Twin Paradox (thorough explanation needed)

[Eli Botkin]

No it's not, not in the "classic twin paradox case"! In that case they both agree on who aged less, and on the ratio of their average rates of aging. In my example they disagree on who aged less, and one's ratio is the inverse of the other's (i.e. if you're considering the ratio of B's elapsed time to A's, A says it's 0.8 while B says it's 1/0.8=1.25).

What do you mean "make the point of time dilation"? Time dilation is normally defined in terms of a ratio of clock time to coordinate time, and in each one's rest frame the other one's clock only elapses 0.8 the amount of coordinate time in that frame. Besides, the "start time" is totally arbitrary, you could imagine both their clocks have been ticking forever from -infinity until the time they meet, and they both read 0 at the moment they meet--in that case how would you define the "time differences" for each one?

JesseM:
Let's look at a classic twin paradox case. A and B start at T=0 and meet when A's clock reads 25. Their relative velocity is 0.6 (where c=1). That means that B turns around when A says that B is at X=7.5. B's clock at turn-around reads 10 (and reads 20 when they meet).For A the dilation ratio will be 20/25 = 0.8.

What is it for B? Well B is aware of A's clock readings from 0 t0 8 and from 17 to 25. As far as B is concerned A instantaneously reset his clock from 8 to 17. So from B's point of view A's clock ticked through 25-9 = 16. Therefore for B the dilation ratio is 16/20=0.8.

And this dilation symmetry makes sense because we expect that each one will see the other clock dilated in the same way. The symmetric two-way time dilation always exists with or without acceleration, but the non-symmetric outcome when they finally meet is a result of only B having that blind-spot on A's worldline.

 

[JesseM]

JesseM:
Let's look at a classic twin paradox case. A and B start at T=0 and meet when A's clock reads 25. Their relative velocity is 0.6 (where c=1). That means that B turns around when A says that B is at X=7.5. B's clock at turn-around reads 10 (and reads 20 when they meet).For A the dilation ratio will be 20/25 = 0.8.

What is it for B? Well B is aware of A's clock readings from 0 t0 8 and from 17 to 25. As far as B is concerned A instantaneously reset his clock from 8 to 17. So from B's point of view A's clock ticked through 25-9 = 16. Therefore for B the dilation ratio is 16/20=0.8.

B can clearly see that A has aged 25 years, not 16 when he returns--proper time is an objective frame-invariant quantity in relativity, there can be no ambiguity about the fact that 25 years of proper time elapse on A's worldline between the event of B departing and the event of B returning. You can't just mix and match results from different inertial frames to define B's "point of view" that way, the phrase "point of view" doesn't even have any well-defined meaning for an observer who doesn't move in a 100% inertial way, there would be many different ways to construct a non-inertial coordinate system where B is at rest and all would be equally valid as far as relativity is concerned. Think of it this way, what if B does not turn around perfectly instantaneously but accelerates in a continuous way to turn around for 1 hour or 1 second or 1 nanosecond, what would you say happens to A's clock from B's "point of view" then?

 

[OnlyMe]

There is a planet, the Earth and another planet, the colony. The two, planets are 20. LY apart and at rest relative to one another. A spaceship takes off from the Earth traveling at 0.5 c, accelerating instantly. And travels directly to the colony planet.

Nothing else exists in this thought experiment.

What an observer on the Earth sees is the colony 20 LY away, holding steady at that distance and the spaceship moving away from the Earth toward the colony, at 0.5 c.

What an observer on the colony sees is the Earth 20 LY away, holding steady at that distance and a spaceship approaching the colony at 0.5 c from the direction of the earth.

What an observer on the spaceship sees is the Earth moving away from the spaceship and away from the direction of the colony at 0.5 c and the colony approaching the spaceship in the direction of the Earth at 0.5 c.

All three see their own clocks as working properly and keeping good time.

An observer in the spaceship cannot observe the distance between the Earth and the colony directly, as they are in front of and behind the spaceship. The observer in the spaceship can only measure the distances of each relative to itself.

In idealized thought-experiments it's assumed that each observer has a grid of rulers and clocks at rest relative to themselves and extending out arbitrarily far, which they use to assign coordinates in their own rest frame--this sort of network is how Einstein originally defined the notion of reference frame.

While this is true the thought experiment as restated above was set up as defining observer perspectives.

Either way, in the spaceship's rest frame the position coordinates of the Earth and station at a single moment of coordinate time will be 17.32 light years apart.

I do not disagree. I should have continued the example to include the time dilated observation. The intent was to demonstrate that with the spaceship essentially moving at a known velocity within the at rest frame of reference of the planets, calculating the proper distance/length between each of the planets and the spaceship was easily demonstrated. Not that this example is unique, just that it lent itself to the purpose.

