# Special relativity and accelerated frames

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1. Sep 15, 2014

### CKH

SR directly addresses only inertial frames. Can SR be applied to analyze accelerated frames?

For example, consider a point on the rim of a rotating disc. I have see an analysis which makes claims by assigning an associated, instantaneous inertial frame comoving with the point at each position. My question is whether this can be used correctly since the frame of the point is always accelerating toward the center of the disc whereas any comoving inertial frame is not accelerating. Can you justify ignoring the acceleration in such an analysis?

2. Sep 15, 2014

### Staff: Mentor

Special relativity works in any flat (that is, no gravity) spacetime, whether you use an inertial frame or not; that's just a choice of coordinates and is handled as is any other coordinate transformation. Rindler coordinates are an example; google wand a serach of this forum will find some decent examples and explanations. (In fact, Einstein considers a rotating frame in his 1905 paper). Most introductory treatments don't cover this material because the math is more complex but doesn't provide much more insight into the fundamental principles.

You should work through the Rindler solution before you take on circular motion; the latter is appreciably more complicated.

The acceleration is just $dv/dt$, so I'm not sure what you mean by "ignore it"; you calculate it, it is what it is, and it tells you what the instantaneous second derivative of the position is. However, when you're working with instantaneous comoving inertial frames ("ICIF" - the concept is used enough to justify its own acronym) there is one potential pitfall: You cannot blithely transfer values from one ICIF to another, just as you cannot blithely transfer values between the different inertial frames of two observers in constant relative motion. We all know to look for this mistake in constant motion case, but it's easy to get it wrong in the accelerated cases because we don't have different observers in the different frames, we have the same observer at different times.

Last edited: Sep 15, 2014
3. Sep 15, 2014

### WannabeNewton

Why focus on SR? We do the same thing in Newtonian mechanics when defining the rest frame of an observer in uniform circular motion around an origin. It's an instantaneously comoving inertial frame so the only thing we care about is that the velocity of the inertial frame matches the velocity of the accelerating observer at the given instant, hence the term "comoving".

The acceleration is a higher order effect, by which I mean it is the second proper time derivative of the trajectory, and in order to analyze its physical effects a single instantaneously comoving inertial frame is not enough. You need a continuous sequence of such frames, one for each instant of time on the clock carried by the accelerating observer. This is how the rest frame of the observer is defined: it is a continuous family of instantaneously comoving inertial frames.

After we have this one-parameter family we can start looking at second order effects like acceleration because we can differentiate velocity given this family. When we only have one instantaneously comoving frame we clearly cannot differentiate velocity and the effects of acceleration cannot be analyzed but that does not mean they are ignored. This is the same in SR and in Newtonian mechanics.

4. Sep 15, 2014

### CKH

I will look up the "Rindler solution". I think I understand your response.

My question is about a problem I see in attempting to derive properties (such as time and distance) in an accelerated frame (relative to a particular IRF) by appealing to a series of ICRFs where we know how time and distance behave relative to that IRF.

The problem I see is that the ICRFs are unlike the accelerated frame at every instant. Although the ICRFs have infinitesimal motion relative to the accelerated frame, they always have a substantial (non-infinitesimal) difference in acceleration. SR itself doesn't say anything about acceleration.

I'm asking: Can you ignore this difference in acceleration when using ICRFs to analyze properties of the accelerated frame? In other words, is it nevertheless valid to analyze an accelerated frame using incremental ICRFs and integratomg? If so, how do you justify ignoring the difference in acceleration in formulating infinitesimals from ICRFs and integrating?

You mention that you have to be careful in transferring values (e.g. in order to do an integration?).

5. Sep 15, 2014

### CKH

Because in Newtonian mechanics time are space are absolute, that is, independent of motion. This is not true in SR. In Newtonian mechanics, we do not have scratch our heads trying to figure out how much time passes on a moving clock while in arbitrary motion.

