Comparing Clock Rates in Accelerated vs Constant Velocity Frames

[Freixas]

@PeroK sent me here based on a discussion starting at https://www.physicsforums.com/threads/some-ideas-on-interstellar-space-travel.1006562/post-6550265.

Summary:

• Someone claimed that an accelerated observer's clock (observer at 1g from Earth to halfway to Alpha Centauri, then -1g the rest of the way) would run slower compared to a clock on Earth (or Alpha Centauri, which appears to be treated as being motionless relative to the Earth).
• @PeroK said that "Time dilation is symmetrical. You could equally say that time is going slower on Earth than in the rocket."
• Later, @PeroK introduced a row of clocks, presumably also motionless with respect to the Earth, placed along the travel path. He said "The crew on the rocket would measure each of these clocks to be running slow as they passed it..."

The gamma solution

One approach to the problem of comparing clock rates would be to just look at the traveler's instantaneous gamma, which is derived from the instantaneous velocity and call it a day. Since the velocity is a relative velocity (Earth vs. rocket), the time dilation is the same from either viewpoint.

The problem is that, at the end of the journey, the traveler and Earth are back in the same inertial frame and the final time tallies are not symmetrical. From each observer's view, the other's clock always ran slower (except at the start and end), yet the traveler's clock is the one with less time.

The "line of simultaneity" solution

When dealing with observers moving with different constant velocity, we can compare clock rates by using lines of simultaneity with respect to each observer's inertial frame. We draw the line of simultaneity for one observer at one point and then one time unit later. These two lines intersect the other's time axis and we read the time interval there. The ratio establishes the rate difference. It is always symmetrical.

I can draw the instantaneous line of simultaneity for an accelerated observer and find the corresponding point on Earth's t axis. I could try to draw a second line of simultaneity one traveler's time unit later to find the rate difference. The problem is that I would be trying to compare two quantities measured relative to two different inertial frames. And the clock rates computed by the traveler with this method would show Earth time sometimes slower and sometimes faster than the traveler's own clock. The Earth observer would come up with the traveler always having a slower rate.

The row of clocks

@PeroK introduced a row of clocks along the travel path, presumably to make it clearer that the time dilation was symmetrical. The idea was for the traveler to compare his clock tick to the tick of the one of the clock he passes.

I am unsure how this brings any clarity to the problem, nor how any method the traveler tried to use to measure the passing clock's tick wouldn't involve measurements that start and end in different inertial frames. Perhaps @PeroK will be willing to walk me through his thinking in this forum.

I may be handicapped by trying to understand the answer using S.R. I suspect this may be more of a G.R. topic. Nothing ventured, nothing gained...

 

[PeroK]

I am unsure how this brings any clarity to the problem, nor how any method the traveler tried to use to measure the passing clock's tick wouldn't involve measurements that start and end in different inertial frames. Perhaps @PeroK will be willing to walk me through his thinking in this forum.

This is where, in general, physics uses the concepts of calculus. It's true that any two measurements must be separated by a finite time interval and that any two such measurements constitute an approximation to the theoretical instantaneous rate. In this case, we have the instantaneous rate $\frac{dt_1}{dt_2}$, where $t_1$ is the time shown on one of the row of clocks and $t_2$ is the time shown on the ship's clock (both as measured from the ship).

In this case the derivative $\frac{dt_1}{dt_2} < 1$, indicating that the ship measures the clock running slow as it passes it. A pair of finite measurements would be $\frac{\Delta t_1}{\Delta t_2}$, which would approximate the instantaneous derivative as $\Delta t_2 \rightarrow 0$.

This concept applies across all of physics and not just SR/GR: we have theoretical differentiable quantities that can be approximated by a pair of finite measurements over some small time interval.

 

[Freixas]

This is where, in general, physics uses the concepts of calculus.

Agreed. Where I still have a problem is with this:

It's true that any two measurements must be separated by a finite time interval and that any two such measurements constitute an approximation to the theoretical instantaneous rate.

If the two measurements that constitute an approximation are nonsensical, then applying calculus to the problem seems to yield more nonsense. I'm not saying that your claim is nonsense—just that I can't see how it avoids being nonsense.

