The Twins Paradox: A Controversial Truth or a Perplexing Paradox?

[yogi]

Jesse - I am trying to follow your logic. D and C are separated by a distance X as measured in the CD frame and A and B are separated by the same number of meters as measured in the AB frame - do I have that right? C and D are in motion v relative to AB and moving to the left - are you saying that when D is abreast of A, C will be abreast of B. Then A launches into motion so there is now no relative motion between A and D.

And then if C is separated from D by x meters - are you saying C will or will not reach B at a time that is definable in each frame relative to the time(s) that are measured by A and D when they meet. I am having a problem here because we have not synced the AB frame with the moving CD frame initially - so could you clarify this before we go on. Thanks

 

[JesseM]

Jesse - I am trying to follow your logic. D and C are separated by a distance X as measured in the CD frame and A and B are separated by the same number of meters as measured in the AB frame - do I have that right?

Yeah, that's right.

C and D are in motion v relative to AB and moving to the left - are you saying that when D is abreast of A, C will be abreast of B.

No. Each frame sees the distance between the other two as Lorentz-contracted--so in B's frame, D will catch up with A before C reaches B, and in D's frame, B catches up with C before A reaches D.

Then A launches into motion so there is now no relative motion between A and D.

Yes, at the moment A and D are at the same location, A instantaneously accelerates so there is no relative motion between A and D. Likewise, at the moment B and C are at the same location, C instantaneously accelerates so there is no relative motion between C and B.

And then if C is separated from D by x meters - are you saying C will or will not reach B at a time that is definable in each frame relative to the time(s) that are measured by A and D when they meet.

Both frames agree that at the moment C reaches B, C changes velocity so it's now at rest relative to B. But they disagree over whether this happens before or after A reached D and A accelerated so it was at rest relative to D.

I am having a problem here because we have not synced the AB frame with the moving CD frame initially - so could you clarify this before we go on. Thanks

There is no way to synchronize two clocks that are moving relative to each other permanently, but what we can do is set the two pairs of clocks either so that C reads the same time as B at the moment they arrive at the same location (and since C then accelerates so it's at rest wrt B, their times will be the same thereafter), or so that A reads the same time as D when they arrive at the same location (in which case A and D will read the same time thereafter).

 

[yogi]

Anyway - maybe another question will clarify your experiment - Taking premise # 1 you state that if A and B clocks read the same upon meeting, then C's clock cannot read the same as B. If the A and B clocks are initally in sync and the C and D clocks are initally in sync - and there is very little time loss in A's coming up to the speed of D, A and B will not have appreciable time difference at the outset - even though they are now in different inertial frames - as A travels toward B that time differential will grow. But if C arrives at B at the almost same time (which of course is undefined universally and must be subjectively measured in each frame) what is the rationale for the impossibility of C and B not registering the same time when they meet - I am missing something I know - so please clarify. Thanks

 

[gonzo]

gonzo -
The situation is very simple - except it doesn't follow. During the major part of his trip A moves at constant velocity - so he is in an inertial frame. His clock should not run at a different rate than B's clock since two identical mechanical devices cannot run at different speeds w/o a mechanical reason. The reason is absent. It has nothing to do with observations in other frames - there are none - only a beginning event (take-off) and an ending event (arrival) These two events exist at the same spatial point in both frames - but the arrival event corresponds to different time in each frame.

What doesn't it follow? The act of changing frames changes a lot of things, whether or not you do it instantaneously. One of these things is lines of simultaneity, and that means which events are in the future or in the past.

One way to think of it is that when one of them changes frames, a bunch of clock ticks in the other frame shift between the past and future as seen by the one in the new frame.

As for needing a mechanical reason for clocks to run differently -- why is that? In different frames, time and space are measured differntly, but there is no paradox or contradiction when you look at the other issues involved like simultaneity.

 

[gonzo]

Yogi, I was looking at your ABCD example above, and one thing you are missing is a reference event. You need an event in spacetime that both frames use for the origin to start counting from.

 

[yogi]

Ok - I posted my second question at the same time you were answering my first - so you are headed in the direction of resolving the problem using the thing I have from the beginning sought to avoid - namely making apparent measurements in the other frame - I know that relativity will always provide a way to manipulate the symbols to coax a solution that relativists will embrace. But as I have tried to stress, it is SR that is being questioned from the standpoint of how the experiments should be interpreted - so what is involved is using those aspects of the theory that have been verified, namely in the present discussion, time differences of high speed particles and satellite and airplane clocks to see, if a mechanism is revealed. Starting with a particular view on contraction to illustrate the validity of SR is a bootstrap argument.

When Einstein says that two clocks intially in sync and at rest in the same frame separted by a distance d - will read differently when brought together - I want to know what it is that is acting upon the clocks that brings about the result - I know the formulas - I derived them in graduate studies long before most of the persons posting on these boards hatched out of the egg.