This is where I see a problem. No instantaneously comoving inertial frame can be said to be perfectly equivalent to the accelerating frame (at that instant) because the acceleration is still different in the two frames, even though they are instantaneous comoving.

So if acceleration has some effect in and of itself, you cannot equate these instantaneous frames.

6. Sep 15, 2014

### pervect

Staff Emeritus
The mathematical techniques needed to deal rigorously with arbitrary coordinate systems (which includes coordinate systems associated with an accelerated observer as a special case) are typically taught along with GR. But they are just mathematical techniques, there isn't any physical content to them, any more than there is physical content in going from cartesian (x,y,z) coordinates to spherical (r, theta, phi) coordinates.

Thus we have the situation that SR + the mathematical techniques needed to rigorously deal with arbitrary coordinate systems can handle accelerated frames.

However, a typical student who has been taught a typical SR course will be lacking the mathematical techniques to deal with arbitrary coordinates in a systematic manner. This falls in the general category of tensor analysis.

As long as the quantities used are tensor quantities, the acceleration of the local frame will not affect the value of any tensor quantities. Therefore it is a perfectly valid technique when one analyzes the situation with tensors.

However, the exact definition of tensors, and also the defintion of local frames (as opposed to global ones) are part of the mathematical apparatus of tensor analysis, something that the typical SR student does have at their disposal.

One can find early treatments of some problem that don't use tensor analysis. Many of these can be useful to students who study and observe the techniques uses. For the student that is always asking "why" and wanting to do things their own way rather than study the literature and observe, if the student gets it wrong, the lack of rigor in the non-tensor analysis may not convice the sutdent. In the worst case the student makes a mistake, isn't convinced by the paper, and goes off on a more-or-less crackpot course of insisting that the published papers are all wrong.

Of course, things aren't always this simple, while published papers have a higher chance of being right than wrong, having gone through some sort of peer review process, there are still plenty of wrong published papers as peer review is far from perfect.

7. Sep 15, 2014

### CKH

Thanks. Seeing that the general problem involves tensors (which I'm not familiar with), can you just answer this simpler question?

Suppose I have an accelerating clock observed from some IRF. Will the instantaneously observed rate of the clock depend only on the instantaneous velocity (using LT) or does the acceleration also enter into the clock's rate (relative to the observer)?

I ask in part because in GR gravitational acceleration is supposed to affect clock rates.

8. Sep 15, 2014

### Staff: Mentor

If the clock is constructed in such a way that the force accelerating it won't disturb it (a pendulum clock would be a bad choice, an atomic clock probably pretty good) the acceleration will not affect it.

There is no gravitational acceleration in GR. A clock deeper in a gravity well will dilate relative to a clock not so deep, but that has nothing to do with acceleration. It's a result of spacetime curvature, which is a property of the spacetime you're in and not the frame (whether inertial or accelerated) that you use to assign coordinates to events in that spacetime.

9. Sep 15, 2014

### CKH

In other words, the LTs depend only on velocity so higher order derivatives are irrelevant. This is now obvious to me.

Hence we can use ICRFs to do path integrals to solve problems involving any motion.

I think I had the impression that acceleration had some effect on clocks from a misunderstanding of GR.

Thanks for pointing out a misunderstanding. I have to learn the right way to look at it.

Einstein points out that an observer in a black box can't tell whether he is on the earth or accelerating at g somewhere far away. That gives the idea that the observer on earth is in fact accelerating wrt to some frame. Would that be the IRF of a free falling object near him?

10. Sep 15, 2014

### Staff: Mentor

Yes.

11. Sep 15, 2014

### pervect

Staff Emeritus
There's a good FAQ on this at http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html, which also points out that my first post was , errr, wrong. It's not just a question of tensor analysis, there is some physical content to the question.

But lets get to the answer. The answer is that the acceleration of an ideal clock does not affect it's ticking rate as observed in an inertial frame.