You have not clearly described a method for computing $\frac{\Delta t_1}{\Delta t_2}$ that avoids making the measurements in two different frames. In the SF&F forum, you said "You could use the transverse Doppler effect, if nothing else. Or, two measurements using an isosceles triangle, which equalises the light travel time."

As far as I could tell, both the transverse and longitudinal Doppler effect corrections assume constant velocity, but perhaps there is a version for correcting for accelerated motion.

I'm not sure what you mean by the isosceles triangle approach. My best guess of your meaning leads me to believe you are combining measurements made in different inertial frames, but I'd prefer to know what you actually meant.

 

[PeroK]

Agreed. Where I still have a problem is with this:

I'm not quite sure whether you're trolling me now. I'll let someone else try to explain that acceleration is not some fundamentally problematic phenomenon that prevents physics being done.

 

[Freixas]

I'm not quite sure whether you're trolling me now.

I'm not trolling you. But since I'm not getting an answer from you, I'd welcome a response from anyone else.

I'll let someone else try to explain that acceleration is not some fundamentally problematic phenomenon that prevents physics being done.

This is your interpretation of my comments—I've never made this claim.

 

[anuttarasammyak]

Is length contraction considered in your thoughts ?

Say at goal rocket clock shows $T_R$ and the Earth IFR clock then and there shows $T_E$.
It takes $T_R/2$ for rocket to reach the middle point which is less than $T_E/2$ of Earth IFR clock then and there because shorter length than start-goal distance set in Earth IFR needs shorter time to reach for same speed of rocket in Earth IFR and the middle point in rocket FR or instantaneous IFR.
$$T_R/2< T_E/2$$
Thus
$$T_R< T_E$$

 

[PeterDonis]

I may be handicapped by trying to understand the answer using S.R.

That's not the problem. The problem is that you are trying to treat two different things as if they were the same thing. That doesn't work.

Here are the two different things, both stated under the assumption that we are working within SR, i.e., flat spacetime, no gravity anywhere:

Thing 1: If you and I are flying past each other, each of us measures the other's clock to be running slow, locally. By that I mean, if we each are emitting light signals at fixed time intervals by our own clocks, and we each measure the time intervals, by our own clocks, at which we receive the other's light signals, and then we correct the latter measurements for light travel time, each of us comes up with a "clock rate" for the other that is slow, and slow by the same factor, namely $\gamma = 1 / \sqrt{1 - v^2}$, where $v$ is our relative velocity.

Note that this can also be viewed as each of us directly observing each other's relativistic Doppler shift factor, which requires no calculation (we each just emit our light signals using the same process that yields light of the same fixed wavelength as measured in our own rest frames, for example by each of us using identically constructed lasers, and then we each measure the shift in wavelength of the other's light), and then correcting that factor for light travel time to get a "clock rate" for the other. But we don't actually have to apply the "light travel time" correction: the observed Doppler shift factor itself is symmetric between the two of us (each of us sees the same Doppler redshift or blueshift in the other's light signals). So we can just as easily consider Thing 1 to be the Doppler shift factor, which is directly observable.

Thing 2: If you and I start out at rest relative to each other, and then we each execute some type of "trip", where we might accelerate, decelerate, etc. in some manner, and then we end up at rest relative to each other, we can compare our elapsed times. This comparison will not show that each of our clocks "ran slow" relative to the other; it will show that one of our clocks had more elapsed time than the other, and we will both agree on which clock that is. (I am leaving out the edge case in which the elapsed times are identical.)

Note that strictly speaking, we can only do this in an invariant manner if we are spatially co-located at both start and end; otherwise we have to depend on a simultaneity convention, since each of us has to know which events on our two worldlines occur at the same time. In your example of this type in the other thread, you were implicitly assuming that we are both at rest in the same global inertial frame at the start and end, and that we used the simultaneity convention of that frame. This frame is picked out physically by the fact that (a) we are both at rest in it at the start and end (and it's the same frame in both cases), and (b) spacetime is flat, so inertial frames can be global. In more general examples those things will not both be true of any frame, and there will be no way to compare "elapsed times" for observers that are not spatially co-located at both start and end.

The important thing to note is that Thing 1 is symmetric between the two of us, while Thing 2 is not--it's asymmetric. That is not a problem or a paradox; it is just a consequence of the fact that these are two different things. It is unfortunate that the same term, "time dilation", is often used in the literature for both of these things. From time to time we try the term "differential aging" for Thing 2 to make clear the difference between the two, but that nomenclature doesn't seem to catch on.