Solutions that depend upon appearances in other frames are unsatisfying - like smoking a cigarette and not inhaling. Any interpretation that depends from Lorentz contraction immediately raises the question of the interpretation to be given to contraction. You may find solice in Eddington's statement: "Contraction is true, but its not really true" but I do not. I would say the reciprocal application of the contraction formulas raises questions at the outset. So in summary I would say, we probably won't be able to find a common turf for meaningful dialog. I do not say you are wrong or that any of my reasons are better - I hate to bring up religeon again - but its sort of like the missionaries that come around once and while and try to give me a Watch Tower. I know from experience the philosophy that separates me from these good folks can never find even a starting point of common ground so... I politely hand back the papar and excuse myself - they leave praying for my salvation - so perhaps jesse - you too should pray for this wayword infidel

 

[JesseM]

Anyway - maybe another question will clarify your experiment - Taking premise # 1 you state that if A and B clocks read the same upon meeting, then C's clock cannot read the same as B. If the A and B clocks are initally in sync and the C and D clocks are initally in sync - and there is very little time loss in A's coming up to the speed of D, A and B will not have appreciable time difference at the outset - even though they are now in different inertial frames - as A travels toward B that time differential will grow. But if C arrives at B at the almost same time (which of course is undefined universally and must be subjectively measured in each frame) what is the rationale for the impossibility of C and B not registering the same time when they meet - I am missing something I know - so please clarify. Thanks

But C won't arrive at B at almost the same time, because of Lorentz contraction. If the CD system is moving at (\sqrt{3}/2)c in the AB frame, then if the distance between A and B is 1 light-year, and the distance between C and D is also 1 light-year in their own rest frame, then in the AB frame the distance between C and D will only be 0.5 light-years. So at the moment D reaches A, C will still be half a light-year away from B's position in the AB frame:

---------C---------D
B------------------A

Also, remember that if C and D were synchronized in their own frame, in the AB frame they will be out of sync--if AB sees the distance between C and D as x (with x=0.5 light years if you use the numbers above), then in the AB frame D will be ahead of C by \gamma (vx/c^2), or \sqrt{3}/2 years using the values of x and v I gave above. So if A and B's clocks read t=0 at the moment things look like the diagram above in the AB frame, then if you set C and D such that D also reads t'=0 at that moment, then C will read t' = -\sqrt{3}/2 at that moment. Since C is traveling at (\sqrt{3}/2)c and it has to travel 0.5 light-years to reach B, it will take an additional 1/\sqrt{3} years to reach B in the AB frame, so the B-clock will read t = 1/\sqrt{3} when C meets it. Meanwhile, the C clock is only ticking at half the rate of B in the AB frame, so it will only have ticked forward by (1/2)*(1/\sqrt{3}) years in this time, so it should read t' = -\sqrt{3}/2 + (1/2)*(1/\sqrt{3}) = -3/(2\sqrt{3}) + 1/(2\sqrt{3}) = -1/\sqrt{3} years when C meets B.

 

[yogi]

gonzo - that was Jesse's example - and I made the same comment. I am fully aware of the shifting hyper planes and a number of other proposed solutions to the twin paradox - these all depend upon observations (in another frame) You can't look at clocks or lines of constant temporal events or whatever and arrive at a physical age difference. That there is a physical age differece we all agree - it comes about because of some physics - remember - according to SR a clock in an inertial frame always runs at the same speed as any other inertial frame - as I previously posted, I believe the difference in the aging lies in the fact that the fame where the distance is measured is not symmetrical with the fame of the clock which is moving. A straight forward application of the principle of the interval invarience shows this to be the case, although relativists balk at this explanation.

Feynman once posed the question of why so many of our physical formulas have the same form - he queried further by asking what one thing they all have in common. He concluded: "It is the space - the framework into which the physics is put." And that is the difference between the two frames - in one frame the interval is compsed only of time - in the other there is a physical change in the spatial coordinates betwen the beginning and ending events - and that in my opinion is where we should look for the mechanism

 

[JesseM]

Ok - I posted my second question at the same time you were answering my first - so you are headed in the direction of resolving the problem using the thing I have from the beginning sought to avoid - namely making apparent measurements in the other frame - I know that relativity will always provide a way to manipulate the symbols to coax a solution that relativists will embrace.

yogi, forget about frames, and think only about physical facts that will be agreed upon by all frames. For example, in the example I gave, it is a physical fact that at the time D and A meet, both their clocks read a time of zero years. (On the other hand, the question of what B and C read at the 'same time' as this event is not a frame-independent physical fact, so let's not even think about it.) Likewise, it is a physical fact that when C and B meet, B reads 1/\sqrt{3} years and C reads -1/\sqrt{3} years. Likewise, we could calculate the time read on all four clocks at the moment they all meet. "Reference frames" help you calculate these numbers, but just think of them as scaffolding, once we have the physical facts of the situation, we can get rid of the scaffolding and just list these facts without any adornment. You do agree that these statements don't depend on reference frames, and that the numbers I've given will be correct, right?

