceleration+SR:

For two intertial frames in relative motion, we can use gamma to describe how each observer measures the other's clock speed. Complete parity. Closely related to relativity of simultaneity. Fine.

Given an inertial frame RF,
and someone moving in a circle AF at velocity v with associated Lorentz gamma,

as far as I understand, RF still use gamma to determine rate of AF's clock?

what will AF use for RF's clock - 1/gamma ?

 

[PAllen]

How do they determine these effects within the frame?
Measure relative clock rates and distance??

They can determine (in principle) that the distance front to back of the rocket expanded by attaching a string that can't stretch; it will break. This is just Bell's spaceship paradox.

They can detect time difference between front and back clocks by exchanging signals.

 

[PAllen]

ok hope this is ok for this same thread. I'm moving in a more basic direction for understanding acceleration+SR:

For two intertial frames in relative motion, we can use gamma to describe how each observer measures the other's clock speed. Complete parity. Closely related to relativity of simultaneity. Fine.

Given an inertial frame RF,
and someone moving in a circle AF at velocity v with associated Lorentz gamma,

as far as I understand, RF still use gamma to determine rate of AF's clock?

what will AF use for RF's clock - 1/gamma ?

[yes, your RF can use gamma for the accelerated object.]

There are different philosophy's on this. To understand anything accelerating observers will measure (including see or photograph), it is simplest just to use any convenient inertial frame. The results of observations are invariant.

To try to come up with a frame for the accelerating observer, you run into the same issues as in GR: there is well defined local accelerated frame, just as there are well defined local frames in GR. However, there is no global frame for an accelerated observer in SR, just as there are no global frames in GR. What you can do, if you insist, is set up a coordinate system in which the accelerated observer remains at fixed coordinate position. Such a coordinate system may not be able to cover all of spacetime. Unfortunately, there are many ways to do this, none preferred. Once you have defined such coordinates (via transform from inertial frame), you can compute the metric in them. Then, using the metric, you can compute time dilation etc. per this coordinate system. It won't be as simple as a constant in place of gamma. The constant gamma results from the fact that the metric in the inertial SR frame is diag(1,-1,-1,-1). With a metric that varies by position and time, you need to integrate contraction of metric with path tangent vectors, instead of having a simple constant.

The up shot of all this is that there is no (preferred) answer to your question (what does the accelerated observer use in place of gamma?). It depends on what coordinate system you set up. On the other hand, let me stress again, if you want to know anything about what the accelerated observer measures or sees, just compute this in any convenient inertial frame.

 

[1977ub]

what does the accelerated observer use in place of gamma?

I'm asking a simpler question than how to define gamma if one is AF.

What RF calls gamma can be used to determine the click speed on AF and its inverse can be used to determine RF's click speed measured from AF.

I mean if RF measures one second while AF is making a revolution, and RF find AF's clock to have moved forward by .5 second, then it's a given that AF will find RFs clock to have clicked twice as fast as his own. I just want to verify it is this simple. I don't see how this can't be true.

 

[PAllen]

I'm asking a simpler question than how to define gamma if one is AF.

What RF calls gamma can be used to determine the click speed on AF and its inverse can be used to determine RF's click speed measured from AF.

I mean if RF measures one second while AF is making a revolution, and RF find AF's clock to have moved forward by .5 second, then it's a given that AF will find RFs clock to have clicked twice as fast as his own. I just want to verify it is this simple. I don't see how this can't be true.

I use AO (accelerated observer) rather than AF, below, because there is no such thing as an accelerated frame (only choices of many possible coordinate systems).

For AO, the behavior they see on clocks in the inertial frame depend on where in the inertial frame they are, and the visual rates vary in time. That is, the observed behavior of inertial clocks will be both position and time dependent. The rates on these clocks averaged over time will show them (per the AO) to fast compared to AO clock. It is true that for pure circular motion at constant speed, the averaged rate seen on the inertial clocks will be gamma (as determined by RF) times faster than AO's clock.

 

[1977ub]

It is true that for pure circular motion at constant speed, the averaged rate seen on the inertial clocks will be gamma (as determined by RF) times faster than AO's clock.

And inescapably, if we simply use a single clock in RF, and AO measures it once per revolution (passing right by it, say) then we can use 1/RF's-gamma to determine the time anti-dilation of RF as perceived by AO. I do understand that this will not apply to all of RF's clocks throughout a revolution. For one thing, AO is moving away from some of them while moving toward others, etc. "On average" AO must be able to use 1/RF's-gamma to determine the average speed of RF's clocks. Every time AO scrapes by clock-RF-0, RF finds AO's clock to have ticked slower by gamma, thus AO must find clock-RF-0 to have ticked faster by 1/RF's-gamma.

 

[PAllen]

And inescapably, if we simply use a single clock in RF, and AO measures it once per revolution (passing right by it, say) then we can use 1/RF's-gamma to determine the time anti-dilation of RF as perceived by AO. I do understand that this will not apply to all of RF's clocks throughout a revolution. For one thing, AO is moving away from some of them while moving toward others, etc. "On average" AO must be able to use 1/RF's-gamma to determine the average speed of RF's clocks. Every time AO scrapes by clock-RF-0, RF finds AO's clock to have ticked slower by gamma, thus AO must find clock-RF-0 to have ticked faster by 1/RF's-gamma.

Agreed. Just don't try to generalize this to other situations without understanding the complexities I described.

 

[1977ub]

Agreed. Just don't try to generalize this to other situations without understanding the complexities I described.

Sure.

1) for uniform linear motion, gamma is purely "subjective" - nobody's clock is moving slower than anyone else's in a meaningful way. it is all wrapped up with planes of simultaneity.

2) for uniform circular motion, gamma gains a clear "objective" meaning - RF can use it to describe *all* of AO's time dilation, and AO can use the inverse to describe *average* time anti-dilation of RF. It is similar to the gravity situation where there are agreed differences in time dilations.

3) The next part is harder for me to work out -
the linearly accelerating observer - how RF and AO can measure their relative clock speeds.

I gather than RF can calculate AO's relative time dilation by simply integrating ever changing gamma with ever changing velocity?

But to ask how AO determines RF's clock speed, this situation is neither the simple 'subjective' or 'objective' case above.

 

[PAllen]

3) The next part is harder for me to work out -
the linearly accelerating observer - how RF and AO can measure their relative clock speeds.

I gather than RF can calculate AO's relative time dilation by simply integrating ever changing gamma with ever changing velocity?

But to ask how AO determines RF's clock speed, this situation is neither the simple 'subjective' or 'objective' case above.

Yes, RF could just integrate gamma(t) over the accelerating path. RF can do this for any path.

For AO, this simple example raises one of the SR concepts difficult to grasp on first encounter. This is the so called Rindler horizon. If you try to ask about how AO would model RF clocks by factoring out light delay using some simultaneity convention, you face the following observation:

RF clocks that AO is accelerating away from appear to red shift until they disappear, at a fixed distance behind the AO. Any RF clock further away cannot be seen or communicate in any way with the AO, so mutual clock comparison is impossible. Note that there is a one way aspect to this (as for all horizons): any RF clock can eventually receive a signal from any point in AO's history; however, there is a last time, for every clock in RF, after which it cannot send any signals to AO.

Visually, you can certainly say the case is more like your (1): any clock at rest in RF and AO's clock each eventually see the other clock freezing and red shifting to infinity.