# Twins paradox explained without forces

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I don't particularly have a problem with that. Use of three observers with regards to the twin paradox has been presented before. It's nothing particularly new.

However to make the OPS scenario complete he should become familiar and apply the transformations. Several of his later posts indicated to me that certain aspects of the Minkowskii symmetry relations with regards to inertial frames vs non inertial frames needed clarification.

Particularly since curved spacetime can lead to some surprising results. Though for this thread I've been assuming his solution is restricted to the Minkowskii or at most the weak field limit.

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18 hours ago, scuddyx said:

The point I am making is that the twin paradox can be explained without resorting to acceleration.

You can explain it purely geometrically - the world line of one of the twins is a geodesic of spacetime, the other one is not. That’s all there is to it. Physically speaking though, a world line not being a geodesic implies the presence of acceleration at some point; but you don’t need to explicitly cast it in those terms.

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I'm not sure what the hoopla is about here.  Scuddyx just says that the time-handoff (all-inertial) scenario explains the relative aging in the twins scenario (of instant turnabout).  That's true.  However, to ignore what happens during twin B's proper acceleration (per B) is to be incomplete in the relativistic explanation of the scenario, because the twin B prediction of the Twin A clock during that phase is never considered.  If one never considers an acceleration phase, one is likely ignorant of a block universe.

Best regards,

Celeritas

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But it isn't called the two clocks paradox or the who's the younger of two people paradox it is called the twins paradox and the fact they are twins has important implications, it implies:

1) they are in close proximity to each other

2) whilst they are in close proximity they are at rest with respect to each other, or both at rest to the same inertial frame

3) whilst they are in close proximity and at rest their respective clocks start simultaneously

4) their clocks tick simultaneously from then on (ignoring insignificant differences)

5) they are simultaneously the same age

if one of those twins were then put on a spaceship and flown around the galaxy at great speed relative to the inertial frame and returned some considerable time later to be with their twin they would once again be in close proximity, be at rest, their clocks would tick simultaneously but one will have ticked less than the others they would simultaneously be different ages. the statement 'they were simultaneously the same age but are now simultaneously different ages' is true

If in a different scenario there is a rogue planet speeding through the galaxy, traveling at a constant speed in a straight line and is heading straight for Earth at 0.9c. It is also true the Earth is heading towards the rogue planet at a constant speed in a straight line and at 0.9c. So there are two inertial frames heading towards each other at 0.9c.

Is there any way we can conclude that an event which occurs in one inertial frame happens simultaneously with an event in the other inertial frame (with the speed and distance involved). Can the birth of a child on Earth be considered simultaneous with the birth of a child on the rogue planet. If the answer is no then the statement 'they were simultaneously the same age but are now simultaneously different ages' is false. They were never simultaneously the same age

If you can conclude that events that occur in two different inertial frames, separated by some vast distance and approaching each other at some great speed, are simultaneous you can agree that two children born are simultaneously the same age. However if they were to fly off to meet each other, accelerating at the same rate for the same time, or, at the instant the planets pass each other you capture a photo (data) of the two individuals they will both be the same age still so the statement 'they were simultaneously the same age but are now simultaneously different ages' is again false. They are not different ages.

So the OPS experiment will explain time dilation and the time dilation will be equivalent to the twins time dilation but in the OPS experiment the statement 'they were simultaneously the same age but are now simultaneously different ages' is false.

and the statement 'they were simultaneously the same age but are now simultaneously different ages' can only be true where acceleration is involved.

so Einstein was right, if he were alive today I'm sure he'd be pleased to know that... (I mock myself)

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10 hours ago, Markus Hanke said:

You can explain it purely geometrically - the world line of one of the twins is a geodesic of spacetime, the other one is not. That’s all there is to it. Physically speaking though, a world line not being a geodesic implies the presence of acceleration at some point; but you don’t need to explicitly cast it in those terms.

I agree. Still, the 3 clocks or 3 siblings experiment OP describes uses only inertial clocks, only geodesics, following the same inertial sections that an instantly accelerating twin passing through the same events would (measuring the same total geometric length). I suppose I'm claiming that you can add the geometric length of two world lines, and get the total length of a world line made by connecting the end of one to the start of another.

8 hours ago, Celeritas said:

I'm not sure what the hoopla is about here.  Scuddyx just says that the time-handoff (all-inertial) scenario explains the relative aging in the twins scenario (of instant turnabout).  That's true.  However, to ignore what happens during twin B's proper acceleration (per B) is to be incomplete in the relativistic explanation of the scenario, because the twin B prediction of the Twin A clock during that phase is never considered.  If one never considers an acceleration phase, one is likely ignorant of a block universe.

I think it is considered. If a twin makes the turnaround, it feels proper acceleration which (in accepted theory) contributes nothing to its proper time, so that feeling won't show up in the ageing equations. In changing inertial frames, its relative simultaneity with events at clock A changes.

