How Time Passes Differently for Twins Traveling at Light Speed

[Ich]

But B only aged less in A's frame. If you look at things in the frame where A was at rest and B was moving towards him, in this frame A aged less between the time B accelerated and the time A and B passed each other. Yogi doesn't accept that either perspective is equally valid, and says there's some objective, frame-independent sense in which B really aged less between the time he accelerated and the time A and B passed, because B was the one who accelerated.

I think yogi is partially right here. When you set up his experiment exactly as he sais, you obtain those 3 events I described. And the distance between 2) and 3) is in an objective, frame-independent sense less than the distance between 1) and 3).
And so yogi is right in claiming that in his experiment the clocks will be out of sync. But this has nothing to do with acceleration - the frame dependent thing is that he said that the clocks have been synchronized in A´s frame.

 

[JesseM]

I think yogi is partially right here. When you set up his experiment exactly as he sais, you obtain those 3 events I described. And the distance between 2) and 3) is in an objective, frame-independent sense less than the distance between 1) and 3).

But there's no frame-independent sense that 2) and 1) happened at the same moment. We could also imagine event 1b): A looks at a special clock designed to show the correct time in B's frame, and sees that it reads time 0. In this case, it is also a frame-independent truth that the time between 1b) and 3) is less than the time between 2) and 3).

And so yogi is right in claiming that in his experiment the clocks will be out of sync. But this has nothing to do with acceleration - the frame dependent thing is that he said that the clocks have been synchronized in A´s frame.

I agree the clocks will be out-of-sync, of course. But the only question of contention here is, is there any frame-independent sense in which B "aged less" than A from the moment he accelerated towards A to the moment he arrived at the same location as A? Is there any sense in which it is somehow less valid to consider this experiment from the point of view of an inertial frame where, after A accelerates, A is at rest and B is moving towards him?

 

[Ich]

Agreed to all:
yogis experiment has a frame-independent outcome because he gave an exact prescription for the definition of event 1) ("simultaneous to event 2) in A´s frame").
And it makes no sense of talking about who aged less if you leave event 1) somewhat undefined. It all reduces (as always) to the definition of simultaneity.

 

[yogi]

Ok - some interesting comments - let me go back Jesse to the thought that I proposed a few weeks ago - and here is where I stand. We start with A and B at rest in an inertial frame - They can be adjacent - it doesn' matter - they are identical clocks. B Starts out North and then swings East, then South in a large half circle passing Altair and then continues South then bends West and returns to A completing a giant circle. As stated, this is a version of the twin paradox - but there is not abrupt turn around - whatever is going on with B's clock is a continuous happening - the circular path taken by B is continuous and B always travels at a velocity v along the tangent to the circle. Now I think everyone will agree that when B returns to A, B's clock reads less (there is a real objective difference between the time logged on B's clock and the longer time accumulated on A's clock) - The question is when and where does the difference occur - The circular path eliminates those explanations that are based upon shifting slopes of world lines, the lost time at turn around, abruptly shifting hyperplanes and the turn around acceleration as being a factor.

So what do we have - Einstein tells us the clocks will no longer be in sync and that the one which took the trip will lag the other by (1/2)t(v/c)^2 (where t is the total time of the trip) Would you argue that B actually has to reach A to experience a real time difference - what if B pulls up in San Fancisco instead of Los Angles where A is waiting - there is still the same time difference essentially - What if B stop on Mars as he is looping back to Earth and then checks his clock and sends a radio signal to A - well, they won't quite agree - but there will be almost as much age difference between A's clock in Los Angeles and B's clock on Mars as if B had continued on to Los Angeles - and so on - if B stops at Alpha Centuri, A will still be much older that B at the time they communicate by radio, but not as much older as if B had completed the trip because the t factor that is appropriate for stopping on alpha centuri is less that the t factor that would correspond to completing the voyage - what I am saying and what I have always said, is that the situation is proportional - B's clock at every point in the journey is falling proportionately behind A's clock - Einstein tells us this in no uncertain terms - there is no ambiguity in what Einstein says in Part 4 of his 1905 paper -

Where the confusion arises is because of the tendency to hark back to the fact that two clocks in motion always see the other guys clock running slow - that is true - but of course we know that both cannot be running slow relative to the other - the difference in the physical aging experiments is that A's frame and B's frame are not the same - B is moving along a path in A's frame that comprises a real length (yes Jesse I said real - same term Einstein used - we now call it a proper length).

