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/

Mike,

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

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.

Just my terminology to distinguish the cases which are in fact quite distinct.

Edgar L. Owen

no Axioms,

First thanks for looking at my site and commenting on it. I'm not quite sure where you are coming from but will try to tease out our differences. We seem to be talking at cross purposes somewhat and you seem to be misinterpreting a number of my statements. Easy to do in relativity if not very carefully stated.

Take your last 3 paragraphs quoting 3 statements of mine.
1. I said " If it follows an inertial path all its constant spacetime distance traveled is through time."
I meant in its own frame where it is at rest. That was assumed but perhaps should have been explicitly stated.

2. I said " Time passes at the same rate on all inertial clocks"
Again I meant in their own frames since in their own frames all their constant motion through spacetime is through time. Normally I mean 'in their own frames' unless I state otherwise.

3. My statement is of course when we ignore signal transit time and red or blue shifts which we normally do when calculating proper times of moving clocks. Ignoring those, two relatively moving clocks do each see each other's clocks ticking slower than their own by the same amount. This is simple time dilation which is well established.

Also you seem to believe the past still somehow exists which leads me to suspect you believe in a block universe in which all past and maybe future states actually exist. I don't agree...

I suspect when we understand how we each use the terminology some of our views are closer than you think.

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.

Mike,

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). The signal transit time doesn't appear in this proper time equation at all and is ignored. This is the standard way to calculate the proper time of a relatively moving clock when the signal transit time is ignored as it usually is. Very simple and correct in all cases. If there is a non-instantaneous acceleration we just integrate this over the path instead.

Edgar L. Owen

Mike,

Also I don't understand all your agonizing over the twins. It's really quite simple.

Ignore the signal transit time. That's an entirely separate issue...

1. In a Minkowski diagram in the earth reference frame use dτi = √(dti2 – dxi2) to calculate the proper times of both legs of the space twin's path, or any point along the path.

2. Use a Lorentz transformation to convert the diagram to the frame of the first leg of the space twin's path. Use the same equation to calculate the apparent proper time of the earth twin from that perspective.

3. Use a Lorentz transformation to convert the diagram to the frame of the second leg of the space twin's path. Use the same equation to calculate the apparent proper time of the earth twin from that perspective.

This method gives both twin's views throughout the whole trip. If you want to add signal transit effects which I don't do, that's a completely separate issue.

Also Caltech? physicist John Baez has a complete animation of all possible effects on his site including the little known Penrose twisting etc. though you have to spend some time with it.

Best,
Edgar L. Owen

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.

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. — Mike Fontenot

I could not make any sense of the new method. The old CADO/CMIF one was worded from a sort of idealistic perspective, but otherwise it didn't seem outright wrong. Can you point me to this simple proof against it? Or to some sort of reason why the standard relativistic view (per Einstein) is unreasonable?

The difference is that her's is actual because she just looks at her comoving clock and reads the time — Edgar L Owen

Both of them are looking at their own comoving clock and reading its time, so this isn't a difference.

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

She is also at a distance from his clock and has a 'perspective view' as you call it. Neither of them is actually there reading the other's clock. Again, no difference.

First thanks for looking at my site and commenting on it. — Edgar L Owen

I only glanced at a few places. Hardly a solid effort to read it all. You seem to hold a sort of dualistic view of mind where the physical universe is a computed virtual reality which is fed real time to a non-physical experiencer elsewhere. The VR has a current state for everything which is sort of updated all at once everywhere for the next 'universal current present moment'.
That seems to be the general gist. Let me know if I didn't get very close.

1. I said " If it follows an inertial path all its constant spacetime distance traveled is through time."
I meant in its own frame where it is at rest. That was assumed but perhaps should have been explicitly stated.

If there is a universal current present moment, then there is a universal location for everything, and if the location of some object changes (in your computer) from one tick to the next, then that object isn't stationary. Yes, I agree that it can be made stationary by selecting a coordinate system with time axis parallel to its worldline, but that frame doesn't correspond to reality in the universe you describe. In such a frame, most moments simultaneous with here and now are in the past or future (have already gone by or have yet to be computed), so the frame aligns the time axis differently than the actual one.

2. I said " Time passes at the same rate on all inertial clocks"
Again I meant in their own frames since in their own frames all their constant motion through spacetime is through time. Normally I mean 'in their own frames' unless I state otherwise.

If I have two clocks in relative motion, in no frame are they both stationary, so I'm not sure how this statement can be satisfied as your modified statement words it. At best you seem to be stating a simple tautology that clocks measure their own proper time, and being inertial isn't required for that.
Under an absolute interpretation such as yours, the concept of the rate of a clock has meaning, but you seem to mix relative and absolute statements without picking a consistent interpretation.
You deny presentism, and yet you talk about time passing at a rate, which is a presentist concept. Under the 4D spacetime interpretation, time isn't something that passes or moves in any way, it's just another dimension with no more present moment than there is a current location (which there is in Fontenot's view, or at least his old CIMF view).

3. My statement is of course when we ignore signal transit time and red or blue shifts which we normally do when calculating proper times of moving clocks.