In the 3-clocks scenario, clocks B and C already have different relative simultaneity with events at A. So we're already accounting for those differences. There's no physical frame-independent difference between B and C, but measurements made in their respective inertial frames, are different. Simply by considering the two inertial frames properly in any relevent equations, everything's accounted.

2 hours ago, between3and26characterslon said:

But it isn't called the two clocks paradox or the who's the younger of two people paradox it is called the twins paradox and the fact they are twins has important implications, it implies:

1) they are in close proximity to each other

2) whilst they are in close proximity they are at rest with respect to each other, or both at rest to the same inertial frame

3) whilst they are in close proximity and at rest their respective clocks start simultaneously

4) their clocks tick simultaneously from then on (ignoring insignificant differences)

5) they are simultaneously the same age

[...]

so Einstein was right, if he were alive today I'm sure he'd be pleased to know that... (I mock myself)

I disagree that the twin paradox implies all those things. If you look at this geometrically, and compare the geometric lengths of two twin's paths, you should be able to see what's important and what's not. If some aspect is not needed in the geometric analysis, then it's not a necessary contribution to the overall and different ageing of the twins. That includes synchronizing clocks, and being at relative rest. Also being twins. All these other aspects illustrate the seemingly paradoxical nature of the different ageing. You could say "the twin paradox" requires some of those things, but "the differential ageing of the twin paradox" does not.

Being in close proximity is important because if you compare the time on two separated clocks, you can get different answers for different observers (inertial frames). The experiment described by OP is okay, because you only need to compare clocks at single locations (events), to predict the result. Those events are the moments and locations of the given clocks passing by each other.

Yes, Einstein was right. I think that a lot of people who say Einstein was wrong, have no idea what he actually said or wrote.

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On 1/25/2020 at 11:43 AM, md65536 said:
On 1/25/2020 at 2:39 AM, Celeritas said:

I'm not sure what the hoopla is about here.  Scuddyx just says that the time-handoff (all-inertial) scenario explains the relative aging in the twins scenario (of instant turnabout).  That's true.  However, to ignore what happens during twin B's proper acceleration (per B) is to be incomplete in the relativistic explanation of the scenario, because the twin B prediction of the Twin A clock during that phase is never considered.  If one never considers an acceleration phase, one is likely ignorant of a block universe.

I think it is considered. If a twin makes the turnaround, it feels proper acceleration which (in accepted theory) contributes nothing to its proper time, so that feeling won't show up in the ageing equations. In changing inertial frames, its relative simultaneity with events at clock A changes.

In the 3-clocks scenario, clocks B and C already have different relative simultaneity with events at A. So we're already accounting for those differences. There's no physical frame-independent difference between B and C, but measurements made in their respective inertial frames, are different. Simply by considering the two inertial frames properly in any relevent equations, everything's accounted.

md65536,

I've been trying to be careful in my response to your prior.  I don't really disagree with what you said there, although I do consider the analysis more complete to consider the A-clock in the B spacetime system during his own turnabout.  I realize that you (yourself) can look at the all inertial Time-Handoff scenario and envision what must happen during (say) a virtually instant twin B turnabout, but many trying to learn the theory may not make that leap (or maybe for some time) if not considering that.  That is, to understand the meaning of a block universe.

In the all-inertial Time-Handoff scenario, if we consider only the 2 inertial legs that coincide with the Twins Instant Turnabout scenario, one can say that B declares A to age by t/gamma = 4/1.25 = 3.2 yr.  C declares A to age by t/gamma = 4/1.25 = 3.2 yr.   Total aging for A per B+C = 3.2 + 3.2 = 6.4 yr.  However, we know that A really ages 10 yr.   That's a difference of 10 - 6.4 = 3.6 yr.  So where does that required extra A-duration come from?  It comes from "the difference in simultaneity" between the B & C inertial frames (which you've pointed out), so the LTs inherently handle that.  In terms of the twin's instant turnabout scenario, it comes from the proper acceleration at B's turnabout, which leads many to argue that it is acceleration that causes A to age more than B.  But this is also to say (once again) that it comes from "the difference in simultaneity" between the inertial frames occupied by B "before vs after" his instant turnabout.

I would submit that it's more about 4-space geometry inherent in SR (as per Minkowski), than anything else.  The geometry is dependent upon the relative motions, the relative motions are dependent upon (existing inertial motions) and any proper accelerations over the defined interval.  All that, shapes the worldlines in spacetime. The jump in A-time that exists between inertial frames (B & C), or during the instant B turnabout, is a function of only the differential in inertial frames (and hence relative velocity), however it requires a proper acceleration for a single observer (twin B) to produce that jump in A-time, because he must rotate his sense of simultaniety to produce the same differential (that exists between the B & C clocks in the Time-Handoff scenario).

