B Gravity vs Acceleration: The Impact on Clocks in Reference Frame R

[calinvass]

Thank you. This explanation at link https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/
Seems very clear up to a point.
The conditiom is, the clocks feel both the same proper acceleration and we set a refference frame F0, where they apear as accelerating at the same time.
We follow the acceleratio curve of the clock in the left hand side ( this is the lower or the rear clock in the direction of motion) then at any point we stop we can draw the coordinates (minkowski spacetime) of a reference frame featuring the left clock at rest FL. When you look on the x-axis to find the second clock, it will apear at point that is on the refference frame F0, further on the x-axis but also at a distance in time because the line of simulataneity (x) is tilted. On the FL reference frame the distance on the x-axis will also be at a greater.

This means the distance between them has increased in FL reference frame but not in F0 reference frame because of different simultaneity lines). Since FL is an accelerated reference frame, the distance between the clocks in this frame will be constantly increasing - constant speed . Now it is becoming a bit harder to follow. This will generate a reduced clock rate observed in the FL reference frame.
However in the F0 reference frame clocks should show the same rates so after the ship stops both clocks should show the same time because the coordinates are tilted always at the same angle. Now it seems to contradict the results in the twins paradox.

 

[PeterDonis]

FL is an accelerated reference frame

No, it isn't. It's an inertial frame in which the chosen spaceship (the one on the left) is momentarily at rest.

This will generate a reduced clock rate observed in the FL reference frame.

You have to be very careful when talking about clock rates and frames in cases like this. Here is a more careful way of stating what is going on:

Pick an event on the worldline of the left (or rear clock) while it is accelerating. Call that event L. We can pick two events on the worldline of the right (or front) clock which are simultaneous with L according to two different inertial frames: event R0, which is simultaneous with L according to frame F0, and event RL, which is simultaneous with L according to frame FL. These are different events.

Call the left clock's elapsed time at event L time T. At event R0, the right clock shows elapsed time equal to T. But at event RL, the right clock shows elapsed time greater than T.

Also, in frame F0, the distance between the two clocks always remains the same--so the distance between events L and R0 is the same as the distance between the clocks before they started accelerating. But the distance, in frame FL, between events L and RL is not the same--it is larger. (This is one way of seeing why the string stretches and eventually breaks.)

after the ship stops both clocks should show the same time

And they will, if they stop at the same time in frame F0. More precisely, if the left clock stops at some event SL, and the right clock stops at event SR0 which is simultaneous with event SL in frame F0, then the clocks will show the same time at those two events. And since the clocks will continue to move at the same speed (since they are no longer accelerating), they will remain synchronized according to frame F0.

But suppose that the left clock stops at some event SL, and the right clock stops at event SRL which is simultaneous with SL in frame FL. Then we have a more complicated situation. The time shown on the left clock at event SL will be less than the time shown on the right clock at event SRL. So in frame FL, the right clock will show more elapsed time than the left clock when they both stop accelerating.

But in frame F0, event SL happens before event SRL; so to see how the clocks end up in frame F0, we have to figure out what the left clock reads at event SLL, which is the event that is simultaneous with event SRL in frame F0. At event SLL, the left clock has been moving inertially for some time after stopping (i.e., in frame F0 the left clock stops accelerating first), so from the viewpoint of frame F0, it is "gaining time" relative to the right clock, which keeps accelerating. How things are when the right clock stops accelerating, at event SRL, will depend on the details, like the original distance between the clocks and how long the left clock accelerates.

Furthermore, in this second scenario, where the right clock stops accelerating at event SRL, the clocks are not moving at the same speed (in any inertial frame), so their synchronization will continue to change (unlike the first scenario, where the right clock stops at event SR0).

 

[calinvass]

Here is the diagram you've described

 

[PeterDonis]

Here is the diagram you've described

Not quite. Event SLL should be on the dotted line going up from event SL--remember, we assumed that the left clock stops accelerating at event L, so it moves inertially after that point, and the dotted line is the inertial worldline of the left clock after event L. Event SLL lies on that worldline.

 

[calinvass]

I was editing the diagram. However you've made some points I didn't take into account. I've judged the acceleration lines as helping curves not the actual path of the clocks. But the mistake was that the horizontal line from SRL should've intersected the dotted blue line not the helping acceleration path ( that path goes beyond c, so it clearly cannot represent the real path.