So yes - I do assert that once two clocks are brought to sync in one inertial frame, the clock which is then moved at velocity v relative thereto will always log less time. Of course, to move, B must change its velocity from zero to v - that necessarily involves an acceleration - but it is only incidental to the real understanding of why there is an objectively real age difference that can, and is measured, in every high speed particle experiment.

And I also claim, based upon the above, that, both A's clock and B's clock will read half as much when B passes Altair as they both do When B completes the trip and lands in Los Angeses - and since we know the real distance to altare as measured in A's frame we can verify this with a radio transmission sent from B as he passes Altare.

 

[JesseM]

Ok - some interesting comments - let me go back Jesse to the thought that I proposed a few weeks ago - and here is where I stand. We start with A and B at rest in an inertial frame - They can be adjacent - it doesn' matter - they are identical clocks. B Starts out North and then swings East, then South in a large half circle passing Altair and then continues South then bends West and returns to A completing a giant circle. As stated, this is a version of the twin paradox - but there is not abrupt turn around - whatever is going on with B's clock is a continuous happening - the circular path taken by B is continuous and B always travels at a velocity v along the tangent to the circle.

Well, in a frame where A is not at rest, B's speed will be constantly changing as he moves around the circle, and so the speed his clock runs in this frame will be constantly changing as well, there will be times when his clock is actually running faster than A's. Of course all frames will get the same answer as to what A and B's clocks will read at the moment they reunite.

Now I think everyone will agree that when B returns to A, B's clock reads less (there is a real objective difference between the time logged on B's clock and the longer time accumulated on A's clock) - The question is when and where does the difference occur - The circular path eliminates those explanations that are based upon shifting slopes of world lines, the lost time at turn around, abruptly shifting hyperplanes and the turn around acceleration as being a factor.

The path integral approach works fine as an explanation here. And like I said in another post, all other explanations are really derivative of the path integral approach. For example, as I said in another post, when people use the explanation that the twin who accelerates is the one who'll be younger, this can be thought of as a consequence of the fact that any non-straight path between two points in spacetime will always have a shorter path integral than a straight (non-accelerating) path between them. Someone also mentioned the "triangle paradox", which is the astonishing fact is that a person walking along the hypotenuse of a right triangle will always walk a shorter distance than the person who walks along the other two edges, despite the fact that they start at the same point and end at the same point. You could explain this by integrating the length of each person's path or just by saying that a straight path is always the shortest one, therefore the person who turned around will always walk a greater length. Both answers are equally correct.

So what do we have - Einstein tells us the clocks will no longer be in sync and that the one which took the trip will lag the other by (1/2)t(v/c)^2 (where t is the total time of the trip) Would you argue that B actually has to reach A to experience a real time difference - what if B pulls up in San Fancisco instead of Los Angles where A is waiting - there is still the same time difference essentially

If there are two clocks in San Francisco and Los Angeles which are in sync in their mutual rest frame, then the most that any other frame can see them out-of-sync by is x/c, where x is the distance between San Fran. and LA in their own rest frame. So, I think this means the maximum disagreement you could have about the amount that A and B are out-of-sync would be 2x/c. But yes, I would say this means there is not a single objective time difference between A and B unless they actually meet at the same location, if they don't then I'd say you can only talk about a time difference of t ± x/c, where t is the time difference in the San Francisco/Los Angeles rest frame. But the actual amount that you add or subtract from t will depend on which inertial frame you're using, and Einstein would say that any inertial frame is equally valid.

what I am saying and what I have always said, is that the situation is proportional - B's clock at every point in the journey is falling proportionately behind A's clock

No, that's not true. In a frame where A is in motion, there will be moments along B's journey where B's velocity in this frame is less than A's, agreed? If so, then the formulas of relativity clearly show that in this frame, it would be A's clock that is ticking slower at that moment.

Einstein tells us this in no uncertain terms - there is no ambiguity in what Einstein says in Part 4 of his 1905 paper -

I just looked over section 4 of the paper, I saw him correctly state that the clock that went in the circle would be behind by (1/2)tv^2/c^2 when they reunited, but nowhere did I see anything about B's clock falling behind A's clock at the same rate throughout the journey. If you think he did, please post the quote you're thinking of.

