Implications of the statement Acceleration is not relative

[stevendaryl]

Prove it. You drew your lines with my rocket in motion. I, in my rocket, have the right to consider myself to be permanently at rest. Prove to me that being non-inertial, yet always at rest, will result in a younger age.

The proof that an inertial path is the one with the longest proper time is essentially the same as the proof that a a straight line is the curve with the shortest length connecting two points. I'll go through both of them in parallel.

Euclidean case
In 2D Euclidean geometry, the formula for the length of a curve connecting two points is given by:

$L = \int \sqrt{1+m^2} dx$

where $m = \dfrac{dy}{dx}$ is the slope of the curve $y(x)$

This formula is good using any Cartesian coordinate system, provided that the curve is never vertical (it breaks down in that case, because the slope becomes infinite).

Finding the path $y(x)$ that makes $L$ an extremum (either a maximum or a minimum), one uses the calculus of variations. The result is that the minimizing or maximizing curve satisfies:

$\dfrac{d}{dx} ( \dfrac{m}{\sqrt{1+m^2}}) = 0$
which has the solution that $m$ is a constant.

So the curve with constant slope is the extremizing curve (the one making the distance either minimal or maximal--we can prove in this case that it is minimal).

Special Relativity case
In Special Relativity, the formula for the proper time of a spacetime path is given by:

$\tau = \int \sqrt{1-\dfrac{v^2}{c^2}} dt$

where $v = \dfrac{dx}{dt}$ is the velocity of the path $x(t)$

This formula is good using any inertial coordinate system.

Finding the path $x(t)$ that makes $\tau$ an extremum (either a maximum or a minimum), one again uses the calculus of variations. In this case, the equation for the extremizing path is:

$\dfrac{d}{dt} ( \dfrac{-\frac{v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}) = 0$
which has the solution that $v$ is a constant.

So the path with constant velocity $v$ is the path that makes the proper time maximal or minimal--we can prove in this case that it is maximal.

 

[harrylin]

I don't agree with this. For instance, the equations show that an object at rest at a constant r in the Schwarzschild vacuum feels a force - and thus is not inertial, nor moving ( relative to the field).

What you here call "at rest" is also what Einstein calls "at rest"; your disagreement seems to be with the definition of acceleration that he used.

Yes, but it will still be non-inertial. Are you saying that in this scenario the rocket feels no acceleration ?

That's correct. In Langevin's "twin" scenario, the space capsule feels no force during the voyage.

 

[harrylin]

[..] Specifically, I mean that SR should not be used to prove that what I earlier called the "second aspect" of the Twin Paradox is not a paradox. That is the claim that when the episode is observed from the "permanently at rest" frame of the rocket, the Earth twin will be younger than the rocket twin.

Indeed, a "permanently at rest" rocket frame in the sense as was meant by objectors means zero acceleration; their objection targeted GR, not SR and Einstein understood this very well. Some people seem to confound that issue with the question if we need GR to describe observations from the accelerating rocket - which is of course not needed, SR is fine for that.

The paradox then is that both twins are "younger than the other", which can't happen in reality.

Ehm no, that basic issue was taken care of in Einstein's 1918 paper, by means of an induced gravitational field. According to 1916GR, one may claim that the firing of the rockets doesn't accelerate the rocket at all but that instead this induces a gravitational field through the universe. That field makes that the stay-at home ages the right amount in comparison with the traveler. However, that solution opened a can of worms that nobody wants - so much, that it has been mostly ducked in the literature.

The problem with treating this aspect of the twin paradox in SR is that the case of the resting rocket cannot be considered. In SR, only observers in inertial frames can consider themselves to be at rest.

Quite so. In SR observers who are accelerating wrt inertial frames may consider themselves to be at rest in an accelerating frame; consequently they cannot consider themselves physically "in rest" in the sense of SR. Its laws of nature for inertial frames do not apply to that frame.

Is it claimed that proper acceleration affects clocks?

Some clocks are affected by such applied forces, but in particular atomic clocks are rather robust. For the typical twin paradox scenario's in which the turn-around only takes a relatively small time this is irrelevant. The assumption that acceleration doesn't affect the clocks is called the clock hypothesis which was probably first brought up in Einstein's 1905 paper.

[..] Unless, of course, it can be shown that the observer in the rocket cannot legitimately consider himself to be permanently at rest.

With respect to that, in my previous post I said, "But always in his mind is that goal, to understand how it is that the rocket can be legitimately understood to be at rest, and the Earth moving."

To which there was this reply: [..]

Harrylin, did you mean to say that the observer in the rocket cannot legitimately consider himself to be permanently at rest? [..]

I understand you to mean "permanently at rest" in the sense of Einstein-1918. There are many objections to make against Einstein's induced gravitational fields. Consider such things as cause and effect, speed of gravitational wave, etc. I have never seen a paper that even attempts to address those issues, let alone solve them.

 

[Dale]

There are many objections to make against Einstein's induced gravitational fields. Consider such things as cause and effect, speed of gravitational wave, etc. I have never seen a paper that even attempts to address those issues, let alone solve them.

Probably because most people don't believe they are issues. The cause is obviously the choice of coordinates, and this meaning of "gravitational fields" does not mathematically produce "gravitational waves" in the usual sense of a wave equation with a characteristic propagation speed.

 

[harrylin]

Probably because most people don't believe they are issues. The cause is obviously the choice of coordinates, and this meaning of "gravitational fields" does not mathematically produce "gravitational waves" in the usual sense of a wave equation with a definite propagation speed.

A change of coordinates isn't a gravitational field. I don't follow Einstein's Machian explanation of invoking a physical, induced gravitational field (for enabling the interpretation that the rocket is constantly "truly in rest"): "all the stars that are in the universe, can be conceived as taking part in bringing forth the gravitational field [..] during the accelerated phases of the coordinate system K' ".

 

[stevendaryl]

A change of coordinates isn't a gravitational field.

