How Time Passes Differently for Twins Traveling at Light Speed

[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.

 

[JesseM]

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

Not exactly, I mean what if you consider things from the perspective of an inertial frame where B is at rest after B accelerates. In this frame, A and B will initially both be traveling at velocity v, then B will accelerate and come to rest, while I'm not sure if you would say this would be true "from B's perspective" (perhaps you would say that both A and B were initially at rest from B's perspective)--as always, it's tricky to talk about the "perspective" of an observer who accelerates at any point during the time period you're considering, it's better to analyze the whole problem from the point of view of a single inertial frame.

Alternately, if you don't want to talk about "frames", you could talk about the perspective of a third observer C who has been traveling at velocity v relative to A since the beginning of the time period we're looking at, who at first sees B traveling at v and then sees B accelerate and come to rest.

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

I think the problem is that when you say "B's frame" you are considering a perspective where A is initially at rest and synchronized with B, but then after B accelerates A is moving with velocity v and running slow. But again, this isn't a valid frame, since acceleration was involved. If you consider things from the perspective of that third observer C who was always traveling at v relative to A, this observer C will initially see A and B traveling together at velocity v, and will see A's clock ahead of B's by vd/c^2. Then B will accelerate and come to rest, while A continues to move at velocity v in the direction of B. A's clock will be ticking slower than B's, but because A's was already ahead, A's clock will still be ahead when A and B meet. C's prediction about how much A's clock will be ahead of B's when they meet will exactly match the prediction made in A's frame, even though in this frame the explanation for why A's clock was ahead will be different.

So, do you agree that the description from C's perspective is just as valid as the description from A's perspective? If not, do you at least agree that all mainstream physicists, including Einstein, would say that the perspective of one inertial frame is just as good as any other?

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.

What if we only start looking at the problem at some time t after B has accelerated? Aren't the initial conditions at t enough to make complete predictions about what will happen in the future, without knowing what happened before t (ie without knowing if it was A or B who accelerated?)

 

[yogi]

Well - here is where I am probably going to differ from what you refer to as mainstream physics - but I think it is not only consistent with SR, it is also in conformity with all experiments that I am aware of . So, from the standpont of a third frame C, if we consider the velocities of A or B or both after they are up to speed relative to C, we would lose what I consider important, namely how did these velocities come about. Now on the other hand, if A, B and C were all at rest at some instant in the past, and all clocks calibrated and brought into sync, and then A or B or both where later accelerated - then the inital conditions are not lost - namely - which clocks were put into motion relative to C. The transforms would work to yield the actual age differences. For example if A had accelerated toward B we could calculate a spacetime interval for A in the C frame and likewise if B had been accelerated (perhaps a different amount of acceleration than A), then B's interval would have both a temporal component and a spatial component as did A's as observed from the C frame. And a straightforward application of Einstein's formula would give the actual age difference between B and A when the meet.

What i think Einstein did in his 1905 part IV which is significantly entitled "Physical Meaning of the Equations..." was to take the transforms to a different level - from their derivation based upon observations between relatively moving frames where each observer infers the other guys clock is running slow - to a full blown Minkowski space time where it becomes crucial to know which clock moved. Admittedly, once the B clock is in motion, everything looks perfectly symmetrical - but as you have pointed out, if the frames are symmetrical, why would there be a real age difference?

This was the point of my post on the cosmological twin paradox introduced by Garth - I don't see how it is possible to sync clocks on the fly - that is in relative motion - you can read a clock of course - as it passes by, but you have no way of knowing whether it is actually running faster or slower - so on any round trip, whether it be a circumnavigation of the universe or once around the earth, you don't have a way of determining whether the other clock is actually running at an identical speed -

As a matter of academic interest - I think it would be a very interesting to set up a PF pole and ask members if they consider the situation of the two clocks initally in sync and then have B move to A, constitutes a real objective age difference or not.

Are you game?

ps - I know Minkowsk came up with the idea after 1905 - but as I read Einstiens's paper, I see it as implicit in his description.

 

[JesseM]

Well - here is where I am probably going to differ from what you refer to as mainstream physics - but I think it is not only consistent with SR, it is also in conformity with all experiments that I am aware of .