Where the confusion arises is because of the tendency to hark back to the fact that two clocks in motion always see the other guys clock running slow - that is true - but of course we know that both cannot be running slow relative to the other - the difference in the physical aging experiments is that A's frame and B's frame are not the same - B is moving along a path in A's frame that comprises a real length (yes Jesse I said real - same term Einstein used - we now call it a proper length).

But a path isn't an object, it can't have a proper length since it doesn't have a rest frame.

So yes - I do assert that once two clocks are brought to sync in one inertial frame, the clock which is then moved at velocity v relative thereto will always log less time.

But what do you mean by "log less time"? I would certainly agree that when the clock that is accelerated meets up with the other one, it will be behind it. But do you agree that in an inertial frame where both clocks start out in motion, and the clock that accelerates actually decreases in velocity, then before it accelerates it will start out significantly behind the other clock, and then it will tick faster after it acelerates, just not by enough to catch up with the other clock when they meet?

And I also claim, based upon the above, that, both A's clock and B's clock will read half as much when B passes Altair

But to ask what A's clock reads "when B passes Altair" depends on your definition of simultaneity. Do you agree that according to Einstein, no inertial frame's definition of simultaneity is preferred over any other's? Do you agree that in a frame where B is at rest while A and Altair are in motion, it would be A's clock that's behind B's clock at the moment B passes Altair?

 

[yogi]

Jesse - I will take your statements one at a time - no - B's velocity does not change - your forgetting what Einstein said - any pologonal path will do - I chose a circle - and Einstein gave the formula (read it again (1/2)t(v/c)^2 - your again trying to confuse the simple unambiguous explanation given by Einstein. As long as his velocity does not change we can use exactly what Einstein prescribed. Velocity does not change for an object that is undergoing centripetal acceleration - there is no change in the energy of the B spaceship - we consider it teathered by a long string with a central anchor point midway between Earth and Altair. I will take up each oint separately.

 

[yogi]

The hypothenus problem is reversed - in SR the temporal leg and the spatial leg are differenced - so the analogy is confusing

 

[JesseM]

Jesse - I will take your statements one at a time - no - B's velocity does not change

Think about this more carefully, yogi. Relativity is not even important here--even in Newtonian physics, if you travel in a circle at a constant speed relative to the center of the circle, then in a frame where the center of the circle is in motion, your speed as you move around the circle will constantly be changing. For example, if you're moving clockwise, so that at the bottom of the circle your velocity is totally to the left and at the top of the circle your velocity is totally to the right, then in a frame where the center of the circle is moving to the right, your speed will be higher at the top of the circle than at the bottom of the circle.

and Einstein gave the formula (read it again (1/2)t(v/c)^2 - your again trying to confuse the simple unambiguous explanation given by Einstein.

Yes, I have already said I agree that when the two clocks reunite at time t, the one that went around the circle will have gotten behind by (1/2)t(v/c)^2. But Einstein does not say that at some earlier time t', like when the clock has only gone around half the circle, the clock will be behind by (1/2)t'(v/c)^2--the amount that it's behind at any given moment before they reunite depends on your choice of reference frames.

 

[yogi]

3rd point - There is no common time between the two frames - but we can still know shat Bis clock reads at any point in his journey - because his spatial path is totally known in advance - it is a measured circle in the A frame which includes a flyby of Altair, Alpha Centuri, Mars, whatever. At any of thiese known objects B can hold up a sign indicating his clock's reading and the transmitter on these remote points will send a signal to A - so even theough there is no common time - A can still know figure out how far behind B's clock is lagging - you can't really believe that they have to get back together to determine that there is a real physical difference at every point in the trip - What if B returns but he is an inch away from A - could they compare --come on.
udes a

 

[JesseM]

The hypothenus problem is reversed - in SR the temporal leg and the spatial leg are differenced - so the analogy is confusing

Just exchange "path of shortest length" with "worldlines of greatest proper time" and the analogy works fine. Remember that in general relativity, objects follow geodesics, which are the worldlines that maximize the proper time, just as a geodesic along a curved spatial surface is the path of minimal spatial distance.