This is where, in general, physics uses the concepts of calculus.

Actually, calculus is irrelevant to the problem the OP appears to me to be having. See above.

 

[PeterDonis]

From each observer's view, the other's clock always ran slower (except at the start and end)

That's not what @PeroK said. The "other clock runs slower" observation can only be made locally; once you, the traveler, have left Earth headed for some distant planet, you and I can't directly measure each other's "clock rates" locally any more, because we're not passing each other.

What @PeroK said is that as you pass each clock placed along your path, you measure its clock rate to be slow compared to yours, locally. But as he also pointed out in the other thread, these clocks do not appear to you to be synchronized, either with my original clock on Earth or with each other. So you cannot derive any global measure of "relative time elapsed" from these local clock measurements. (This point was also made in the previous thread.)

we can compare clock rates by using lines of simultaneity with respect to each observer's inertial frame.

The problem with this is that, in general, you cannot construct a single non-inertial frame from these "lines of simultaneity", because they will cross each other, which means you will end up assigning different "times" to the same events on Earth's worldline. That doesn't make sense.

 

[Freixas]

Thanks for stepping in, Peter.

The problem with this is that, in general, you cannot construct a single non-inertial frame from these "lines of simultaneity", because they will cross each other, which means you will end up assigning different "times" to the same events on Earth's worldline. That doesn't make sense.

Ah, yes, that sounds familiar. I think Dale has a favorite paper on the topic.

Note that this can also be viewed as each of us directly observing each other's relativistic Doppler shift factor, which requires no calculation (we each just emit our light signals using the same process that yields light of the same fixed wavelength as measured in our own rest frames, for example by each of us using identically constructed lasers, and then we each measure the shift in wavelength of the other's light), and then correcting that factor for light travel time to get a "clock rate" for the other. But we don't actually have to apply the "light travel time" correction: the observed Doppler shift factor itself is symmetric between the two of us (each of us sees the same Doppler redshift or blueshift in the other's light signals). So we can just as easily consider Thing 1 to be the Doppler shift factor, which is directly observable.

I'm still trying to digest this bit.

With constant velocity, I get a constant Doppler shift. The example I learned was: two observers with constant velocity beam a video of their clocks to each other. They compare the rate of time on the video to their own clocks, adjusting for the Doppler shift.

But now one observer is accelerating. The Doppler frequency shift at the start of a tick is now not the same as at the end of the tick. This sounds similar to trying to make measurement in different inertial frames. Or are you saying that we ignore the clock tick and that the the instantaneous Doppler shift is enough?

What @PeroK said is that as you pass each clock placed along your path, you measure its clock rate to be slow compared to yours, locally.

I'm having a problem understanding why the distance of the clock is significant. If the clocks are on my path, then I am co-located at one instant in time, where everyone agrees on simultaneity. After that, even 1 nanosecond later, I am neither co-located, nor even in the same inertial frame as when I was, which would seem to be the case with a clock anywhere (e.g. on Earth).

OK, technically, I am never exactly co-located, so maybe we're ignoring very small distances and not worrying about very small inertial frame changes?

 

[PeterDonis]

are you saying that we ignore the clock tick and that the the instantaneous Doppler shift is enough?

Yes. In the limit, if you want to be precise, you can have each observer continously emitting light signals towards the other instead of intermittent pulses (e.g., by each one shining a continuous laser at the other). Then it is indeed a continuous wavelength (or frequency) comparison. Of course, as you say, if either observer is accelerating the shift will be changing continuously, but that's fine, you can still identify the instantaneous shift at any instant. And as long as you're just using local measurements, you don't have to worry about any other issues. See further comments on that below.

I'm having a problem understanding why the distance of the clock is significant.

