Implications of the statement "Acceleration is not relative"

Mentz114

It is a well known fact that an inertial worldline between two events has a larger proper interval than any non-inertial worldline connecting the events. I'll see if I can find a proof.

See section 6.6 of this lecture

http://physics.ucsd.edu/students/cou...61.14jan11.pdf

DaleSpam

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.

DaleSpam

Sure it can. See here:
http://math.ucr.edu/home/baez/physic...eleration.html

You can always make a coordinate system where any given object is permanently at rest. You just have to be very detailed about your specification of the coordinate system since there is no "standard" meaning.

You also cannot apply formulas derived for inertial frames in non-inertial frames. They are both legitimate, but not equivalent.

stevendaryl

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:

[itex]L = \int \sqrt{1+m^2} dx[/itex]

where [itex]m = \dfrac{dy}{dx}[/itex] is the slope of the curve [itex]y(x)[/itex]

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

[itex]\dfrac{d}{dx} ( \dfrac{m}{\sqrt{1+m^2}}) = 0[/itex]
which has the solution that [itex]m[/itex] 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:

[itex]\tau = \int \sqrt{1-\dfrac{v^2}{c^2}} dt[/itex]

where [itex]v = \dfrac{dx}{dt}[/itex] is the velocity of the path [itex]x(t)[/itex]

This formula is good using any inertial coordinate system.

Finding the path [itex]x(t)[/itex] that makes [itex]\tau[/itex] 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:

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

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

harrylin

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

harrylin

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.
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.
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.

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.
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.

DaleSpam

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

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

It depends on how you define "gravitational field". If you drop an object, and it follows a trajectory [itex]h = -\frac{1}{2}g t^2[/itex] for some constant [itex]g[/itex], then many people would call [itex]g[/itex] 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 [itex]g=0[/itex] (no gravitational field), then a change of coordinates to accelerated coordinates can make [itex]g[/itex] nonzero.

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

DaleSpam

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 prefered 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

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

DaleSpam

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.

In the same discussion he 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

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

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

GregAshmore

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/physic...eleration.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

How about this one:



from this thread.

GregAshmore

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.

DaleSpam

See post 80.

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.

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.