 

[yogi]

Your 4th point is not even possible given the initial conditions - there is only one velocity involved - and this is where we differ - you want to say that because part of the journey (where B passes Altair) A will believe the velocity is different - or when B is heading home - its a negative velocity - no - it makes no difference - these are observational aspects that would lead to apparent differences where each clock observes the other to be running slow - but what we are talking about is object age difference - any pologonal path (Again I quote Einstein) and that means the velocity is constant during the entire trip but the trip can wander all over the place -

 

[JesseM]

3rd point - There is no common time between the two frames - but we can still know shat Bis clock reads at any point in his journey

Sure we can. But if we know B's clock reads time t when it's at a particular point on the circle, the only way to say what A's clock reads "at the same time" that B is at that position is to pick a reference frame. Are you denying that different frames would disagree about this according to the rules of relativity? If not, are you denying that Einstein would say that no frame's definition of simultaneity can be preferred over any other?

because his spatial path is totally known in advance - it is a measured circle in the A frame which includes a flyby of Altair, Alpha Centuri, Mars, whatever. At any of thiese known objects B can hold up a sign indicating his clock's reading and the transmitter on these remote points will send a signal to A - so even theough there is no common time - A can still know figure out how far behind B's clock is lagging

How far it is lagging IN WHOSE FRAME?

- you can't really believe that they have to get back together to determine that there is a real physical difference at every point in the trip - What if B returns but he is an inch away from A - could they compare --come on.
udes a

If B was an inch away from x, then the quantity 2x/c, which is the upper bound on the amount that different frames can disagree about how far the clocks are out-of-sync, would be a very tiny amount--about 0.00000000017 seconds. Still, as long as they are not at exactly the same position, there will be at least some disagreement between different frames about how far they are out-of-sync, and the greater the distance, the greater the potential disagreement. Again, do you deny that different inertial reference frames in relativity will disagree about the amount that two separated clocks are out-of-sync? Do you deny that relativity (and Einstein) say that no inertial reference frame is preferred over any other?

 

[JesseM]

Your 4th point is not even possible given the initial conditions - there is only one velocity involved

If you really believe that an object moving at constant speed in a circle in one frame would be moving at a constant speed in other frames, then you don't even understand Newtonian mechanics.

- and this is where we differ - you want to say that because part of the journey (where B passes Altair) A will believe the velocity is different

I never said anything of the sort, obviously the speed is constant in A's frame. But there's no reason we have to analyze the problem in A's frame, is there? You'd get the same answer for how far B is behind A at the moment they reuinite no matter which frame you used.

 

[yogi]

5th point - A path certainly can have a proper length because it is defined in the A frame - I have described a path that B will follow - it is an itenery that A and B work out while they are both at rest in A frame - and all points are measured in the A frame - just because it is a circle doesn't mean it doesn't have a length

 

[JesseM]

5th point - A path certainly can have a proper length because it is defined in the A frame

Nonsense, you're misusing the term "proper length". Proper length is the length of an object in its own rest frame, not just whatever frame we choose to define its length in for the purpose of writing down a problem. I could say that in A's frame there's a rod moving at 0.866c that's 1 meter long, but that's not its proper length, its proper length is 2 meters because that's the length in its own rest frame. A path has no rest frame, and again, the frame we arbitrarily choose to use when writing down a particular problem has jack squat to do with "proper length".

 

[yogi]

I don't understand your 6th point - but to the extent I do - The situation is only defined per Einstein when the two clocks are moving together or stationary and brought into sync - whether one slows or not is of no moment - there is just an acceleration in a different direction - we only refere to the situation when they were last in sync and take all readings from there

 

[JesseM]

I don't understand your 6th point

What sentences in my post are you calling my "6th point"? I've said this before, but I would prefer that you would use the "quote" function instead of just responding to what you think my points were.

- but to the extent I do - The situation is only defined per Einstein when the two clocks are moving together or stationary and brought into sync - whether one slows or not is of no moment - there is just an acceleration in a different direction - we only refere to the situation when they were last in sync and take all readings from there

You'll have to be more specific, I don't understand what a single one of these phrases ('only defined per Einstein', 'whether one slows or not is of no moment', 'an acceleration in a different direction', 'take all readings from there') refers to or means.

