Can an Infinitely Stiff Rod Defy the Speed of Light in Relativity?

[?????]

There has been so much interest lately in objects going faster than the speed of light, I though I would propose a little thought experiment.

An observer is standing stationary relative to Object A, which is 1x10E10 meters away.

Since there is no relativistic limit on acceleration, suppose the observer accelerates to velocity 2.85 x 10E8 m/s in 0.1 seconds.

At this velocity, v/c = .9524 and (1-(v/c)^2)^.5 = .305. Assume c = 3 x 10E8 m/s.

The instant that the accelerating observer becomes stationary in an inertial reference frame going at velocity 2.85 x 10E8 m/s, Object A is now approximately 3.05 x 10E9 meters away (length contraction). I know Object A is actually closer than that, but I am not factoring this in because the trick I am using doesn't need the distance to be any shorter than 3.05 x 10E9.

Object A has traveled 6.95 x 10E9 meters in 0.1 seconds...or 6.95 x 10E10 m/s.

 

[HallsofIvy]

No, it hasn't because one measurement, the initial distance, and the other, the distance after the acceleration, were measured in different frames of reference. You cannot compare them.

 

[Dickfore]

First of all, if the motion of the observer is uniformly accelerated relative to object A, then the distance covered by the moving observer in those 0.1 seconds would be:

$$s = \frac{a t^{2}}{2} = \frac{v_{f} t}{2} = \frac{2.85 \times 10^{8} \, \frac{ \mathrm{m} }{ \mathrm{s} } \times 0.1 \mathrm{s} }{2} = 1.425 \times 10^{7} \, \mathrm{m}$$

This distance is smaller than the inital distance, so the observer had not flown by object A while accelerating.

Second of all, it is true that the distance is contracted by a Lorentz factor of $\frac{1}{\gamma}$, but time is also contracted by the same factor $\frac{1}{\gamma}$, since the proper time is relative to the moving observer and that is the shortest time.

Thus, the speed that the observer would deduce for object A, after he had started moving with constant velocity is:
$$\frac{\frac{1}{\gamma} \, (L - s)}{\frac{1}{\gamma} \, \frac{L - s}{v}} = v$$

in the opposite direction of the direction motion of the observer to the object A.

 

[Dickfore]

Notice that the proper time according to the moving observer while they are accelerating takes:

$$\tau = \frac{t_{0}}{2} \left[ \frac{\arcsin{\beta}}{\beta} + \sqrt{1 - \beta^{2}}\right], \ \beta = v_{f}/c$$

For $\beta = v_{f}/c = (2.85 \times 10^{8} \, \mathrm{m \, s}^{-1})/(3.00 \times 10^{8} \, \mathrm{m \, s}^{-1}) = 0.950$, the factor is:

$$\tau = 37.9 \, t_{0}$$

 

[?????]

No, it hasn't because one measurement, the initial distance, and the other, the distance after the acceleration, were measured in different frames of reference. You cannot compare them.

Yes, the measurements are made in two different frames of reference. Does this solve the puzzle?

Suppose the observer and Object A are fastened to the end of a pole and they undergo the same acceleration as a unit. When they arrive in the moving reference frame, they will now be 1 x 10E10 meters apart. They started out 3.05 x 10E9 meters apart as seen by this moving frame and now have increase their spacing by 6.95 x 10E9 meters in a time of around .3 seconds.

 

[ghwellsjr]

Yes, the measurements are made in two different frames of reference. Does this solve the puzzle?

Combining measurements made in two different Frames of Reference is a good way to create puzzles, not solve them. You need to describe and analyze your entire scenario using just one FoR. The idea that each observer and object "owns" their own rest FoR is just wrong. It's OK to analyze your scenario from each of these FoRs, but you need to do them one at a time and use the Lorentz Transform, not guess-work, to get from one frame to the other.

 

[?????]

Notice that the proper time according to the moving observer while they are accelerating takes:

$$\tau = \frac{t_{0}}{2} \left[ \frac{\arcsin{\beta}}{\beta} + \sqrt{1 - \beta^{2}}\right], \ \beta = v_{f}/c$$

For $\beta = v_{f}/c = (2.85 \times 10^{8} \, \mathrm{m \, s}^{-1})/(3.00 \times 10^{8} \, \mathrm{m \, s}^{-1}) = 0.950$, the factor is:

$$\tau = 37.9 \, t_{0}$$

I am not familiar with this equation. In my original post above, I assume that the .1 second is the time seen by the accelerating observer. The inertial references frames will therefore see an acceleration time of .328 seconds. The way this is calculated is given in the attachment to this post.

 

[Dickfore]

I will not download any .doc files. If you can export it in PDF, fine. Otherwise, I'm afraid I cannot open it.