For some experimental tests of this see http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis

I believe there are some remarks about this issue in the textbook "Gravitation" as well. I recall reading them but couldn't find the exact section to refresh my memory. My recollection is that the authors noted that even the enormous acceleration due to the strong nuclear force between protons in the atomic nucleus had no effects on timekeeping.

Note, however, that while acceleration of a clock does not affect it's rate as observed by an inertial observer, if we accelerate both the clock AND the observer, the result is no longer true. If we insist that our observer remain inertial, however, we can say that acceleration does not affect the rates of clocks.

12. Sep 16, 2014

### CKH

I'm not sure how that case is different. Naively, at any instant, the accelerating observer can assign a rest ICRF to himself and another ICRF to the clock, measure the velocity of the clock and then calculate the clock rate (as seen by the observer) at that instant using the LT. So, it shouldn't matter that the observer is accelerating.

Can you explain what you mean?

13. Sep 16, 2014

### jbriggs444

The synchronization convention in use by the observer's ICRF at one instant and by the ICRF at the next instant is constantly changing due to the acceleration. This affects the sequence of time coordinates ascribed to the ticks of the observed clock.

The observed clock's "tick rate" in the accelerating frame is judged based on the delta between the time coordinates ascribed for each tick.

Last edited: Sep 16, 2014
14. Sep 17, 2014

### CKH

Yes, but 1) the clock hypothesis is not violated, and 2) the apparent changes in the behavior of the clock versus an observer in an IRF are entirely due to the observer's acceleration, right? Moreover, the observer knows his own acceleration vector at every instant. So, it seems that he can still calculate the clock's behavior without violating the assumption that the clock's rate is independent of it's own acceleration (the clock hypothesis), just like an observer in an IRF.

It's just harder to analyze, right?

Say at $t_0$ (in the observer's frame) the observer selects his instantaneous ICRF as a base IRF. He can thereafter measure his own motion relative to that base IRF, using an accelerometer. Can't he then transform his measurements of the clock's path and rate to that IRF and realize that all variations in the clock rate are due to both his motion and the clock's motion in that base IRF?

From that perspective, I don't understand why we should claim that there is anything special about the case where the observer is accelerating versus when he is in an IRF, except that the calculations are different because the observer's velocity is changing.

15. Sep 17, 2014

### pervect

Staff Emeritus
If you are careful to always use inertial frames in your analysis (as you did with your MCIF), then there is in fact no problem. So the caution would be unnecessary.

In the FAQ I quoted earlier, what I was trying to say would be addressed by the following section.

A minor note of clarification. The astronauts who can't see outside are applying the equivalence principle when they say that they think they are accelerating at 1g in deep space. They are saying that their observations should be "equivalent". This is fairly obvious from context but wasn't explicitly stated.

16. Sep 18, 2014

### CKH

OK, I think I understand you correctly now. An observer's acceleration means his velocity is changing, so what he measures about some independent clock is affected by his own motion. What threw me off was this:

If I understand correctly, whether acceleration of a clock (which is absolute) affects the rate of the clock is not observer dependent; it is just a fact that acceleration doesn't affect the clock (only relative velocity alters the clock's rate from any observer's point of view). An accelerating observer who thinks that the clock's own acceleration does affect the observed clock rate, is mistaken and hasn't analyzed the problem correctly.

Do you agree?

Your response above seemed to make some exception to the clock hypothesis, but i don't think that was your intention.

You have left me wondering about the issue of clock rates in a declining gravitational field such as at higher altitudes above the earth. A clock farther from the earth (e.g. on a tower) runs faster than one on the earth? It is true that their accelerations are different, but that shouldn't matter. Moreover we know that the difference in rates is constant. So the only SR conclusion I can see is that ICRFs of the clocks are in constant motion relative to one another.

Perhaps this constant relative motion is the same as the velocity of an object dropped from the tower when it reaches the earth?