This applies whether the observer is moving relative to the object or the object is moving relative to the observer.

Do you understand that this is a completely meaningless distinction? In relativity there is no difference between "A is moving relative to B" and "B is moving relative to A"

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

If there is an observer in both the moving frame of reference and the stationary frame

Do you understand that "moving" and "stationary" can only be defined relative to a choice of reference frame, and that all reference frames are equally valid in SR?

I am pretty sure the thought experiment included sufficient information to know which frame of reference was in motion. One at rest frame of reference, one moving frame of reference and a total of three observer dependent "perspectives".

Yes, all reference frames are equally valid. SR also includes the means to reconcile observed differences between two frames of reference in uniform rectilinear motion relative to one another.

While in motion the spaceship is length contracted, but the distance between the planets is not.

Relative to what frame? In the spaceship's frame the spaceship is not in motion, it's stationary, and therefore its length is not contracted while the distance between planets is contracted since they are the ones in motion in this frame. And this frame's perspective is every bit as valid as the perspective of the frame where the planets are stationary and the spaceship is in motion. Do you disagree?

No and yes. This gets to the heart of my intent. The difference between proper distance/ length and length contracted observation of distance/length. Proper distance/length is the same for all observers. Where a moving frame of reference is involved the proper distance/length appear length contracted. As long as the relative velocity is known, the Lorentz transformations provide the means to calculate the proper distance/length from the observed contracted distance/length.

This is not a new or unique discussion. There are theoreticians on both sides of the issue. My position is that the proper distance/length is the real distance/length.

In my original post I excluded velocity dependent time dilation and length contraction.

If you assume two planets, moving uniformly with respects to one other through space

Moving with respect to one another, or with respect to some observer's frame of reference? Obviously if they are moving with respect to one another the distance between them is changing!

This was misstated. It should have been, "at rest relative to one another and moving uniformly in space...".

the planets can be length contracted while in motion, the distance between them remains constant. It is not length contracted.

Distance is neither an object.., matter nor energy. Though it may appear length contracted under some circumstances, distance itself does not move and so cannot be length contracted, in a way similar to rods and spaceships.

Both of these can be viewed as similar to, Bell's spaceship paradox. I don't think you will find it in a textbook, but a description of the thought experiment should be available on the net.

Very briefly, it involves two spaceships that begin at rest with a string stretched tight connecting them. They accelerate uniformly such that the distance between them remains the same when observed from the "rest" frame of reference from which they began. Does the string break as it is length contracted? There were and are some very bright theorists on both sides. The CERN theory group decided that the string would not break. I am not yet sure but I like their answer.

I did not raise this example earlier, because it deals with the length contraction of the string, an object and that carries the conversation further than was my intent.

I do understand SR and the math involved. I did not and do not believe that math is necessary to present the perspective.

 

[pervect]

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

Well, relativity says that one cannot determine which frame of reference is actually moving.
Thus, as a consequence, the whole twin paradox is "meaningless" - i.e. it's a false paradox. It's only a paradox if one has the belief (which is inconsistent with relativity) that one can determine which frame of reference is "actually" moving.

One would hope we could wrap this long thread up at this point - my intuition says that that's not going to happen. But it really is that simple!

 

[Eli Botkin]

B can clearly see that A has aged 25 years, not 16 when he returns--proper time is an objective frame-invariant quantity in relativity, there can be no ambiguity about the fact that 25 years of proper time elapse on A's worldline between the event of B departing and the event of B returning. You can't just mix and match results from different inertial frames to define B's "point of view" that way, the phrase "point of view" doesn't even have any well-defined meaning for an observer who doesn't move in a 100% inertial way, there would be many different ways to construct a non-inertial coordinate system where B is at rest and all would be equally valid as far as relativity is concerned. Think of it this way, what if B does not turn around perfectly instantaneously but accelerates in a continuous way to turn around for 1 hour or 1 second or 1 nanosecond, what would you say happens to A's clock from B's "point of view" then?

JesseM:
I don't disagree that "B can clearly see that A has aged 25 years." I was addressing the point of how to compute the dilation ratio (DR). For even in the classical twin paradox problem it should turn out that every observer "sees" a moving clock as running more slowly than his own.

If the acceleration is continuous, then instead of being a constant the DR would be a function having value 1 when the relative velocity passes through zero. Though I haven't yet made the computation, I expect that both DR functions would have the same average value.

Jesse, I appreciate your insightful replies. They've made me think more. But I don't have more to offer so I'll sign off on this thread. Thanks.