If the observer had accelerated with constant accelerating relative to a co-moving inertial reference frame at any instant, then, the acceleration relative to a fixed inertial reference frame is:

$$a \equiv \frac{d v}{d t} = \frac{d \left( \frac{v' + V}{1 + \frac{v' V}{c^{2}}} \right)}{d \left( \frac{t' + \frac{V x'}{c^{2}}}{\sqrt{1 - \frac{V^{2}}{c^{2}}}}\right)} = \frac{d v'}{d t'} \frac{(1 - V^2/c^2)^{3/2}}{(1 + v' V / c^2)^3}$$

In the co-moving frame $v' = 0$, $a' = d v'/d t' = a_{0}$, and $V = v$, so:
$$a = \frac{d v}{d t} = a_{0} \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}$$

If we want to find the derivative w.r.t. proper time, we need to remember that:
$$dt = \frac{d \tau}{\sqrt{1 - v^2/c^2}}$$
so:
$$\frac{d v}{d \tau} = \frac{d v}{d t} \frac{d t}{d \tau} = a_{0} \left( 1 - \frac{v^{2}}{c^{2}} \right)$$

This equation may be integrated:
$$d \tau = \frac{d v}{a_{0} (1 - v^2/c^2)}$$
$$\tau = \frac{1}{a_{0}} \, \int_{0}^{v}{d \bar{v} \frac{1}{1 - \bar{v}^2/c^2}}$$
The intergral over velocities is most easily done if we make the hyperbolic trigonometric substitution:
$$\frac{\bar{v}}{c} = \tanh{p} \Rightarrow d \bar{v} = c \, \mathrm{sech}^{2}{p} \, dp, 1 - \bar{v}^2/c^2 = 1 - \tanh^{2}{p} = \mathrm{sech}^{2}{p}$$
and we have:
$$\tau = \frac{c}{a_{0}} \, \tanh^{-1}{ \left( \frac{v}{c} \right)} \Rightarrow v = c \tanh{\left( \frac{a_{0} \tau}{c} \right)}$$

The time elapsed by the stationary frame is:
$$dt = \frac{d \tau}{\sqrt{1 - \frac{v^2}{c^2}}} = \cosh{\left( \frac{a _{0} \tau}{c} \right)} \, d\tau \Rightarrow t = \frac{c}{a_{0}} \, \sinh{\left( \frac{a_{0} \tau}{c} \right)}$$
[/tex]

The distance traveled during this acceleration is:
$$dx = v \, dt = c \tanh{\left(\frac{a_{0} \, \tau}{c} \right)} \, \cosh{\left( \frac{a_{0} \tau}{c} \right)} \, d\tau = c \, \sinh{\left( \frac{a_{0} \tau}{c} \right)} \, d\tau$$

$$x = \frac{c^{2}}{a_{0}} \left[\cosh{\left( \frac{a_{0} \tau}{c} \right)} - 1 \right]$$

Eliminating $a_{0}$ from the relation between velocity and proper time, we get the following expressions for the elapsed time and distance traveled relative to the fixed frame:
$$t = \tau \frac{\beta}{\sqrt{1 - \beta^{2}} \, \tanh^{-1}{\beta}}$$

$$s = \frac{c \tau}{\tanh^{-1}{\beta}} \, \left[ \frac{1}{\sqrt{1 - \beta^{2}}} - 1 \right]$$

Taking $\beta = 0.950$ and $\tau = 0.1 \, \mathrm{s}$, we get:
$$t = 0.166 \, \mathrm{s}$$

$$s = 3.61 \times 10^{7} \, \mathrm{m}$$

 

[?????]

I will not download any .doc files. If you can export it in PDF, fine. Otherwise, I'm afraid I cannot open it.

Interesting. I think we agree most of the way through. I lose you when it comes time to integrate velocity. See Attached.

 

[ghwellsjr]

The details of how the acceleration takes place have nothing to do with this problem. It won't matter whether the acceleration is constant over the 0.1 second interval or all occurring at any instant of time during that interval. All that matters is the difference in speed between the beginning of the interval and the end of the interval.

You need to pay attention to posts #2 and #6.

 

[?????]

The details of how the acceleration takes place have nothing to do with this problem. It won't matter whether the acceleration is constant over the 0.1 second interval or all occurring at any instant of time during that interval. All that matters is the difference in speed between the beginning of the interval and the end of the interval.

You need to pay attention to posts #2 and #6.

My thought experiments here are examples of one of my favorite types of thought experiments. For those readers who suspected reference frames are involved in the analysis, you get a gold star. I have seen some confusion in a number of posts throughout this website on what a reference frame is. So, I would analyse the experiment the following way:

A reference frame is nothing more than a point of view from which to do an analysis (or a series of calculations). Object A is an element in the stationary and moving reference frames - in fact, it is an element of all reference frames in the universe. Special Relativity is totally concerned with converting results from one reference frame to another, so I would disagree with saying that you can't compare calculations in two reference frames. Of course you can. So I will. Assume the distance traveled by the accelerating observer is not significant to the overall dimensions of this problem. Let's look at the problem from the stationary reference frame. The observer starts off 1 x 10E10 meters from Object A and ends up 1 X 10E10 meters away after the acceleration. Object A never breaks the speed of light. Looking at it from the moving reference frame, the observer starts off 3.05 x 10E9 meters from Object A and ends up 3.05 x 10E9 meters from Object A. There never was a puzzle. The trick I used was to word the experiment as though something happened, but that thing never did happen.