If so, go back to the question at the end of post #87, and note that the parts in bold are only saying things about such frame-independent physical facts:

Now, in terms of how D and A's times relate when they meet, and how C and B's times relate when they meet, you have two choices:

1. When D meets A, their clocks read the same time, and read the same time thereafter. This means C's clock will not read the same time as B's when they meet.
2. When C meets B, their clocks read the same time, and read the same time thereafter. This means A's clock will not read the same time as D's when they meet.

You can't have both, but do you agree there's no reason to prefer one over the other, since if you pick #1 the situation in B's frame looks exactly how the situation in D's frame looks if you pick #2, and vice versa? (this also means that if you pick #1, and when all four meet A and D's clocks read time x while C's reads y and B's reads z, then if you pick #2, you know that when all four meet C and B's clocks will read time x while A's reads y and D's reads z--once again, totally reciprocal).

So, do you agree with this last statement in bold, which has nothing to do with "reference frames" at all?

But as I have tried to stress, it is SR that is being questioned from the standpoint of how the experiments should be interpreted

But there is no "interpretation" in physical facts like the time that two clocks read at the moment they meet at a single point in space, agreed? "Interpretation" only arises when we start making statements that depend on reference frame, like what C read "at the same time" that A and D were meeting.

 

[yogi]

Jesse - My view of the ABCD problem is this - First - I do not believe in actual space contraction - While it is true that each frame measures space as contracted in the other frame - there is not a real contraction in the sense envisioned by Lorentz - this notion appears to be a carryover from Lorentz Ether theory... and while measurments made in the other frame are real in the same sense of all measurements, they do not correctly depict things in the other frame. So from my point of view, ABC and D can all be brought into sync - the distance between A and B is x and the distance between C and D is x - so even though A would calculate the distance x in the CD frame is different from the x measured in his own frame - A and D could be coincident and brought into sync and C and B can be brought into sync when they are coincident and all 4 clocks will read the same number (if they could instantantly communicate with one another) .. But since they can't, each clock could record the time at which they are synced on a permanent memory to be later compared).

In your last post - why is it a physical fact that when C meets B the time shift is
1/(3)^1/2. Doesn't this follow again from your premise re the reality of contraction.



Permit me to go back to the two clock set up - A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets A. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B - the two times are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

 

[JesseM]

Jesse - My view of the ABCD problem is this - First - I do not believe in actual space contraction - While it is true that each frame measures space as contracted in the other frame - there is not a real contraction in the sense envisioned by Lorentz - this notion appears to be a carryover from Lorentz Ether theory... and while measurments made in the other frame are real in the same sense of all measurements, they do not correctly depict things in the other frame. So from my point of view, ABC and D can all be brought into sync - the distance between A and B is x and the distance between C and D is x - so even though A would calculate the distance x in the CD frame is different from the x measured in his own frame - A and D could be coincident and brought into sync and C and B can be brought into sync when they are coincident and all 4 clocks will read the same number (if they could instantantly communicate with one another) .. But since they can't, each clock could record the time at which they are synced on a permanent memory to be later compared).

In your last post - why is it a physical fact that when C meets B the time shift is
1/(3)^1/2.

Actually the time shift is 2/(3)^1/2, since B reads 1/(3)^1/2 and A reads -1/(3)^1/2.

Doesn't this follow again from your premise re the reality of contraction.

It follows from the fact that the equations of the laws of physics have the mathematical property of Lorentz-invariance, and that both observers "synchronize" their two clocks by turning on a light at the midpoint between the two clocks and setting the clocks to read the same time at the moment the light reaches them. Are you saying you're not just disagreeing with the "interpretation" of relativity, but with the actual physical predictions of relativity? You think that even if they both use the "synchronization" procedure described above, it is possible for A and D to read the same time when they meet and for C and B to read the same time when they meet? Do you think the equations that correspond to the most fundamental known laws actually give incorrect predictions about physical events, and that the "true" fundamental laws would not be Lorentz-invariant?

Permit me to go back to the two clock set up - A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets A. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B - the two times are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If A resets his clock to match that of E at the moment he passes it, and E is in sync with B in the BE rest frame (which means they aren't in sync in A's frame) then of course when A reaches B his clock will read a smaller time than B's. On the other hand, if there is a clock F at a constant distance from A and in sync with it in the AF rest frame, and B resets his clock to read the same time as F when he passes it, then B will read a smaller time than A when A and B meet. Once again, the situation is completely symmetrical.

 

[CronoSpark]

Light travels at a constant speed.
We percieve light at a constant rate.
Light rebounds from the clocks and reaches the reciever at a certain time.
When both clocks are going away from each other, we percieve a slower FPS.
When both clocks are traveling towards each other, we percieve a higher FPS.
I do not think there is a paradox. Both twins should be the same age in their own frames of reference for the whole trip. When B is x distance away and is stationary, then A will see B being x/c time younger. When B reaches Earth again, both should be the same age.

