I garbled the answer for the circular case. Sorry. I should have said the linear acceleration causes the traveling twin and the home twin to disagree about their respective ages when they are separated. But in the circular motion case, they agree about their respective ages, even while they are separated.

I do the analysis for the circular motion case in the lower, older portion of my webpage:

https://sites.google.com/site/cadoequation/cado-reference-frame

If you scroll down far enough, you will get past my recent work on my new simultaneity method, and get to my old work on the co-moving-inertial-frames simultaneity method , which I called "the CADO frame". The CADO frame features a "CADO Equation", that is NORMALLY written

CADO_T = CADO_H - v * L,

where the asterisk just denotes multiplication of two scalars. But in Section 12, on the CADO Equation for 2 or 3 Spatial Dimensions, the equation becomes

CADO_T = CADO_H - v "dot" L,

where v and L are now vectors. In the circular case, v and L are perpendicular vectors, and so their dot-product is zero. Therefore CADO_T (the age of the home twin, according to the traveler) is equal to CADO_H (the age of the home twin, according to the home twin), so they agree about their two ages.

The trouble with this, as you would find if you watch the other video, is that the same "paradox" is observed in a (thought) experiment in which acceleration plays no part. — tim wood

There are two "red herring" examples that claim to prove that acceleration doesn't cause the time difference in the twins' ages at the reunion.

One is the example that uses three perpetually-inertial observers: the home twin, and two unrelated people. The fist unrelated person takes the place of the traveler on the outbound leg, and the second one takes the place of the traveler on the inbound leg. The latter is younger than the home twin at the "reunion" by the same amount as the twins in the original scenario. The fallacy is that in the revised case, no one is surprised at the result, so there is no paradox to resolve.

The second red herring is the case where the traveling twin circles the home twin, at a high constant speed. When he returns, she isn't older. But it's not hard to show that whenever the motion is perpendicular to the line connecting the two twins (which is always is, in the circular case), their rates of ageing will be equal.

I have a question for you. I was told acceleration or the application of force on the male twin is what solves the twin paradox. Please explain. — Michael Lee

Everyone knows what is observed at the reunion: The home twin is older than the traveling twin. The controversy is over the traveler's conclusions about what the home twin's current age is during the trip. There are at least five different answers to that question, with no consensus, even 100-plus years after special relativity was discovered. One answer is that simultaneity at a distance is a completely meaningless concept. The other four answers say simultaneity at a distance isn't meaningless, but disagree about how the home twin's current age varies during the trip. Acceleration by the traveler IS the cause of the age difference at the reunion, despite the various arguments to the contrary. For details, see my webpage:

https://sites.google.com/site/cadoequation/cado-reference-frame

Would like to get your responses to my responses to you from those other posts yesterday..... — Edgar L Owen

Too busy right now with work on my new simultaneity method. Sorry.

Sorry ... I don't know why it did that to you. I've been banned for life on Physics Forum (since more than 10 years ago), so at least you didn't get bashed that bad!

I could not make any sense of the new method. — noAxioms

I might be able to help you understand my simultaneity method a little by just saying that my new method and the proof both share a focus on determining by how much the home twin (she) ages while the pulse she sends the traveling twin (him) is in transit. My method basically says that when that pulse is partially in the left half of the Minkowski diagram (with the time axis plotted horizontally), and partially in the right half of the Minkowski diagram, that her ageing during the transit in the left half needs to be determined by the perpetually-inertial observer in the left half of the diagram, and likewise for the right half. Those two components of her ageing during the pulse's transit are then added together to get the accelerating twin's conclusion about her total ageing during the transit. Hope that helps a little.

I could not make any sense of the new method. — noAxioms

The most important result I got was the proof that the CMIF simultaneity method is incorrect. If my proof is correct, that's a BIG deal, because the CMIF method has been the most-used simultaneity method for many decades. Since I also rejected the only two other simultaneity methods that I'm aware of (because they are non-causal), that left NO known simultaneity methods at all (for an accelerating observer). But immediately after I discovered that proof, I also saw a way to reasonably define a new simultaneity method, and I fleshed that out over several months. It's not quick or easy to describe, so I'm not surprised that you've had trouble understanding it. But it turned out to have some remarkably nice features (like the linear middle portion of the age correspondence diagram), and that has given me some confidence that it may be THE correct simultaneity method for an accelerating observer. Either way, it's the only game in town now.