Having said that, the discussion on reference frames shows that I completely bungled getting my real trick to work. Here's what I mean:

The observer starts the experiment 1 x 10E10 meters from Object A and ends up 3.05 x 10E9 meters from Object A. Doesn't this imply that an accelerated reference frame can see a speed faster than the speed of light? The answer: no. Just because the observer is an accelerating reference frame doesn't mean that the laws of physics don't apply. One main foundation of Special Relativity is that the laws of physics apply in all reference frames - and I see no justification for restricting that notion to just inertial frames.

What happened? The key idea in analyzing the experiment from the accelerated reference frame is "failure of simultaneity at a distance." The trick I was trying to use in this example is that I have never found a text that coherently addresses how Special Relativity relates to accelerated reference frames. Certainly, simultaneity would be a challenge for any analysis of acceleration. Some might say that SR only deals with inertial frames. I would say that this was true 100 years ago, but don't you think it's time we moved beyond that limitation?

The same thought applies to my post #5. Although all the analysis in that experiment is done from the moving reference frame, the observer and Object A cannot accelerate simultaneously as seen by this frame, even if they are linked together with a pole. Once again, there's no standard way to deal with accelerating objects as view by moving reference frames.

The two experiments I cite are pretty simple experiments. Their solution should be standard textbook stuff. It appears to me that there is a giant chunk of science missing from current SR theory.

 

[ghwellsjr]

Accelerated reference frames are trickery. That's why you can't find any texts that deal with them.

In your scenario, the observer has accelerated to 0.9524c in 0.1 seconds. Even if the entire acceleration occurred instantly at time zero and actually went to almost c and then dropped back down to 0.9524c after 0.1 seconds, the farthest the observer could have traveled is 3 x 10E7 m and would be no closer to Object A than 1000 x 10E7 - 3 x 10E7 = 997 x 10E7. He's barely moved. I don't know where you're getting these other numbers from.

Look at it this way: it would take light 33.3 seconds to travel 1 x 10E10 meters. In a tenth of a second, it has only gone 1/333 of the distance.

 

[PeterDonis]

The trick I was trying to use in this example is that I have never found a text that coherently addresses how Special Relativity relates to accelerated reference frames.

It would seem that you haven't looked very hard. Take a look at the Wikipedia page on non-inertial reference frames:

http://en.wikipedia.org/wiki/Non-inertial_reference_frame

Note particularly the list of references at the bottom.

Some might say that SR only deals with inertial frames. I would say that this was true 100 years ago, but don't you think it's time we moved beyond that limitation?

Yes. We have. See above. Also try this page on the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

The same thought applies to my post #5. Although all the analysis in that experiment is done from the moving reference frame, the observer and Object A cannot accelerate simultaneously as seen by this frame, even if they are linked together with a pole.

This is just the Bell Spaceship Paradox in another guise. See here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

Or Google it. Or search this forum for previous threads on that topic.

 

[?????]

PeterDonis

Let me further explain my post #11 by reviewing the links you gave. The Usenet Physics FAQ discussion: I see that acceleration is discussed, including the twin paradox, General Relativity and rotating objects. Even some simple equations are given. But this discussion doesn't approach a solution procedure when I ask if an accelerating observer can see Object A go faster than the speed of light (my post #1). The point I was trying to make in post #11 - shouldn't such a simple question have a relatively simple answer?

As I look at the Bell Spaceship Paradox, my feelings are only amplified. Is this really the best solution to such a simple example? This is just more complicated than I feel should be necessary. How would this technique answer my main question from post #1?

I understand that you may prefer the solution techniques presented in the links you gave. This is fine. I am always hoping for a more visual, more intuitive solution technique (dare I say it? - even a more Newtonian feeling solution technique). This is just my personal preference.

 

[PeterDonis]

The point I was trying to make in post #11 - shouldn't such a simple question have a relatively simple answer?

It does: "no".

Bear in mind that my links were not directed at the specific problem you posed, because you admitted that you posed it not to seek an answer but to make a claim that SR somehow does not deal with acceleration. (In post #11 you admit as much, and you give the correct answer to your own question, which is the one I just gave above.) My post was addressed to that claim, which is a very general one, and which I was arguing was false: SR has plenty of methods for dealing with acceleration, but apparently you are not familiar with them. That's ok, but there's no need for trickery to elicit information.

I am always hoping for a more visual, more intuitive solution technique (dare I say it? - even a more Newtonian feeling solution technique). This is just my personal preference.

Have you tried drawing a spacetime diagram of your proposed scenario? That's a pretty basic visual technique, which I use a lot. Plus, it forces you to confront and resolve any ambiguities or vagueness in your formulation of the problem. If you're not familiar with them, I would recommend a basic relativity text, for example Taylor-Wheeler's Spacetime Physics.