 

[JesseM]

Light travels at a constant speed.
We percieve light at a constant rate.
Light rebounds from the clocks and reaches the reciever at a certain time.
When both clocks are going away from each other, we percieve a slower FPS.
When both clocks are traveling towards each other, we percieve a higher FPS.

What's "FPS"?

I do not think there is a paradox. Both twins should be the same age in their own frames of reference for the whole trip. When B is x distance away and is stationary, then A will see B being x/c time younger. When B reaches Earth again, both should be the same age.

Not according to relativity, no. And everything about relativity follows from the assumptions that the laws of physics should work the same in every inertial reference frame, and that the speed of light should travel at the same speed in every inertial reference frame. So if you disagree with relativity's prediction that the twin who departs and then returns will be younger than the twin whose velocity never changed, then you must disagree with one of these two assumptions.

 

[yogi]

Hi Jesse - In the ABCD problem, the issue for me is not the validity of the LT - it is the interpretation - in each frame the separation between the end points is x - if relative motion means actual contraction then there is no coincidence between B and C when A and B are concurrent, and vice versa. That is to me an interpretional issue - not a rebuke of Einstein's theory. Let me again regress to the simplier question of whether you find a difference in the situation where two separated objects are in at rest in the same frame and in sync and a third clock enters the picture flying by one of the at rest clock to copy its reading. I see no way the transforms can be interpreted to predict a different age for the newcomer clock when it continues on to meet the other at rest clock.

 

[yogi]

jesse - to recap: A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets B. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B but separated from B by a distance d- the two times (A and E) are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If A arrives at B having aged less than B after he has set his clock to correspond with E, how could that happen since in this situation we have no way of knowing whether E and B are moving relative to a space defined by a distance d in the A frame or a distance defined in the EB frame. There is symmetry in this case. Contrary wise if E takes off and flies parallel to A to arrive at B coincident with A. In this case the symmetry is broken because E is traveling in a space defined by distance measured in the B frame - so E's clock will read less than B's when E meets B.

 

[JesseM]

Hi Jesse - In the ABCD problem, the issue for me is not the validity of the LT - it is the interpretation - in each frame the separation between the end points is x

Ok, let's not say anything about the distance between them, since "distance" is not an objective physical concept. Let's just state the following objective facts: if a light is turned on at the midpoint of B and A before A accelerates, then A's clock and B's clock will read the same time at the moment the light hits them; likewise for C and D before C accelerates. And before any acceleration happens, if A aims his telescope at B, he will see B's clock 1 year behind the clock he is carrying, and vice versa; same goes for what happens when C aims his telescope at D or vice versa. Given these conditions, do you agree with the numbers I gave for what times C and B will read when they meet, given that A and B both read a time of 0 years when they meet? Do you agree that the situation is "symmetrical" in the sense I described at the end of post #87, so if you instead set things up so C and B read 0 years when they meet, D will read 1/\sqrt{3} and A will read -1/\sqrt{3} when A and D meet, and likewise the situation is also symmetrical when all four meet?

 

[JesseM]

jesse - to recap: A and B are two clocks brought to sync in a common frame in which they are both at rest - A accelerates to v and travels at v until he meets B. Any path will do. A and B could initailly have been side by side and A buys a round trip ticket - or they could have been originally separated - the time differential doesn't distinguish between the two situations. Do you think it is possible to have the same result if A were already in motion relative to B? For example let's assume A is an incoming rocket with velocity v that passes near a clock E that is in sync with B and at rest in the frame of B but separated from B by a distance d- the two times (A and E) are compared as they pass (e.g., A resets his clock on the fly to correspond with that E -- A continues on toward B w/o changing his velocity v - when A arrives at B what will he report as to his trip time from E to B?

If t_E is the time that E read as A passed it, and t_B is the time that B read when A met it, then the time that A measures between passing E and meeting B will be \sqrt{1 - v^2/c^2}(t_B - t_E).

If A arrives at B having aged less than B after he has set his clock to correspond with E, how could that happen since in this situation we have no way of knowing whether E and B are moving relative to a space defined by a distance d in the A frame or a distance defined in the EB frame.

A hasn't "aged less", his clock just reads a lesser time than B's. From A's point of view, the reason for this is that B's clock is always ahead of E's clock. (If I put a clock in my hallway that's one hour behind the clock in my study, and it takes you only 15 seconds to walk through my hallway and into my study, am I justified in saying that you've aged 59 minutes and 45 seconds less then me when we meet?) If you want to stick to objective physical facts that don't depend on your choice of reference frame, the question of "who has aged less" is meaningless for two observers moving at constant velocity, just like the question of which of two meter sticks in relative motion "really shorter". The only situation where you can talk objectively about which of two observers has aged less is where they first compare clocks at a single location, then move apart, then come back together to compare clocks again.