OK. That's definitely not how you worded it the first time, where you talked about what's being viewed and not what's being computed.

Ignoring those, two relatively moving clocks do each see each other's clocks ticking slower than their own by the same amount. This is simple time dilation which is well established.

Assuming a relativist interpretation and assuming they're inertial, agree. It seems not to be true in your VR universe where a moving observer should compute his own clock as running slow because he's not stationary (not at the same location in the simulation from one moment to the next). Using his own frame is wrong in that situation because that frame doesn't represent the universal frame.

Also you seem to believe the past still somehow exists which leads me to suspect you believe in a block universe in which all past and maybe future states actually exist. I don't agree...

This is what I mean by mixing interpretations. You explicitly deny the block interpretation (that's fine), but then say you're not a presentist, which is what's left. If you deny the block, then you must deny any frame that has past and future events being simultaneous with some current event, a contradiction if you deny the reality of such events.

I didn't look to see how much of these contradictions are in the books since you said the SR parts were under revision. I scanned pretty much none of that. I looked a bit at consciousness, entanglement and dark matter, just sort of random places to click.

The underlying solution to the apparent contradictions you mention is the notion of two kinds of time.

Processor time which proceeds at the same rate throughout the universe in each tick of which the entire universe is recomputed including the computation of the allocation of the constant identical total distance traveled through spacetime of every object between distance in time and distance in space. The result is the universe as the present moment surface of a cosmic hypersphere in which everything is at the same processor time but objects have different proper times depending on how much spatial distance they have traveled along their own world lines in their own frames,

Within this present moment surface, frames view other frames from the perspective of their different coordinate systems and calculate differed relative values to the space and time values of clocks in relative motion to their own.

Edgar L. Owen

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.

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.

It's what they observe. Or maybe you're right and everyone else - and the world itself! - is wrong. Or maybe any curious high-school student - or middle schooler these days - knows more about any kind of relativity than you. But the remedy is always only a few clicks away: why not try it?

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.

Well done for not engaging with what I said.

Has time passed more slowly for the apple in the fridge?

If 'no' (and obviously the answer is 'no'), what's the difference between that case and the twin case?

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. — Mike Fontenot

And I have. I'm a newbie there.

This, on the other hand, is philosophy:

The underlying solution to the apparent contradictions you mention is the notion of two kinds of time.

Processor time which proceeds at the same rate throughout the universe in each tick of which the entire universe is recomputed including the computation of the allocation of the constant identical total distance traveled through spacetime of every object between distance in time and distance in space. The result is the universe as the present moment surface of a cosmic hypersphere in which everything is at the same processor time but objects have different proper times depending on how much spatial distance they have traveled along their own world lines in their own frames, — Edgar L Owen

You lost me there. In their own frames (a block concept), they travel no spatial distance at all, by definition. You continue to mix philosophy of time interpretations.

Also, why a hypersphere? Are you saying space is curved and finite and wraps around if you go far enough? Glorified (non-local) computational balloon analogy?

Within this present moment surface, frames view other frames from the perspective of their different coordinate systems and calculate differed relative values to the space and time values of clocks in relative motion to their own.

OK, 'present moment surface.' is consistent with what you're saying, but then other frames do not correspond to this surface, but rather to hyperplanes tilted one way or another so that only along the 2D plane of intersection are simultaneous events 'actual' (part of the hypersurface. Why would you consider such a frame valid if most of consists of nonexistent events.

For example, Bob (the guy traveling) just before turning around at age 10 figures that Alice (the stay at home sister) is 5 years old, but her 5th birthday is actually a past event and in reality (assuming she's the stationary one) she's 20 years old, or more precisely, the universal computer is computing Bob's 10th birthday at the same time as it is computing Alice's 20th. The 5th birthday is long gone and it is a mistake for Bob to compute her age using a coordinate system where he is stationary. The fact of the matter is, he's very much not stationary, but moving like a bat outa hell.

The block interpretation has no problem with Bob's frame since all those events are equally real, and hence any frame is as good as another. This is why the block interpretation (which had been around since before Einstein) suddenly became more intuitive.

- - -

Has time passed more slowly for the apple in the fridge?

If 'no' (and obviously the answer is 'no'), what's the difference between that case and the twin case? — Bartricks

The difference is that relative to the apple in the fridge, the apple on the table still rots faster. With the twins, in the frame of either, it is the other one that rots more slowly.

So, when the twins meet back up, they're both older than each other?

They don't meet back up if they stay stationary in different frames

But they can meet back up, yes? And if they do, are they both older than each other?

No, they can't. One would have to accelerate and become stationary in a different frame. The other twin ages slower in that frame as well, but in that new frame, the twin back home is already much older than the twin turning around.

BTW, the apple example is more like gravitational dilation. An apple on the top floor of a building rots faster than one on ground floor, all else being equal.

I am unclear why they can't meet up. If one is travelling through space and the other is stationary, why can't they meet up?

BTW, the apple example is more like gravitational dilation. An apple on the top floor of a building rots faster than one on ground floor, all else being equal. — noAxioms

We're agreed, though, that the apple in the fridge isn't travelling through time slower though, yes? Time doesn't run slower in fridges, or faster at the tops of buildings, presumably.