 

[yogi]

Your 7th point hits the nail on the head as to where we disagree - Yes as to there being no preferred frame - and Yes as to the question as to What B would measure as to A and Altair - B would at say that both A's clock and the clock on Altair are both running slower than his - but this is an apparent effect - in fact by the same line of reasoning B would get all the way back home and if he made a number of observations on the trip considering his own frame as stationary - every one of those observations made by B would yield the same result - of course he would be very certain that his brother A was aging less - but here is the difference Einstein claims when he explains the physical consequences of the transforms - what he says in Part 4 - there is a real age difference that accumulates in accordance with how far you have traveled in the frame where the two clocks were initailly at rest and in sync - when you add in this factor - the intervals are still equal, but the components that make up the interval for the traveler (B) are different that the single component (time) that comprises the totality of the interval for the A clock. Age differences are real and you don't have to turn around to measure them - The interval for B has a spatial and temporal component, the interval for A has only a temporal component.

 

[yogi]

Here is another way to sharpen the issues - Einstein tells us that a clock at the equator will run slower than a clock at the North Pole (actually there are oblateness issues that cancel - but we will ignor that) - now suppose we have a high tower on the North Pole and we put a clock J on top - we launch another clock Q in polar orbit at the same height as the top of the tower. Since Q and J are at the same height we do not need to consider GR effects - so according to Einstein the orbiting clock Q will run slower - but if Q considers himself stationary for a short time each time he passes J, will he not measure J's clock to be slower? - and will not J measure Q's clock to be slower? - but in actuality it is Q that is ageing less (if you don't believe this you need to speak with some GPS engineers).

 

[JesseM]

Your 7th point hits the nail on the head as to where we disagree

Arrgghh! Please quote my post, I don't know what "point" you're replying to!

Yes as to there being no preferred frame - and Yes as to the question as to What B would measure as to A and Altair - B would at say that both A's clock and the clock on Altair are both running slower than his - but this is an apparent effect - in fact by the same line of reasoning B would get all the way back home and if he made a number of observations on the trip considering his own frame as stationary - every one of those observations made by B would yield the same result

I don't remember saying anything about what things would look like from B's perspective, since B doesn't even have a valid inertial frame. Maybe I did say something like that, but I don't feel like going back and rereading all my posts--again, please quote when responding so I can have some idea of the context.

And no, as a matter of fact I wouldn't say that B would see A's cock running slower throughout the journey. If you're talking about what B would actually "see" using light-signals, at times he would see A's clock running slower and at other times he'd see it running faster. If you're talking about how fast A's clock would be running at any given time in B's coordinate system, this depends on how you define B's coordinate system, since B is not moving inertially I don't think there's any "standard" way of doing it.

When I talked about looking at the same problems in different frames, I meant looking at things from the frame of a separate inertial observer (let's call him 'C') moving at constant velocity relative to A (and remember, 'constant velocity' means both constant speed and constant direction, so B doesn't count even though his speed relative to A is constant).

of course he would be very certain that his brother A was aging less

No he wouldn't. Do you understand that the rules of special relativity do not apply to non-inertial coordinate systems?

but here is the difference Einstein claims when he explains the physical consequences of the transforms - what he says in Part 4 - there is a real age difference that accumulates in accordance with how far you have traveled in the frame where the two clocks were initailly at rest and in sync

If you're saying that Einstein ever said that one inertial reference frame's view of a problem is more physically real than any other's, then you are totally and utterly confused about the most basic ideas of relativity.

 

[JesseM]

Here is another way to sharpen the issues - Einstein tells us that a clock at the equator will run slower than a clock at the North Pole (actually there are oblateness issues that cancel - but we will ignor that) - now suppose we have a high tower on the North Pole and we put a clock J on top - we launch another clock Q in polar orbit at the same height as the top of the tower. Since Q and J are at the same height we do not need to consider GR effects - so according to Einstein the orbiting clock Q will run slower - but if Q considers himself stationary for a short time each time he passes J, will he not measure J's clock to be slower?

NO. Q IS NOT IN AN INERTIAL REFERENCE FRAME, RULES OF SR LIKE TIME DILATION ONLY WORK IN INERTIAL REFERENCE FRAMES.

and will not J measure Q's clock to be slower?

Yup, and J is in an inertial reference frame.

but in actuality it is Q that is ageing less (if you don't believe this you need to speak with some GPS engineers).