But that doesn't seem to be working very well. Where is the symmetry of clock rates here? Do observers in the tower and on earth agree on which clock is ticking faster?...

I need to sleep. If you have any hints I'd appreciate them.

17. Sep 18, 2014

### pervect

Staff Emeritus
A quick summary of my position: proper time is observer independent, and coordinate time isn't. It is perhaps unclear what is meant by "the rate of the clock" without talking about this distinction. I make certain unstated assumptions by what I mean, but you, as a reader , may be making different unstated assumptions :(. So I think we need to disentangle the two sorts of time.

Proper time is the sort of time an idealized clock measure. As a non-idealized realization of proper time, we can use the NIST (National institute of standards) defiintion of the second. The second is defined as a certain number of a specified atomic transition regardless of any of these other factors such as where you are. what the gravitational potential is. how fast you are moving, or how much you are accelerating. The proper second, or its realization via the NIST standard second, are independent of the observer.

Coordinate time is something that we build up on top of proper time. To define coordinate time, we need to define the notion of simultaneity. And as Einstein's train thought experiment shows (I hope you are familiar with this - I don't want to digress to explain it, but let me know if explanation is needed) the notion of simultaneity is observer dependent. This observer dependence creeps into the notion of coordinate time, which is also observer dependent.

Given a reference clock, and a notion of simultaneity, we can define a coordinate time for any specified event covered by our coordinate system as the proper time reading on our reference clock which is simultaneous with our specified event.

If two clocks are not at the same location in space-time, we need to use coordinate time to compare them, because we need the notion of simultaneity to define what "now" is for the distant clock. This means that observer dependence is introduced (subtly) when we talk about "observers" and/or "frames". The observer or frame has a proper clock (which is independent of the observer), but it also has a notion of simultaneity, which is dependent on the observer.

If we use only proper time, there isn't any way to meaningfully compare clocks at different locations, we are lacking the necessary definition of simultaneity. Using proper time we can only say "all clocks tick at 1 second per second", which we operationally define as xxxx transitions of the yyyyy atomic transition, so there isn't any concept of clocks having "different rates".

Thus, when we talk about comparing clocks, I implicitly assume that we are talking about comparing them via the observer dependent coordinate time.

Defintely not my intention, I hope this new response clarifies things.

If you have an actual gravitational field, like that of the Earth, there isn't any coordinate system that represents "an inertial observer". So it's unclear how to analyze the problem without using relativity, because the prescription "use an inertial observer" doesn't have any clear meaning in GR.

Assuming, then that we are willing to use GR, if we wish to be precise, we also need to specify exactly what coordinate system we are using to compare clocks. I will use the TAI coordinate time standard, also known as "atomic time", as described in the wiki http://en.wikipedia.org/wiki/International_Atomic_Time, without going into the details of why this choice is a good one, and how many/most other common choices will yield equivalent results.

Then, using the TAI coordinate time standard, higher clocks definitely tick faster than the lower clocks. In GR the important issue for coordinate clock rate is the gravitational potential. Clocks deeper in a gravity well tick slower than clocks higher in a gravity well according to TAI time.

If you recast the problem using the principle of equivalence to an accelerating spaceship in flat space-time, you do not have to use GR. Inertial observers are clearly available, and you can use them to solve the problem without getting involved in GR at all.

18. Sep 18, 2014

### CKH

Now I understand better what you meant about accelerating observers. Proper time has no relativity of rates and is by definition independent of motion. Observation of rate changes happens only for observers not moving with the clock. The clock hypothesis has to do with other observers of the clock. However, observed acceleration is absolute for all observers (because they can measure the effects of their own acceleration) and therefore the clock hypothesis is indeed observer independent, but the hypothesis is a statement about observers. Maybe I'm rambling here.

Perhaps we can think about accelerated frames by imagining that we have an initial inertial frame in which to synchronize clocks. (It seems difficult to do so in an accelerating frame.) Then as that frame accelerates the Coordinate clocks of the frame remain synchronized. (But not in rotation because the acceleration is not uniform in the frame.)