You can't have it both ways, sometimes rejecting ideas which depend on reference frames like Lorentz contraction, at other times wanting to talk in terms of language that has no frame-independent meaning, like which of two observers moving at constant velocity has "aged less".

There is symmetry in this case.

No, this is not symmetrical, because you have two separated clocks at rest and synchronized in B's frame, while A has only one clock. To make it symmetrical, you'd have to add another clock F which is at rest relative to A, and synchronized with A in A's rest frame. Then it's clear that you can either arrange things so that A synchronizes his clock with E as he passes it, or that B synchronizes his clock with F as he passes it--and whichever arbitrary choice you make, the one that synchronized his clock with the clock at rest in the other's frame will read a smaller time when A and B meet. Do you agree?

 

[gonzo]

Yogi, it might help if you didn't think about inertial frames as having any existence of their own. As I understand it, they are just artificial constructs for humans to look at things

Frames don't exist, only events in spacetime exist, whether that event is a tick of a clock, or two particles being in the same place. We humans can then try and measure things about these events, using various metrics. It turns out that these metrics differ in different inertial frames, though the spacetime interval is always the same. The events are still real, still happen, and aren't affected by our way of measuring them.

Taylor and Wheeler's book has a good analogy of two people making a map of a city, using slightly different directions for North. Using the city center as the origin, they will assign different East-West and North-South distances to the same point in the city. That point still exists in the same place, and that point doesn't really care about how it is being measured, but they will still have different numbers for it. However, straight distance from the city center will be the same for both cartographers.

In your example in #115 of A flying between E and B which are stationary and in sync, assuming no acceleration, you are looking at it in a funny way as I see it. There are simply two events. Event one is when A arrives at E, and event two is when A arrives at B. You can even take event one to be the agreed upon origin.

These are two events that have nothing to do with any inertial frames. However, we can if we want measure the time and distance between these events. If this measurement is done in different inertial frames, then the interval between these events will be same.

Luckily, this is a simple example since if we look at the inertial frame following A, the distance is 0 between the two events, so the the Interval is equal to the Time.

Assuming everyone sets their origin to when A passes E, then you can compare measurements when A passes B. And yes, A's clock and B's clock will show different elapsed times, just like they will show different distances. For A, there was zero distance between the two events. They both happened at A. For B there was the distance BE between the two events. Different distance means different times. So what?

I don't really see what you mean by "symetrical" in this case.

 

[yogi]

Jesse: From your post 117: "The only situation where you can talk objectively about which of two observers has aged less is where they first compare clocks at a single location, then move apart, then come back together to compare clocks again."

Simply not true - look again at what I have quoted from the 1905 paper - A and B are not together - they are at rest separated by a distance d, brought to sync - and then A moves to meet B. There is now an age difference. A's clock will have recorded less time than B's. Seems we have already established this about 10 times.

 

[yogi]

Let us assume Yogi and Jesse are on orbiting spacecraft - I am flying around the equator east to west and you are flying west to east. Each time we pass each other we check the other guys clock using two clocks in our own spacecraft - You will say poor Yogi - not only is he slow to understand SR - his clock is running slow also. And I will look out my window and measure the rate of your clock as it flies by and say - Jesse is to busy posting on the forums to set his clock rate up to speed. Gonzo is at the North pole - at the same height above the Earth as you and I - he will see both our clocks running slow. Now each time we pass I will see your clock running slow and you will see mine running slow - but a funny thing happens -neither of us accumulates any age difference relative to the other - our situation is entirely reciprocal. So after many orbits we decide to land and we do so bu detouring to the North Pole - we compare clocks - Jesse and yogi's logged times will be the same - but Gonso's clock will read more than Jesse's and Yogi's Or do you have a different conclusion?

 

[JesseM]

Jesse: From your post 117: "The only situation where you can talk objectively about which of two observers has aged less is where they first compare clocks at a single location, then move apart, then come back together to compare clocks again."

Simply not true - look again at what I have quoted from the 1905 paper - A and B are not together - they are at rest separated by a distance d, brought to sync - and then A moves to meet B. There is now an age difference. A's clock will have recorded less time than B's. Seems we have already established this about 10 times.

Can you provide that quote again? I've forgotten which post of yours included it. I am confident that Einstein would not say that in that situation A has objectively "aged less" than B in a frame-independent sense, if you think I'm wrong please show a specific quotation where he says something to that effect.

Let us assume Yogi and Jesse are on orbiting spacecraft - I am flying around the equator east to west and you are flying west to east. Each time we pass each other we check the other guys clock using two clocks in our own spacecraft - You will say poor Yogi - not only is he slow to understand SR - his clock is running slow also. And I will look out my window and measure the rate of your clock as it flies by and say - Jesse is to busy posting on the forums to set his clock rate up to speed.