I never said one is traveling through space and the other isn't. I said each was stationary relative to a different inertial reference frame. You're using absolute phrasing. The situation is quite different given an absolute interpretation.

The original set up is that one is stationary and the other travels away at a constant speed, yes? And they turn around and travel back, yes?

And the 'paradox' is, well, what? That one will 'think' that they have experienced less time than the other, and the other will think the reverse, yes?

And that's somehow supposed to tell us something about time?

I mean, they can't actually both be older than each other, yes? So they've endured for the same amount of time. It is just they don't think this, based on their experiences.

Just as, by analogy, the apple in the fridge, though it appears not to have endured as much time as the apple on the sideboard, has, in fact, endured the same amount of time.

It would be monstrously silly to conclude that appearances are accurate and that time slows down in fridges. And so it would surely be every bit as silly to conclude that the appearances the twins are subject to are accurate.

No. The original setup is that the two move apart from each other. There is no designation of either of them being the stationary one.

Under absolute terms like that, perhaps the guy with the rocket is initially the stationary one and it is Earth that is moving quickly. The answer still comes out the same.

Doesn't matter - same point applies. They both speed away and then come back together, yes?

THey won't both be older than each other, will they? So, what's the point?

is it that under such a scenario two people will find themselves with contradictory but equally well epistemically justified beliefs about how much time has elapsed?

You seem to only be able to think in absolute terms. That works as well.

The simplest case then is Earth is arbitrarily designated as stationary and Bob moves fast the whole time and thus ages less because physical processes slow down if they're not stationary. That's actually pretty simple, and the twins thing isn't a paradox at all. Reference frames don't come into play at all with this interpretation. Why don't you go with it?

Doesn't matter - same point applies. They both speed away and then come back together, yes? — Bartricks

The apples are just sitting there, not in relative motion. One in the fridge (which works by retarding chemical reactions, not dilating time) and one not.

THey won't both be older than each other, will they? So, what's the point?

The apples stay the same age. One just rots quicker.

If the apples move fast, then yes, one actually gets older than the other. This has been demonstrated conclusively with small fast objects that decay at very known rates.

The simplest case then is Earth is arbitrarily designated as stationary and Bob moves fast the whole time and thus ages less because physical processes slow down if they're not stationary. That's actually pretty simple, and the twins thing isn't a paradox at all. Reference frames don't come into play at all with this interpretation. Why don't you go with it? — noAxioms

Er, I did. And then you said both move.

What I want to know is why physicists think it tells us something interesting about time. Because it seems to me to tell us nothing more than my fridge/apple example.

For instance, here's another variation: one twin travels from the earth and the other stays put. Twin one thinks "hm, my twin is getting smaller and smaller than me". Whereas the other twin - twin two - thinks "hm, my twin is getting smaller and smaller than me".

Are they both getting smaller than each other? No, obviously not. But they both have equally justified beliefs that one is getting smaller than the other.

If the apples move fast, then yes, one actually gets older than the other. This has been demonstrated conclusively with small fast objects that decay at very known rates. — noAxioms

Hahaha, so you DO think time goes more slowly in fridges?! It has been demonstrated conclusively that apples decay more slowly in fridges.

The apple in the fridge on the sideboard does not 'age' faster than the one in the fridge. They are both the same age. One is just more shrivelled than the other. Processes have happened in one faster than they have in the other.

Er, I did. And then you said both move. — Bartricks

Motion by definition is relative. I said they both move relative to each other. The absolute way would be relative to some implied absolute frame, so one just says 'this one is moving and that one isn't'. I have a hard time thinking in such terms, as I said, you should go with it.

What I want to know is why physicists think it tells us something interesting about time. Because it seems to me to tell us nothing more than my fridge/apple example.

In absolute interpretation, light speed is not frame independent. That's where it becomes complicated. Has nothing to do with apples. How can I measure how long light takes to cross the room if I don't know if my clock is dilated or running full tilt? For one, no clock in reality, even if stationary and compensating for gravity, runs at full speed. They've never computed how slow Earth clocks run, which sort of puts a dent in the claim that there is such a rate.
Translation, as an absolutist, I have no idea how long it takes the apple to rot since my clock doesn't measure time accurately at all.

For instance, here's another variation: one twin travels from the earth and the other stays put. Twin one thinks "hm, my twin is getting smaller and smaller than me". Whereas the other twin - twin two - thinks "hm, my twin is getting smaller and smaller than me".

Are they both getting smaller than each other? No, obviously not. But they both have equally justified beliefs that one is getting smaller than the other.

Not justified. An object at twice the distance occupies a quarter of my visual field. They didn't take that into account if either concludes that the other is actually getting smaller.
Similarly, an approaching clock appears to run faster but nobody who knows their physics actually concludes it is. That's just blue shift.

Has time passed more slowly for the apple in the fridge?

If 'no' (and obviously the answer is 'no'), what's the difference between that case and the twin case? — Bartricks

I have to take you question seriously. Watch this:

https://www.youtube.com/watch?v=noaGNuQCW8A