# Accelerated obserevers in special (not general!) theory of relativity

## Main Question or Discussion Point

Hi, folks!

I am interested for the following question:

The original Einstein's formulation of the special theory of relativity does not permit consideration of the accelerated observers, but this limitation is not essential because we can consider the special relativity as limit case of the general relativity by substituing gravitational constant by 0. Is it possible to consider accelerated observers in the special relativity without mathematical language of the general relativity (manifolds, tensors etc.)?

My idea is introducing following additional axiom:

Acceleration is relativistic irrelevant.

Let we consider two birds A and B, flying nonlinearly and nonuniformly with the motion laws A(t) and B(t) in inertial system S (of the ground), where A(t) and B(t) are indefinitely differentiable vector functions of the time. How the bird A see the bird B.

We will fix a moment $$t_0'$$ of the bird's A clock and compute $$t_0$$ such that $$t_0'=\int_0^{t_0}\sqrt{1-(A'(t)/c)^2}\,dt$$. Then we compute the vector $$v=A'(t_0)$$ and fix the Innertial system S' such that S' has constant velocity $$v$$ relative to system S. Then we transform coordinates $$(A(t_0),t_0)$$ of bird A to coordinates $$(a',t_0')$$ in system S' and coordinates $$(B(t_1),t_1)$$ of bird B to coordinates $$(b',t_1')$$. We compute moment $$t_1$$ using condition $$t_1'=t_0'$$, and position $$b'$$ using computed moment $$t_1$$. Then, intensity of vector $$b'-a'$$ is distance bitween two birds measured by bird A when it scan the time $$t'$$ from it's clock.

Is it correct?

Related Special and General Relativity News on Phys.org
Special relativity can deal with accelerations just fine, you don't need GR for that at all. And it's not quite as simple as you think, because accelerations bend space-time quite a bit, even with "only" special relativity.

Remember how simultaneity depends on speed? Clocks run at different rates, lengths are different, etc... Well, imagine the changes when you accelerate.

Suppose you want to fly to a star which, before you start accelerating towards it, is 100 light-years away. If you accelerate to 0.8 c, the star will now only be 60 light years away due to length contraction. This means that, if you accelerate from 0 to 0.8 c in less than a year, the star will have moved more than 40 light years closer during that time. In other words, in your accelerating frame of reference, the star is moving at over 40 times the speed of light. This is quite possible since your reference frame is not inertial.

Also, clocks on the star will, again in your frame of reference, have moved forward by about two centuries (while clocks at the same distance behind you will actually have turned backwards). You don't actually see this happening, you just infer it because before and after the acceleration you see images of the same events while light will now have been under way for a longer time (only traveling at 0.2 c relative to the star), which means these same events must have happened longer ago.

(edit: I originally said "5 decades" and "0.4", but I had messed up by switching two figures around)

These effects are proportional to distance, so you can easily make the clocks on a distant quasar move back and forth a few seconds by simply jumping around in your living room. Funny but true. Of course other people in the room will say you're crazy because for them the quasar's clocks will just be ticking at their normal, slow rate. Until they get up for a cup of coffee.

This plays a role in the twin paradox: if you are the twin that moves away and comes back again, you will find your brother's watch has been ticking slower during the whole time you were away, except during the time you accelerated back toward him, which is precisely when his clock suddenly jumped forward, making you the one who turns out to have aged less when you return.

So I wouldn't say accelerations are irrelevant in special relativity at all!

General relativity just says gravity is equivalent to acceleration, so it causes the same kind of effects. Only, since the acceleration caused by gravity becomes smaller at a distance, the effects won't be quite as dramatic. But it does turn out to make clocks run at different rates depending on how high you are in the field of gravity.

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jtbell
Mentor
These effects are proportional to distance, so you can easily make the clocks on a distant quasar move back and forth a few seconds by simply jumping around in your living room. Funny but true. Of course other people in the room will say you're crazy because for them the quasar's clocks will just be ticking at their normal, slow rate. Until they get up for a cup of coffee.
And the (very hardy) people living on the quasar will say you're all crazy because they never experience anything unusual happening to them, no matter how you all jump around. Hurkyl
Staff Emeritus
Gold Member
Nedeljko: Do you understand how classical, non-relativistic mechanics is able to deal with accelerating frames of reference? You can use exactly the same methods for SR....