 

[calinvass]

 

[PeterDonis]

the mistake was that the horizontal line from SRL should've intersected the dotted blue line not the helping acceleration path

Yes.

that path goes beyond c

It shouldn't if you generated it correctly. You need to use the $\cosh$ and $\sinh$ functions.

 

[calinvass]

Yes, I understand. I've used constant Newtonian acceleration in flat spacetime but that is not the trajectory in frame F0.

 

[David Lewis]

It's not an illusion, and it's not symmetrical like kinetic time dilation.

Please explain what you mean by not symmetrical.

 

[Nugatory]

Please explain what you mean by not symmetrical.

Gravitational time dilation is asymmetrical in the sense that if B's clock is running slow relative to A's clock as observed by A, it is also running slow (or equivalently, A's clock is running faster) as observed by B. In physical terms: If B is deeper in a gravity well than A, and they exchange light signals from identically constructed (so the emission frequencies are the same) light sources B will receive a light signal that is blueshifted to a higher frequency than his own source, while A will receive a light signal that is redshifted to a lower frequency than his own source.

The speed-based time dilation is symmetrical: as observed by A, B's clock is running slow relative to A's clock but as observed by B, A's clock is running slow relative to B's clock. If they exchange light signals they will both receive either blue-shifted or red-shifted signals according to whether they are approaching one or another or moving apart.

 

[David Lewis]

Many thanks. Is it correct to say gravitational time dilation is asymmetric under interchange of observers?

 

[stevendaryl]

Many thanks. Is it correct to say gravitational time dilation is asymmetric under interchange of observers?

Yes, it's asymmetric.

But it's actually a little complicated as to what "gravitational time dilation" means. What's definitely true is that if you use a curvilinear, noninertial coordinate system, then in general the value of \frac{d \tau}{dt} (where \tau is the elapsed time on a standard clock, and t is coordinate time) will be position-dependent, as well as velocity-dependent, while in SR, using inertial Cartesian coordinates, it is only velocity-dependent.

 

[David Lewis]

So a satellite clock, having higher gravitational potential, would appear to tick faster than an Earth clock, and (assuming the satellite is moving with respect to the Earth clock) SR time dilation would make the satellite clock appear to tick slower than the Earth clock. I suppose you would add the two effects together to arrive at the net dT/dt.

 

[Nugatory]

So a satellite clock, having higher gravitational potential, would appear to tick faster than an Earth clock, and (assuming the satellite is moving with respect to the Earth clock) SR time dilation would make the satellite clock appear to tick slower than the Earth clock. I suppose you would add the two effects together to arrive at the net dT/dt.

Yes, and we have several older threads that show how to do this calculation. Getting it right is kinda important for the GPS system...

 

[calinvass]

There is also frame dragging effect thus the orbiting rotation direction or whether it is geostationary should influence the difference in tick rates.

 

[cianfa72]

if we have two observers inside a rocket accelerating in flat spacetime ("accelerating" meaning "experiencing proper acceleration"), at rest relative to each other (as measured by them exchanging round-trip light signals and seeing that the round-trip travel time by each of their clocks remains constant), the observer at the bottom of the rocket will be time dilated (clock running slower) relative to the observer at the top (as measured by noting the elapsed time on both of their clocks between two successive round-trip light signals).

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Furthermore if in the above procedure each observer sends/encodes in the reflected light signal the reading of his own clock (when the signal is reflected back), the receiving observer can evaluate the difference between the encoded times and the readings of his own clock between the receipts of two successive round-trip light signals.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Is the above correct ? Thank you.

 

[PeroK]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Furthermore if in the above procedure each observer sends/encodes in the reflected light signal the reading of his own clock (when the signal is reflected back), the receiving observer can evaluate the difference between the encoded times and the readings of his own clock between the receipts of two successive round-trip light signals.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Is the above correct ? Thank you.

You could do it that way. Alternatively, each observer sends out a signal once per second, say. They each receive their own signals back at one second intervals. But, receive the other's signals with a longer or shorter time interval.

 

[cianfa72]

Alternatively, each observer sends out a signal once per second, say.

yes, once per second according their own clocks.

They each receive their own signals back at one second intervals. But, receive the other's signals with a longer or shorter time interval.

Again, as measured by their own clocks.

 

[jartsa]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Quite the opposite. All inertial observers say that there is some relative motion. They might call it "length contraction motion". I am guessing that "we" are inertial observers.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Yes, according to those non-inertial people the bottom people are time dilated. Different inertial observers have different opinions about the time dilation.