Only in my instantaneous rest frame at any given moment can I say your clock is running at \sqrt{1 - v^2/c^2} times the rate of mine, but I can't integrate \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dv to find the elapsed time on your clock as my clock ticks from t_0 to t_1, since I am not in an inertial reference frame. Also, if we actually watched each other's clocks using light instead of calculating the rate of the clocks in our instantaneous rest frame, we would sometimes see each other's clocks ticking faster, sometimes slower.

Gonzo is at the North pole - at the same height above the Earth as you and I - he will see both our clocks running slow. Now each time we pass I will see your clock running slow and you will see mine running slow - but a funny thing happens -neither of us accumulates any age difference relative to the other - our situation is entirely reciprocal. So after many orbits we decide to land and we do so bu detouring to the North Pole - we compare clocks - Jesse and yogi's logged times will be the same - but Gonso's clock will read more than Jesse's and Yogi's Or do you have a different conclusion?

Yes, you're correct. But gonzo is the only one in an inertial reference frame here, so all of us must use his frame to calculate the actual time elapsed on all three of our clocks when we meet (assuming we're doing the calculations in SR, not GR).

 

[yogi]

Jesse - take a look at post #55 re your Q.

As far as you and I and Gonzo - none of us is in a truly inertial frame - we are accelerating due to the Earth's rotation and Gonzo is accelerating because of the Earth orbit around the Sun. Its just a matter of degree. Even if the measurments are perfect, you can still do the experiment - just assume you kink your orbit slightly each pass so that for an instant your path is straight. The apparent slowing of my clock will be detected by two clocks spaced apart in your spaceship.

 

[JesseM]

yogi, the quote you provided in post #55 was:

"If at points A and B there are stationary clocks which viewed in the stationary system, are synchronous, and if clock A is moved with the velocity v along the line AB to B then on its arrival the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)t(v/c)^2 ..."

There is absolutely nothing there about A "aging less" in an objective, frame-independent sense, in fact there is nothing about the relative rates of the clocks at all, all he says is that the clock A will be behind the clock B "on its arrival", which of course I agree with. The claim that A has objectively aged less, despite the fact that in A's frame it is B who has aged less, is totally ridiculous and conflicts with the basic assumptions of relativity, so I am sure Einstein never suggested such a thing.

Let me ask you this: do you agree that what's important in determining who has aged less when A and B meet is only the time intervals recorded by each frame, not the actual times on each clock when they meet? Suppose at B and E are synchronized in their own rest frame, and A is moving towards them at constant velocity, first passing E and then meeting B, and we don't assume that A synchronizes his clock with E at the moment they pass. Instead, at the moment A passes E, E's clock reads t_{EA} and A's clock reads some different time T_{AE}, and at the moment A reaches B's location, B's clock reads t_{BA} and A's clock reads T_{AB}. Do you agree that even if T_{AB} > t_{BA}, we can still say that A has aged less in the BE frame since (t_{BA} - t_{EA}) > (T_{AB} - T_{AE})? If so, are you claiming that not only would this mean that A aged less in the BE frame, it would mean that A aged less in a frame-independent sense? Please answer these questions yes or no.

 

[yogi]

Jesse --We have an entirely different interpretation of what Einstein has said - The only significant thing is Einstein's lack of explanation of why one of the two clocks objectively aged less - A or B. This is the whole point I have been trying (obviously w/o success), to get across. Einstein didn't explain it -but as I keep saying there is asymmetry in the one way twin problem because once A and B are in sync in the same rest frame, one must move so that they eventually meet - and the one that moves will have logged less time on his clock. If it is A that moves (by accelerating) it is of no consequence that in A's frame, B's clock appears to be running slow while they are in relative motion - that is the "apparenency" that always occurs when making measurments in a relatively moving frame -

if both A and B accelerated equally, A and B while in relative motion, would each believe the other clock is slow, but because there is symmetry in this example, there would be no real age difference when they compared clocks upon meeting. Einstein's whole point of part 4 of his 1905 paper was to express the physical meaning of the transforms - real measurable age difference - this is the transition from illusory observations to reality. He stated the result, but he didn't explain it.

.

With regard to the non-inertiality of the frames in my orbiting spaceship description - I think it is insignificant. These are all thought experiments - we don't need to get bogged down with the fact that there is some slight curvature to the orbits during the sampling period - in one sense you can consider our spaceships to be non accelerating since they are in orbit, there are no G forces - and in another sense you can rely upon the fact that it is not necessary to take into account the curvature in GPS - we get the correct offset straight away from the relative velocity between the clock in the Earth centered reference frame and the satellite clock(s) via the Lorentz transforms - ergo it is not a general relativity problem.

 

[yogi]

Jesse - Your question - No - I don't think there is adequate information to make any conclusions about actual age difference - at least as I understand your question, I would have to know whether the relative velocity between A and the EB frame came about by acceleration of the EB frame or A.