I agree that "twin B's feeling his own inertia" is not the driving factor for quantification of B's lesser aging (it's velocity), but it is certainly true that the twin who properly accelerates is the one who ages less.  Besides changing direction, he must produce that frame-differential that exists in the Time Handoff scenario (that produces the A-time jump, so A can age the most), and only a proper acceleration by B can attain that.

Best regards,

Celeritas

Edited by Celeritas

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That all sounds great. Your carefulness is appreciated. If we were all careful, we could remove anything that isn't agreeable, like opinions or interpretations.

If instead of "In terms of the twin's instant turnabout scenario, it comes from the proper acceleration at B's instant turnabout.", we said "...it corresponds with B's proper acceleration", then I can agree.

Sure, it's helpful to consider the experiment from different points of view and different measurements, but when you speak of the time at clock A, when B passes C, then you're comparing distant clocks, and you have to deal with the caveats of that.

Then if you want an even closer to complete understanding of it, it is useful to know the reason that we can say the BC event is simultaneous in B's frame with an event at A at time 3.2 years on A's clock, but is simultaneous in C's frame with an event at A at time 6.8 years on A's clock. It comes from the definition of time that Einstein gives in his 1905 paper that establishes special relativity, in the section translated as "Definition of Simultaneity". As Einstein wrote (in Relativity: The Special and General Theory):

Quote

... it assumes absolutely nothing about light. There is only one demand to be made
of the definition of simultaneity, namely, that in every real case it must supply
us with an empirical decision as to whether or not the conception that has to be
defined is fulfilled. That my definition satisfies this demand is indisputable.
That light requires the same time to traverse the path A -> M [the midpoint between A & B] as for the
path B -> M is in reality neither a supposition nor a hypothesis about the
physical nature of light, but a stipulation which I can make of my own freewill
in order to arrive at a definition of simultaneity.

Anyway, whether the definition of simultaneity has a physical basis or is merely a chosen convention, is something that can be argued. It doesn't matter though. Especially if A's distant clock is never compared to a local clock, the definition of simultaneity can be completely avoided. I also believe that a closer-to-complete understanding of relativity involves being aware of the definition of simultaneity and knowing when that definition doesn't matter (eg. for invariants).

1 hour ago, Celeritas said:

When you go from the B POV to the C POV at BC flyby, the A clock's time-readout instantly jumps from 3.2 yr to 6.8 yr, because A must age 10 yr but must also age by 3.2 y over the BC → CA leg (10-3.2=6.8 yr).

A's clock "jumping ahead" is not something that B or C can instantly detect. It's something that's seen gradually over time as C approaches.

Actually, a Doppler analysis is easy here because with a speed of .6 c, the relativistic Doppler factor is 2. On its 4-year journey to meet C, B sees A's clock appear to tick at half its rate. It watches A age 2 years. On its 4-year journey to meet A, C sees A's clock appear to tick at twice its rate. It watches A age 8 years. Together they see the full 10 years of A's ageing. When B and C pass, or when a twin instantly turns around, no ageing of A appears to happen then. The instantaneous jump in A's coordinate time, is not something that is measured at that time.

Edited by md65536

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On 1/26/2020 at 5:21 PM, md65536 said:
On 1/26/2020 at 5:21 PM, md65536 said:

A's clock "jumping ahead" is not something that B or C can instantly detect. It's something that's seen gradually over time as C approaches.

Actually, a Doppler analysis is easy here because with a speed of .6 c, the relativistic Doppler factor is 2. On its 4-year journey to meet C, B sees A's clock appear to tick at half its rate. It watches A age 2 years. On its 4-year journey to meet A, C sees A's clock appear to tick at twice its rate. It watches A age 8 years. Together they see the full 10 years of A's ageing. When B and C pass, or when a twin instantly turns around, no ageing of A appears to happen then. The instantaneous jump in A's coordinate time, is not something that is measured at that time.

Agreed, receipt of light signals from A shows he exists continguously over the interval. But I was speaking strictly of what exists in the spacetime system, not the receipt of light signals.  The jump in A clock time (not per A himself) is something that must exist, because spacetime systems define what exists now across the all of 3-space for any moment in time.  It's not that twin A himself skips that 3.6 yr A-duration, only that it becomes missing due to the instant change in POV at B turnabout (or B/C flyby).  The Block universe is required to explain said A-time jump.

Best regards,

Celeritas

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So ultimately, the relative aging is inherent in the geometry of the spacetime diagram and it's worldlines.  It's the very same for the Time Handoff and Twins scenarios, given the turnabout is instant and hence ignorable. However, all this really means is that twin B must execute proper acceleration(s) to the precise tune that produces the outbound and inbound legs of the Time Handoff scenario.  So the frame relations (LTs) dictate the outcome, but to attain those frame relations, twin B (not A) must execute the accomodating proper acceleration(s).  So while there exists a geometry in the abstract that produces the relative aging presented by the Time Handoff scenario, twin B must attain that geometry by his own proper acceleration(s).