It depends on how you define "gravitational field". If you drop an object, and it follows a trajectory $h = -\frac{1}{2}g t^2$ for some constant $g$, then many people would call $g$ the "acceleration due to gravity" or the "gravitational field". But it certainly is affected by a coordinate change. It can be made to vanish by careful choice of coordinates. If $g=0$ (no gravitational field), then a change of coordinates to accelerated coordinates can make $g$ nonzero.

The equivalence principle is about this notion of gravitational field: there is no difference (other than the variation of $g$ with location) between the effects of a $g$ due to gravity and a $g$ due to the use of accelerated cooridnates.

 

[Dale]

A change of coordinates isn't a gravitational field.

Remember that there is more than one possible meaning of the term "gravitational field" in GR. In the sense that Einsetin meant it (also my preferred sense) the "gravitational field" is the Christoffel symbols, and the Christoffel symbols do in fact change under a change of coordinates. So a change of coordinates does cause a gravitational field in the sense Einstein used the term.

[EDIT: stevendaryl made essentially the same point, but faster!]

 

[harrylin]

It depends on how you define "gravitational field". If you drop an object, and it follows a trajectory $h = -\frac{1}{2}g t^2$ for some constant $g$, then many people would call $g$ the "acceleration due to gravity" or the "gravitational field". But it certainly is affected by a coordinate change. It can be made to vanish by careful choice of coordinates. If $g=0$ (no gravitational field), then a change of coordinates to accelerated coordinates can make $g$ nonzero.

The equivalence principle is about this notion of gravitational field: there is no difference (other than the variation of $g$ with location) between the effects of a $g$ due to gravity and a $g$ due to the use of accelerated cooridnates.

A change in coordinates can of course produce a fictitious gravitational field; but if you don't consider this a real, physical field, then you side with the objector ("Critic") of early GR. Einstein countered that "the accelerated coordinate systems cannot be called upon as real causes for the field". As I elaborated, the issue here is with Einstein's Machian explanation of a gravitational field that is induced by the distant stars.

[..] a change of coordinates does cause a gravitational field in the sense Einstein used the term. [..]

As you see here above, Einstein stated just the contrary.

 

[Dale]

A change in coordinates can of course produce a fictitious gravitational field; but if you don't consider this a real, physical field, then you side with the objector ("Critic") of early GR.

I don't worry too much about "real" or "fictitious" except where it is part of standard terminology (i.e. "real numbers" or "fictitious forces"), and I won't take either side of such a debate.

Einstein countered that "the accelerated coordinate systems cannot be called upon as real causes for the field". As I elaborated, the issue here is with Einstein's Machian explanation of a gravitational field that is induced by the distant stars.

As you see here above, Einstein stated just the contrary.

In the http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativityhe also said "the gravitational field in a space-time point is still not a quantity that is independent of coordinate choice; thus the gravitational field at a certain place does not correspond to something 'physically real', but in connection with other data it does. Therefore one can neither say, that the gravitational field in a certain place is something 'real', nor that it is 'merely fictitious'." He also said in the same discussion "Rather than distinguishing between 'real' and 'unreal' we want to more clearly distinguish between quantities that are inherent in the physical system as such (independent from the choice of coordinate system), and quantities that depend on the coordinate system." and "the distinction real - unreal is hardly helpful".

My point remains, that the "gravitational field" he refers to by "A gravitational field appears, that is directed towards the negative x-axis. Clock U1 is accelerated in free fall, until it has reached velocity v. An external force acts upon clock U2, preventing it from being set in motion by the gravitational field. When the clock U1 has reached velocity v the gravitational field disappears" is the field from the Christoffel symbols and it is entirely determined by the choice of coordinates.

I have not read all of Einsteins writings, but I believe that this is the meaning he usually attributes to the term. If you believe that he refers to something besides the Christoffel symbols then please be explicit about what mathematical term you think he intends and why.

 

[Mentz114]

What you here call "at rest" is also what Einstein calls "at rest"; your disagreement seems to be with the definition of acceleration that he used.

I don't know what you are talking about. The case I gave refers to proper acceleration which is unambiguous.

That's correct. In Langevin's "twin" scenario, the space capsule feels no force during the voyage.

I still don't know what you mean. You are moving the gaoalposts and and being slippery, because you're wrong.

 

[GregAshmore]

I did. I suspect that you didn't bother to read it, but stop acting as though the objection were summarily dismissed and no answer were given when you simply haven't bothered to read the answer given.

Yesterday I skimmed some posts, believing that I had "the gist" of the argument, without stopping to digest all the details. It's possible I missed your "gist", which would be sloppy of me. However, I don't think I missed what I was looking, for. I don't recall seeing it in posts prior to yesterday, either.

Before I say what I was looking for, I will accept a good deal of the criticism in recent posts. I have made some overly broad and not well thought out statements, resulting in errors. I realized that as I went through the mental exercise of constructing the spacetime diagram for the twin paradox, rather than accepting it as a finished product.

In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed.

Every SR solution that I have seen has the rocket in motion. Certainly, in every SR spacetime diagram I have seen, the rocket is in motion; it is the reversing motion of the rocket which produces the knee in the rocket's worldline, and thus the difference in proper times. The article discussing acceleration in special relativity also speaks of the accelerating object as in motion. (http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html)

I, in my rocket, claim that I am not in motion. I have been sitting in my rocket in the same position throughout the episode. You claim that I will be younger than my twin at the end of the episode. As proof, you present to me a diagram that shows me in motion. I reject it. I categorically deny that the diagram applies to me. The diagram shows me in motion; I have not moved.

I do not deny that my experience has been non-inertial. I do not deny that the experience of my twin has been inertial. I do not even deny that I have experienced "proper acceleration", because you have told me that I can experience proper acceleration while remaining motionless.

I deny that I have been in motion. Therefore, I insist on a solution that does not put me in motion.

What would a diagram of such a solution look like? There would have to be a point representing my position and time at the beginning of the episode, another point representing my position and time at the end of the episode, and a (possibly curvilinear) path that connects the two points. Whatever the shape of the path, the value of the position coordinate must not vary from the beginning to the end of the path, because I do not move.