@michelcolman

Thank you for your fast and detail answer, but it seems that you are not readed my post carefuly.

Remember how simultaneity depends on speed? Clocks run at different rates, lengths are different, etc...
I know for all these effects and when I implement my idea to formulas for concrete cases, I find that the length contraction and the time dilatation are consequences of obtained formulas and I find nothing from your post that is not consequence of my formulas. Yes, if I accelerate to 0.8c I will see star as more near because length contraction and I can be near the star in arbitrary short my proper time because time dilatation, and I return from my journey to Earth for example 1 second older, but my frends from the Earth will be dead (because the time measured from the Earth must be longer than 200 years). I know it.

For example, if the bird A moved from the moment $$t_0$$ to the moment $$t_1$$ of the inertial system S (of the ground), then the adequate proper time of bird A is $$\int_{t_0}^{t_1}\sqrt{1-\left(\frac{A'(t)}{c}\right)^2}\,dt$$. Is it correct?

If the bird A measured it's proper time $$t'$$, we can compute corresponding "earth's" time $$t_0$$ by solving the equation $$t'=\int_0^{t_0}\sqrt{1-\left(\frac{A'(t)}{c}\right)^2}\,dt$$. I thinked that it is the easier part of my considerations.

Does anybody can show me what is mjy concrete mistake. The best is constructing an example where we obtain different results.

JesseM
Special relativity can deal with accelerations just fine, you don't need GR for that at all. And it's not quite as simple as you think, because accelerations bend space-time quite a bit, even with "only" special relativity.

Remember how simultaneity depends on speed? Clocks run at different rates, lengths are different, etc... Well, imagine the changes when you accelerate.

Suppose you want to fly to a star which, before you start accelerating towards it, is 100 light-years away. If you accelerate to 0.8 c, the star will now only be 60 light years away due to length contraction. This means that, if you accelerate from 0 to 0.8 c in less than a year, the star will have moved more than 40 light years closer during that time. In other words, in your accelerating frame of reference, the star is moving at over 40 times the speed of light. This is quite possible since your reference frame is not inertial.

Also, clocks on the star will, again in your frame of reference, have moved forward by about five decades (while clocks at the same distance behind you will actually have turned backwards). You don't actually see this happening, you just infer it because before and after the acceleration you see images of the same events while light will now have been under way for a longer time (only traveling at 0.4 c relative to the star), which means these same events must have happened longer ago.

These effects are proportional to distance, so you can easily make the clocks on a distant quasar move back and forth a few seconds by simply jumping around in your living room. Funny but true. Of course other people in the room will say you're crazy because for them the quasar's clocks will just be ticking at their normal, slow rate. Until they get up for a cup of coffee.

This plays a role in the twin paradox: if you are the twin that moves away and comes back again, you will find your brother's watch has been ticking slower during the whole time you were away, except during the time you accelerated back toward him, which is precisely when his clock suddenly jumped forward, making you the one who turns out to have aged less when you return.

So I wouldn't say accelerations are irrelevant in special relativity at all!
This is a little misleading, because you're making it sound like there is a single "correct" way for an accelerating observer to define simultaneity at each point on his worldline. It's certainly true that you can construct a "rest frame" for the accelerated observer in such a way that his definition of simultaneity at each moment will coincide with that of the inertial frame where he's instantaneously at rest (I think Rindler coordinates, which work as a rest frame for an observer experiencing constant proper acceleration, would have this property), and in that case everything you say is correct, but you don't have to construct the accelerating observer's "frame" in this way. Unlike for inertial observers, there is not really a "standard" convention for defining the rest frame of an accelerating observer, and thus no standard simultaneity convention either (and remember that simultaneity is just that, a convention, the question of what events are simultaneous from a given observer's 'point of view' is not really a physical question at all).

I can not belive that discussion started by me is so interessant for other members of the forum.