 

[ghwellsjr]

If we want to be precise, we should use metronome for measuring time intervals and rods for measuring distance intervals and then use clock for measuring accumulated time and odometer for measuring accumulated distance.

Hmm, interesting. You are of course correct, but metronomes have such a strong association with music and the acoustic output that I have never heard it used in a physics context.

Well then, you never read this post:

The point I was trying to make is that time and distance do not behave the same way in SR. The fact that the time difference persists while the length difference does not (quite aside from how it happens "physically") underscores that difference.

But time and distance both persist, you only think they don't because you are making an invalid comparison of a clock to a ruler. Time dilation does not directly affect the time on a clock, it directly affects the tick rate of a clock and then the clock integrates (or counts) the ticks to keep track of elapsed time.

To get similar behavior, we should use a metronome (which does not count ticks) and a ruler. Take them both on a high speed trip, during which the metonome slows down and the ruler contracts, and when we come back to the starting point, the ruler is the same length as one that did not take the trip and the metronome ticks at the same rate as one that did not take the trip.

Now if you want to get similar behavior to a clock, you need an odometer. This will integrate distance traveled just like the clock integrates time. And you could have a speedometer which calculates divides the (contracted) distance traveled by the (dilated) time.

For example, let's suppose that we take a vehicle with a clock, an odometer and a speedometer. We accelerate the vehicle to 0.6c and take it on a round trip for 50 years according to the starting frame. It's speedometer will read 0.6c and from the point of view of an observer that was stationary with the vehicle before it left, the vehicle's speed is also 0.6c. The gamma factor at this speed is 1.25 which means the clock will be running slow by 1/1.25 according to the rest frame. Its lengths along the direction of motion will also be contracted to 1/1.25 of what the observer in the rest frame sees.

So in our example, the vehicle will take 50/1.25 or 40 years to make the complete trip and this is what will be indicated on its clock. Similarly, the distance traveled according to the rest frame is 0.6*50 or 30 light-years. But according to the on-board odometer, it has traveled 24 light-years. And the speedometer will have read 24/40 = 0.6c during the trip.

 

[bobc2]

Well then, you never read this post:

Very nice way to explain that, ghwellsjr.

 

[bobc2]

Well, relativity says that one cannot determine which frame of reference is actually moving.
Thus, as a consequence, the whole twin paradox is "meaningless" - i.e. it's a false paradox. It's only a paradox if one has the belief (which is inconsistent with relativity) that one can determine which frame of reference is "actually" moving.

pervect, the twin paradox is not about which one moves with constant velocity. It's about which twin accelerates. Relativity definitely does not say that you cannot determine which twin is accelerating. More to the point, it's about which twin took the shorter world line path through spacetime from the event of the departure to the event of the reunion. That can certainly be determined--and any observer can figure that one out (no matter what reference frame you wish to use).

By the way the reference frame does not move. Just pick a reference frame and then analyze the motion of some object (or observer) with respect to that reference frame.

 

[IsometricPion]

Both of these can be viewed as similar to, Bell's spaceship paradox. I don't think you will find it in a textbook, but a description of the thought experiment should be available on the net.

I found it and its solution (that the string breaks) on wikipedia http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox" .

It is only meaningless if you are not able to determine which frame of reference is actually moving. If it were meaningless the whole twin paradox would be meaningless, as it requires, some means of determining which twin "travels".

It is not much more difficult to analyse the twin paradox using an inertial frame in which both planets are in motion. The phrase "actually moving" does not have meaning in the context of SR (since it implies there is some coordinate independent notion of (spatial) velocity, which there isn't).

The difference between proper distance/ length and length contracted observation of distance/length. Proper distance/length is the same for all observers. Where a moving frame of reference is involved the proper distance/length appear length contracted. As long as the relative velocity is known, the Lorentz transformations provide the means to calculate the proper distance/length from the observed contracted distance/length.

This is not a new or unique discussion. There are theoreticians on both sides of the issue. My position is that the proper distance/length is the real distance/length.

This was recently discussed on the linked page of https://www.physicsforums.com/showthread.php?t=484405&page=7" thread. As was said there, distance has a well known definition which is not frame invariant. The proper length between two points in space-time is frame invariant and corresponds to the distance between the two points in only one frame. In the frame in which the spaceship is at rest in, e.g., its outbound journey, the proper length defined by the path connecting points simultaneous in frame in which Earth is at rest does have the same value as the distance measured in Earth's rest frame, but it is not a distance (since it has a temporal component).

 

[JesseM]

pervect, the twin paradox is not about which one moves with constant velocity. It's about which twin accelerates.

How are the two different? If one moves at constant velocity and one doesn't, that's exactly equivalent to saying one accelerates and one doesn't.