 

[?????]

An exact calculation of the speed of Object A relative to the accelerating observer (as defined in my Post #1 above). See attached. I have changed the acceleration period in the attachment to mean that the acceleration of 0.1 seconds is measured relative to the stationary reference frame. This gives a more extreme example of Post #1 experiment and provides a better representation of what is going on.

 

[PeterDonis]

See attached.

A couple of questions/comments:

(1) You say "the accelerated reference frame", and you talk about an observer "at xo in the accelerated reference frame". What do you mean by "accelerated reference frame" and "xo"? Do you mean "the inertial frame moving at velocity $\beta$, in which the accelerated observer is at rest once he stops accelerating"? In which case the location xo would be defined as the spatial location of object A at the instant, in that frame, that the acceleration stops? Or do you mean something else? (As I noted before, drawing a spacetime diagram of the scenario, instead of just the spatial diagrams you drew, forces you to confront and resolve these types of ambiguities in your formulation of the problem.)

(2) You seem to be implying that, if you can calculate a velocity greater than c using anyone's "frame", it's a problem. That's not the case. The rule in SR that "nothing can go faster than light", applied as you're applying it, only holds in inertial frames. The more general form of the rule, which holds even in curved spacetime and for objects in arbitrary states of motion, is simply that all objects with non-zero rest mass must travel on timelike worldlines (i.e., their worldlines must lie within the light cones at each event they pass through). In your example, that appears to be true by hypothesis, so there's no problem.

 

[PeterDonis]

An exact calculation of the speed of Object A relative to the accelerating observer (as defined in my Post #1 above).

You might also want to check out the following page on the relativistic rocket equation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

Looking at the formulas there, I'm not sure all of them in your posted calculation match up, but some of your terminology is a little different. Anyway, it's a useful check.

 

[?????]

A couple of questions/comments:

(1) You say "the accelerated reference frame", and you talk about an observer "at xo in the accelerated reference frame". What do you mean by "accelerated reference frame" and "xo"? Do you mean "the inertial frame moving at velocity $\beta$, in which the accelerated observer is at rest once he stops accelerating"? In which case the location xo would be defined as the spatial location of object A at the instant, in that frame, that the acceleration stops? Or do you mean something else? (As I noted before, drawing a spacetime diagram of the scenario, instead of just the spatial diagrams you drew, forces you to confront and resolve these types of ambiguities in your formulation of the problem.)

Assume the accelerated observer is at the origin of a massless, infinitely stiff framework (which still is subject to length contraction, naturally). When he accelerates, the framework goes with him.

I'm not sure I understand your second question. My calculation shows that even for the extreme acceleration shown, Object A is not near the speed of light. I am (vaguely) aware of the hypothesis you state and agree with it.

 

[PeterDonis]

Assume the accelerated observer is at the origin of a massless, infinitely stiff framework (which still is subject to length contraction, naturally). When he accelerates, the framework goes with him.

There's no such thing. The stiffness of materials is limited by special relativity: one way of stating the limit is that the speed of sound in the material can't exceed the speed of light. Your prescription would require an infinite speed of sound, so the information that a certain part of the framework has started to move could propagate infinitely fast; that's the only way the framework could maintain its structure as it accelerates. That's not physically possible, even in principle, so it's not allowed even in a thought experiment. So this definition of "accelerated frame" won't work.

I'm not sure I understand your second question. My calculation shows that even for the extreme acceleration shown, Object A is not near the speed of light. I am (vaguely) aware of the hypothesis you state and agree with it.

It looked to me like at least one of the calculated velocities you quoted in your PDF was greater than 3 x 10^8 meters per second. I may have misread. If you agree with what I stated about worldlines staying within the light cones, that's good.

 

[?????]

1. Yes, you are quite right. I agree with your statement "the inertial frame moving at velocity β, in which the accelerated observer is at rest once he stops accelerating? In which case the location xo would be defined as the spatial location of object A at the instant, in that frame, that the acceleration stops?" I become a bit uncomfortable explaining to myself how these other observers in the accelerating frame can actually follow the original accelerating guy at the origin of the frame without some kind of connection. Are they in his frame or are they in their own separate co-accelerating frames? Or, does it matter?

The question you raise has actually occupied much of my time recently. To me, it all comes down to "speed of signals". An infinitely stiff structure would allow signals to instantly go from the origin to some far location faster than the speed of light. I do not like that idea, even if does connect all the accelerating observers together in the same "frame". But I am also uncomfortable saying that signals traveling at the speed of light is a property of spacetime - implied by world lines, gravity waves, LIGO and all that. I accept the current wisdom on that issue, but I do have an interesting experiment that questions if signals traveling at the speed of light violate the Law of Conservation of Momentum. I will present this experiment in a future post.

