New Arguments In A Possible Proof That Negative Ageing Doesn't Occur In Special Relativity

[Neddy Bate]

Mike,

Somewhere online you should be able to find a GR solution to the SR twin 'paradox' using pseudo-gravity, and the equivalence principle, rather than just SR. Gravitational potential depends on distance, so it should give the same answer, I would think.

But why? You already know how to draw Minkowski diagrams, so just look at the rules for drawing lines of simultaneity:

When the velocity changes from v=+0.866 to v=-0.866 the simultaneity lines should look symmetrical, showing that there is no absurd and inconsistent asymmetry as you have described above.

Another way is to simply look at the SR equations rather than starting with the GR equations and having to work backwards from there.

Here is a simple SR equation:
t = γ(t' + (vx' / c²))
The above equation is the given SR relationship between a time in an unprimed frame (t) and the time in a primed frame (t') at various locations along the x' axis.

For simplicity, choose the time t'=0 and use units where c=1:
t = γ(0 + (vx' / 1))
t = γvx'

This demonstrates that for the case t'=0, the times (t) are directly related to their location as measured by the x' axis. For a constant v, clearly t varies with x'. Gamma is always positive, but velocity can be positive or negative. So for t'=0 and a positive velocity, t is increasingly positive on the positive x' axis, and increasingly negative on the negative x' axis. But for t'=0 and a negative velocity, that is reversed, so t is increasingly positive on the negative x' axis, and increasingly negative on the positive x' axis. The asymmetry you describe is not there, so you have made a mistake someplace.

(Note: negative t values here only mean that they were an earlier time than the time assigned t=0.)

 

[Ssssssss]

Ugh I screwed up where the factors of \(c\) go and ran out of edit time so I'm gonna ask for my previous post to be deleted. The rest of this is an exact copy except with the dimensional analysis done correctly.

No, at the beginning of the scenario, right before he accelerates, they are separated by the distance "d" lightyears. It is unspecified, but it is non-zero. Regardless of how big you make "d", her age doesn't decrease (according to him) when he accelerates away from her, but her age suddenly increases by a large amount (according to him) when he accelerates toward her.

Then your maths is wrong. Let your observer be at rest at \(x=c^2/A\) in an inertial frame until time 0 at which time he starts accelerating with constant proper acceleration \(A\) (the initial \(x\) value makes the maths a bit neater but you can use any value you like) so before the acceleration starts his worldline is \(x=c^2/A\) and after it is \(x^2-c^2t^2=c^4/A^2\). During the acceleration phase we can write \[\frac{dx}{dt}=\frac{Act}{\sqrt{A^2t^2+c^2}}=\frac{c^2t}{x}\]which means that we can write a four vector tangent to the line as \((cdt/dt,dx/dt)=(c,c^2t/x)\) which isn't the four velocity because it isn't normalised but is parallel to it. A vector perpendicular to this in the Minkowski sense (remember that this is \(v_tu_t-v_xu_x-v_yu_y-v_zu_z=0\) for two vectors \(u\) and \(v\)) is \((c,x/t)\) and this must lie parallel to the observer's instantaneous inertial rest frame's spatial planes. So the equation of the MCIF spatial plane of the accelerating observer at time \(t=T\) must have \(dx/dt=X/T\) where \(X\) is the x position of the accelerating observer at that time and it must pass through the point \(t=T,x=X\) and it's obvious that this is satisfied by \(x=(X/T)t\) or rearranging into a more usual form for drawing on a Minkowski diagram \(t=(T/X)x\).

Let's see what this means. It means that if you have an observer who is inertial in this frame and has a positive \(x\) coordinate then if we consider the accelerating observer's MCIF simultaneity planes at two times then the later plane passes through the inertial worldline later so both time coordinates are increasing so the inertial observer's coordinate age according to the accelerating one is increasing. It's only when we get to negative \(x\) coordinates that we get the intersections in the reverse order because every one of the accelerating observer's MCIF simultaneity planes goes through the origin but it's not surprising that the coordinates go bad when you extend them through that coordinate singularity at the origin. That's your bad mapping choices not a problem with physics. We also don't care because our accelerating observer can never receive information from there until he stops accelerating at which point he gets parallel MCIF spatial planes again although there's still a dodgy region where your coordinates don't work.