@JesseM

I understand your remark, but remember that scientific theory must be refudable, and that essential for scientific theory are measurable phenomens. If we compute any value different way using different definitions, but we have same observable predictions, then we have two images of the same scientific theory.

Consider we an airplane. Airplane is at airport in the city M, it starts acceleration, after any time it flies linearly and uniformely, and then it slows to land at airport in city N. Pilot and passanger (at distance $$d$$ behind pilot) has instruments for measuring accelerations. At different positions in same airplane pilot and passanger have different maximum of acceleration during accelerating from airport in the city M. This proposition is observable and it must be decidable by theory and independent on used definitions.

HallsofIvy
Homework Helper
I can not belive that discussion started by me is so interessant for other members of the forum.

@JesseM

I understand your remark, but remember that scientific theory must be refudable, and that essential for scientific theory are measurable phenomens. If we compute any value different way using different definitions, but we have same observable predictions, then we have two images of the same scientific theory.

Consider we an airplane. Airplane is at airport in the city M, it starts acceleration, after any time it flies linearly and uniformely, and then it slows to land at airport in city N. Pilot and passanger (at distance $$d$$ behind pilot) has instruments for measuring accelerations. At different positions in same airplane pilot and passanger have different maximum of acceleration during accelerating from airport in the city M.
Can you give any support for that statement?

This proposition is observable and it must be decidable by theory and independent on used definitions.
Where has it been observed?

JesseM
@JesseM

I understand your remark, but remember that scientific theory must be refudable, and that essential for scientific theory are measurable phenomens. If we compute any value different way using different definitions, but we have same observable predictions, then we have two images of the same scientific theory.
I agree, different coordinate systems don't represent different scientific theories, they all make the same predictions about localized events (like what two clocks will read when they cross paths, or what an accelerometer reads at the moment a clock next to it shows a certain time) and frame-invariant quantities (like rest mass or the spacetime interval).
Nedeljko said:
Consider we an airplane. Airplane is at airport in the city M, it starts acceleration, after any time it flies linearly and uniformely, and then it slows to land at airport in city N. Pilot and passanger (at distance $$d$$ behind pilot) has instruments for measuring accelerations. At different positions in same airplane pilot and passanger have different maximum of acceleration during accelerating from airport in the city M. This proposition is observable and it must be decidable by theory and independent on used definitions.
Sure, accelerometers, which measure proper acceleration (the instantaneous acceleration in an object's inertial rest frame at that moment, which is proportional to the G-forces experienced), will always be predicted to give the same readings in every coordinate system, so if that's all you're saying I agree (HallsofIvy seems to think you're saying something else but I don't read your statement as anything more than a comment about proper acceleration being frame-invariant and measurable by accelerometers, let me know if I'm misreading you). But my response to michelcolman was specifically about the issue of how accelerating observers define simultaneity with distant events, which depends completely on their choice of coordinate system, it's not a frame-invariant quantity like proper acceleration is.

Can you give any support for that statement?
Hint: Proper distance between pilot and passanger is constant, but because motion, same distance measured from the ground is shorter (contraction of length). In inertial system of ground, passanger must be faster than pilot because decreasing distance. I can give complete computation for any concrete case.

Where has it been observed?
This is the imagined experiment, not real. I am talkig about my prediction of value those does not depend on chosed set of definitions. This is observable value - it has same value in all images of the same scientific theory.

JesseM
Nedeljko, you're just talking about Born rigid acceleration, right? The fact that observers experiencing such acceleration (in which the distance between any two observers remains constant in their inertial rest frame at every moment) will feel different proper accelerations is also discussed on this page:
We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. In principle, these ships could be physically connected together by ladders, allowing passengers to move between them. Although each ship would have a different proper acceleration, the spacing between them would remain constant as far as each of them was concerned.

@michelcolman
For example, if the bird A moved from the moment $$t_0$$ to the moment $$t_1$$ of the inertial system S (of the ground), then the adequate proper time of bird A is $$\int_{t_0}^{t_1}\sqrt{1-\left(\frac{A'(t)}{c}\right)^2}\,dt$$. Is it correct?