Because in that case the acceleration of one observer in itself has asymmetric effects. If the "traveling" observer accelerates, the Doppler shift he observers in the light from the other observer changes instantly for him. But the other observer does not see any change in the first observer's Doppler shift until a light travel time has passed--i.e., until the light signals from the acceleration period have reached him. (In the standard twin paradox, this is called the "Doppler shift explanation" in the Usenet Physics FAQ twin paradox article that frequently gets referenced in PF threads.) So now the Doppler shift is neither Thing 1 nor Thing 2; it's yet another different thing, Thing 3, and you would have to analyze from scratch how it behaves and what, if anything, it tells you. Why go through all of that when you don't have to?

technically, I am never exactly co-located

Yes, but by "local" we just mean "close enough that all light travel times between observers are negligible compared to other time scales in the problem". For example, human conscious perception works on a time scale of roughly 10 to 100 milliseconds. So "local" for human perceptions of things is roughly within 3,000 to 30,000 kilometers (10 to 100 milli-light-seconds). By contrast, if the "observers" are computers that can distinguish changes on nanosecond time scales, then "locally" means "within a foot or so".

 

[Dale]

@PeroK sent me here based on a discussion starting at https://www.physicsforums.com/threads/some-ideas-on-interstellar-space-travel.1006562/post-6550265.

Summary:

• Someone claimed that an accelerated observer's clock (observer at 1g from Earth to halfway to Alpha Centauri, then -1g the rest of the way) would run slower compared to a clock on Earth (or Alpha Centauri, which appears to be treated as being motionless relative to the Earth).
• @PeroK said that "Time dilation is symmetrical. You could equally say that time is going slower on Earth than in the rocket."
• Later, @PeroK introduced a row of clocks, presumably also motionless with respect to the Earth, placed along the travel path. He said "The crew on the rocket would measure each of these clocks to be running slow as they passed it..."

The gamma solution

One approach to the problem of comparing clock rates would be to just look at the traveler's instantaneous gamma, which is derived from the instantaneous velocity and call it a day. Since the velocity is a relative velocity (Earth vs. rocket), the time dilation is the same from either viewpoint.

The problem is that, at the end of the journey, the traveler and Earth are back in the same inertial frame and the final time tallies are not symmetrical. From each observer's view, the other's clock always ran slower (except at the start and end), yet the traveler's clock is the one with less time.

The "line of simultaneity" solution

When dealing with observers moving with different constant velocity, we can compare clock rates by using lines of simultaneity with respect to each observer's inertial frame. We draw the line of simultaneity for one observer at one point and then one time unit later. These two lines intersect the other's time axis and we read the time interval there. The ratio establishes the rate difference. It is always symmetrical.

I can draw the instantaneous line of simultaneity for an accelerated observer and find the corresponding point on Earth's t axis. I could try to draw a second line of simultaneity one traveler's time unit later to find the rate difference. The problem is that I would be trying to compare two quantities measured relative to two different inertial frames. And the clock rates computed by the traveler with this method would show Earth time sometimes slower and sometimes faster than the traveler's own clock. The Earth observer would come up with the traveler always having a slower rate.

The row of clocks

@PeroK introduced a row of clocks along the travel path, presumably to make it clearer that the time dilation was symmetrical. The idea was for the traveler to compare his clock tick to the tick of the one of the clock he passes.

I am unsure how this brings any clarity to the problem, nor how any method the traveler tried to use to measure the passing clock's tick wouldn't involve measurements that start and end in different inertial frames. Perhaps @PeroK will be willing to walk me through his thinking in this forum.

I may be handicapped by trying to understand the answer using S.R. I suspect this may be more of a G.R. topic. Nothing ventured, nothing gained...

Sorry, what is the specific question or problem?

 

[Ibix]

Another way of conceptualising this is to consider the tick rates of two inertial clocks in relative motion. One clock is part of @PeroK's family of clocks and the other just happens to have the same instantaneous speed as you. The instantaneous tick rate of your clock must match that of the inertial clock co-moving with you unless acceleration affects the tick rates of clocks, which it doesn't (the "clock hypothesis", which has been tested to something like $10^{18}\mathrm{g}$ IIRC).

 

[cianfa72]

Another way of conceptualising this is to consider the tick rates of two inertial clocks in relative motion. One clock is part of @PeroK's family of clocks and the other just happens to have the same instantaneous speed as you.

The family of @PeroK clocks are actually at rest and Einstein synchronized each other in the global inertial frame in which they are at rest (we assumed a flat spacetime since the beginning). So they actually show the coordinate time $t$ of that global inertial frame.

 

[Freixas]

Sorry, what is the specific question or problem?

Hi, Dale, no worries, Peter understood what I was getting at.

 

[Dale]

Ok, we will close it then so that nobody else bothers. Private conversations can be done by PM