2. One of the velocities that I calculated in my post was greater than the speed of light. But this is not the velocity of the object. I think of it this way: A single observer cannot measure velocity. He can only measure and event (coordinates in space or clock readings). After the experiment, all the observers get together, compare notes and calculate the velocity. So they all have an input into what the true velocity is. Therefore, I took a shortcut and averaged all the inputs (there were only two of them in my calculation) for the final velocity. I suppose you could add a number of observers between xo and L, then calculate all their clock readings and determine the Object A velocity as it moves from observer to observer. I have not done this, but I feel somewhat confident that Object A will still not go faster that c.

 

[ghwellsjr]

...I think of it this way: A single observer cannot measure velocity. He can only measure an event (coordinates in space or clock readings). After the experiment, all the observers get together, compare notes and calculate the velocity. So they all have an input into what the true velocity is. Therefore, I took a shortcut and averaged all the inputs (there were only two of them in my calculation) for the final velocity. I suppose you could add a number of observers between xo and L, then calculate all their clock readings and determine the Object A velocity as it moves from observer to observer. I have not done this, but I feel somewhat confident that Object A will still not go faster that c.

You have been told that you can't compare measurements made from different frames and yet you say:

A reference frame is nothing more than a point of view from which to do an analysis (or a series of calculations). Object A is an element in the stationary and moving reference frames - in fact, it is an element of all reference frames in the universe. Special Relativity is totally concerned with converting results from one reference frame to another, so I would disagree with saying that you can't compare calculations in two reference frames. Of course you can. So I will.

I think you are missing the point with regard to what a Frame of Reference is. It appears to me that you are thinking that a FoR represents what different observers see and measure of the universe around them and that is why it makes sense to you to have more observers, each with their own FoR, and for them to take those different observations and measurements and combine them by averaging to get a better measurement of what's really going on.

What you are missing is that each FoR is a different definition of the space and time in the universe, it's not a different measurement and it has no bearing on what observers in that FoR will observe or measure. The distance between two events in one frame can be different than in another frame. The time interval between two events in one frame can be different in another frame. A speed in one frame can be different in another frame. All because you are using different definitions of spatial distances and time.

Once you have set up a scenario according to your arbitrarily chosen reference frame, like you did in your first post, you have all the information you need to calculate any speed, distance, time interval, or whatever according to the definition provided by you in the description of your scenario. You cannot derive any more information by incorporating additional reference frames, supposedly for additional observers, to get a better understanding of what is true. You have defined all the truth in your original scenario.

It's OK, of course, to take your original scenario defined according to your chosen Frame of Reference and transform all the significant events into a different Frame of Reference using the Lorentz Transform, not guesswork, to see what it looks like to you in another FoR, but this will not have any bearing on what your observers will see or measure.

This is why, by the way, so-called accelerating frames are trickery. Special Relativity describes how to define an inertial Frame of Reference but it has no guidelines, no protocols, no traditions, no suggestions, no rules, no hints, no demands on which inertial Frame of Reference you should use. That's the whole point of SR, all inertial frames are equally valid and equivalent. But when you think in terms of each observer having his own rest frame, you are now thinking that SR is suggesting or demanding that that is what you are supposed to do. But you have to realize that there are an infinite number of rest frames for every observer and SR doesn't prescribe which one is preferred because none of them is. So when you think in terms of an observer starting out at rest in one FoR, accelerating and then coming to rest in another FoR, you are thinking that this is what SR demands, which it is not. You can come to all kinds of strange conclusions when you do this. Now when you also think that the accelerating observer is also at rest in a non-inertial FoR (really a series of inertial FoRs), you are implying again that there is just one such way to do this but there are an infinite number of FoRs at each instant of his acceleration in which he is at rest and SR offers no guidelines on which one to use, it's up to you. Which means it's all trickery.

 

[PeterDonis]

What you are missing is that each FoR is a different definition of the space and time in the universe

This is a very important point! I was trying to figure out how to say something like this, and you nailed it.

But when you think in terms of each observer having his own rest frame, you are now thinking that SR is suggesting or demanding that that is what you are supposed to do. But you have to realize that there are an infinite number of rest frames for every observer and SR doesn't prescribe which one is preferred because none of them is.

However, I think one has to be careful phrasing statements like this. If an observer never accelerates (i.e., he remains inertial--weightless--for his entire existence), then there *is* a single inertial frame in which he is always at rest. In the standard usage of the term "rest frame", there *would* be only one unique rest frame for that observer, not an infinite number.

Acceleration complicates things because if an observer is accelerating, then at each event on his worldline, he is at rest in a *different* inertial frame. (I'm assuming for now that spacetime is flat, so the notion of global inertial frames makes sense, and we can meaningfully talk about the same inertial frame at different events. Curved spacetime adds a new set of complications.) However, it is still true that, at each event, there is only *one* "instantaneous rest frame" for the observer if by that one means an inertial frame in which the observer is, for an instant, at rest (the standard term is "momentarily comoving inertial frame", or MCIF). There are not an infinite number of "rest frames" in this sense at a single event. The infinite number only comes in because the MCIF changes from event to event, and there are an infinite number of events on the observer's worldline.