So is it twin B's proper acceleration(s) that produce the relative aging?  If B doesn't properly accelerate, he never ages differently from A.  If A properly accelerates while B does not, A ages less than B (so that's out).  Twin B's proper accelerations must produce all the velocities required to attain the Time Handoff worldlines (and hence its related relative aging result), which in this case are instantaneous accelerations.  Velocities are governed by kinematic acceleration, and kinematic accelerations are governed by proper accelerations in any local spacetime.

I think its fair to say that the twin's relative aging is quantified by their relative velocity, and the locations of the related events in spacetime.  But I also figure it's fair to say that twin B's proper acceleration is what makes it all happen, and to the precise tune that it does.

This is kinda like the chicken vs the egg deal.

Best regards,

Celeritas

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On 1/27/2020 at 11:40 PM, Celeritas said:

But I also figure it's fair to say that twin B's proper acceleration is what makes it all happen, and to the precise tune that it does.

"Makes it all happen" is too vague to be meaningful. What's "all"? B's acceleration doesn't determine A's ageing. One might as well say "B's proper acceleration is the magic that makes it work" and "magic" means the part that doesn't show up in the maths, or it's the answer to the "Why?" questions that aren't satisfied with knowing the "what".

I don't know why people look for such explanations anyway. We agree everything that's "what" is there in the geometry, and it's there in the 3-clock variation when nothing physical accelerates. Does there need to be more than that?

I think it's fair to say that twin B's velocity relative to A is affected by its proper acceleration. That's about all that's needed to explain acceleration's role in the twin paradox.

If B undergoes the same proper acceleration but at A's location, there's no change in the coordinate time at A. So someone else might say "It's distance that makes it all happen!" Then someone else might suggest that if B instead orbits A at a very small distance but at 0.6c, you get the same ageing as in OP's experiment, but with B undergoing constant proper acceleration. So neither distance nor proper acceleration alone is making it all happen here. You can scratch a hole in your head trying to figure it out how one or the other is making it all happen. But if all you have is relative velocity and time, you can calculate it.

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md65536,

The question that is always asked is this ...  what causes twin B to age less than twin A?

In the defined Twins scenario, if twin B does not properly accelerate, he cannot age differently than A.  If B does properly accelerate, he always ages less.  Did B's own proper acceleration cause him to age less?  Indeed it did, but this does not address the extent of relative aging, ie why the worldlines geomtery ended up precisely as it did.  How much younger is B?  Well then you need to run the LTs, which use velocity.  The fact that the LTs use velocity, does not change the fact that all velocities (and hence worldline geometry) are completely governed by all the proper accelerations over the interval.  This is inescapable.

So I completely agree with you, that the worldlines geometry is everything when it comes to quantifying relative aging.  Inherent in that geometry, are all the relative velocities and relative accelerations over the interval.  And indeed, the LTs use velocity.

In regards to proper acceleration, it's not just about the feeling of an applied force (eg B experiencing equal and opposite forces applied to himself, eg a compression, with no net motion of himself wrt A).  It's about what an accelerometer would measure (in collective) during applied force(s).  Wrt twin B in the usual twins scenario, we usually assume a single point acceleration applied unto twin B, ie twin B changes in his own state of motion.  B properly accelerates.

Best regards,

Celeritas

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md65536,

So I've been trying to minimize the wording here ...

The question is ... Why does twin B age less?

One can say twin B ages less than twin A because B ages 8y over the interval, while A ages 10y.  This is presented by the geometry of the worldlines. IOW, either twin may run the LTs to predict the aging of the other, and he with the lower result ages the least.  Also (as you've said), each twin could just compare clocks upon return and the same result arises (so, no prediction of the other's clock necessary).  However, the only reason that works out is because of the assumed relativistic effects that take place (length contracton), and so its not as though relativity (specifically the LTs) are ignored in that limited consideration. For example ...

One can also say this ...

Twin B must TURN in spacetime to record the separation between A/B departure and B-turnabout as a moving contracted length (L'=L/gamma), and by definition Twin A (and earth) records that separation at its stationary proper length (L, the longest recordable length for that separation).  A & B always hold the same relative speed for the outbound and inbound legs, so he who records that separation the shortest, ages the least.  And, that be twin B.  That's why B ages less. His "turning in spacetime" causes that to happen, ie his own proper acceleration.  It does not matter that B's proper acceleration is instant.  It only matters what the result of those proper accelerations produce ... the resultant specific relative velocities (used by the LTs).  Twin B must TURN in spacetime, and that answers the question.  However, if we wish to predict the relative aging, we must quantify the relative time over the interval, and so the LTs must be run (which use velocities).  We then find that A ages 10y while B ages 8y, from which it is easy enough to state that B ages the least (because the LTs require it).

