Age-Changes Caused By Instantaneous Velocity-Changes

[PAllen]

What is in dispute, is whether the curve CADO_T(t) should be determined using my reference frame for the traveler (I'll refer to it as "the {MSIRF(t)} frame"}, or using PassionFlower's frame (I'll refer to it as "the PF frame") ... or perhaps some other alternative (Doby & Gull (?)).

Actually, what is in dispute is that virtually everyone except you claims this is not a question of physics at all; it is a question of convention, convenience, and taste.

 

[Passionflower]

Do you detect any problem with PassionFlower's frame?

At this stage I think it is not very fruitful to argue as you appear to be too involved with the formulas you use and I fear you lost objectivity.

What is called 'Passionflower's frame' by the way is a rock solid concept worked out by Minguizzi, there are three interesting papers on this:

Differential aging from acceleration, an explicit formula (2004)
http://arxiv.org/abs/physics/0411233

Towards a closed differential aging formula in special relativity (2006)
http://arxiv.org/abs/gr-qc/0611076

Relativity principles in 1+1 dimensions and differential aging reversal (2006)
http://arxiv.org/abs/physics/0412010

 

[Mentz114]

The Minguizzi paper is great. It's a breath of fresh air that he ( and the other authors cited) uses the term 'differential ageing'. Please, can we ban the phrase 'twin paradox' forever from this forum.

 

[matheinste]

The Minguizzi paper is great. It's a breath of fresh air that he ( and the other authors cited) uses the term 'differential ageing'. Please, can we ban the phrase 'twin paradox' forever from this forum.

You have a seconder for that proposal.

Matheinste.

 

[Dale]

Actually, what is in dispute is that virtually everyone except you claims this is not a question of physics at all; it is a question of convention, convenience, and taste.

Yes, that is correct. Nobody has a real problem with CADO (particularly not in regions of spacetime where the CADO only goes forward). The only disagreement is with Mike's misunderstanding that his preference for CADO is more than merely a personal preference, but a physical requirement.

The CADO_H(t) curve is then a straight line, of slope 2. ...
The CADO_T(t) curve is a straight line, of slope 1/2. ...
For PassionFlower's frame, the CADO_T_PF(t) curve is a straight line, of slope 1.

Yes, the different synchronization conventions yield different results for which ages of Tom are synchronized with which ages of Sue. There is nothing surprising about that.

So, it is time for you to address my challenges now.

 

[Mike_Fontenot]

[CORRECTION]:
My posting below is incorrect. I failed to answer my own previous question correctly (about how the PF frame for Jerry compares to the PF frame for Tom).

I'll post my corrected version shortly.
[END CORRECTION]

I'm surprised that no one has pointed out the obvious problem with PassionFlower's reference frame for Tom ... the readers of this thread must be half asleep.

Here's the obvious problem with PasssionFlower's frame (a frame which has also been endorsed by DaleSpam):

Using PassionFlower's frame (the "PF" frame), we want to know what Tom and Jerry each conclude about Sue's current age, relative to their own. And we are focusing attention on only the first 20 years of Tom's and Jerry's lives (when they are co-located, and identical in essentially every way).

There is a PF frame for Tom, in which Tom is forever stationary at the spatial origin. There is also a PF frame for Jerry, in which Jerry is forever stationary at the spatial origin. We want to know how these two frames compare, in their description of Sue's current age, for the first 20 years of Tom's and Jerry's lives.

The PF frame for Tom says that, during the first 20 years of his life, Tom concludes that Sue is always his SAME age. In contrast, the PF frame for Jerry says that Jerry will always conclude that Sue is HALF his age. [The above sentence is incorrect ... see my next posting for the correct PF frame for Jerry]

So, even though NOTHING distinguishes Tom and Jerry during the first 20 years of their lives, the two PF frames say that Tom and Jerry come to different conclusions about Sue's current age, purely because Tom accelerates when he is 20, whereas Jerry never accelerates.