It is maintained that SR can be used to solve this problem. I will attempt to draw the spacetime diagram. For convenience, I'll make my time and position axes orthogonal, in the usual manner with time positive toward the top of the page. And as usual I will set my starting position and time at the origin.

Because I do not move, my worldline must be parallel to the time axis; in this case it will be coincident with the time axis. I realize that this is the worldline of an inertial object. I am not inertial. It doesn't seem right that my worldline should be inertial, but that's how it must be, because I do not move.

Now to draw the earth. The Earth is moving in my frame, and inertial in its own frame. I cannot find any worldline that will satisfy both conditions, and also meet me at the end of my worldline. I cannot complete the spacetime diagram.

Being unable to represent in a spacetime diagram the perfectly legitimate scenario of myself at rest throughout the episode, and the Earth reversing, I conclude that SR is not suitable for solving the problem.

I may be wrong in my conclusion, as has been claimed. If so, I would like to see a SR solution which has me in one position throughout the episode.

 

[ghwellsjr]

I would like to see a SR solution which has me in one position throughout the episode.

How about this one:

from this thread.

 

[GregAshmore]

How about this one:

from this thread.

I haven't read it yet. I was just coming on to see if I could save myself from the same mistake I've made before: seeing the thing in my head without checking it on paper. Some people learn slowly.

 

[Dale]

However, I don't think I missed what I was looking, for. I don't recall seeing it in posts prior to yesterday, either.
...
In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution.
...
What would a diagram of such a solution look like?
...
I may be wrong in my conclusion, as has been claimed. If so, I would like to see a SR solution which has me in one position throughout the episode

See post 80.

Because I do not move, my worldline must be parallel to the time axis; in this case it will be coincident with the time axis. I realize that this is the worldline of an inertial object. I am not inertial. It doesn't seem right that my worldline should be inertial, but that's how it must be, because I do not move.

Don't forget, your frame is non-inertial so straight lines do not correspond to inertial worldlines. Inertial worldlines are geodesics, which is not the same thing as a straight line when you are using non-inertial coordinates.

Another, more accurate, way to say it is that your coordinate lines are bent. So lines of constant coordinates are not straight lines and straight lines don't have constant coordinates. A similar thing happens, e.g. in polar coordinates. Since the coordinates are curved the equation r=mθ+b does not represent a straight line.

Being unable to represent in a spacetime diagram the perfectly legitimate scenario of myself at rest throughout the episode, and the Earth reversing, I conclude that SR is not suitable for solving the problem.

Your inability to solve this reflects your own personal limitation, not a limitation of SR. It has been made abundantly clear to you that SR is not limited in this way.

 

[GregAshmore]

See post 80.

I have downloaded the paper.

Don't forget, your frame is non-inertial so straight lines do not correspond to inertial worldlines. Inertial worldlines are geodesics, which is not the same thing as a straight line when you are using non-inertial coordinates.

Another, more accurate, way to say it is that your coordinate lines are bent. So lines of constant coordinates are not straight lines and straight lines don't have constant coordinates. A similar thing happens, e.g. in polar coordinates. Since the coordinates are curved the equation r=mθ+b does not represent a straight line.

I'll have to work on this to understand it.

Your inability to solve this reflects your own personal limitation, not a limitation of SR.

I fully expect to find that you are correct in this.

It has been made abundantly clear to you that SR is not limited in this way.

Well, no, it hasn't been made clear. It ought to be clear, I'm sure. The fact that it isn't clear is much more a factor of my response to what has been said than a factor of the content.

I have the impression that some of the contributors on this forum are teachers by trade, so what follows may be of interest. If not, no need to read further.

I've been trying to figure out why I have had so much trouble learning relativity. In particular, I have never had the experience of repeatedly thinking that I understand a subject, only to discover that I am profoundly wrong.

One reason, no doubt, is the bizarre premises that we are called on to accept. However, that was much more of a stumbling block at the beginning than it is now. At this point, I can "suspend disbelief" and treat the problem as an exercise in abstract logic. The "truth" or "reality" of the premises can be evaluated later.

So perhaps I'm just not good at abstract logic. Maybe. I'm sure I'm no Einstein, at any rate. But I'm not profoundly stupid, either. So how do I repeatedly find myself in the position of being profoundly wrong?

I had an "aha" moment on this a couple of weeks ago, which was reinforced and clarified last night. It has to do with my pattern of learning.

The entire subject of relativity is completely hands-off for me. I'll never see, much less operate, a particle accelerator. I learn new things all the time in my work, but in every case I can test my understanding of what ought to happen against what actually happens when I act on my understanding.

In addition, many aspects of relativity are hypothetical (hands-off) for everyone, at least in our lifetimes. We'll never travel at relativistic speeds. So none of us have the opportunity to directly test our understanding by experiment. (We have indirect experimental evidence to support what is predicted; that's not the same thing as making the prediction come to pass.)

I have made the mistake of thinking that "unable (in practice) to test by experiment" means "completely unable to verify". In my usual method of learning, I form a mental picture of what ought to happen, then I test it by experiment. Because I am unable to test, I have been in the habit of stopping after forming the mental picture.

In my work, I may do calculations after forming a mental image and before conducting an experiment. (I nearly always did when I designed machinery; I rarely do now that I work in a larger company and only write software.) The calculations are viewed as a means of avoiding failure in the experiment; they are never seen as a verification of anything. The calculations are never an end in themselves; they are a means of getting to the end, which is a functioning piece of equipment.

In relativity, for someone in my situation, the calculations are both the verification and the deliverable. That's what finally penetrated my thick skull last night. I can no more put something on this forum without verifying it by calculation than I can deliver an untested product to a customer.

Now maybe we'll see fewer dumb statements by me on the forum.

I did find this interesting, for perspective. Errors are never acceptable. But if even professionals have trouble, I should not be surprised if I have trouble, too.