 

[JesseM]

Jesse --We have an entirely different interpretation of what Einstein has said

No, your "interpretation" cannot possibly be valid, unless you imagine that Einstein himself disbelieved the principle that each reference frame's perspective was equally valid, and one would have to be grossly ignorant to assert such a thing, given everything else he wrote.

The only significant thing is Einstein's lack of explanation of why one of the two clocks objectively aged less - A or B.

No clock has objectively aged less, that's utter nonsense. I notice you didn't answer my question earlier, do you agree with my statement about the time intervals being all that matter? In other words, if E reads 2:00 and A reads 4:00 when they pass, and then A reads 5:00 and B reads 4:00 when they meet, do you agree that it doesn't matter that A has a greater time, all that really matters is that A saw an interval of 1 hour while the interval was 2 hours in the EB frame? And if you do agree on this, would you also say that this proves that A was objectively aging slower as he approached B? Please address this question.

if both A and B accelerated equally, A and B while in relative motion, would each believe the other clock is slow, but because there is symmetry in this example, there would be no real age difference when they compared clocks upon meeting. Einstein's whole point of part 4 of his 1905 paper was to express the physical meaning of the transforms - real measurable age difference - this is the transition from illusory observations to reality. He stated the result, but he didn't explain it.

Einstein made very clear that different frames define simultaneity differently--this was the whole point of section 1--so obviously that's how he would explain why this situation is symmetrical, because in A's frame B did not read the same time as E, B was in fact ahead of E, so the fact that B was ahead of A when they met is perfectly consistent with the idea that B was ticking more slowly than A in A's frame. If you don't like this answer, fine, but it would be foolish to deny that Einstein and other relativists would see it this way.

With regard to the non-inertiality of the frames in my orbiting spaceship description - I think it is insignificant. These are all thought experiments - we don't need to get bogged down with the fact that there is some slight curvature to the orbits during the sampling period

Slight curvature? They make a complete circle during the sampling period! If you were just looking at a very short time in which the curved orbit was close to a straight line, that'd be one thing, but a complete circle is about as far as a straight line as you can get.

in one sense you can consider our spaceships to be non accelerating since they are in orbit, there are no G forces

Sure there are, there will be centrifugal force on board the orbiting ship. edit: I just remembered that in orbit, the centrifugal force will be equal and opposite to the gravitational force...and of course the equivalence principle says that in an arbitrarily small region of spacetime, the laws of physics will look the same for an observer in free-fall as they do for one moving inertially. But if you want each observer to keep track of the movement of the other observer at moments other than the one where they are passing each other in orbit, then their coordinate systems cannot be arbitrarily small, and there will be effects which distinguish their coordinate system from an inertial one, like tidal forces and the fact that light seems to go faster in one direction than the other.

and in another sense you can rely upon the fact that it is not necessary to take into account the curvature in GPS - we get the correct offset straight away from the relative velocity between the clock in the Earth centered reference frame and the satellite clock(s) via the Lorentz transforms - ergo it is not a general relativity problem.

Nonsense, the GPS satellites certainly take into account the fact that clocks on the Earth's surface are not moving inertially--where are you getting your information? I suggest you take a look at http://relativity.livingreviews.org/Articles/lrr-2003-1/ on relativity and the GPS satellites...for example, look at section 2 (Reference Frames and the Sagnac Effect), where they show the time required for light to travel a certain path in a rotating reference frame (equation 7), and then say:

Observers fixed on the earth, who were unaware of Earth rotation, would use just \int d{\sigma}^{'} /c for synchronizing their clock network. Observers at rest in the underlying inertial frame would say that this leads to significant path-dependent inconsistencies, which are proportional to the projected area encompassed by the path.

 

[JesseM]

Jesse - Your question - No - I don't think there is adequate information to make any conclusions about actual age difference - at least as I understand your question, I would have to know whether the relative velocity between A and the EB frame came about by acceleration of the EB frame or A.

OK, thanks for addressing this question--but are you asserting that the history of which accelerated most recently is somehow relevant? In other words, if A is at first at rest relative to B and E and then accelerates towards them, this would lead you to a different conclusion about whether A had "objectively" aged less than B as it passed from E to B than if A had been moving at constant velocity towards the position of E and B since before E and B were moved (accelerated) from Earth to their current position in space? What if A last accelerated a million years ago while E and B last accelerated a million and one years ago, would this lead you to a different conclusion about who objectively aged less than if E and B last accelerated a million years ago while A last accelerated a million and one years ago?

 

[yogi]

Jesse:

The issue is not which frame accelerated most recently - it is simply this - two clocks in one frame are brought in sync - one is moved. It will be found to be out of sync with the one which did not move - and the amount of the discrepency is given in Part 4 of the 1905 paper.

With regard to the counter orbiting spacecraft - you again want to impose conditions that are not required - I do not have to wait for a complete orbit to determine that the other clock is running slow - I simply place two clocks at opposite ends of my spaceship, measure the distance between them L , and then use my two clocks to measure how long it takes your clock to travel the distance L between my two clocks. Based upon the time difference recorded by my clocks and the time recorded your clock - I will always believe that your clock is running slow.