So as I have said several times you are wrong. The MCIF age of a person within \(c^2/A\) of your accelerating observer will monotonically increase and it only decreases at larger distances where the coordinate system is ill-defined and you are wrong to be using it. If you want a coordinate system that covers all of spacetime for an arbitrarily but not eternally accelerating observer then I'd recommend Dolby and Gull's radar coordinates: https://arxiv.org/abs/gr-qc/0104077.

Edit: if you pick a less contrived initial x value than I did then you will get a few more constants in the maths that just move the point where all the accelerating observer's MCIF spatial planes cross but will not otherwise change anything else about the argument.

 

[Neddy Bate]

...so how could it produce an erroneous result when its acceleration equivalent (using the equivalence principle) is used? Is it possibly because the gravitational time dilation equation is not intended to be used with gravity being switched on and off, and therefore we can't use it to describe what happens when the observer switches his acceleration on and off?

I do think there may be a problem in assuming that the equivalence principle can be used in this way. I have seen a similar thing done in clever explanations for the twin 'paradox' where they make an analogy to an imaginary gravitational field that switches on over the whole universe during the times when the rocket's engines burn. This explanation has the rocket remaining 'stationary' while everything else in the universe goes into free fall in that gravitational field.

Then, at the turnaround, they use the large distance between the rocket and the home-twin to explain that the home twin's age will advance much more than the rocket's, due to the home-twin being so high up in the gravitational field. But if the rocket then fires its engine in the opposite direction, making a quick double-turnaround, this analogy fails to explain how the home-twin's age could possibly go back to being young again as it was before. Yet that is what a proper SR analysis of the quick double-turnaround would say should be the case.

So, I do think that this kind of use of the equivalence principle is more of a parlor trick than a proper use of physics. GR should reduce to SR in the case where there is no gravity, but that does not mean that gravity can be substituted for acceleration any time one wishes, and still expect to get the same results as pure SR would give.

 

[Ssssssss]

This explanation has the rocket remaining 'stationary' while everything else in the universe goes into free fall in that gravitational field.

Agree. And that means that the inertial observer is changing altitude and velocity in the implied gravitational field and you need to work out quite carefully what the potential actually is and it certainly isn't the uniform field Mike_Fontenot is using because that doesn't give the right redshift.

 

[Ssssssss]

But if the rocket then fires its engine in the opposite direction, making a quick double-turnaround, this analogy fails to explain how the home-twin's age could possibly go back to being young again as it was before.

It's just one twin aging slowly then the other one aging slowly I think and the exact result is open to interpretation anyway unless one twin does another acceleration so they can meet up.

How do you post images on mobile? Or is that desktop only?

 

[Neddy Bate]

It's just one twin aging slowly then the other one aging slowly I think and the exact result is open to interpretation anyway unless one twin does another acceleration so they can meet up.

How do you post images on mobile? Or is that desktop only?

I upload my image to imgur.com and then post the automatically-assigned address of the image between image tags like this:

[img]https://i.imgur.com/Kodfr.jpg[/img]

I'm not sure if its different on mobile.

imgur.com doesn't require any sign in, and even a clipboard image can be cut & pasted into their upload window (or you can drag & drop, upload file, etc.).

https://imgur.com/upload

 

[Ssssssss]

Thanks Neddy. I managed to do the imgur thing so if I haven't screwed the rest of that up here's a Minkowski diagram of the situation I described in #83 above

The axes are unlabelled but time is vertical as usual and the blue line is the accelerating observer and the yellow line is the asymptote of his worldline and the dotted lines are instantaneous "now" planes at easy to draw intervals rather than an even spacing in blue's proper time or anything. However you can easily see that an observer to the right of the origin will encounter these planes in the same order as the accelerating observer and one passing through the origin will encounter multiple planes simultaneously (a coordinate singularity) and an observer to the left will encounter the planes in reverse order (as well as the not shown planes from the inertial phase) but the map isn't valid there due to the coordinate singularities.

 

[Ssssssss]

Here's the situation with an acceleration and deceleration following each other.

The blue line is the accelerating observer and the green line is an inertial observer and the dotted lines are boundaries between the MCIF sense of "during" the initial inertial phase and each acceleration phase and the final inertial phase but this time there are two regions where these coordinates don't work due to overlaps which I've hatched in red and also I've sketched on the direction of "down" for the pseudogravitational fields as light blue arrows and you can see that the green line is initially "below" the blue line so "ages slowly due to time dilation" and then is "above" it so "ages fast due to time dilation". I don't like this approach because if you want numerical answers you need to determine both the pseudoaltitude and speed of the green line in the accelerating frame and calculate the combined time dilation at each instant and integrate which isn't easy so I'd rather work in the inertial frame or come up with some better coordinates that don't overlap.