From the paper referenced in post #80:

The path through this confusion existed already in Einstein’s original paper[9], and was popularised by Bondi in his work on ‘k-calculus’. It lies in the correct application of ‘radar time’ (referred to as ‘Marzke-Wheeler Coordinates’ in Pauri et al.[10]). This concept is not new. Indeed Bohm[1] and D’Inverno[2] both devote a whole chapter to k-calculus, and use ‘radar time’ (not under that name) to derive the hypersurfaces of simultaneity of inertial observers. However, both authors then apply this definition wrongly to the traveling twin.

 

[stevendaryl]

In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed.

Once you understand how things work in inertial coordinates, then how they work in any other coordinate system (such as one in which the traveling twin is always at "rest") is just an application of calculus. So you are demanding that someone demonstrate a calculus exercise to you?

Okay, if it really will make you happy, I will post such a demonstration.

 

[stevendaryl]

Accelerated Rocket in Inertial Coordinates

Let's choose the zero for $x$ and $t$ to make the mathematics as simple as possible.

So assume that one twin is at rest at some location $x=L$, in inertial coordinates. The other twin starts off at location $x=L_1$ at time $t = -t_1$, travels in the $-x$ direction until he reaches $x=L_0$ at time $t=0$, and then travels in the $+x$ direction until he reaches $x=L$ again at time $t=+t_1$. The time origin is chosen so that his journey is symmetric in time about the point $t=0$. The mathematical description of the traveling twin's path is:

$x(t) = \sqrt{(L_0)^2 + c^2 t^2}$

Time $t_1$ is chosen so that $(t_1)^2 = \dfrac{L_1^2 - L_0^2}{c^2}$

The velocity of the traveling twin at any time is given by:

$v(t) = \dfrac{c^2 t}{\sqrt{(L_0)^2 + c^2 t^2}}$

and the time-dilation factor $\sqrt{1-\dfrac{v^2}{c^2}}$ is given by:

$\sqrt{1-\dfrac{v^2}{c^2}} = \dfrac{L_0}{\sqrt{(L_0)^2 + c^2 t^2}}$

The elapsed time for the traveling twin is given by:
$\tau = \int \sqrt{1-\dfrac{v^2}{c^2}} dt$

I'm not going to do the integral, but you can see that the integrand is less than 1, so the result will certainly be less than $\int dt = 2 t_1$. So the traveling twin will be younger.

I'm going to make another post where I describe this same situation from the point of view of the traveling twin.

 

[stevendaryl]

Accelerated Rocket in Accelerated Coordinates

In inertial coordinates, the path of the traveling twin is given by:
$x(t) = \sqrt{L_0^2 + c^2 t^2}$

Now, let's switch to a coordinate system in which the traveling twin is at rest, by making the transformation:

$X = \sqrt{x^2 - c^2 t^2}$
$T = \dfrac{L_0}{c} arctanh(\dfrac{ct}{x})$

where $arctanh$ is the inverse of the hyberbolic tangent function.

In terms of these coordinates, the path of the stay-at-home twin is given by:

$X(T) = L_1\ sech(\dfrac{cT}{L_0})$

where $sech$ is the hyperbolic secant function.

To see that this is reasonable, you can look at the Taylor series expansion: $sech(\theta) = 1 - \frac{1}{2} \theta^2 + \ldots$. So for small values of $T$, we have:

$X(T) = L_1(1 - \frac{1}{2} \dfrac{c^2T^2}{L_0^2} + \ldots)$
$= L_1 - \frac{1}{2} g_1 T^2 + \ldots$

where $g_1 = \dfrac{c^2 L_1}{L_0^2}$

That is the path of an object that starts off at $X = L_1$ at time $T=0$ and falls at the acceleration rate of $g_1$.

So in these coordinates, the traveling twin is always at the location $X = L_0$, while the stay-at-home is at $X = L_0$ at some point (at some time prior to $T=0$, rises to $X=L_1$ at time $T=0$, and then falls back to $X=L_0$ at some later time.

In terms of the coordinates $X,T$, the elapsed time $d\tau$ for a traveling clock is given by:

$d\tau = \sqrt{\dfrac{X^2}{L_0^2} - \dfrac{V^2}{c^2}} dT$

where $V = \dfrac{dX}{dT}$

Note the difference with the formula for an inertial frame: $d\tau$ not only on the velocity, but on the position $X$.

For the traveling twin, who is always at $X = L_0$, his proper time is:
$d\tau = dT$

For the stay-at-home twin, whose position $X$ is always greater than or equal to $L_0$, the first term $\dfrac{X^2}{L_0^2} > 1$. If you work out the details, you will find that $d\tau = \sqrt{\dfrac{X^2}{L_0^2} - \dfrac{V^2}{c^2}} dT > dT$

So in accelerated coordinates, it's also the case that the stay-at-home twin ages more than traveling twin.

 

[PeterDonis]

In Langevin's "twin" scenario, the space capsule feels no force during the voyage.

IIRC this is only because spacetime in Langevin's scenario has a different topology than standard Minkowski spacetime; it is spatially a cylinder in the x direction instead of an infinite line. The "traveling" twin goes around the cylinder whereas the "stay at home" twin does not. The two paths belong to different topological classes; you can't continuously deform one into the other.

In this kind of situation you can have multiple free-fall paths between the same pair of events with different proper times; but each free-fall path has maximal proper time compared to all non-free-fall paths within the same topological class. (There are other topological classes possible as well; for example, there could be another free-fall twin that went around the cylinder twice, etc.) For example, an accelerating "twin" that didn't go around the cylinder would have less elapsed proper time than the free-fall stay-at-home twin (but not necessarily less than the free-fall traveling twin); and an accelerating "twin" that *did* go around the cylinder would have less elapsed proper time than the free-fall traveling twin.

 

[Dale]

I'll have to work on this to understand it.

I think that this is probably one of the key topics of modern relativity.

The key concepts of relativity, both special and general, are geometrical (Minkowski geometry for SR and pseudo-Riemannian geometry for GR).

Just like you can take a piece of paper and draw geometrical figures and discuss many things, such as lengths and angles, without ever setting up a coordinate system. The same thing is possible in relativity. The "piece of paper" is spacetime which has the geometrical structure of a manifold. The "geometrical figures" are worldlines, events, vectors, etc. that represent the motion of objects, collisions, energy-momentum, etc.