There is no objective sense in which Q is aging less at all points on his orbit, although he is aging less over the course of one complete orbit. Suppose that in J's frame, Q is moving to the right at 20,000 miles per hour when he passes the North Pole, and to the left at 20,000 miles per hour when he passes the South Pole. Now consider another observer K who is moving inertially to the right at 5,000 miles per hour in J's frame, so that in K's frame J is moving to the left at 5,000 miles per hour. This means that in K's frame, the satellite Q is moving to the right at 15,000 miles per hour when it crosses the North pole, and it's moving to the left at 25,000 miles per hour when it crosses the South Pole. So you can see that in K's frame, when Q crosses the North Pole its velocity is actually less than J's velocity at that moment, so Q's clock will be ticking faster at that moment. But at other points in the orbit K will see Q's clock ticking slower, and if you do a path integral of the rate that J and Q's clock tick over the course of one complete orbit from the perspective of K's frame, you will see that overall Q does get further and further behind J each time they reunite.

The point is, again, every inertial frame is equally valid in special relativity, so you can look at this problem from either K's or J's perspective here.

 

[yogi]

Jesse - There is definitely an objective sense to the fact that Q is ageing less at every point in his orbit - we monitor satellite clocks continuously from different ground stations and if the Height is factored out, an uncorrected GPS satellite clock will continually fall behind the ground observer clock - Are you really saying that, relative to the North Pole clock J, Q will be seen it leading and then lagging and that an opposite result would occur at the South Pole... I shutter to think of the state of GPS technology if you had been project engineer.

AS to the fact that the Earth is not an inertial system - it is true - but the experiment can always be adjusted to eliminate the slight curvature when making measurments - For example, in the traveling clock example, B can actually straighten out his curve at any number of destinations (if the curvature bothers you so much) where he wants to take note of the clocks in A frame - so long as his velocity remains at v, the observations of B will indicate that clocks at every point in the A frame are running show - they will not be in sync, but that is not the issue - each individual clock will be seen to be running slow.

And yes - one inertial frame is presumed to be as good as another - but that is where you are missing my point - the individual components of the interval are not the same - and that is why, when relativly moving observers only have one piece of data, namely relative velocity, they will both conclude the other clock is running slow (but this is impossible - its an observational fallacy). But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest. But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.

And so has Yogi

 

[Ich]

- they will not be in sync, but that is not the issue -

that is exactly the issue. A contiuous shift of sync is nothing less than an observed accelerating/decelerating of A´s clock, as seen by B.

- the individual components of the interval are not the same -

SR is all about the individual components of the interval being different from frame to frame.

But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest. But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.

The only permanent thing about those age differences is the way you set up your Gedankenexperimente. You always choose A´s frame as the one where you measure time differences. And you always have in mind that you would let B 'stop' at any point of his journey, i.e. you would match velocities with A and then there would be a residual time difference. That´s true, at least if you would B slowly travel back to A to compare clocks.
But A´s frame is NOT the one and only frame where physically meaningful things happe. If you set up your Gedankenexperiment differently (eg let A join B´s frame), the outcome woul obviously be different, too.

 

[JesseM]

Jesse - There is definitely an objective sense to the fact that Q is ageing less at every point in his orbit - we monitor satellite clocks continuously from different ground stations and if the Height is factored out, an uncorrected GPS satellite clock will continually fall behind the ground observer clock - Are you really saying that, relative to the North Pole clock J, Q will be seen it leading and then lagging and that an opposite result would occur at the South Pole...

No, I'm not saying that. Relative to the north pole clock J, Q is always moving at exactly 20,000 miles per hour, so it is always slowed down by the same amount. But relative to the clock K, which is moving inertially at 15,000 mph relative to J (I messed up the numbers in my example last time, if K only saw J moving at 5,000 mph like I said earlier, then K would always see Q moving faster than J), Q's speed varies from between 5,000 mph at the North Pole and 35,000 mph at the South Pole. So at the North Pole K sees Q moving at 5,000 mph to the right and J moving at 15,000 mph to the left, therefore in K's frame at this moment, J's clock is ticking slower than Q's. Do you disagree with this? Do you disagree that K's inertial frame is just as valid for working out the problem as J's inertial frame?