But how does anyone KNOW, when Tom is (say) 10 years old, that he actually WILL choose to accelerate when he is 20? What if, when they are 20, Tom and Jerry decide to flip a coin, to decide which of them accelerates? Maybe it will be Jerry, not Tom, that actually ends up accelerating. Or maybe neither of them will accelerate. Or maybe both of them will accelerate. Who can KNOW, when those twins are 10 years old, what they will choose to do when they are 20? Apparently, the PF frames know.

The PF frames are obviously NON-CAUSAL.

Such a situation would probably be considered quite reasonable, by a mystic. But no physicist worth his salt could possibly take such an absurd situation seriously.

Mike Fontenot

 

[Mike_Fontenot]

OK, here's my correction to my previous post:

The PF frame for Jerry, is exactly the same as the PF frame for Tom, during their first 20 years of life. (The two frames DO differ after that point in their lives.)

Both PF frames (for the first 20 years) say that Sue is the SAME age as Tom and Jerry.

So my problem with the PF frame is that, for a perpetually inertial observer (Jerry), it produces a different result that the standard Lorentz frame. I have already argued, in another thread, that any frame other than the standard Lorentz frame, for a perpetually inertial observer, is invalid. Here's the link to my post that discusses that issue:

https://www.physicsforums.com/showpost.php?p=2978931&postcount=75

Mike Fontenot

 

[Mike_Fontenot]

The incorrect statements that I recently made about Passionflower's frame (specifically, that Sue's age in his frame depends on what the observer does in the distant future), may actually be correct for the Dolby & Gull frame. I'm not knowledgeable about that frame, but Fredrik seemed to express the opinion, in a previous thread, that Dolby & Gull DOES behave as I mistakenly originally thought PassionFlower's frame behaved. Here is a link to Fredrik's post in that thread:

https://www.physicsforums.com/showpost.php?p=2811379&postcount=30 .

If Fredrik's interpretation of Dolby & Gull's frame is correct, then my original (incorrect) objections about PassionFlower's frame DO apply to Dolby & Gull's frame ... i.e., that Dolby & Gull's frame is NON-CAUSAL.

(DaleSpam originally planned to provide a description of the simultaneity plots for both PassionFlower's frame and for Dolby & Gull's frame, for the specific example I described earlier in this thread. I saw DaleSpam's description for PassionFlower's frame, but if he ever posted it for Dolby & Gull's frame, I must have missed it.)

Mike Fontenot

 

[Passionflower]

I'm surprised that no one has pointed out the obvious problem with PassionFlower's reference frame for Tom ... the readers of this thread must be half asleep.

Frankly I am no longer interested in arguing, since as I wrote before I fear you lost objectivity. Based on this I suggest you attempt to remove yourself a bit from 'your' CADO equations and try to find some more objectivity.

A topic on differential aging as described by Minguzzi would be very interesting and helpful but if every other posting has the acronym CADO more than three times I for one will not be interested in participating.

 

[Dale]

The PF frames are obviously NON-CAUSAL.

Such a situation would probably be considered quite reasonable, by a mystic. But no physicist worth his salt could possibly take such an absurd situation seriously.

I have never heard the term "non-causal" applied to a reference frame. You will have to define it clearly and unambiguously (not your strong point) and demonstrate why a "non-causal" reference frame is invalid (i.e. makes incorrect predictions about physical experiments). You can call something a nasty-sounding name like "non-causal" and imply some sort of guilt by association by linking it with mystics, but any "physicist worth his salt" would avoid falling for your logical fallacies and ask you to define your terms and demonstrate your claim.

In addition, once you fix the origin of the "PF" frame the lines of simultaneity are fixed regardless of any future or past accelerations, and the D&G method only assigns coordinates to events in the past light cone of the observer, so it is hard for me to imagine any definition of "non-causal" that would make sense. If anything the CADO seems more likely to be "non-causal" since it assigns new coordinates to events outside of the past light-cone as the observer accelerates.