 

[Neddy Bate]

Thanks Neddy. I managed to do the imgur thing so if I haven't screwed the rest of that up here's a Minkowski diagram of the situation I described in #83 above

The axes are unlabelled but time is vertical as usual and the blue line is the accelerating observer and the yellow line is the asymptote of his worldline and the dotted lines are instantaneous "now" planes at easy to draw intervals rather than an even spacing in blue's proper time or anything. However you can easily see that an observer to the right of the origin will encounter these planes in the same order as the accelerating observer and one passing through the origin will encounter multiple planes simultaneously (a coordinate singularity) and an observer to the left will encounter the planes in reverse order (as well as the not shown planes from the inertial phase) but the map isn't valid there due to the coordinate singularities.

Oh, that's interesting, now I understand exactly what you meant by coordinate singularities.

Below is a diagram of the "double turnaround" that I was talking about before. The one on the left is the single turnaround, and the one on the right is the double turnaround.

On the left is the standard twin scenario, where the traveler says the home twin gets much older during the turnaround. She goes from 10 years old to 70 years old while the traveler is 20 years old, according to his simultaneity change.

On the right is the same thing, except very shortly after the turnaround, he does a second turnaround. So in this case, she goes from 10 years old to 70 during the first turnaround, but after the second turnaround she is much less than 70, according to him. That type of case is where Mike_Fontenot's "negative aging" idea comes from.

You might say that this is invalid, because of a coordinate singularity, but what is the use of simultaneity lines in SR if not this? Surely the traveler can use them to assess her age throughout his entire journey, and absolutely no problems result from his doing so, correct?

 

[Mike_Fontenot]

[...]
The MCIF age of a person within \(c^2/A\) of your accelerating observer will monotonically increase and it only decreases at larger distances where the coordinate system is ill-defined and you are wrong to be using it.
[...]

Where did you get the above information? Is it in Taylor and Wheeler?

 

[phyti]

Mike;

fig.1
A and B have no relative motion.
At t0, B moves relative to A with a gradual acceleration to velocity v, and sends a light signal to A.
If B moves toward A (path 1), he will receive the return signal early compared to path 0.
The interval between signals will be shorter, B's perception will be, the A-clock rate is faster.
If B moves away from A (path 2), he will receive the return signal later compared to path 0.
The interval between signals will be longer, B's perception will be, the A-clock rate is slower.
fig.2
At t3, after attaining v, B maintains an inertial frame.
Comparing return times of fig.2 with fig.1, the short intervals are shorter and the long intervals are longer. These are examples of doppler shift, a perception of changing frequencies/clock rates, but not aging.
The A-clock has not changed, since a change in B's motion cannot affect distant clocks.

Also included a pdf where the acceleration is removed for an SR comparison.

 

[Ssssssss]

Where did you get the above information? Is it in Taylor and Wheeler?

I do believe that I just derived where the simultaneity planes cross in the post you quoted from and you are obviously aware of the problems with non-diffeomorphic maps because you tried to claim it wasn't an issue in SR.

 

[Ssssssss]

what is the use of simultaneity lines in SR if not this?

The smartarse relativist's answer is that they aren't useful for anything and we only care about light cones and worldlines and there is truth in that which you touched on when you said "no problems result from his doing so". A more nuanced answer is that you are correct that it doesn't matter what age the inertial observer is now because what now means elsewhere in space is something you can choose but as I said to Mike_Fontenot earlier that is not the only thing we use coordinates for. Look at this Minkowski diagram.

It's a standard twin paradox with Earth in blue and traveller in red and I've added in pink and purple the boundaries of "during" the in and out legs according to the traveller's MCIF and I've also added two asteroids in light and dark green and my question is do they hit Earth (in 1d the answer is yes but let's pretend they are slightly out of the plane and have some small velocity in that direction). With the MCIF chart you cannot predict the trajectory of the light green asteroid once it crosses the pink line because the chart doesn't cover the left wedge between the pink and purple lines so you cannot define a trajectory in that region so you cannot predict where the asteroid is and you cannot predict the trajectory of the dark green asteroid because where it crosses the pink line isn't where it crosses the purple line and you have a discontinuous trajectory. You can of course use the global inertial frame to fix both of these issues but that's the point that you need the global chart to understand and fix the issues with your MCIF chart. Actually if you don't use an instantaneous turnaround the problems on the left mostly go away because the now planes sweep across the region they skip in this diagram but you always have issues on the right with multiple coverage. This is a lot more of an issue in GR where it's possible that you don't have a global coordinate system and it can be tricky to work out where different local coordinate systems overlap or don't meet whereas in SR you can always use a global inertial frame to paper over any problems.