So, it's not as though I disagree with anything that's been said thus far.  I just think the usual debates on this matter result from slightly different interpretations of the question ... who ages the least and why?  Is the water boiling because its 212 degrees Fahrenheit, or because heat was first applied to reach that temperature?

Best regards,

Celeritas

Edited by Celeritas

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In my opinion both twins age in accordance to their locality. With $m_i=m_g$ applying at each locale regardless of other observers.

The paradox arises with an invalid assumption of an at rest frame.

In essence the difference between the different twin aging rates depends on the relative velocity and subsequently factoring in the acceleration changes to those relative velocities

Even under constant velocity if the two twins have a difference in constant velocity regardless of acceleration they will age at different rates.

Now let's consider the following thought experiment. Remove all signals from A to B. Twin A has a different constant velocity than B. Those two twins will age differently regardless of signal exchange or acceleration change.

In essence through $m_i=m_g$ the difference between aging rates is identical to placing two clocks at different gravitational potentials. Ie one at sea level and one at the top of Mount Everest. Neither accelerates and you don't require signal exchange until you compare clocks at some later time.

To put it another way the paradox only arises on the SR treatment but doesn't arise in the "all reference frames are inertial" under the GR treatment.

Edited by Mordred

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3 hours ago, Celeritas said:

Twin B must TURN in spacetime to record the separation between A/B departure and B-turnabout as a moving contracted length (L'=L/gamma),

No, that's false. The twin that turns at B-turnaround (aka BC event) measures the same contracted length between AB and BC that OP's inertial clock B measures between the same events. The distance between AB and BC is 3 light years according to A, and 2.4 light years according to twin B or clock B. If you have a marker at BC that is stationary in A's frame (so that the proper length between A and the marker is the length measured in A's frame), then the marker approaches B (twin or clock) at a speed of 0.6c for 4 of B's years, traveling 2.4 light years in outbound B's frame.

3 hours ago, Celeritas said:

That's why B ages less. His "turning in spacetime" causes that to happen, ie his own proper acceleration.

Bringing this back to OP's post, I'll agree that your twin B's turn around causes it to follow the shorter world line made up of OP's B and C between the 3 events where clocks pass.

3 hours ago, Celeritas said:

Is the water boiling because its 212 degrees Fahrenheit, or because heat was first applied to reach that temperature?

Well let's consider that analogy. You have two twins A and B, and the only differences between them is that B turns around and ages less. In the absence of some other form of time dilation (GR), having B turn is necessary to have it age less. Therefore B's turnaround is the cause of it ageing less.

Now say you have 2 glasses of water, A and B, and a kettle on the stove, and you pour B into the kettle and it boils. The only difference is that B was the one that was poured and it was the one that boiled. In absence of some other way to heat it, pouring B into the kettle was necessary to have it boil. Therefore, pouring B is the cause of it boiling.

I suppose it's possible to interpret that as a true statement. But in isolation it is just misleading. Pouring water causes it to boil. Proper acceleration causes differential ageing.

Then you do the equations and you find that the world line AB to BC to AC is shorter than AB to AC, and that the energy used by the kettle heated the water, and the pouring doesn't factor into the maths. Someone points out that if the water started out in the kettle, it would boil in the same time as if it was poured in. Then someone else says that involves hidden pouring.

Edit: Maybe I'm dumbing down the argument too much. You're arguing that the change in coordinate time at A's clock when B turns around, is something real (because the LT says it is?) and I'd argue that's just a coordinate effect that disappears with a different system of coordinates or without a definition of simultaneity. Equivalently, you're arguing that Einstein's definition of simultaneity must be physically real? Or, back to OP's experiment, could we say the debate is about whether clock A is physically different to clock B vs. clock C besides how it appears differently to different observers? Would you argue that when twin B turns around (or if clock B were to turn into clock C), that causes a physical change in A? I say it doesn't, that the observed differences are only relative and depend only on the observer.

2 hours ago, Mordred said:

Even under constant velocity if the two twins have a difference in constant velocity regardless of acceleration they will age at different rates.

Now let's consider the following thought experiment. Remove all signals from A to B. Twin A has a different constant velocity than B. Those two twins will age differently regardless of signal exchange or acceleration change.

If they're both inertial and their velocities are different, does that just mean that you're choosing a frame of reference where they have those velocites? Otherwise, regardless of the direction of their different velocities, if B's speed is v in A's frame, then A's speed is v in B's frame. The only way to have the twins age the same over a time t is to choose an inertial frame of reference where A's speed is the same as B's speed, right? (Of course there's no inertial frame where A is inertial and B turns to reunite with A, and they have the same speed relative to the observer the whole time.)