AS to the fact that the Earth is not an inertial system - it is true

I never said that the Earth was not an inertial system, over short time intervals it's acceptable to say that the center of the Earth (and the poles) are moving inertially. I just said that Q is not an inertial system.

but the experiment can always be adjusted to eliminate the slight curvature when making measurments - For example, in the traveling clock example, B can actually straighten out his curve at any number of destinations (if the curvature bothers you so much) where he wants to take note of the clocks in A frame

That won't help your argument at all, because the point is that in looking at the entire trip from start to finish, you still can't treat B as if he's in an inertial frame and thus sees A's clock ticking slower throughout the trip. If you analyze the entire problem from within a single inertial frame, at best B can only be at rest in this frame during one of the line segments that make up his trip.

so long as his velocity remains at v, the observations of B will indicate that clocks at every point in the A frame are running show

Only if you switch inertial frames in the middle of the problem, but that will involve complexities like B's surface of simultaneity suddenly swinging around, and at those moments A's clock may not be running slow, but instead leaping forward very quickly into the future. You can't just assume that the normal formulas of relativity like time dilation apply to the perspective of an observer who does not move inertially throughout his trip, like B.

So again, it's better to analyze the entire problem from start to finish from within a single inertial frame. Then you really can say that at any moment, the rate a clock is ticking is determined solely by its velocity in that frame. And once again, there are perfectly valid inertial frames where B's velocity is slower than A's at certain points in the circle it makes, or where Q's velocity is slower than J's at certain points in its orbit.

And yes - one inertial frame is presumed to be as good as another - but that is where you are missing my point - the individual components of the interval are not the same

I don't know what you mean by "the individual components of the interval".

and that is why, when relativly moving observers only have one piece of data, namely relative velocity, they will both conclude the other clock is running slow (but this is impossible - its an observational fallacy).

Do you also think it's "impossible" that two observers in Newtonian mechanics could each see the other observer's velocity as greater than their own, from the perspective of their respective rest frames? If not, I don't see why two observers who each see the other's clock running slower is any more problematic.

But when there is more data (as Einstein gives you in his physical description in part 4 - namely a rest frame in which both clocks are synced

As long as the clocks start out at the same location, they are synced in all frames.

and in which real (proper) distances are measured, the spacetime interval for the traveling clock has a spatial component that is a proper distance measured in the original frame where both clocks were initially at rest.

You are making nonsense of the term "proper distance". Einstein would never have said a path can have a proper length, and neither would any other relativist. Go on, ask some others on this board if you don't believe me.

But not vice versa (there is no spatial component in the interval of the stay at home clock) - and that is why age differences are real and permanent, and Albert has told you so.

And so has Yogi

Complete nonsense. Einstein would have said it would be just as valid (though less simple mathematically) to analyze the problem from the perspective of a different inertial frame where A is in motion at constant velocity and B's velocity varies as he moves away from A, and if you analyze the problem in this frame you will get exactly the same answer for how far B will have fallen behind A when they reunite, (1/2)t(v/c)^2. That is why the age difference is real, because you will get exactly the same answer for the age difference when they reunite regardless of what frame you use.

Do you agree that we would get the same answer for the age difference when A and B reunite if we used a different frame, and that Einstein and all relativists would say that it would be just as valid to analyze the problem from some other frame besides A's rest frame? Please answer this question yes or no.

 

[yogi]

yes or no. You are hung up on the idea that the two clocks must return to the same point for there to be real time dilation - and the fact that the moving clock B cannot be changing course in the space defined by the rest frame of A.

As always, our discussions usually turn out to be about as gratifying as shoveling smoke - and I am sure you are equally frustrated.

 

[JesseM]

yes or no.

Cute, yogi. Now could you answer the question for real?

You are hung up on the idea that the two clocks must return to the same point for there to be real time dilation

I'm not "hung up" on this, I'm just telling you something that all relativists, including Einstein, would agree on. If you have an alternate theory, then feel free to present it, but you should stop making the obviously false claim that Einstein's paper supports your position.

and the fact that the moving clock B cannot be changing course in the space defined by the rest frame of A.

You mean that we can't apply the standard formulas of relativity to B's perspective if B is moving non-inertially? Once again, this is something that all relativists would agree on. If you doubt me on this, please ask some other people on this board, or email some physics professors, you'll see that I'm not just giving you my own interpretation here.

As always, our discussions usually turn out to be about as gratifying as shoveling smoke - and I am sure you are equally frustrated.

If you would actually address my specific questions, and use the QUOTE function so that I know what comments you're responding to, these discussions would be a lot less frustrating for me. A good place to start would be the yes-or-no question about whether different inertial frames are equally valid, and whether they'll give the same answer for the time delay when the two clocks reunite.

 

[Huckleberry]

Maybe this will help you understand what would happen if you brought the twins back together after moving them around at near light speed. If I'm wrong I'm sure someone will correct me.