If the bird A measured it's proper time $$t'$$, we can compute corresponding "earth's" time $$t_0$$ by solving the equation $$t'=\int_0^{t_0}\sqrt{1-\left(\frac{A'(t)}{c}\right)^2}\,dt$$. I thinked that it is the easier part of my considerations.

Does anybody can show me what is mjy concrete mistake. The best is constructing an example where we obtain different results.
Wouldn't those equations indicate that the bird's time and the earth's time had both moved slower than the other? So when you put the clocks together, they would both have to be behind the other, which is impossible?

I may have misunderstood what you are writing, but it seems like you are considering the whole surface of the earth as a reference, which is a bit difficult because time at different places on the surface will be moving at a different rate as observed by the bird. You have to compare with a clock at the point of departure/arrival of the bird. If the bird flies away from its perch and then turns around to come back, the clock next to the perch will have suddenly jumped forward during the turn. At first sight, I don't see that happening in your equations.

This is a little misleading, because you're making it sound like there is a single "correct" way for an accelerating observer to define simultaneity at each point on his worldline. It's certainly true that you can construct a "rest frame" for the accelerated observer in such a way that his definition of simultaneity at each moment will coincide with that of the inertial frame where he's instantaneously at rest (I think Rindler coordinates, which work as a rest frame for an observer experiencing constant proper acceleration, would have this property), and in that case everything you say is correct, but you don't have to construct the accelerating observer's "frame" in this way. Unlike for inertial observers, there is not really a "standard" convention for defining the rest frame of an accelerating observer, and thus no standard simultaneity convention either (and remember that simultaneity is just that, a convention, the question of what events are simultaneous from a given observer's 'point of view' is not really a physical question at all).
That is very true, indeed. My accelerating reference frame merely corresponded to the reference frame that would be used by an observer at the same speed who was not accelerating (so the transition is smooth when the acceleration starts/stops). It seemed to be a logical choice, but anyone is at liberty to pick any other kind of reference frame as long as he/she applies the required corrections.

Wouldn't those equations indicate that the bird's time and the earth's time had both moved slower than the other? So when you put the clocks together, they would both have to be behind the other, which is impossible?
$$A(t)$$ is the motion law of bird A in inertial system S (of the ground). In other systems we must calculate the bird's motion law.

$$A(t)$$ is the motion law of bird A in inertial system S (of the ground). In other systems we must calculate the bird's motion law.
If you were saying that acceleration of other objects doesn't matter to an unaccelerating observer, I think you are correct. So from an earth-bound observer's point of view, the birds can accelerate as much as they like, their clocks will only be affected by their speeds, not their accelerations.

The birds, however, will see clocks going back and forth around them all the time as they accelerate in different directions.

tiny-tim
Homework Helper
The original Einstein's formulation of the special theory of relativity does not permit consideration of the accelerated observers, but this limitation is not essential because we can consider the special relativity as limit case of the general relativity by substituing gravitational constant by 0. Is it possible to consider accelerated observers in the special relativity without mathematical language of the general relativity (manifolds, tensors etc.)?
Hi Nedeljko! Special relativity can be combined with the equivalence principle, and it does give many of the results "of general relativity".

For example, SR plus EP gives correct results for both geodetic and gravitomagnetic precession (ie, rotation of an observer's frame), at least in the "weak" case …

see, for example, K. Nordtvedt's "Special Relativity Equivalence Principle", pp. 71-94 in the unpromisingly-titled book "Gravitation: From the Hubble length to the Planck length" (2005).

The equivalence principle is only a local principle, and therefore its combination with special relativity does not give correct global results, such as for precession of the perihelion of Mercury …

this is why Einstein needed to invent general relativity. I attached my computation in the attachment. The $$\LaTeX$$ source is in the attached zip file. My question is: Is this computation correct or not? If it is incorrect, why?

tiny-tim

Special theory of relativity does not include the gravitation. By this reason, the only prediction Mercury's motion law by the special theory of relativity is the straight line, but not a bounded path. By this reason, the general theory of relativity is introduced because including the gravity.