Edited by md65536

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I'm going to reference a particular lesson from Mathius Blau General relativity.

Quote

light is deflected by a gravitational field just like material objects; clocks run slower in a gravitational field than in the absence of gravity. To see the inevitability of the first assertion, imagine a lightray entering the rocket / elevator in Figure 1 horizontally through a window on the left hand side and exiting again at the same height through a window on the right. Now imagine, as in Figure 2, accelerating the elevator upwards. Then clearly the lightray that enters on the left will exit at a lower point of the elevator on the right because the elevator is accelerating upwards. By the equivalence principle one should observe exactly the same thing in a constant gravitational field (Figure 3). It follows that in a gravitational field the lightray is bent downwards, i.e. it experiences a downward acceleration with the (locally constant) gravitational acceleration g

See page 26.

In essence inertial mass can also be referred to as acceleration mass. (Resistance to acceleration)

You will want to look at the previous pages as well. A more complex GR treatment is that a gravitational field is defined by the energy momentum tensor in terms of flux.

Edited by Mordred

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4 hours ago, md65536 said:

No, that's false. The twin that turns at B-turnaround (aka BC event) measures the same contracted length between AB and BC that OP's inertial clock B measures between the same events. The distance between AB and BC is 3 light years according to A, and 2.4 light years according to twin B or clock B. If you have a marker at BC that is stationary in A's frame (so that the proper length between A and the marker is the length measured in A's frame), then the marker approaches B (twin or clock) at a speed of 0.6c for 4 of B's years, traveling 2.4 light years in outbound B's frame.

Bringing this back to OP's post, I'll agree that your twin B's turn around causes it to follow the shorter world line made up of OP's B and C between the 3 events where clocks pass.

Hmm, not sure what you are talking about there md65536.  I did not say that twin B would measure the outbound & inbound separation-lengths differently than B or C.  Your response seems to imply you thought I said that. I said that twin B must properly accelerate, ie TURN in spacetime (which means "transition inertial frames") to age less than twin A.  That causes the separation between the events to length-contract, as that length then goes into motion per twin B.  Shorter length, he gets there quicker.  Similarly on return, so twin B travels a shorter duration, and hence a shorter worldline.  So when I said ... TURN (in spacetime), I meant "transition inertial frames".

Far as the Turnabout event goes (at the halfway point), it is defined (unrealistically) as instantaneous.  This means that the twin A-clock would (in the B spacetime system) swing from 2.4 ly away, thru 5 ly away, then to 2.4 ly away, instantly.  The Twin A clock's readout would (in the B spacetime system) swing from 3.2y-A, thru 5y-A, then to 6.8y-A instantly.  Of course, the receipt of light signals (a different thing) would not reveal any loss in A-clock time over the interval, per the doppler effect you pointed out prior.

4 hours ago, md65536 said:

Edit: Maybe I'm dumbing down the argument too much. You're arguing that the change in coordinate time at A's clock when B turns around, is something real (because the LT says it is?) and I'd argue that's just a coordinate effect that disappears with a different system of coordinates or without a definition of simultaneity. Equivalently, you're arguing that Einstein's definition of simultaneity must be physically real?

Or, back to OP's experiment, could we say the debate is about whether clock A is physically different to clock B vs. clock C besides how it appears differently to different observers? Would you argue that when twin B turns around (or if clock B were to turn into clock C), that causes a physical change in A? I say it doesn't, that the observed differences are only relative and depend only on the observer.

Relativity is an accepted theory, and it uses the Einstein convention for simultaneity.  The 1-way speed of light equals the 2-way speed of light.  I agree, that is a convention.  However, it may very well be true to mother nature.  Occam's Razor suggests we should assume it is, as everything works out far easier.  However, no matter what the true simultaneity of mother nature is (even if 1-way <> 2-way), would not a proper acceleration cause similar results? ... ie, swings in remote clock "location and time-readout" in the spacetime system of he who transitions inertial frames.

At face value, SR requires that all moments in time coexist.  It's a fused spacetime "continuum".  Time is laid out in the continuum as inches are laid out on the ruler.  So the rapid change in A clock readout per B during his turnabout, is physical in the sense that twin A advances along his own worldline in a spacetime continuum (in only B's system), due only to a change in the twin B POV during his turnabout. What must happen in the B spacetime system (per LTs) has no effect on twin A's own experience ... ie time passes as usual per A.

I agree with the OP in that the accrual of proper time is the very same for the Twins and Time Handoff scenario.  That's because the geometry of the worldlines are the very same.  But that's because the twin B proper-accelerations are considered instantaneous (impossible) and are hence ignorable, since no observer accrues proper time over that period.  The matter regarding how A exists in the B spacetime system during B's turnabout is a separate thing from the OP, however it should be considered for a complete relativistic analysis.