In this geometrical approach, the twin scenario is simply a triangle, and the fact that the traveling twin is younger is simply the triangle inequality for Minkowski geometry. In a coordinate-independent sense, the traveler's worldline is bent, and that in turn implies that his worldline is necessarily shorter as a direct consequence of the Minkowski geometry.

Now, on top of that underlying geometry, you can optionally add coordinates. Coordinates are simply a mapping between points in the manifold (events in spacetime) and points in R4. The mapping must be smooth and invertible, but little else, so there is considerable freedom in choosing the mapping. It is possible to choose a mapping which maps straight lines in spacetime to straight lines in R4, such mappings are called inertial frames.

It is also possible to choose a mapping which maps bent lines in spacetime to straight lines in R4, a non-inertial frame. Such a mapping does nothing to alter the underlying geometry. The bent lines are still bent in a coordinate-independent geometrical sense, but because it simplifies the representation in R4 it can still be useful on occasion in order to simplify calculations.

Because the mapping is invertible, in many ways it doesn't matter if you are talking about points in the manifold or points in R4. So you talk about things being "at rest" based on R4, and things happening "simultaneously" based on R4, and many other things. However, it is occasionally important to remember the underlying geometry.

I hope this helps.

Well, no, it hasn't been made clear. It ought to be clear, I'm sure.

It has, in fact, been made clear to you that SR can handle the twins paradox. What obviously hasn't been made clear to you is why. Your repetition of bald assertions that have already been contradicted is unhelpful. It wastes your time in repeating it and it wastes our time in repeating our responses. It also irritates those (maybe only me) who feel like their well-considered and helpful responses have been completely ignored by you.

You have been provided explanations and references addressing the topic, which clearly didn't "do it" for you. Read the explanations and references and point out the specific things that you don't understand or don't agree with. Then we can help clarify and make some progress.

 

[TrickyDicky]

Just like you can take a piece of paper and draw geometrical figures and discuss many things, such as lengths and angles, without ever setting up a coordinate system. The same thing is possible in relativity. The "piece of paper" is spacetime which has the geometrical structure of a manifold.

If this is so I wonder why rigorous definitions of manifolds are based on charts (that define local coordinate systems). IOW and simplifying: any object (that is of course also a topological space with certain topological features that make it well behaved) that can be "charted" is a manifold.
So the "piece of paper" must have the property that you can set up a coordinate system on it, that is its defining property if you want to call it a manifold. The fact that "you don't have to" is obviously relying on the fact that it is implicit in the definition, so I always have trouble understanding the insistence on banning charts.

 

[TrickyDicky]

Addressing the OP more specifically, maybe he is finding difficult to understand why if there is no two notions of velocity coexisting (an absolute and a relative velocity) in relativity, one does have both absolute and relative acceleration. While one justifies easily the first case because in relativity we have a spacetime rather than separate space and time and therefore one doesn't have any absolute space wrt which define an absolute velocity, this doesn't work in the case of acceleration and ultimately it seems one has to settle with "because this is just the way it is".
This isn't something that is very often clarified.

 

[stevendaryl]

Addressing the OP more specifically, maybe he is finding difficult to understand why if there is no two notions of velocity coexisting (an absolute and a relative velocity) in relativity, one does have both absolute and relative acceleration. While one justifies easily the first case because in relativity we have a spacetime rather than separate space and time and therefore one doesn't have any absolute space wrt which define an absolute velocity, this doesn't work in the case of acceleration and ultimately it seems one has to settle with "because this is just the way it is".
This isn't something that is very often clarified.

I don't think that things are that dissimilar when it comes to velocity and acceleration.

You can define a 4-velocity $V^\mu$ by $V^\mu = \dfrac{d}{d \tau} x^\mu$ and you can similarly define a 4-acceleration $A^\mu$. Neither is more absolute than the other. However, you can always choose coordinates so that the spatial components of $V^\mu$ are all zero, but you can't always do that for the spatial components of $A^\mu$

 

[TrickyDicky]

However, you can always choose coordinates so that the spatial components of $V^\mu$ are all zero, but you can't always do that for the spatial components of $A^\mu$

Right there, that's what I mean. This is the asymetry that is not so easy to explain. And maybe the OP naively thinks that if there is an absolute acceleration it would imply a rate of change of absolute velocity but that can't be because there is no such a thing as absolute velocity in relativity.
At this point I guess I should wait for the OP to confirm if this gets any close to his line of thought.

 

[harrylin]

[..] My point remains, that the "gravitational field" he refers to by "A gravitational field appears, that is directed towards the negative x-axis. Clock U1 is accelerated in free fall, until it has reached velocity v. An external force acts upon clock U2, preventing it from being set in motion by the gravitational field. When the clock U1 has reached velocity v the gravitational field disappears" is the field from the Christoffel symbols and it is entirely determined by the choice of coordinates. [..]

Once more, Einstein's comment related to your point was:
"To be sure, the accelerated coordinate systems cannot be called upon as real causes for the field, an opinion that a jocular critic saw fit to attribute to me on one occasion. But all the stars that are in the universe, can be conceived as taking part in bringing forth the gravitational field; because during the accelerated phases of the coordinate system K' they are accelerated relative to the latter and thereby can induce a gravitational field, similar to how electric charges in accelerated motion can induce an electric field."

please be explicit about what mathematical term you think he intends and why.

I think that Einstein was clear enough, and I was explicit in my clarification - he gave a physical explanation which I don't copy and to which you turned a blind eye. That shows that you also don't copy it; please stop trying to turn your agreement with me in an argument about something else.

 

[harrylin]

[..] In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed. [..] The article discussing acceleration in special relativity also speaks of the accelerating object as in motion.

Indeed, SR presents solutions for the physical consideration of a traveler who changes velocity; SR isn't made for the view of a traveler who is constantly at rest, such that the universe is bouncing around while the traveler doesn't accelerate.