What is consistent with SR? The idea that B was objectively aging slower after it accelerated towards A, even though in C's frame it would be A that was aging slower? Or do you disagree that in C's rest frame, A would be aging slower?

So, from the standpont of a third frame C, if we consider the velocities of A or B or both after they are up to speed relative to C,

What does "up to speed relative to C" mean? C initially sees A and B moving at velocity v, then B accelerates and comes to rest in C's frame.

we would lose what I consider important, namely how did these velocities come about.

Velocities relative to what? Are you suggesting some concept of absolute velocity here? Why can't C and A have been moving at a constant velocity v relative to one another since the beginning of time? If C and A are just taken as the origins of inertial coordinate systems rather than real physical objects (after all, in relativity you are free to use 'reference frames' which no physical object is permanently at rest in), then the very definition of "inertial reference frames" demands that these two origins always move at constant velocity relative to one another.

Now on the other hand, if A, B and C were all at rest at some instant in the past

Nope, that's not the scenario I'm asking about. Again, the whole concept of an "inertial reference frame" presupposes that different frames are moving at constant velocity relative to one another for all time--if you want to consider what things look like from the perspective of a physical object which accelerates, you are not talking about inertial reference frames. Please don't duck the question--I'm asking whether the perspectives of different inertial reference frames are equally valid, not the perspectives of different physical observers who don't always move inertially.

By the way, are you saying that if our initial conditions at time t show A and B moving towards each other at constant velocity v, then in order to decide who is aging less we must know who accelerated last before time t? This is in clear contradiction of what you said on this thread, when I asked you:

Let me put it this way--do you agree that if we know the complete set of initial conditions at some time t in a particular frame (the position and velocity of both objects at t, the time on their own clocks, etc.) and we want to make some predictions about what will happen later, then since the laws of relativity are completely deterministic, these initial conditions at t are sufficient to make a unique prediction about the future? Do you agree that what happened before t is irrelevant, including the question of which of two clocks accelerated before t?

You replied "Yes - i will agree to that." So unless you misunderstood, or unless you are going back and changing your answer, that means if at some time t we see A and C moving towards each other at constant velocity, then the question of whether A or C accelerated at some time before t is not relevant to deciding any physical question, including the question of who is aging slower.

and all clocks calibrated and brought into sync, and then A or B or both where later accelerated - then the inital conditions are not lost - namely - which clocks were put into motion relative to C.

A was not put into motion relative to C during the period we are considering. A and C have been moving at constant velocity relative to each other since the beginning of the time period we are considering. If you like, you can consider them to have been moving at constant velocity relative to one another since the beginning of time, although as I said above this shouldn't be necessary if you agree that we don't need to know anything about times before our initial conditions to answer the question of who ages slower.

What i think Einstein did in his 1905 part IV which is significantly entitled "Physical Meaning of the Equations..." was to take the transforms to a different level - from their derivation based upon observations between relatively moving frames where each observer infers the other guys clock is running slow - to a full blown Minkowski space time where it becomes crucial to know which clock moved.

Well, that's just ignorant. Anyone who understands relativity knows that different reference frames disagree about which of two clocks is ticking slower, and that all frames are equally valid, the question of which clock accelerated in the past does not force you to choose one frame over another. This symmetry of different reference frames is a very basic part of "Minksowski space time", if you think that the question of which object "moved" (accelerated) is relevant to deciding which inertial reference frame to use, you're badly confused about how Minkowski space time works. The point of the twin paradox is not that one inertial frame is favored over another depending on who accelerated, it's just that you can't put the perspective of the non-inertial twin on the same footing as an inertial reference frame.

Admittedly, once the B clock is in motion, everything looks perfectly symmetrical - but as you have pointed out, if the frames are symmetrical, why would there be a real age difference?

How many times do I have to repeat this? The reason there is a real age difference is because different frames define simultaneity differently. Do you not believe me that C's frame will make exactly the same prediction as A's frame about what A and B will read when they meet, in spite of the fact that C's frame says A is ticking slower as they approach each other and A's frame says B is ticking slower as they approach each other?