I'll just accept the effects of a gravitational gradient for now. I am not prepared to understand the mathematics of GR.

BTW, is there a name for uniformly accelerated frames?

19. Sep 18, 2014

### Staff: Mentor

No, they don't. More precisely, there is *no* way to have a frame in which all of the following are true:

(1) There is a family of observers all of whom are "at rest" in the frame (i.e., they are all at rest relative to each other);

(2) All of the observers have nonzero proper acceleration;

(3) All of the observers's clocks stay synchronized.

This is true even if all of the observers are accelerating linearly in the same direction. The (non-inertial) coordinates in which a family of such observers is at rest are called Rindler coordinates; see here:

http://en.wikipedia.org/wiki/Rindler_coordinates

Note that, although all of the observers are at rest in Rindler coordinates, their proper accelerations are not all the same; the ones "higher up" (i.e., further along in the direction of acceleration) have smaller proper accelerations than the ones "lower down". Also, the rates of their clocks are different: the clocks "higher up" run faster than the clocks "lower down". (This is similar qualitatively to what happens in a gravity well, but the variation of clock rate with "height" is different.) This is why their clocks can't stay synchronized.

20. Sep 20, 2014

### CKH

By proper acceleration, I suppose you mean the acceleration that an observer measures with an accelerometer (e.g. by measuring the instantaneous acceleration of a body to which no forces are applied)?

Perhaps this is tangential, but I ask myself, in the case of a rigid body undergoing uniform acceleration at some point in the body, are the proper accelerations of all points in the body constant and equal?

From the point of view of some IRF, the body contracts continuously in the direction of acceleration because it is rigid in its rest frame. So it seems that in an IRF, the coordinate accelerations of all points in a rigid body cannot be the same. (If the velocities in the IRF were the same at all points in the body at all moments, the body could not contract.)

Taking acceleration as absolute (in any IRF), this implies that the proper accelerations cannot be uniform throughout the body. (These differences in proper accelerations are very small for accelerations like g.)

Since there are different velocities among the particles in the body (due to increasing contraction) in the IRF, their clocks go out of sync as well in the IRF (very slightly and only along the direction of the acceleration).

Then, what happens in the Rindler frame of a particle in this accelerating rigid body? If we take an ICFR for that Rindler frame, we are in an instantaneous rest frame for the body. So, no contraction is seen and the clocks remain synchronized. This seems rather nice because if we are in an accelerating spaceship we don't see any distortions of shape nor in the synchronization of clocks (I think). However we can measure a very slight difference in the proper accelerations of the nose of the ship versus the tail! Very strange.

---

In the other case, there is a group of independent objects with the same proper accelerations. Viewed from an IRF (S), the objects have coordinates at some time t0. Because the proper accelerations are equal, the coordinate accelerations are also equal in the IRF (S). Thus the group of objects moves as a rigid configuration (is translated) in the IRF (without any contraction of the group) and their velocities are uniform in IRF (S) at any instant. The clocks on the objects remain in sync by symmetry in S.

Now consider one of these bodies in it's Rindler frame and refer back to the IRF (S) above. Take an ICRF in the Rindler frame at a moment when it is not at rest with S. From this ICRF for the Rindler frame, we see the group of objects as contracted because we are moving relative to S. In the ICRF frame the contraction increases with time as the configuration moves faster. So, the objects are moving because the group is contracting in any ICRF and therefore the clocks aren't synchronized in the Rindler frame.

The two cases of rigid body acceleration versus a uniformly accelerating rigid group appear complementary. It's seems that the changes in proper acceleration in the one case and clock desynchronization in the second case are "second order" effects which are very small except at extreme accelerations.

I've tried to think this through in a qualitative sense without doing all the calculations. Have I've gone off the tracks?