I, in my rocket, claim that I am not in motion. I have been sitting in my rocket in the same position throughout the episode. You claim that I will be younger than my twin at the end of the episode. As proof, you present to me a diagram that shows me in motion. I reject it. I categorically deny that the diagram applies to me. The diagram shows me in motion; I have not moved. [..]

An SR diagram does not address your issue. Einstein's 1918 paper does; however few people accept his Machian solution.

I would like to see a SR solution which has me in one position throughout the episode.

A non-moving traveler isn't addressed in SR. SR can only provide a mapping to rocket, such that the description is from the view of the accelerating rocket.

 

[Dale]

I think that Einstein was clear enough, and I was explicit in my clarification - he gave a physical explanation which I don't copy and to which you turned a blind eye.

I wasn't asking for a physical explanation, I was asking for clarification about its definition. What is meant by the term "gravitational field"? Until it is known what is meant by the term it is nonsense to even talk about providing a physical explanation for it. If I give you a physical explanation for a flubnubitz without defining the term, have I actually told you anything? E.g. "All the stars that are in the universe, can be conceived as taking part in bringing forth the flubnubitz."

You certainly were not explicit at all about what you believe he means by the term "gravitational field" even when directly asked for clarification, and you seem to disagree with Einstein's use of the term although you quote him. If you wish to clarify what specifically you believe Einsetin refers to by the term "gravitational field" then we can continue the discussion.

I don't understand your reluctance to clarify your position. Surely by now you realize how easily misunderstandings can arise in online forums. A request for clarification should always be taken seriously and complied with willingly.

 

[stevendaryl]

Once more, Einstein's comment related to your point was:
"To be sure, the accelerated coordinate systems cannot be called upon as real causes for the field, an opinion that a jocular critic saw fit to attribute to me on one occasion. But all the stars that are in the universe, can be conceived as taking part in bringing forth the gravitational field; because during the accelerated phases of the coordinate system K' they are accelerated relative to the latter and thereby can induce a gravitational field, similar to how electric charges in accelerated motion can induce an electric field."

I would say that there are several aspects of gravity:

1. For each point in spacetime, and for each possible initial velocity, there is a unique inertial path, the freefall path, or geodesic. These geodesics are physical, and are affected by the presence of matter and energy.
2. In general, geodesics that start close together and parallel do not remain close together and parallel as you follow them. This is a manifestation of spacetime curvature.
3. If one takes a path through spacetime that does not follow a geodesic, there will be resistance--it requires the application of a physical force to depart from geodesic motion.
4. If one uses a coordinate system in which coordinate axes are not geodesics, then the path of an object with no physical forces acting on it will appear "curved", meaning that the components of velocity in this coordinate system are not constant.

The first 3 aspects of gravity don't involve coordinates at all, and only mention physical forces, not fictitious forces. The 4th aspect to me is what the meaning of "fictitious forces" are, which is that the coordinate flows are not geodesics. The terminology "coordinate flow" is made up by me; I'm not sure what the technical term is, or if there is one. But in a similar way that specifying an initial point in spacetime, together with an initial velocity, determines a path through spacetime--the geodesic, specifying an initial point in spacetime, together with an initial velocity, determines a different path through spacetime, the coordinate flow, which is the solution to the component equation:

$\dfrac{dV^\mu}{d\lambda} = 0$

(where $\lambda$ is the affine parameter for the path).

This would be a geodesic, if it were flat spacetime and inertial cartesian coordinates were used.

I think that Einstein was clear enough

Actually, I don't think he was very clear, mainly because he is using the words "gravitational field" (or something in German, more likely) without giving a precise definition of what it means.

 

[Dale]

The 4th aspect to me is what the meaning of "fictitious forces" are, which is that the coordinate flows are not geodesics. The terminology "coordinate flow" is made up by me; I'm not sure what the technical term is, or if there is one.

I think that the technical term would be "integral curves of the coordinate basis", but it certainly isn't a commonly used term. "Coordinate flows" sounds nice.

 

[my_wan]

I just wanted to add a few words for GregAshmore concerning the distinction between coordinate and proper acceleration in the form of a thought experiment, in the hopes it might help make it more obvious.

I take it you understand the relative nature of velocity, but when two inertial objects have a constant velocity with respect to each other it cannot be said which is 'really' moving and which is at rest. It depends on which coordinates you choose. If one hits the gas to speed up, not only does the occupant feel the force, but no matter which coordinates you choose every observer will agree that its speed is changing. Hence anybody with a good understanding of relativity can calculate the force felt by the occupants. This is the consequence of proper acceleration. However, depending on the coordinate choice of the observer, not everybody will agree on how much or in what direction it is accelerating with respect to their coordinates. The acceleration is absolute, but how much acceleration occurs in relation to a particular coordinate choice is relative. Some observers will even say they are slowing down, which also requires acceleration. This is referred to as coordinate acceleration.

To illustrate imagine sitting in a seat and tossing a rock straight up and catching it when it falls. That rock traces out a straight up and down line in your coordinates. Now imagine your seat is the back seat of a car moving at a constant 100 kph. The guy on the side of the road sees you toss the rock down the road and and the car carries you down with it to catch it. As far as the laws of physics are concerned the straight up and down path and the curved path up over the road are just as real. There are no 'real' paths in that sense, and the classical aether failed because it essentially sought to establish which path was real. It is effectively like an American arguing with the Chinese over which way is really up. The laws of physics allows everybody to agree on what is accelerating, though not necessarily by how much or what direction, but does not allow everybody to agree on what is moving at some constant velocity verses at rest.

Gravity turns this relationship on its head. Note the force felt when accelerating. When on the surface of a gravitational mass, like Earth, the principle of equivalence tells us this weight we feel is the same force we feel when accelerating. Now if you jump off a roof then while accelerating toward the ground you feel no force, effectively weightless. Hence, in your frame of reference you are at rest, i.e., not accelerating, but the Earth is accelerating toward you. Yet everybody in the Universe can agree that your speed is changing, even if not by how much. In this case proper gravity is absolute, while how much gravity and coordinate paths are relative. So in this case gravity is absolute, but the coordinate acceleration due to gravity is relative.

