Implications of the statement "Acceleration is not relative"

GregAshmore

I have downloaded the paper.

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

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

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:

stevendaryl

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 [itex]x[/itex] and [itex]t[/itex] to make the mathematics as simple as possible.

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

[itex]x(t) = \sqrt{(L_0)^2 + c^2 t^2}[/itex]

Time [itex]t_1[/itex] is chosen so that [itex](t_1)^2 = \dfrac{L_1^2 - L_0^2}{c^2}[/itex]

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

[itex]v(t) = \dfrac{c^2 t}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex]

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

[itex]\sqrt{1-\dfrac{v^2}{c^2}} = \dfrac{L_0}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex]

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

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 [itex]\int dt = 2 t_1[/itex]. 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:
[itex]x(t) = \sqrt{L_0^2 + c^2 t^2}[/itex]

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

[itex]X = \sqrt{x^2 - c^2 t^2}[/itex]
[itex]T = \dfrac{L_0}{c} arctanh(\dfrac{ct}{x})[/itex]

where [itex]arctanh[/itex] is the inverse of the hyberbolic tangent function.

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

[itex]X(T) = L_1\ sech(\dfrac{cT}{L_0})[/itex]

where [itex]sech[/itex] is the hyperbolic secant function.

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

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

where [itex]g_1 = \dfrac{c^2 L_1}{L_0^2}[/itex]

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

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

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

[itex]d\tau = \sqrt{\dfrac{X^2}{L_0^2} - \dfrac{V^2}{c^2}} dT[/itex]

where [itex]V = \dfrac{dX}{dT}[/itex]

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

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

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

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

PeterDonis

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.

DaleSpam

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 travelling 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.

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

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 wich 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

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

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

TrickyDicky

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

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 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

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

DaleSpam

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

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:

[itex]\dfrac{dV^\mu}{d\lambda} = 0[/itex]

(where [itex]\lambda[/itex] is the affine parameter for the path).

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

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.

DaleSpam

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.

DaleSpam

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