Can you point me to this simple proof against it? — noAxioms

Yes. It is near the end of Section 7 of my webpage:

https://sites.google.com/site/cadoequation/cado-reference-frame

Or to some sort of reason why the standard relativistic view (per Einstein) is unreasonable? — noAxioms

What do you mean by "the standard relativistic view (per Einstein)"?

It would be best to continue this discussion on the SciForums forum, rather than here, because this is a philosophy forum, and the above is a physic issue.

There need be no signal received to calculate the apparent age of a relatively moving clock. It's a simple calculation in a Minkowski diagram dτ = √(dt^2 – dx^2). — Edgar L Owen

Please show how to use that equation to compute the age of the home twin, according to the traveling twin, immediately before and immediately after the traveler's instantaneous velocity change. Do the case with gamma = 2.0, and with the turnaround at 20 years of age for the traveler, and 40 years of age for the home twin (according to the home twin).

Also, it would be better to continue this discussion on the SciForums forum rather than here, because this is a physics issue, not a philosophy issue.

Normally (in most but not all methods) the signal transit time is ignored when the proper time of a moving clock is computed. — Edgar L Owen

I don't understand that comment. It's obvious that the image received is an old, out-of-date image, so it should never be considered to be showing the current age of the distant person.

And, sure both perspectives are correct in their frames. The difference is that her's is actual because she just looks at her comoving clock and reads the time, while his is apparent or observational because he from a distance in a different state of motion has a perspective view. He's not actually there reading her time on her clock. — Edgar L Owen

Understood.

There are some philosophers on this forum who are also fairly knowledgeable about special relativity and the twin paradox. Those of you in that category might be interested in some discussion going on in the SciForums physics forum. Until recently, I've been a proponent of the co-moving inertial frames (CMIF) simultaneity method, but I recently discovered a simple proof that shows that the CMIF method is incorrect. And I also defined a new simultaneity method. Here's a link to that forum:

http://www.sciforums.com/forums/physics-math.33/

It's what he computes or would see disregarding the time lag of light traveling between them. — Edgar L Owen

I would say that it's what he computes when he properly takes the image transit time into account.

It's her apparent age in his reference frame as opposed to her actual age in her own reference frame.

I wouldn't use the term "actual" for her perspective, and I wouldn't use the term "apparent" for his perspective. They each are equally correct in their perspectives. They disagree, but neither one is more correct than the other.

Yes but of course this is only how the space twin VIEWS the aging of his earth twin. That doesn't affect the actual aging rate of the earth twin in the least which goes on completely as normal unaffected by how anyone else views it. — Edgar L Owen

That's true, provided by "views" you mean what the traveling twin CONCLUDES about the home twin's current age. His conclusion about her current age is NOT what he sees on a TV screen showing her, because that image is "out of date" ... if they are separated by many light years, it took many years for that image to travel from her to him. I suspect you already know this. It's best to not refer to his "view" when you mean what he concludes, or what he computes her current age to be.

If he instantaneously changes his velocity from +v to zero, he will conclude that she instantaneously ages by half the amount she would have aged if he has gone from +v to -v. Then, during that 10 minutes at zero velocity, they would be ageing at the same rate. And during the change from zero velocity to -v, she would instantaneously age by the same amount of ageing that happened when going from +v to zero velocity. I.e., in the case where her she instantaneously ages by 60 years in the standard twin paradox, in your example, she would instantaneously age by 30 years at the beginning of the 10 minutes, then age at his rate for 10 minutes, and then age by another 30 years at the end of the 10 minutes. That is the result for the co-moving inertial frames simultaneity method.