Now reconsider the twin paradox. If two observers experience the same amount of acceleration X time, then neither one will age any faster than the other. If two spaceship pass each other at a constant velocity, such that t=0 is mutually defined at the point of closest approach, then each will observe the others clock going slower than their own. To resolve this you accelerate your ship to catch up to the other, which requires applying an absolute force, i.e., proper acceleration. It doesn't matter whether you accelerate a little and take longer to catch up, or a lot to catch up faster, the effect is the same. You are the one that experienced a proper acceleration, hence your clock will be the one that appears to have slowed when you catch up. If you both apply the same proper acceleration to catch up to each other then no time dilation will be apparent under the definition given by t=0. Time is as fluid as the path of the rock in the back seat of the car, but like the rocks coordinate path it must transform from one coordinate choice to another by a well defined set of rules. Others have done an excellent job of articulating these quantitative rules here.

 

[Dale]

If one hits the gas to speed up, not only does the occupant feel the force, but no matter which coordinates you choose every observer will agree that its speed is changing.

Just a point of clarification. This should be "no matter which inertial coordinates you choose".

 

[my_wan]

Just a point of clarification. This should be "no matter which inertial coordinates you choose".

Yep, it even looks awkward stated the way I did now that you pointed it out.

 

[harrylin]

I just wanted to add a few words for GregAshmore [..]
Note the force felt when accelerating. When on the surface of a gravitational mass, like Earth, the principle of equivalence tells us this weight we feel is the same force we feel when accelerating. Now if you jump off a roof then while accelerating toward the ground you feel no force, effectively weightless. Hence, in your frame of reference you are at rest, i.e., not accelerating, but the Earth is accelerating toward you. Yet everybody in the Universe can agree that your speed is changing, even if not by how much. [..] So in this case gravity is absolute, but the coordinate acceleration due to gravity is relative.

Thanks for bringing that up, as it is exactly that modern argument that 1916GR denies; and I had the impression that GregAshmore noticed that point, that it's basically that issue that he discovered. Einstein tried to relativise acceleration by relativising gravitation, so that it's a matter of free opinion if a rocket accelerates or not. Nowadays few people accept that view.

Now reconsider the twin paradox. If two observers experience the same amount of acceleration X time, then neither one will age any faster than the other.

That argument fails in the first version by Langevin, see my earlier remarks as well as elaborations by others.

 

[PAllen]

Gravity turns this relationship on its head. Note the force felt when accelerating. When on the surface of a gravitational mass, like Earth, the principle of equivalence tells us this weight we feel is the same force we feel when accelerating. Now if you jump off a roof then while accelerating toward the ground you feel no force, effectively weightless. Hence, in your frame of reference you are at rest, i.e., not accelerating, but the Earth is accelerating toward you. Yet everybody in the Universe can agree that your speed is changing, even if not by how much. In this case proper gravity is absolute, while how much gravity and coordinate paths are relative. So in this case gravity is absolute, but the coordinate acceleration due to gravity is relative.

Not quite everyone in the universe: someone who jumped off the roof next to you would not agree. As for someone far away, in GR, the whole concept of relative velocity at a distance is fundamentally ambiguous because you can't compare vectors at a distance in curved spacetime. So this argument is not so clear cut.

I would agree that gravity is absolute for a different reason: tidal 'forces', physically; curvature geometrically. Tidal forces are detectable in a small region.

 

[GregAshmore]

I decided that I ought to do the calculations that I should have done Friday night, before reading any of the posts or papers referenced since then. Here's my best shot at the Twin Paradox. Later this evening, or maybe tomorrow night, I'll see how what I did compares with your suggestions.

I can't say that I did this without any outside influences since Friday night. I did see that the image that George referenced looks like a spacetime diagram. That may have triggered some thoughts about coordinates.

I do all the calculations for the spacetime diagram but did not include an image of it. You all know what it looks like.

Solution of the Twin Paradox - to the degree possible knowing only the Lorentz transform and the usage of the spacetime diagram.

Given:
G1. Earth and rocket are both at rest at same position. Earth clock and rocket clock are synchronised.
G2. At time 0.0, rocket fires a pulse.
G3. Earth and rocket separate at relative velocity 0.8c.
G4. At distance 10 units from Earth, as measured in Earth frame, rocket fires a pulse.
G5. Earth and rocket approach at relative velocity -0.8c.
G6. Upon reaching Earth, rocket fires a pulse, coming to rest on Earth.
G7. Gravitational effects of mass are to be ignored.

Questions:
Q1. Both the Earth and the rocket claim to be at rest throughout the episode. Can both make good on this claim?

Q2. What are the clock readings on Earth and rocket at G6?

Q3. Are the clock readings calculated for Q2 unambiguously unique?

Solution:
The questions are with regard to kinematics only: positions and times. With one exception noted later, the dynamics of the episode need not be considered.

It will also be assumed that acceleration does not affect clocks. This assumption, together with G7, allows special relativity to be used in the attempt at a solution.

It will also be assumed that the acceleration from rest to velocity V is instantaneous. Any effect assumption this might have on the calculated clock readings will be ignored for the purposes of this exercise.

The Earth is at the origin of its own coordinate system. Likewise, the rocket is at the origin of its own coordinate system. By G1, the origins are coincident at the start of the episode. For convenience, the X axes of the two systems are colinear, and the relative velocity is along that axis, with positive V in the positive X direction.

Four spacetime events will be considered.
Event A: Corresponds to G2. Time and position are zero in both the Earth frame and the rocket frame. Velocity of the rocket frame is V. For convenience, the axes are set so that positive V is along coordinates ; positive V is made to be along the for convenience . For the moment, the Earth frame will be shown with orthogonal the velocity will be shown "to the right" on the spacetime diagram,

Event B: Corresponds to G4, on the worldline of the Earth.

Event C: Corresponds to G4, on the worldline of the rocket.

Event D: Corresponds to G6. The worldlines of the Earth and rocket meet here, and become colinear.

Notation:
T represents time.
X represents position.

Events and frames are represented by lower case letters; event followed by frame.
e represents Earth frame.
r represents rocket frame.