 

[DrGreg]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Quite the opposite. All inertial observers say that there is some relative motion. They might call it "length contraction motion". I am guessing that "we" are inertial observers

If "we" are inertial observers, you are right, but I suspect what cianfa72 meant to say was "each observer claims the other is at rest relative to themself", and, in that case, that is true.

EDIT: Just to clarify possibly ambiguous language:

• Observer A claims that Observer B is at rest relative to Observer A
• Observer B claims that Observer A is at rest relative to Observer B

Both these claims are true; they can be taken as a definition of what "at rest relative to a non-inertial observer" means.

 

[PeroK]

If "we" are inertial observers, you are right, but I suspect what cianfa72 meant to say was "each observer claims the other is at rest relative to themself", and, in that case, that is true.

That's what we all meant! Starting with:

The precise way of putting it is this: if we have two observers inside a rocket accelerating in flat spacetime ("accelerating" meaning "experiencing proper acceleration"), at rest relative to each other (as measured by them exchanging round-trip light signals and seeing that the round-trip travel time by each of their clocks remains constant), the observer at the bottom of the rocket will be time dilated (clock running slower) relative to the observer at the top (as measured by noting the elapsed time on both of their clocks between two successive round-trip light signals).

 

[jartsa]

EDIT: Just to clarify possibly ambiguous language:

• Observer A claims that Observer B is at rest relative to Observer A
• Observer B claims that Observer A is at rest relative to Observer B

Both these claims are true; they can be taken as a definition of what "at rest relative to a non-inertial observer" means.

Last sentence is ambiguous. It should be said like this: Both these claims are true; they can be taken as a definition of what "at rest relative to a non-inertial observer, according to the observer himself" means.

Relative to what, and according to who.

 

[DrGreg]

Last sentence is ambiguous. It should be said like this: Both these claims are true; they can be taken as a definition of what "at rest relative to a non-inertial observer, according to the observer himself" means.

Relative to what, and according to who.

I think in relativity, when you say "A is at rest relative to B" it is automatically assumed it's "...according to B". Or in other words, it means "A is at rest in B's coordinate system". I think it's a bad idea to say "A is at rest relative to B according to C". I'm not even sure it has a well-defined meaning. It would be better to say "A and B have the same velocity relative to C", if that's what you mean, or "A and B have a constant separation as measured by C". The details depend on exactly how C chooses to do the measurement. It may also depend on whether spacetime is curved or flat. ("Same velocity" doesn't always make sense.)

 

[PeterDonis]

I think in relativity, when you say "A is at rest relative to B" it is automatically assumed it's "...according to B".

Actually, I think the proper meaning of "at rest relative to" should be that it is an invariant: the best way to say it is "A and B are at rest relative to each other", and this is to be taken as meaning that, for example, round-trip light signals between them have a constant round-trip travel time, according to either of their clocks, which is an invariant, independent of any choice of coordinates. That is the meaning @cianfa72 was using, and I think it's correct.

 

[PeterDonis]

All inertial observers say that there is some relative motion.

In coordinate terms, yes. But what invariant indicates "relative motion" in this case? Note that, per my previous post just now, there is an invariant (timing of round-trip light signals between A and B) that says there is no "relative motion".

 

[jartsa]

In coordinate terms, yes. But what invariant indicates "relative motion" in this case? Note that, per my previous post just now, there is an invariant (timing of round-trip light signals between A and B) that says there is no "relative motion".

Velocity reading on an inertial navigation system bolted on the rocket floor minus velocity reading on an inertial navigation system bolted on the rocket ceiling. That number is the same in all frames.

 

[Ibix]

Velocity reading on an inertial navigation system bolted on the rocket floor minus velocity reading on an inertial navigation system bolted on the rocket ceiling. That number is the same in all frames.

You need to compare readings taken at the same time, though. What simultaneity condition are you using? You can get growing, shrinking, or steady distances depending on your choice.

 

[cianfa72]

Velocity reading on an inertial navigation system bolted on the rocket floor

Btw, what does a such inertial navigation system actually measure ?

 

[Ibix]

Btw, what does a such inertial navigation system actually measure ?

The proper acceleration and possibly rotation rates with respect to the Fermi-normal axes, all as functions of the instrument's proper time, and integrals thereof.