I don't see how it is possible to sync clocks on the fly - that is in relative motion - you can read a clock of course - as it passes by, but you have no way of knowing whether it is actually running faster or slower

What do you mean by "sync"? Do you mean to make sure they're reading the same time? That's impossible if they're in relative motion, since they will be ticking at different rates. Or if by "sync" you just mean determining what one clock reads "at the same moment" that another clock reads a given time, like "when clock A read 2:00, clock B read 4:00" (ie the issue of simultaneity), then each frame can do this either using light-signals (taking into account the time the light took to reach you), or by using local readings on a network of clocks which are synchronized in that frame (the clocks are defined to be synchronized in this frame if light emitted at the midpoint of two clocks hits both at the same time according to their readings). This is the procedure that Einstein outlined in his paper, so even if you disagree with it, if you disagree that this is how Einstein defined simultaneity then you're just being ignorant.

As a matter of academic interest - I think it would be a very interesting to set up a PF pole and ask members if they consider the situation of the two clocks initally in sync and then have B move to A, constitutes a real objective age difference or not.

Are you game?

Only if both the scenario and the question are phrased with enough detail so people don't misunderstand the question. For example, I might phrase it like this:

Suppose you have two clocks A and B which are initially at rest relative to one another and synchronized in their own frame, then B accelerates towards A and moves toward it at velocity v until they meet. Do you think it is objectively true that B was aging more slowly as it approached A, even though it is possible to view this situation from a frame where A and B are both initially moving at velocity v, then B comes to rest and A continues to move towards it at velocity v until they meet, so that in this frame A would be ticking more slowly after B comes to rest?

Of how about this:

Suppose two clocks start out at rest relative to one another and synchronized in their own frame, then one accelerates briefly and after that they move towards one another at constant velocity. If different inertial reference frames disagree about which clock is ticking slower as they move together, can we determine which frame is "objectively" correct by checking which clock was the one that accelerated?

 

[yogi]

Briefly - rather than go into all your misinterpretations as to what I have said - if you know the complete history of the two clocks - then yes, the predictions as to which is ticking faster, if either, is determined. By complete history you would have to know more than just the fact that the two frames have relative motion.

You can find recogonized authors that interpret Einstein's statement as real one way time dilation (age difference). John D Barrow for example explains the reality based upon the difference in the energy that can be attributed to the change in velocity - anyway - the primary experiments that are available to show time dilation are those dealing with high speed particles, H and K round the world airplane journeys, GPS, all of which are easily explained by taking Einstein's explanation at face value.

So let us see if we can agree upon a correct way to state the poll. I object to the words "objectively correct" I propose the following:

Einstein makes the following statement in part 4 of his 1905 paper under the Heading Physical meaning of the Equations in respect to moving Rigid bodies and moving clocks:

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

Question for members - is this a real age difference or would the two clocks actually read the same when compared?

 

[Ich]

We´ve been through this - The time difference is real, it is the difference in proper time since the clocks of A and B have been synchronized in frame A.
The synchronization is of course frame dependent, and so is the time difference if you chose tho synchronize in a different frame.

 

[JesseM]

You can find recogonized authors that interpret Einstein's statement as real one way time dilation (age difference). John D Barrow for example explains the reality based upon the difference in the energy that can be attributed to the change in velocity - anyway - the primary experiments that are available to show time dilation are those dealing with high speed particles, H and K round the world airplane journeys, GPS, all of which are easily explained by taking Einstein's explanation at face value.

No, you are simply misinterpreting them, as usual.

Einstein makes the following statement in part 4 of his 1905 paper under the Heading Physical meaning of the Equations in respect to moving Rigid bodies and moving clocks:

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

Question for members - is this a real age difference or would the two clocks actually read the same when compared?

yogi, this is a totally dishonest strawman on your part--you know perfectly well that I agree the clock A would be behind B when they meet, I just disagree that this means A was "aging slower" during its journey.