In any case, I see that you are still trying to change the subject from the fact that you have failed to answer my repeated challenge. It makes me wonder what you are trying to hide. I suspect that you know full well either that your position is wrong or that you are incapable of doing the math required to prove your position right.

 

[Mike_Fontenot]

In my response to PAllen's post,

https://www.physicsforums.com/showpost.php?p=2962578&postcount=56 ,

specifically the quote I gave from his posting:

> [...]
> Imagine the 'instant turnaround twin' is looking through a telescope at an
> image of clock with the 'stationary twin'.
> [...]
> Passing the point of instant turnaround the turning twin sees:
> [...]
> - clock appears further away
> [...] ,

I said:

As long as the MAGNITUDE of the velocity stays the same before and after the velocity change (as it does in this example), then the apparent size of the image is exactly the same immediately after and immediately before the instantaneous velocity change. The distance to the home twin, according to the traveler, is L/gamma, where L is the distance according to the home twin. Gamma has the same value for v = +0.866 as it does for v = -0.866.

My last two sentences were correct, but my first sentence may well be incorrect ... my apologies to PAllen. I have never investigated how the two-dimensional image of some distant object (when viewed through a telescope) would APPEAR to an accelerating observer ... it's a question that I consider to be a distraction from more important issues, and of minor importance. It is the TRUE current distance (according to the traveler) to the distant object, that I think is of fundamental importance, not the APPARENT distance.

The image being viewed through a telescope is an OLD image, and the solid angle subtended by the object gives an incorrect indication of the CURRENT distance to the object.

This issue came up in another thread. Here are two postings of mine from that thread:

https://www.physicsforums.com/showpost.php?p=2987503&postcount=24

and

https://www.physicsforums.com/showpost.php?p=2988591&postcount=30 .

Mike Fontenot

 

[PAllen]

In my response to PAllen's post,

https://www.physicsforums.com/showpost.php?p=2962578&postcount=56 ,

specifically the quote I gave from his posting:

> [...]
> Imagine the 'instant turnaround twin' is looking through a telescope at an
> image of clock with the 'stationary twin'.
> [...]
> Passing the point of instant turnaround the turning twin sees:
> [...]
> - clock appears further away
> [...] ,

I said:



My last two sentences were correct, but my first sentence may well be incorrect ... my apologies to PAllen. I have never investigated how the two-dimensional image of some distant object (when viewed through a telescope) would APPEAR to an accelerating observer ... it's a question that I consider to be a distraction from more important issues, and of minor importance. It is the TRUE current distance (according to the traveler) to the distant object, that I think is of fundamental importance, not the APPARENT distance.

The image being viewed through a telescope is an OLD image, and the solid angle subtended by the object gives an incorrect indication of the CURRENT distance to the object.

This issue came up in another thread. Here are two postings of mine from that thread:

https://www.physicsforums.com/showpost.php?p=2987503&postcount=24

and

https://www.physicsforums.com/showpost.php?p=2988591&postcount=30 .

Mike Fontenot

May I ask how you propose to know the true current distance? Since what you see now, the other twin may have blown up, or started moving rapidly in a random direction. The length that has the property you claim (doesn't change before and after turnaround) would have to be defined as follows:
---
assuming the other twin continues to move at that same speed as I last saw (measured e.g. by their redshift), continuing until my 'now', then I can claim this distance doesn't change before and after turnaround (noting that my definition of 'now' for distant points just changed a lot as I shifted frames).
---

To me, that's not physics, it's just an exercise in "let's pretend".

 

[terrence777]

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[Mike_Fontenot]

[...]
In my response to PAllen's post,

https://www.physicsforums.com/showpost.php?p=2962578&postcount=56 ,

I said:

[...]
As long as the MAGNITUDE of the velocity stays the same before and after the velocity change (as it does in this example), then the apparent size of the image is exactly the same immediately after and immediately before the instantaneous velocity change. The distance to the home twin, according to the traveler, is L/gamma, where L is the distance according to the home twin. Gamma has the same value for v = +0.866 as it does for v = -0.866.
[...]