So when you say this:

there are an infinite number of FoRs at each instant of his acceleration in which he is at rest and SR offers no guidelines on which one to use, it's up to you.

I think what you mean to say is not that there are an infinite number of MCIFs at a single event, but only that, if we abandon the attempt to construct a single *inertial* global "rest frame" for an accelerating observer (because it can't be done), and instead just try to construct *some* global "frame" in which the accelerating observer's worldline is always at the origin (so in a sense he is "at rest" in it), there are an infinite number of ways of doing that; or, even if we just try to agree on what the instantaneous "frame" looks like at a single event, for an accelerating observer, if we abandon the restriction that it must be an MCIF, there are an infinite number of ways of doing *that*. I agree with all this; I just think it's important to be precise about what we are talking about when we talk about frames of reference, since that term is used in different and incompatible ways.

 

[PeterDonis]

I agree with your statement "the inertial frame moving at velocity β, in which the accelerated observer is at rest once he stops accelerating? In which case the location xo would be defined as the spatial location of object A at the instant, in that frame, that the acceleration stops?"

Yes, you could use this definition of "the location xo". But it's important to understand that that definition makes sense only relative to that particular inertial frame. You can't, for example, take the distance xo in the "end" inertial frame (the one in which the accelerated observer is at rest after he stops accelerating), mark it off from the observer's location in the "start" inertial frame (the one in which he was at rest before he started accelerating), and call that "location xo" in the start frame. More precisely, you can do that, but it won't have any physical meaning; there is no natural "connection" between those two locations.

I become a bit uncomfortable explaining to myself how these other observers in the accelerating frame can actually follow the original accelerating guy at the origin of the frame without some kind of connection.

They can't. Either they have to wait until the information about his acceleration gets to them, or they have to all have some kind of prearranged scheme for starting their accelerations in order to maintain their relative positions. But even that is complicated; Google "Born rigid acceleration" if you want to get an idea of the issues involved.

Are they in his frame or are they in their own separate co-accelerating frames? Or, does it matter?

As ghwellsjr's post made clear, you appear to be thinking of "frame" the wrong way. There are ways for the accelerating observers to all use a common "frame", meaning a common convention for assigning space and time coordinates to events. But that's a choice; there's nothing in the physics that puts them into a particular "frame".

But I am also uncomfortable saying that signals traveling at the speed of light is a property of spacetime - implied by world lines, gravity waves, LIGO and all that.

Don't get hung up on the term "speed of light". The property of spacetime is only that there is an invariant speed: if any object moves at that speed relative to one observer, it moves at that speed relative to all observers (more precisely, relative to all observers who see each other as moving at less than that speed). The fact that light moves at the invariant speed is a separate physical fact about light, not a property of spacetime.

 

[ghwellsjr]

However, I think one has to be careful phrasing statements like this. If an observer never accelerates (i.e., he remains inertial--weightless--for his entire existence), then there *is* a single inertial frame in which he is always at rest. In the standard usage of the term "rest frame", there *would* be only one unique rest frame for that observer, not an infinite number.

Let's take, for example, the scenario from the Original Post:

"An observer is standing stationary relative to Object A, which is 1x10E10 meters away."​

In the interest of brevity, I will describe the coordinates of the Observer and Object A as [t,x] since all motion is along one line and I will use units of seconds for time and units of billions of meters for distance. Here are the events for six of the infinite number of Frames of Reference that we could use to describe this starting point:

FoR#    Observer    Object A
  1      [0,0]       [0,10]
  2      [0,0]       [0,-10]
  3      [0,10]      [0,0]
  4      [0,-10]     [0,0]
  5      [0,-5]      [0,5]
  6      [0,5]       [0,-5]

Note that all of these have a starting time of zero, but it would be just as valid to use any other number, including a negative number for the time.

Now which one, if any, of these is the "unique rest frame for that observer" according to "the standard usage of the term 'rest frame'"? And where is this standard defined?

 

[?????]

You have been told that you can't compare measurements made from different frames and yet you say:

I think you are missing the point with regard to what a Frame of Reference is. It appears to me that you are thinking that a FoR represents what different observers see and measure of the universe around them and that is why it makes sense to you to have more observers, each with their own FoR, and for them to take those different observations and measurements and combine them by averaging to get a better measurement of what's really going on.

What you are missing is that each FoR is a different definition of the space and time in the universe, it's not a different measurement and it has no bearing on what observers in that FoR will observe or measure. The distance between two events in one frame can be different than in another frame. The time interval between two events in one frame can be different in another frame. A speed in one frame can be different in another frame. All because you are using different definitions of spatial distances and time.

Once you have set up a scenario according to your arbitrarily chosen reference frame, like you did in your first post, you have all the information you need to calculate any speed, distance, time interval, or whatever according to the definition provided by you in the description of your scenario. You cannot derive any more information by incorporating additional reference frames, supposedly for additional observers, to get a better understanding of what is true. You have defined all the truth in your original scenario.