Clocks in relative motion do record different absolute times whenever they have been synced at rest and one clock is put into motion - all evidenced by H and F airplane experiments, GPS, and the extended lifetime of high speed muons and pions that are created in the Earth reference frame and subsequently move relative thereto.

I don't really think we can learn anything further from each other - this thread has gone on way too long as it is.

 

[JesseM]

Jesse:

The issue is not which frame accelerated most recently - it is simply this - two clocks in one frame are brought in sync - one is moved. It will be found to be out of sync with the one which did not move - and the amount of the discrepency is given in Part 4 of the 1905 paper.

What does "moved" mean? Does it mean "accelerated", or does it just mean that A is moving relative to the rest frame of B and E? If acceleration isn't relevant, why did you say "at least as I understand your question, I would have to know whether the relative velocity between A and the EB frame came about by acceleration of the EB frame or A"? If acceleration is relevant, yet it doesn't matter who was accelerated most recently, then in every possible experiment both A and EB will have been accelerated at some point in their past, it's not like there are any clocks in the universe which have been traveling at constant velocity since the Big Bang. So what does this lead you to conclude about which one was "moved", and which clock is "really" ticking slower?

With regard to the counter orbiting spacecraft - you again want to impose conditions that are not required - I do not have to wait for a complete orbit to determine that the other clock is running slow - I simply place two clocks at opposite ends of my spaceship, measure the distance between them L , and then use my two clocks to measure how long it takes your clock to travel the distance L between my two clocks. Based upon the time difference recorded by my clocks and the time recorded your clock - I will always believe that your clock is running slow.

In this case, you're right that the curvature of the orbit will be negligible, as long as your ship is small. But then what was the point of bringing an orbit into the picture at all? Why not just imagine that your ship is moving inertially, and mine is too?

Finally, what does this have to do with the claim that there is an absolute truth about which clock was ticking slower? Don't you agree that if my ship has two clocks on either end, and I look at the time on my clocks as one of your clocks passes both, I will say that your clock ran slow?

Clocks in relative motion do record different absolute times whenever they have been synced at rest and one clock is put into motion - all evidenced by H and F airplane experiments, GPS, and the extended lifetime of high speed muons and pions that are created in the Earth reference frame and subsequently move relative thereto.

None of these experiments show evidence for an "absolute" truth about which clock is running slow, all of them are perfectly consistent with the idea that each inertial frame sees clocks in other frames running slow, and that there is no way to settle which frame's point of view is the correct one. If you think Einstein would have disagreed, then you are completely misunderstanding his paper.

I don't really think we can learn anything further from each other - this thread has gone on way too long as it is.

I think that if you will clarify some of the ambiguities in your arguments, like what you mean by "moved" and why you think there is an absolute truth about which of two clocks is running slow, then this argument could still end up being a productive one.

 

[JesseM]

Following up some comments from the Cosmological Twin Paradox thread:

Jesse. Moved means - they are initially brought to rest in one frame - and one of the clocks is accelerated - we know which is accelerated because some agency is required to bring about a velocity change of one clock only. But the acceleration has nil to do with the reading of the clocks

If it has nil to do with the reading of the clocks, then why did you bring it up? Do you agree that in the cosmological twin paradox, if the twins are originally both on Earth (which we assume is moving inertially) and then one twin accelerates briefly and then flies away from the Earth at constant velocity, the twin who accelerated may have aged more rather than less when the two twins meet again, since the Earth may not be at rest in the preferred coordinate system defined by the topology of the universe? Do you also agree that in SR, the question of which of two clocks was "moved" is irrelevant to the question of which clock "really" aged less in a particular time-interval?

Let us see if the cosmological twin problem can be localized - say that we have two clocks A and B in orbit about the Earth - Clock A was built on Alpha and was in sync with all alpha clocks before it was launched a million years ago and eventually captured by the Earth's G field into an east-west circular orbit. Clock B was built on Earth and was in sync with Earth clocks before being launched into a west-east circular Earth orbit. As they pass each other every 2 hours - will one or the other of these clocks appear to be gaining time?

Why did you bother to say that B was built on Alpha Centauri and A was built on Earth? This is totally irrelevant to the problem. Again, past history doesn't matter in relativity, including the question of which one accelerated (or which one 'moved' in your terminology). So all that matters is that we have two clocks, one orbiting west-east and the other orbiting east-west, if we know their specific velocities and the size of their orbit this will tell us whether one clock has gained any time from one moment they pass to another, and if so how much. If they are both orbiting at the same speed in the rest frame of the Earth's center, then neither will gain time; if they are orbiting at different speeds relative to the center (which would be true if they were orbiting at the same speed relative to the surface), then the one that's orbiting faster in the rest frame of the Earth's center will get show less elapsed time between successive passings.