My last two sentences were correct, but my first sentence may well be incorrect ... my apologies to PAllen. I have never investigated how the two-dimensional image of some distant object (when viewed through a telescope) would APPEAR to an accelerating observer ... it's a question that I consider to be a distraction from more important issues, and of minor importance. It is the TRUE current distance (according to the traveler) to the distant object, that I think is of fundamental importance, not the APPARENT distance.

The image being viewed through a telescope is an OLD image, and the solid angle subtended by the object gives an incorrect indication of the CURRENT distance to the object.
[...]

I decided to do an analysis of the angle subtended by the diameter of the (assumed inertial) earth, at some instant when two perpetually inertial observers (with velocities of beta = +0.866 and beta = -0.866, with respect to the earth) happen to be momentarily co-located, far from the earth. In particular, I wanted to determine that subtended angle, according to each of the three inertial frames (the Earth frame, and the two "traveling" observers' frames). Those three frames disagree about the angle. The analysis is elementary, but a bit tricky. Like most calculations in special relativity, it is easy to get wrong.

Here's a specific result from the equation I derived (which relates the subtended angles, according to the three reference frames):

Suppose the subtended angle, according to the Earth frame, is 10 degrees.

Specifically, we draw a diagram (in the Earth frame) with a straight line between the center of the Earth and the point of co-location of the two traveling observers. That straight line lies along the direction of relative motion of the travelers with respect to the earth. We arbitrarily choose any Earth diameter that is perpendicular to the above straight line. Then, we draw a straight line between each end of that Earth diameter, to the point of co-location. The angle between those latter two lines is taken to be 10 degrees.)

The inertial traveler, moving AWAY from the Earth at speed 0.866c, will measure the angle subtended by the Earth diameter to be about 33 degrees. The inertial traveler, moving TOWARD the Earth at speed 0.866c, will measure a subtended angle of about 2.7 degrees.

So I definitely WAS wrong when I said that the outbound and inbound inertial observers would see the SAME size telescopic image at the instant when they are co-located. (If they receive a TV image, at the instant of co-location, giving the home twin's age at the time of image transmission, they each WILL receive exactly the same TV image. I had originally thought that the image through a telescope would likewise be the same for both travelers, but that's not the case.)

But even though the travelers see different images through the telescope at the instant of co-location, the CURRENT separation of the travelers, from the earth, at that instant of co-location, WILL be the same distance, according to both travelers' inertial reference frames: they will each agree that the separation is HALF what the Earth inertial reference frame says it is (because gamma = 2 for v = +-0.866). This result follows from the Lorentz equations, or (much easier and quicker) from either the length-contraction result, or else the time-dilation result combined with "velocity reciprocity".

If the two travelers try to DIRECTLY infer their CURRENT distance to the Earth from the angle subtended by the image of the Earth in their telescopes, they will be badly mislead. The image that each of them sees through the telescope is an old, out-of-date image ... it shows them what the Earth WOULD have looked like at some earlier time in their pasts, if light speed were infinite. For the outbound traveler, he was CLOSER to the Earth in his past, so he sees an image that is LARGER than it would be if he could see the Earth as it currently is. The inbound traveler was FARTHER from the Earth in his past, and so he sees an image which is SMALLER than it would be if he could see the Earth as it currently is.

If the travelers want to deduce the separation from the telescopic image, the old, out-of-date, directly deduced separation CAN be ADJUSTED to give the correct CURRENT separation. If they do that adjustment correctly, they will get a separation which agrees with the Lorentz equations (or with the length-contraction result). But that's a very tortuous way to get the same result that can be obtained much easier and quicker using length-contraction.

Mike Fontenot

 

[Mike_Fontenot]

CORRECTION: The angle I computed in my previous posting was the angle subtended by the RADIUS of the earth, NOT the DIAMETER of the earth. Sorry for the confusion.

Mike Fontenot