I wish use my idea for accelerated systems without gravity effects. Have I this way the same results as in general relativity with gravitational constant equal to 0.

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tiny-tim
Homework Helper
Ken Nordtvedt

Special theory of relativity does not include the gravitation. By this reason, the only prediction Mercury's motion law by the special theory of relativity is the straight line, but not a bounded path. By this reason, the general theory of relativity is introduced because including the gravity.
Well, http://en.wikipedia.org/wiki/Ken_Nordtvedt" [Broken] ) would disagree with you.

Special relativity does not exclude gravity …

so you can add in the equivalence principle …

and if you do, special relativity then gives correct local gravitational results.

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I am confused now. Is it true or not that the special relativity is limit case of the general relativity with G=0?

tiny-tim
Homework Helper
I am confused now. Is it true or not that the special relativity is limit case of the general relativity with G=0?
I'm also confused isn't G just the conversion factor which converts mass to an equivalent distance?

(so, for example, the Schwarzschild radius RM of a mass M is a length, but can be written, in terms of mass, 2GM)

May be we have different understandig (definitions) of the special theory of relativity. I consider the special relativity as the limit case of the general relativity with G=0. What definition you use? Relativistic theory in the flat spacetime?

Can anybody tell me is the computation attached at the first post of this page correct or not and why?

JesseM
May be we have different understandig (definitions) of the special theory of relativity. I consider the special relativity as the limit case of the general relativity with G=0. What definition you use? Relativistic theory in the flat spacetime?
I think the special theory is usually understood as the limit when you zoom in on an arbitrarily small neighborhood of a point in curved spacetime--this is the basis for the equivalence principle--or the limit as the curvature of spacetime (and thus the tidal forces) goes to zero, which happens naturally in the limit as the size of the patch of spacetime you're looking at approaches zero.

Fredrik
Staff Emeritus
Gold Member
Can anybody tell me is the computation attached at the first post of this page correct or not and why?
It isn't a great idea to attach the calculation in a pdf file. It's too hard to include bits of the math in quotes. I'll just comment on what you did in #1.

Let we consider two birds A and B, flying nonlinearly and nonuniformly with the motion laws A(t) and B(t) in inertial system S (of the ground), where A(t) and B(t) are indefinitely differentiable vector functions of the time. How the bird A see the bird B.

We will fix a moment $$t_0'$$ of the bird's A clock and compute $$t_0$$ such that $$t_0'=\int_0^{t_0}\sqrt{1-(A'(t)/c)^2}\,dt$$.
So far so good. Note that t0 is just the time coordinate in frame S of some event on A's world line.

Then we compute the vector $$v=A'(t_0)$$...
This isn't a vector (when we're considering 1+1-dimensional spacetime), but you already know that.

...and fix the Innertial system S' such that S' has constant velocity $$v$$ relative to system S. Then we transform coordinates $$(A(t_0),t_0)$$ of bird A to coordinates $$(a',t_0')$$ in system S' and coordinates $$(B(t_1),t_1)$$ of bird B to coordinates $$(b',t_1')$$.
This makes S' a co-moving inertial frame (co-moving with A at time t0 in frame S). That's fine, but your presentation is a bit weird because you talk about transforming a point on B's world line before you specify which point, but I see that you're doing that next.

We compute moment $$t_1$$ using condition $$t_1'=t_0'$$, and position $$b'$$ using computed moment $$t_1$$. Then, intensity of vector $$b'-a'$$ is distance bitween two birds measured by bird A when it scan the time $$t'$$ from it's clock.
You have chosen to consider two events (one on A's world line and one on B's world line) that are simultaneous in S', which is the frame that's co-moving with A, so yes, b'-a' is the distance in S' between the birds at time $t_0'$. But before you say that this is what A measures you should think about how he measures it. If he e.g. measures the time it takes to bounce a radar signal off the other bird, the result isn't going to be b'-a'. The coordinate distance in the co-moving inertial frame only agrees with the radar distance in the limit where the size of the region of spacetime where the measurement is performed goes to zero.

Fredrik
Staff Emeritus