Best regards,

Celeritas

Edited by Celeritas

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19 hours ago, md65536 said:

Edit: Maybe I'm dumbing down the argument too much. You're arguing that the change in coordinate time at A's clock when B turns around, is something real (because the LT says it is?) and I'd argue that's just a coordinate effect that disappears with a different system of coordinates or without a definition of simultaneity.

md65336,

Maybe we should further address this point, regarding how real a coordinate time is.

Lorentz considered coordinate time a mathematical artifact that arose during the derivation of his LTs.  He called it "local time", to distinguish it from something "real".  Einstein said that what Lorentz called local time, "was just time".

Let's consider the ABC TIme Handoff scenario.  At B/C flyby, B declares the A-clock to then read 3.2y-A but C declares the A clock to then read 6.8y-A.  Are both correct, 1 correct, or neither correct?  Are they all correct, but their predictions something less than real?  Does a clock truly read what it's hands then display?

The eventual receipt of light signals will reveal that all are correct, and their predictions (per LTs) represent the "real time" as experienced by the remotely located A-clock, per the observer.  A clock is as old as it's time readout then displays, and we might assume that all clocks start ticking from zero at construction/birth.  The fact that observers disagree, does not lead that LT coordinate predictions are something less than real.  The fact that coordinate time differentials lessen with lesser range, and vanish when co-located with the observers (eg B&C at B/C flyby), does not lead that LT coordinate predictions are something less than real.

The problem is ... what convention of synchronicity should twin B use during his proper accelerations?  The Einstein convention does not work, because he's no longer inertial.  What we do assume, is that LTs predict reality for inertial POVs within any locally-flat-spacetime far from gravity source.

All that said, the Time Handoff and Twins scenarios produce the same result, because the geometry is the same. due to the specific scenario specifications and restrictions.  The Twins scenario is reduced to inertial purely segments that match the Time Handoff scenario, because zero duration during proper accelerations are zero and hence ignorable (wrt the accrual of proper time).  Again, the proper accelerations "are what causes" twin B to attain that geometry.  As such, they play as much a role in attaining that result, IMO, as does the quantification of relative time per the LTs over the inertial phases of flight.

Best regards,

Celeritas

Edited by Celeritas

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Coordinate time must be an artifact of the metric as a change in coordinates can alter the results. For example if you define a vector for time under the Minkowskii metric and change to polar coordinates the magnitude and direction of the ct (time interval) is not invariant under coordinate change. Though if you use the applicable four vectors or one forms they are invariant under coordinate change.  However coordinate time itself as a vector would follow the same problems as a Euclidean vector under coordinate change in so far as not being invariant.

Hence proper time defined by the four vectors ate invariant while coordinate time is not.

Recall time is given units of length by defining the interval ct.

To add to this an observer in his own reference frame will think time runs normally that is his coordinate time however another observer will see his time differently. Proper time is the only invariant time.

Edited by Mordred

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Celeritas,

I don't want to argue this. I can't prove simultaneity is merely a convention and you can't prove that it's physically real.

Einstein's convention does work in the case of the accelerating twin, because there is always a momentary comoving inertial reference frame that it can use. Even in the case of instant acceleration, you can let the twin instantly sweep through all velocities from its outbound to its inbound velocities, and you can instantly sweep through all planes of simultaneity between outbound and inbound. Whether or not there's any physical meaning to that doesn't even matter, because whether you do it that way or not, you're not going to predict any different results.

The LT uses Einstein's definition of simultaneity to establish the time at different locations within an inertial frame. I don't understand what you mean when you say the LT is real but Einstein's definition of simultaneity is a convention.

When B and C meet (or when B turns around), the time at A is relative. There is no one "then" that you're asking about when you say "Does a clock truly read what it's hands then display?" The clock at A is correct at all times through its journey. Also you talk about what the clock at A "then displays". But you know that B and C when they meet see the same time appearing on A's clock, in accordance with the relativistic Doppler effect. That they disagree on when "then at A" is, can come down to the fact that they disagree on how long it has taken light from A to reach them at that moment (because in this situation, B and C each have A moving at the same relative speed in different directions, and they agree on the rate at which A's clock ticks in accordance with SR). They measure at that moment that the clock at A appears to display the same time, but they can say that the time at A is "really" a different time at that moment because of the "time" it has taken light to reach them. BUT the "time" it has taken light to reach them is based on the definitions that Einstein uses, which are the same that establish whether two distant events are simultaneous or not. So the relative time at A, according to some distant observer, is "really" what the LT says it is, exactly to the extent that Einstein's simultaneity definition is "real". Whether one believes it is or not, it is established by definition, not by physical measurement.

19 minutes ago, Mordred said:

Hence proper time defined by the four vectors ate invariant while coordinate time is not.