Let's say A and B start out 1 light year apart in their rest frame, and synchronized in this frame, then at the moment that both clocks read 2005, A accelerates to 0.5c and moves at constant velocity towards B. In this frame, A will reach B in (1 light year)/(0.5c) = 2 years, in 2007. But A was ticking at (1 - 0.5^2) = 0.866 of B's rate during this trip, so when it meets B it will only read 2005 + 2*0.866 = 2006.73.
So, in B's rest frame:
amount of time that B ages after A accelerates: 2 years
amount of time that A ages after A accelerates: 1.73 years

But in the frame of an inertial observer C who sees A and B initially moving at 0.5c, then sees A decelerate until it's at rest, things look a bit different. In this frame, A and B were initially not synchronized--instead, A and B were initially out-of-sync, with B ahead of A by 0.5 years. So if A read 2005 at the moment it decelerated, B already read 2005.5 at that moment. After that, B was moving towards A at 0.5c, and in this frame the distance between them was only 0.866 light years, so it takes (0.866 light years)/(0.5c) = 1.73 years for B to reach A. Since A was at rest during this time, its clock will elapse 1.73 years during this time, so it will read 2005 + 1.73 = 2006.73 when B arrives. But B is only ticking at 0.866 the normal rate during this time, so itwill tick 1.73*0.866 = 1.5 years in this time, and since it read 2005.5 at the moment A accelerated, it will read 2005.5 + 1.5 = 2007 at the moment it reaches A.
So, in C's rest frame:
amount of time that B ages after A decelerates: 1.5 years
amount of time that A ages after A decelerates: 1.73 years

So, both frames predict that B reads 2007 and A reads 2006.73 at the moment they meet, but in B's rest frame it was A who was aging slower after A accelerated, and in C's frame it was B who was aging slower after A decelerated. The question is, are both these descriptions equally valid? Am I correct in understanding that you would say both descriptions are not equally valid, yogi, but that the description in B's frame is somehow more correct than the description in C's frame? I promise that any relativist, including John D Barrow, would say both descriptions are equally valid.

 

[yogi]

Jesse - you are weasel wording your way around again - you have previously taken the position that there can be no objectively real age difference in the situation I quoted in post 64.

 

[yogi]

Jesse - Here is one of my previous statements:
Originally Posted by yogi
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.

And your response:

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.

Did I use the word "correct" or preferred or universal -no as to each

I consistently said that when B moves to A and the two clocks are compared in the frame in which they were synced - A will have logged more time than B - there is a one way age difference, real and objective.

 

[JesseM]

Jesse - you are weasel wording your way around again - you have previously taken the position that there can be no objectively real age difference in the situation I quoted in post 64.

No, I explicitly said in many posts (I can quote some if you like) that in the scenario where the two clocks are initially at rest relative to each other and synchronized in their rest frame, and then one accelerates towards the other, then the one that accelerated will show a lesser time reading. My position has always been that although this clock shows a lesster time reading when they meet, that does not mean that the clock that accelerated has "aged less" between the time it accelerated and the time they meet. This is clear in the numerical example I give in my last post, where A accelerates towards B, and all frames agree that when they meet, A will read 2006.73 while B will read 2007. But you see that in B's frame, A ages 1.73 years and B ages 2 years between the time A accelerates and the time they meet; but in C's frame, A ages 1.73 years but B only ages 1.5 years between the time A accelerates and the time they meet. Are you saying that the description of the situation from C's perspective is any less valid than B's, or aren't you? Please give me a simple straight answer to this question.

I consistently said that when B moves to A and the two clocks are compared in the frame in which they were synced - A will have logged more time than B - there is a one way age difference, real and objective.

"Logged more time" since when? In C's frame, B only logs 1.5 years after A accelerates while A logs 1.73 years, but B still comes out ahead because A read 2005 at the moment it accelerated while B already read 2005.5 at the same moment.

To pin down what you mean here, let's imagine that A and B were out-of-sync in their own rest frame--at the same moment B read 2005, A already read 3164. Then A accelerates, and when they come together, B reads 2007 and A reads 3165.73. Would you say that A has "logged more time" simply because it reads a greater time when they meet, ignoring the question of whether they were in sync to begin with? Or would you say that the question of whether they were initially in sync is relevant to the question of which clock "logged more time"?

 

[yogi]

Don't add other factors - I am saying that, in the example I proposed as a PF poll question in post 64, between a first spacetime event defined as A and B at rest in the same frame and both set to read zero in the frame in which they are at rest, and a second spacetime event (B's subsequent arrival at A consequent to a short acceleration period followed by constant relative velocity v thereafter), then, if A and B both stop their respective clocks and the hands viewed, the clocks will read different (For example A could read 50 minutes and B could read 40 minutes).