[...] in general relatively moving observers each view the time on each other's clocks ticking slower than their own. — Edgar L Owen

That's a special relativity result. The twin "paradox" doesn't require (or profit from) general relativity. And that result applies ONLY to perpetually inertial observers. An accelerating observer in the twin paradox scenario will find his home twin will be older than him at their reunion. So sometimes during the trip, he MUST conclude that she is ageing FASTER than he is. The co-moving inertial frames (CMIF) simultaneity method says that her age instantaneously increases by a large amount during his instantaneous velocity change at the turnaround. My new simultaneity method says that her age doesn't change at all during his turnaround, but that it does increase linearly for years after his turnaround, at a rate greater than his own rate of ageing. There are also two other simultaneity methods (Dolby and Gull, and Minguizzi) that give an increased ageing rate for her, without an instantaneous increase.

For details, see my webpage:

https://sites.google.com/site/cadoequation/cado-reference-frame

Please remember that my reason for posting here is to get opinions from the philosophers on this forum on my "intuitive" philosophical comments. I'm not here to teach, or debate, or defend, my views on special relativity and the twin paradox.

I would agree that her heartbeats continually in his absence, so she must always have a well defined biological age. — Devans99

Thanks.

I wonder if anything could be deduced by a constant, mutual, radio broadcast of each other's heartbeats to each other. That would allow verification of the existence of the other twin and also act as sort of body clock by which they might be able to judge each others relative speeds of ageing.

Maybe even quantum entanglement could be used to transmit, instantaneously this time, the beat of each other's hearts. — Devans99

Your first paragraph doesn't work, because the distance between them is constantly changing, and so the actual period of the heartbeats is distorted by the varying travel times of the messages. The same thing happens if she continuously transmits a TV image of her, holding a sign that gives her exact age. When he properly allows for her ageing during the transit of the message, he gets the correct current age for her. And that is the same answer that can be obtained analytically from the Lorentz equations.

Your second paragraph doesn't work, because special relativity and quantum mechanics are mutually inconsistent ... neither theory recognizes the legitimacy of the other. Einstein never accepted quantum mechanics.

But NOW, I want to know what you and other philosophers think of my intuitive philosophical reasoning about the question of whether simultaneity at a distance is meaningful or meaningless. THAT'S why I posted here. This isn't the appropriate place for me to either teach special relativity or defend special relativity.

"

I am not sure I understand the problem. Is it that situations can arise in which one person, Jill, has apparent evidence that James has aged more slowly than she, and James will have apparent evidence that Jill has aged more slowly than he?

If so, that as it stands is not really a problem as such, for we can simply say that one of them is mistaken. They may both be equally justified in their beliefs, nevertheless, one of them is incorrect. — Bartricks

No, they are actually BOTH correct. And the evidence that they each have is valid evidence fro THEMSELVES, but it contradicts the evidence of the other person. That's strange, but it can't be shown to lead to an actual logical inconsistency.

She actually gets younger, according to him.

"For the traveler traveling away or the observer staying behind their is no difference in their relative speed so they age at the same rate."

As I've already said, special relativity says that if they are perpetually inertial (and moving at a non-zero speed), they will each conclude that the other is ageing more slowly. It's time now to address my philosophical comments. I'm not here to teach or defend special relativity. I want to know what philosophers on this forum think about my philosophical comments.

"Where do you get that from. You and are 'perpetually-inertial observers' and do not age differently."

My statement applied to the case where we are moving at non-zero speed. I forgot to state that. Sorry.

I forgot to respond to your other two points.

For circular motion by the traveler, the two twins Do agree with one another. Both say the twin at the center of the circle is ageing faster than the twin who is moving in a circle. The standard twin paradox scenario assumes that their relative motion is one-dimensional (although it CAN be extended to two of three spatial dimensions).

As to your last point, the undisputed results of special relativity HAVE been experimentally confirmed in lots of ways. We have never been able to check the most dramatic predictions by accelerating large objects (like humans) to relative speeds that are large fractions of the speed of light, and probably never will ... it simply requires WAY too much energy.

Special relativity says that, for two perpetually-inertial observers (meaning that they have never accelerated, and never will accelerate), they EACH will conclude that the other is ageing more slowly. (I believe that all physicists believe that). So those two inertial observers DON'T agree with each other. But special relativity says that they are BOTH CORRECT, even though they disagree with each other! And it's impossible prove that an actually INCONSISTENCY results in that situation. To what extent the above can be extended to also apply to an observer who has accelerated at some point in the past, or may accelerate at some point in the future, is the source of disagreement among physicists.