That's why the results of the twin paradox and OP's experiment should be incontrovertible; they only require the comparison of proper times.

I've been debating the meaning of coordinate time and that's not helpful. The outcome of the twin paradox is as certain as SR is, and doesn't depend on how distant clocks relate, about which I'm not going to agree with others.

Edited by md65536

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SR has the capability of handling the twin paradox. Of that I have zero doubt however it requires the correct examination to handle assymmetric relations instead of strictly the symmetric Lorentz transforms. Quite frankly every solution involves showing the assymetry.

Perfect example is the assymetry due to acceleration.  That is one of the primary reason why I mentioned that much earlier in this thread to the OP.

Coordinate time can get tricky. One of the reasons tensors uses one forms is that Euclidean vectors are not invariant under coordinate transforms. (Well I should be more precise the compoments of a vector are not invariant, though few readers are familiar wirh vector components) The use of four vectors applies the Lorentz transforms to preserve Lorentz invariance. A good GR Introductory textbook should always dedicate a section to one forms. (One forms take a vector and returns a scalar )

The Minkowskii metric itself uses the inner product of two vectors. Hence it is coordinate dependent (Cartesian)

One can define the Minkowskii metric by $\mu\cdot\nu=\nu\cdot\mu$ which describes symmetric and orthogonal vector relations defined by the inner products under the Kronecker delta.  However the function of a metric is to provide the mapping of a vector to its one form. Ie a tensor or rank (m,n) contains the one forms (m) and vectors (n). However now I can quarantee I have gotten too complex for the average reader lol.

To put in simpler terms coordinate time has a coordinate basis of a vector.

Edited by Mordred

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17 minutes ago, Mordred said:

SR has the capability of handling the twin paradox. Of that I have zero doubt however it requires the correct examination to handle assymmetric relations instead of strictly the symmetric Lorentz transforms. Quite frankly every solution involves showing the assymetry.

Perfect example is the assymetry due to acceleration.

Agreed. If twins are symmetric, they can't have aged differently at an event through which they both pass (or ever, in a reference frame in which they're always symmetric).

Another example of asymmetry is OP's experiment, where the asymmetry is not due to acceleration.

I agree, in the twin paradox class of experiments, twin B's acceleration is the cause of the asymmetry. In other experiments, acceleration doesn't cause asymmetry (eg. if two twins both accelerate), in others asymmetry doesn't cause a difference in path length (you can contrive two twins to have aged the same amount when they meet), in others asymmetry has a cause other than acceleration.

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Correct the main gist of any valid solution is to show assymetry from the Lorentz transforms

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4 hours ago, Mordred said:

Correct the main gist of any valid solution is to show assymetry from the Lorentz transforms

Or you can simply compare the geometric length of their world lines (given shared events to begin and end at), and be done with it

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Lol true enough it's still fun to study different examinations as a lot of lessons can be learned.

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md65536,

So, our discussion was regarding whether acceleration plays any role in the relative aging outcome.  Everyone agrees that the LTs quantify relative time by the use of velocity, everything is inherent in the geometry of the worldlines, and it is ultimately about the length of the worldlines over the interval.

Camp 1 ... says proper acceleration plays no role in the relative aging in the twins scenario.

Camp 2 ... says (it does, ie) all relative velocities used by the LTs are the very result of the initial velocity and all subsequent twin B proper accelerations.

The twins must be co-located at the ends of the interval to produce a relative aging that's invariant.  It certainly takes twin B's proper accelerations, to make that happen.  The fact that we reduce the periods of acceleration to zero, does not change that.  It's about "the transitioning of inertial frames". Yet, no one disputes that the relative aging is quantified by the use of only velocities per the LTs.

Regarding coordinate time, we call it coordinate time because observers disagree per POV, not because the LTs produce coordinate times that are merey apparent or even potentially incorrect. Is it possible they are incorrect?  Yes.  However, we assume the Einstein clock synchronization procedure is true to nature and hence correct, per Occam's Razor (until some future test proves otherwise).  So at B/C flyby, we assume the A clock really does then exist at 3.2y-A per B, and the A-clock really does then exists at 6.8y-A per C.  Ie that that's true to reality. If a future test proves 1-way <> 2-way, well, then we'll be changing the LTs to accomodate.

Certainly, the accrual of proper time always dictates the relative aging result, instant accelerations or not.  Yet the use of coordinate time in the relative aging prediction does not alter the outcome, although it is a less convenient calculation.

One more ... at B/C flyby, C then holds A to exist at 6.8y-A.  We call that coordinate time, because other observers (eg B at B/C flyby) can disagree.  Yet, 6.8y is the proper time of the moving A-clock "when it displays 6.8y".

Best regards,

Celeritas

Edited by Celeritas
type-O

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