 

[phyti]

S7;

The paper "On Radar Time...", Dolby & Gull, presents a fantasy of simultaneity. Dealing only with the x axis of simultaneity (aos), the observer has to establish a simulated system of simultaneity relative to his ref. frame.
fig.1
Two clocks c1 and c2 are synchronized to a master clock midway between the two, using em signals, per the SR convention. The graphic shows this process for A as viewed by U, resulting in the green aos. Constant velocity segments are used in place of curved segments for clarity.
fig.2
An intermediate signal during the change of v results in out of synch clocks.
1. The aos is an imaginary mathematical convention.
2. The spatial extent of the aos is determined by the separation of the clocks.
3. If A changes speed as at t1, he has to resynchronize his clocks.
Based on these facts, the rotation of the aos, and its extent across space is incorrect as typically shown in publications.
The synch convention applies to a single inertial frame and there is no such thing for the universe of objects with a wide range of motions. In the case of curvilinear motion, there is no time interval with constant velocity (speed, direction, or both), to perform the synchronization, beyond a local approximation.
As an example, the local GPS system requires daily corrections.

 

[Mike_Fontenot]

The paper "On Radar Time...", Dolby & Gull, presents a fantasy of simultaneity.

If you want a coordinate system that covers all of spacetime for an arbitrarily but not eternally accelerating observer then I'd recommend Dolby and Gull's radar coordinates: https://arxiv.org/abs/gr-qc/0104077.

I've never been able to get a straight answer from Steve Gull as to what the ACD ("Age Correspondence Diagram") looks like for his "Radar Method" in the case of the standard twin paradox scenario (with an instantaneous velocity change at the turnaround).

The "Age Correspondence Diagram" is just a plot of the traveling twin's (his) age on the horizontal axis, and the home twin's (her) age, according to him, on the vertical axis. (It gives her age, according to him, at each instant of his life). For example, for the CMIF simultaneity method, the ACD starts out from the origin, and goes upward in a straight line of slope = 1/gamma (where gamma is greater than zero and less that 0.5), until it reaches the turnaround point. (gamma is equal to 1/sqrt(1 - v*v)). Then, when he instantaneously changes his velocity from +v to -v, the plot goes vertically upward (by an amount that depends on v and also on their distance apart). Finally, the plot goes upward to the right, with a slope of gamma again. That segment ends at their reunion, and their two ages there must match what her much simpler calculations give.

So, my question (for both of you) is, what do you think the Dolby and Gull method gives for the ACD in the above scenario? Steve Gull never would tell me. I THINK (but don't now for sure) that his method gives an ACD that starts out like the CMIF diagram, but part way to the turnaround, it changes to a steeper (but not vertical) straight line, and continues that line until part way back from the turnaround, and then finishes with a slope of gamma, like the CMIF diagram does.

What do YOU (each of you) think the ACD for the Dolby&Gull method looks like?

Regardless of what Gull's ACD looks like, the biggest problem with his method is that it is NON-CAUSAL ... a portion of the ACD before the turnaround is reached depends on what the traveler decides to do at the turnaround, and that's not known for sure until he gets there. That's a disqualifier for me.

 

[Mike_Fontenot]

I do believe that I just derived where the simultaneity planes cross in the post you quoted from and you are obviously aware of the problems with non-diffeomorphic maps because you tried to claim it wasn't an issue in SR.

I would still like to either SEE your derivation of this, or else be given a reference of where it's been derived:

"The MCIF age of a person within c2/A of your accelerating observer will monotonically increase and it only decreases at larger distances."

I would think that Taylor and Wheeler would have derived that result. I suspect they did, and if so, I doubt that they said the decreasing age portion was illegitimate. After I saw it in your post, I ran an old program that I wrote about 25 years or so ago, which was based on the information in Taylor and Wheeler about how to handle piecewise-constant accelerations. It confirmed that there is no negative ageing until their separation is greater than 1/A, where A is the acceleration in ly/y/y. Until then, her age slowly increases. I had never realized before that that happens.