Twin 1 blasts off to Alpha Centauri 4 light years distant leaving twin 2 standing at the launch pad. Twin 1 would perceive almost no passage of time. He has been traveling along with the same light that was reflected by the Earth. So when he arrives at his destination he pulls out his ultra powerful telescope and looks back at twin 2 standing on the launchpad while the smoke clears. They appear to be the same age.

What twin 1 observes is not reality. Well, it's real enough, but it is light that has been traveling for 4 years from the time it was reflected off twin 2. In reality, while twin 1 is looking at twin 2 on the launch pad, four years has actually passed for twin 2. Twin 1's consciousness would experience no time and as he looked back at Earth his senses would tell him likewise.

Twin 1 turns around and goes home at the same speed he left. He would be running into all the light that has been reflecting from the Earth at an accelerated pace. In 4 light years distance he would experience 8 years worth of reflected light, the 4 from the time he left and 4 more until he returns. Would his consciousness match his senses in this scenario? If it does then both of the twins would be 8 years older than they were when they first seperated.

Twin 2 waits 4 years and looks for twin 1 at Alpha Centauri, but doesn't find him there. He sees him about halfway there. After almost 4 more years he looks again and finds twin 1 has finally arrived and is looking at some point in space where the launch pad was eight years ago. Twin 1 looks like he hasn't aged at all in eight years. A few moments later twin 1 arrives at the launch pad unexpectedly and now both twins are the same age again.

This assumes that just because a person is moving does not guarantee that they have a positive velocity. I imagine their actual velocity would be a vector based on the center of gravity for the universe and taking into account regional irregularities. Any motion requires a 4th dimension.

Ok, you can crucify me now but please be kind enough to educate me instead of just insulting me.

What was the question?
Huck

 

[yogi]

All right Jesse - I will try to say it in a way that conveys what Einstien said as I see it unambiguously - let's keep it real simple. A and B at rest in the same frame separated by a distance d. Clocks brought to sync. B starts moving toward A at relative velocity v (a short acceleration to get up to speed - then cruises at constant relative v). Einstein says when B arrives the clocks will no longer read the same - B's clock lags (has logged less time - is younger - whatever you want to call it).

I am saying that this is a real difference - it is not an observational apparency - it is a real objective difference in the two clocks.

I think your question is - what if you consider things from B's perspective - once B is in motion, according to the transforms and the fact that B is now in an inertial frame - why can't he say that A's clock is running slow. He will measure it to be running slower if he considers himself at rest and A moving toward him in his own frame.

But if B draws this conclusion, it cannot be a reality. There is only apparent symmetry. When the two clocks arrive, we cannot have a result where B's clock reads less than A's while at the same time A's clock reads less than B's. So if you try to do the problem in the frame of B you would get a different answer if you fail to include the initial conditions But if you take into account the inital conditions (namely - how did the relative motion come about) you will get the correct answer in any frame in which the motions are transformed. So my answer to your question is both yes and no

Huck - didn't mean to ignor your post - Jesse and I have this love affair going (sort of like Roy and the tiger) and so its hard to switch to a new problem when we can't get by first base on this one.

 

[Huckleberry]

It's quite alright yogi. I was kind of hoping my post would be ignored anyway. What I wrote feels right, but I have the impression that it is dreadfully wrong and I've made a fool of myself by writing it. I was expecting fire and brimstone and haven't seen it yet. I'm thankful for that. I'm fairly new to this site and don't know what to expect yet.

Huck

 

[jtbell]

What I wrote feels right, but I have the impression that it is dreadfully wrong and I've made a fool of myself by writing it.

Actually, it's mostly more or less correct as a qualitative description, assuming the traveling twin's speed is high enough. The only place you trip up is at the very end of the journey:

A few moments later twin 1 arrives at the launch pad unexpectedly and now both twins are the same age again.

Twin 1 does return home close on the heels of the light by which twin 2 sees him at Alpha Centauri; but they are not the same age when they meet again. Furthermore, if they watch each other continuously through telescopes during the trip, they will see each other aging in such a way that just before twin 2 arrives home, both of them will expect twin 2 to be younger.

For a detailed numeric example using a somewhat lower speed than you're imagining, see posting number 3 in this thread:

https://www.physicsforums.com/showthread.php?t=69214

You might try re-calculating the numbers using, say, v = 0.99c and see what you get.