Example: Tbe represents the time at event B in the Earth frame.

Times are given as the distance that light travels in one unit of time. T = ct.
With this unit of time, and with velocity given as constant factor of c (v = Vc), the Lorentz transforms have the form:
X' = g(X - VT)
T' = g(T - VX)
where g = 1 / Sqrt(1 - V^2)

With V = 0.8, g = 1.667


Calculate time in Earth frame at Events B and C.
By the statement of G4, events B and C are simultaneous in the Earth frame. Measurements of distance in a frame are by definition taken at a single instant in the frame.

Time in the Earth frame at Events B and C is distance from Earth to rocket (as measured in Earth frame) divided by relative velocity.
Tbe = Tce = 10 / V = 10 / 0.8 = 12.5.


Calculate coordinates at Event B.
Xbe = 0.0 (Earth is inertial; Xae = 0.0; position in an inertial frame does not change with time.)
Tbe = 12.5 (As calculated above.)

Xbr = 1.667 * (0.0 - (0.8 * 12.5)) = -16.67
Tbr = 1.667 * (12.5 - (0.8 * 0.0)) = 20.83


Calculate coordinates at Event C.
Xce = 10.0 (By G4)
Tce = 12.5 (As calculated above.)

Xcr = 1.667 * (10.0 - (0.8 * 12.5)) = 0.0
Tcr = 1.667 * (12.5 - (0.8 * 10.0)) = 7.5

Note that calculation of Xcr confirms what is already known. Xcr must be zero because Xar = 0.0 and the rocket is inertial to this point.


Earth or rocket must change frames at velocity reversal.
The reversal of velocity in G4 must be represented by a change of frame in the spacetime diagram. Without a change of frame, the worldlines of Earth and rocket can never meet.
Either the Earth or the rocket, or both, must change frames.
The Earth cannot change frames: No unbalanced force acts on it; it is inertial.
The rocket must change frames: It is acted on by an unbalanced force; it is not inertial.


Setting up the new rocket frame.
The rocket will be in its new frame during the approach in G5. The rocket approach frame will be represented by the addition of the lower case 'p' to the notation.

The rocket must be assigned position and time coordinates in its approach frame. At the start of an exercise, coordinate values may assigned at will, due to the linearity of the Lorentz transform. In this case, the approach frame comes into play at an event in an ongoing episode, at Event C. To maintain correspondence with the physical reality, and taking into account the assumption of instantaneous acceleration, the coordinates of the rocket in the approach frame at Event C must match the coordinates of the rocket in the original separation frame at Event C.


Transformation from the rocket approach frame to the Earth frame.
The Lorentz transformation equations were derived with the origins of the two frames coincident. Therefore, Event C must be treated as a local origin for the purposes of transformation from frame to frame. Event coordinates relative to the local origin are transformed from frame to frame, as shown in the following equations.

In these equations, replace the underscore with the symbol for the event to be transformed.

To transform from the Earth frame to the rocket approach frame:
X_rp = g((X_e - Xce) - V(T_e - Tce)) + Xcrp
T_rp = g((T_e - Tce) - V(X_e - Xce)) + Tcrp

To transform from the rocket approach frame to the Earth frame:
X_e = g((X_rp - Xcrp) + V(T_rp - Tcrp)) + Xce
T_e = g((T_rp - Tcrp) + V(X_rp - Xcrp)) + Tce

As discussed above,
Xce = 10.0
Tce = 12.5

Xcrp = 0.0
Tcrp = 7.5

Calculate the coordinates of Event B in the rocket approach frame.
Xbrp = g((Xbe - Xce) - V(Tbe - Tce)) + Xcrp
Xbrp = 1.667((0.0 - 10.0) - (-0.8)(12.5 - 12.5)) + 0
Xbrp = -16.67 (Same as Xbr)

Tbrp = g((Tbe - Tce) - V(Xbe - Xce)) + Tcrp
Tbrp = 1.667((12.5 - 12.5) - (-0.8)(0.0 - 10.0)) + 7.5
Tbrp = 1.667(0 - (-0.8)(-10.0)) + 7.5
Tbrp = 1.667(0 - 8.0) + 7.5
Tbrp = 1.667(-8.0) + 7.5
Tbrp = -5.83 (Compare 20.83 for Tbr)

Calculate the coordinates of Event D.
Time for approach is same as time for separation. (Relative velocity is the same.)

Xde = 0.0
Tde = Tbe + Tbe = 25.0

Xdrp = g((Xde - Xce) - V(Tde - Tce)) + Xcrp
Xdrp = 1.667((0.0 - 10.0) - (-0.8)(25.0 - 12.5)) + 0.0
Xdrp = 1.667(-10.0 - (-10.0)) + 0.0
Xdrp = 0.0 (Confirms inertial behavior of rocket from Event C to Event D: Xcrp = 0.0)

Tdrp = g((Tde - Tce) - V(Xde - Xce)) + Tcrp
Tdrp = 1.667((25.0 - 12.5) - (-0.8)(0.0 - 10.0)) + 7.5
Tdrp = 1.667(12.5 - (-0.8)(-10.0)) + 7.5
Tdrp = 1.667(12.5 - 8.0) + 7.5
Tdrp = 7.5 + 7.5
Tdrp = 15.0 (Confirms approach time equals separation time in rocket frame.)


Answers to questions:
Q1. Both the Earth and the rocket claim to be at rest throughout the episode. Can both make good on this claim?
Yes, kinematically. Earth and rocket X coordinates are 0.0 throughout.
Whether this makes physical sense dynamically is unknown, given limited knowledge noted above.

Q2. What are the clock readings on Earth and rocket at G6?
The Earth clock reads 25.0.
The rocket clock reads 15.0.

Q3. Are the clock readings calculated for Q2 unambiguously unique?
Yes. There is only one way to construct the spacetime diagram, due to the unique non-inertial behavior of the rocket.
Visually, the Earth and rocket experiences are symmetric. Each sees the other move away and return. Nevertheless, the rocket is unambiguously younger than the Earth at reunion.