If you take the case where the acceleration A approaches infinity, but occurs for only an infinitesimal time, you get the idealization of an instantaneous velocity change. I.e., the acceleration is a Dirac delta function. In that case, 1/A is essentially zero, so the duration of the slow positive ageing period goes to zero, and the negative ageing starts essentially immediately and is instantaneous. That simplification is very useful, and it is used frequently in analyzing twin paradox scenarios, with both directions of the velocity change, giving both positive and negative instantaneous age changes. But you contend that the instantaneous age REDUCTIONS are somehow illegitimate, whereas the instantaneous age INCREASES are legitimate.

In your opinion, WHAT happens when the traveler (he) makes two back-to-back instantaneous velocity changes of alternating directions, with no time lapse in between? I.e., suppose his velocity has been +v for a while (they are getting farther apart), and he instantaneously changes it to -v. THEN, he immediately changes his velocity to +v. What is the overall result of that, in terms of the ACD (Age Correspondence Diagram)?

 

[Mike_Fontenot]

... it doesn't matter what age the inertial observer is now because what now means elsewhere in space is something you can choose ...

For a perpetually-inertial observer, there is only one correct answer to the question "What is the current age of that distant person right now, according to you (the perpetually-inertial observer)?". It is the answer given by his array of synchronized clocks and "helper observers" stationed with those clocks. ANY different answer will contradict those direct measurements.

So, given the above, is it reasonable to tell an observer who sometimes accelerates that for him, there ISN'T one single correct answer to the above question? I think it would be absurd to contend that. There IS one and only one answer to that original question for an accelerating observer. We just don't know for sure what that answer is.

The reason I am so interested in the equation obtained from the equivalence-principle version of the gravitational time dilation equation, with it's array of clocks moving with the accelerating observer, is that for the first time, it gives the accelerated observer an array of clocks similar to what the perpetually-inertial observer has. It has the potential of giving him a certain answer to that original question about the current age of the distant person. But so far, I haven't been able to make any sense out of it.

 

[phyti]

forum;

If we ignore the means of B following the discontinuous speed profile from 0 to 10, the simplest solution is to count the periodic light signals.
At .6c, gamma is 1.25.
At reunion:
The A-clock sent 10, all received by B.
The B-clock sent 8 (not shown), all received by A.
They both agree, B is younger than A.

 

[phyti]

Neddy;

The green aos points to At=3.2 which corresponds to Bt=4.0, time dilation of .8 in agreement with v=.6. The problem, B does not continue along the magenta segment, but reverses direction at Bt=4.0. He intercepts the blue return signal early at Bt= 4.6, reducing the round trip time by 1.8. Without any adjustments, (4.6+1.6)/2 = 3.1 corresponds to At=3.2, a revised aos. B's perception is A's clock is running faster. Another case of B changing his motion after emitting a light signal, which alters his time of events.

It's my opinion that the triangle was intended to represent the departure and return of a 2nd observer without further detail, in the same manner as Einstein did in his paper, and as a child draws a stick figure to show a human form.

 

[Neddy Bate]

With the MCIF chart you cannot predict the trajectory of the light green asteroid once it crosses the pink line because the chart doesn't cover the left wedge between the pink and purple lines so you cannot define a trajectory in that region so you cannot predict where the asteroid is and you cannot predict the trajectory of the dark green asteroid because where it crosses the pink line isn't where it crosses the purple line and you have a discontinuous trajectory. You can of course use the global inertial frame to fix both of these issues but that's the point that you need the global chart to understand and fix the issues with your MCIF chart.

The way I see it is that before the turnaround, the traveler is just in an an inertial frame. So, he has the full chart of that inertial frame at his disposal which he can use to predict the asteroid's path. After all, before the turnaround, he might not even know that the captain of his spaceship has a plan to turnaround. Then, after the turn around, the traveler is once again in an inertial frame, and he can use the full chart of that inertial frame as well. Perhaps that is what you mean by, "using the global chart?"

Actually if you don't use an instantaneous turnaround the problems on the left mostly go away because the now planes sweep across the region they skip in this diagram but you always have issues on the right with multiple coverage.

I still don't see any problems. The time of impact (or near miss) of the asteroid can always be expressed in the coordinates of any of the inertial frames. The instantaneous turnaround does create an apparent discontinuity that is not really there, but even that doesn't matter if he just specifies which inertial frame he is using, rather than trying to stitch them into one.