It's OK, of course, to take your original scenario defined according to your chosen Frame of Reference and transform all the significant events into a different Frame of Reference using the Lorentz Transform, not guesswork, to see what it looks like to you in another FoR, but this will not have any bearing on what your observers will see or measure.

This is why, by the way, so-called accelerating frames are trickery. Special Relativity describes how to define an inertial Frame of Reference but it has no guidelines, no protocols, no traditions, no suggestions, no rules, no hints, no demands on which inertial Frame of Reference you should use. That's the whole point of SR, all inertial frames are equally valid and equivalent. But when you think in terms of each observer having his own rest frame, you are now thinking that SR is suggesting or demanding that that is what you are supposed to do. But you have to realize that there are an infinite number of rest frames for every observer and SR doesn't prescribe which one is preferred because none of them is. So when you think in terms of an observer starting out at rest in one FoR, accelerating and then coming to rest in another FoR, you are thinking that this is what SR demands, which it is not. You can come to all kinds of strange conclusions when you do this. Now when you also think that the accelerating observer is also at rest in a non-inertial FoR (really a series of inertial FoRs), you are implying again that there is just one such way to do this but there are an infinite number of FoRs at each instant of his acceleration in which he is at rest and SR offers no guidelines on which one to use, it's up to you. Which means it's all trickery.

A good comment. Going back to my original post, the question I was really asking has to do with how you define velocity. Any accelerating observer will have various far off points in the universe length-contracting towards him. Is this length contraction actually velocity? Is Object A moving during the acceleration (due to length contraction)? It starts off at one relative location, and after the acceleration, ends up at another relative location. Not only that, but both of these relative locations are some distance away from the accelerating observer, at the start and the end of the acceleration. Because Object A is "way over there", does that mean that it's velocity is not relative to the accelerating observer?

So, the discussion has developed into one about reference frames. Because Object A is "way over there", this is a relevant point. The only way to determine what "way over there" means is to define a reference frame so that you can go measure events surrounding Object A. I agree with what you are saying about reference frames.

The real question for me is "Where is Object A before and after the acceleration?" Let's propose another experiment. An observer is at the origin of an inertial reference frame and Object A is located at coordinate xo and is traveling toward the observer at speed .9524c. Now you can calculate many things about the Object A position relative to the observer without any problems. Isn't this the same experiment as my acceleration experiment after the acceleration is over? Saying that reference frame is important is certainly something that should be discussed. But Object A is still at position xo after the acceleration, no matter how you define the reference frame. The whole point of my original post was to ask a question about whether length contraction is actually "velocity" for an accelerating observer.

 

[PeterDonis]

Going back to my original post, the question I was really asking has to do with how you define velocity.

For accelerating observers, there is no well-behaved definition of "velocity" for distant objects. Only within a small local region of spacetime around a given event, i.e., in a MCIF, can you apply a workable concept of "velocity", and that, of course, only works for objects that are within that small local region of spacetime.

The real question for me is "Where is Object A before and after the acceleration?"

And since you've said you agree with what's been said about reference frames, you should realize that there is no well-defined answer to this question. Now here's another question: what actual physical phenomenon depends on having a well-defined answer to this question? The answer is, none. In other words, the fact that Object A has no well-defined "position" to an accelerating observer has no actual physical consequences, so it's OK.

Let's propose another experiment. An observer is at the origin of an inertial reference frame and Object A is located at coordinate xo and is traveling toward the observer at speed .9524c. Now you can calculate many things about the Object A position relative to the observer without any problems.

Yes, as long as spacetime is flat. If spacetime is curved, as I noted before, even a freely falling observer will run into problems trying to define a global "reference frame" that behaves the way your intuition says it ought to. But we don't have to go into that here.

Isn't this the same experiment as my acceleration experiment after the acceleration is over?

Not if you try to assign a meaning to the measurements in that "ending" inertial reference frame, that applies to the observer while he is still accelerating.

The whole point of my original post was to ask a question about whether length contraction is actually "velocity" for an accelerating observer.

The answer to this question is "no". And as I noted above, since there are no actual physical consequences to this "length contraction", that answer causes no problems.

 

[pervect]

Suppose you're in the end phase of braking after a relatiavistic approach. Here are some semi-rhetorical questions to think about.

1) Would a physically meaningful definition of velocity ever say that the "braking" maneuver actually causes your destination to move away, so it gets more distant?>

2) Does the proposed definition result in your destination getting further away as you brake? (To avoid being to vague, I'll drop a big hint and suggest that it does, though I don't have time to get into the details).

3) Is the proposed definition supported by any textbooks? (Actually, I'm not sure if the textbooks talk about this sort of basic issue. I'm not sure if it's an omission, or through caution. Maybe someone can come up with a textbook reference that does talk about it.)

4) Is the proposed definition based on geometric objects, or is it coordinate dependent?

5) Is it really the best definition to use, or are the results so screwy that you'd be better off with a different one?

 

[ghwellsjr]

A good comment. Going back to my original post, the question I was really asking has to do with how you define velocity. Any accelerating observer will have various far off points in the universe length-contracting towards him. Is this length contraction actually velocity?

No.

Is Object A moving during the acceleration (due to length contraction)?

Not in your original frame of reference and it's not changing its motion in your final frame of reference.

It starts off at one relative location, and after the acceleration, ends up at another relative location. Not only that, but both of these relative locations are some distance away from the accelerating observer, at the start and the end of the acceleration. Because Object A is "way over there", does that mean that it's velocity is not relative to the accelerating observer?

Object A has no velocity as a result of describing its position as defined by two different Frames of Reference. Stick with one FoR. It may have a velocity in some reference frames, but not because of the acceleration of something else.

So, the discussion has developed into one about reference frames. Because Object A is "way over there", this is a relevant point. The only way to determine what "way over there" means is to define a reference frame so that you can go measure events surrounding Object A. I agree with what you are saying about reference frames.

The only reason we are discussing reference frames is because you brought them up in your first post. Your observer doesn't need to use any reference frame to determine how far away Object A is. He can measure it by sending a radar signal to it and seeing how long it takes for the echo to get back to him. If it's 10 billion meters away, it will take 66.666 seconds for the radar signal to make the round trip. When we use reference frames, we don't measure things, we calculate them. Reference frames can help us determine what the observers in our scenarios might measure, but these observers cannot take advantage of the knowledge we have that the distance is 10 billion meters, at least not legitimately.

The real question for me is "Where is Object A before and after the acceleration?"

It's in the same place it always was in the initial inertial reference frame and it is traveling at a constant speed in the final inertial reference frame, both before, during and after the acceleration. Let's look at these two frames in a little more detail:

In the initial inertial reference frame, you had the observer and Object A at rest with each other and separated by 10 billion meters which is 33.333 light seconds. Then, at some point in time, you had the observer accelerate toward Object A and achieve a speed of 0.9524c. Object A has not moved. Since the distance is 33.333 light seconds away according to the frame, it will take 33.333/0.9524 seconds or 35 seconds to make the trip.

However, according to the observer's clock which is now time dilated by a factor of 0.305, it will take 10.67 seconds. Also, the length contracted distance is 10.166 light seconds and the speed will calculate out to be the same value.

Now let us look at the final inertial reference frame. To be rigorous, we should use the Lorentz Transform to establish the coordinates of the events in the new frame but since it's simple, I will just tell you the salient points (which are quite similar to the initial frame). First off, this final frame is traveling at 0.9524c with respect to the initial one. This means that distances will be length contracted along the direction of motion by the factor of 0.305. That means the initial distance between the observer and Object A is 10.166 light seconds. Remember, they have both been traveling this distance apart for some time.

Now, all of a sudden, well, actually over the course of fraction of a second, the observer comes to a stop in this reference frame while Object A continues to move in the reference frame toward the observer at 0.9524c. Note that there has been no sudden jumping around of Object A, it's a smooth transition in this frame just like in the initial frame. The length contracted distance between the observer and Object A merely starts getting shorter than it was before. We can now use the numbers from the initial frame for the observer to apply to Object A in the final frame to see that it moves 10.166 light seconds in 10.67 seconds.

One way for the observer to calculate his speed relative to Object A during his trip instead of waiting until the trip is over is to use Relativistic Doppler by comparing the tick rate of a clock co-located with Object A compared to his own clock. Since we know his speed, we can calculate the Relativistic Doppler Factor and that tells us what the observer will measure and from that, he can determine his speed.

Let's propose another experiment. An observer is at the origin of an inertial reference frame and Object A is located at coordinate xo and is traveling toward the observer at speed .9524c. Now you can calculate many things about the Object A position relative to the observer without any problems. Isn't this the same experiment as my acceleration experiment after the acceleration is over? Saying that reference frame is important is certainly something that should be discussed. But Object A is still at position xo after the acceleration, no matter how you define the reference frame. The whole point of my original post was to ask a question about whether length contraction is actually "velocity" for an accelerating observer.

This is confusing because you say that Object A is located at coordinate xo and is traveling and then later you say Object A is still at position xo after the observer accelerates implying that Object A is not traveling.

Anyway, I hope my previous comments will clarify the issue so you won't need another experiment.

 

[?????]

PeterDonis

Originally Posted by ?
Assume the accelerated observer is at the origin of a massless, infinitely stiff framework (which still is subject to length contraction, naturally). When he accelerates, the framework goes with him.

Your Reply:
There's no such thing. The stiffness of materials is limited by special relativity: one way of stating the limit is that the speed of sound in the material can't exceed the speed of light. Your prescription would require an infinite speed of sound, so the information that a certain part of the framework has started to move could propagate infinitely fast; that's the only way the framework could maintain its structure as it accelerates. That's not physically possible, even in principle, so it's not allowed even in a thought experiment. So this definition of "accelerated frame" won't work.

I don't agree with your statement. Assuming a massless, infinitely stiff reference framework has no effect on the calculation of the experiment. See attachment.