By the way, pervect doesn't post that much any more but I remember from when he did he was one of the most knowledgeable posters here, if you think he's getting some basic aspect of relativity wrong chances are you're misunderstanding what he's trying to say.

 

[GrayGhost]

JesseM:
I don't disagree that "B can clearly see that A has aged 25 years." I was addressing the point of how to compute the dilation ratio (DR). For even in the classical twin paradox problem it should turn out that every observer "sees" a moving clock as running more slowly than his own.

Eli,

My understanding is that twin B can see the twin A clock ticking faster than his own "while B himself is non-inertial". He has to, for otherwise he could not agree with twin A on their relative aging when they are again co-located. The faster ticking is not the result or proper time speeding up, but only a change in the twin B POV due to frame transitioning, which causes the A-clock to advance faster along its worldline (per B).

GrayGhost

 

[Dale]

Well then, you never read this post:

Yes, I don't know how I missed that one, since I was an active participant on that thread. Good post!

 

[bobc2]

How are the two different? If one moves at constant velocity and one doesn't, that's exactly equivalent to saying one accelerates and one doesn't.

You can tell the difference when the twins reunite and compare proper distances traveled.

By the way, pervect doesn't post that much any more but I remember from when he did he was one of the most knowledgeable posters here, if you think he's getting some basic aspect of relativity wrong chances are you're misunderstanding what he's trying to say.

Either he was not thinking clearly at the moment or I was not. I do that a lot, so I'm probably the culprit (I get a little distracted when my boss starts giving me those quizzical looks--I try not to post very often from my desk). But thanks for the heads-up.

 

[Dale]

I realize you can claim that all frames of reference are equally valid. However, you are talking about every individual inertial frame being as valid as ever other inertial frame. Each one of the traveling twin's frames are as valid as all of the others in the accelerating sequence. So, does that necessarily mean that all of those frames collectively (yielding the bazzar faster-than-light result) are as valid as any individual inertial frame? I don't think so, but I'll need to go back and review Rindler and a couple of others to see if they comment on that aspect. Maybe bcrowell, JesseM, or DaleSpam already have the answer.

The collective frames do have some problems, most notably that there are events in the spacetime where this approach assigns multiple sets of coordinates. This violates one of the defining properties of a coordinate system. However, you can fix this simply by saying that the regions where there is ambiguity are excluded from the coordinate system. This makes your system a bit "choppy" and you cannot analyze physics in areas that are not covered by the chart, but similar things happen in the Rindler and most other non-inertial coordinate systems.

If you do it right, it is OK, but most people using this approach don't recognize that there are hidden pitfalls, don't know that there are equally valid alternatives, and don't realize that there is no standard convention.

 

[SeventhSigma]

So what does it mean if, say, the universe is a hypersphere and going in one direction means you come out the other? e.g. zooming off Earth and then arriving back at Earth without turning around?

 

[GrayGhost]

So what does it mean if, say, the universe is a hypersphere and going in one direction means you come out the other? e.g. zooming off Earth and then arriving back at Earth without turning around?

It does seem to be the case, assuming the universe is a closed unbounded system. Now as to whether the Earth still exists, has yet to begin existing, or what the date-time happens to be upon your arrival, is the big question.

GrayGhost

 

[Gulli]

One handy formula to know when thinking about the relativity of simultaneity is this: if two clocks are synchronized and a distance D apart in their own rest frame, then that means that in the frame of an observer who sees the clocks moving at speed v (parallel to the axis between them), then at a single moment in the observer's frame the clock at the rear will show a time that's ahead of the clock at the front by an amount vD/c^2. So in this example, since the Earth clock and the colony clock are synchronized and a distance of 20 light-years apart in their own frame, then for the spaceman who sees them moving at 0.5c, at any given moment in his frame the two clocks must be out-of-sync by (0.5c)(20 light-years)/c^2 = 10 years, with the rear colony clock ahead of the front Earth clock by that amount. So, at the same moment the Earth clock reads 600,000 AD, the colony clock must read 600,010 AD, according to the definition of simultaneity in the spaceman's rest frame.[\QUOTE]

I'm sorry to bother you again but I'm still confused by that 10 year head start: I mean, I sort of understand why that would happen (it's similar to the Andromeda paradox, which doesn't seem like much of a paradox to me, after all, both observers on Earth still see the same event through a telescope and that event happened in the past anyway), but I don't understand why it doesn't happen the other way around. Why doesn't the colony see the spaceship with a (0.5c)*(17.32LY)/c^2 = 8,66 year head start? And why does the D stand for the distance in the rest frame of a point, does "a point" even have a rest frame?

 

[JesseM]

You can tell the difference when the twins reunite and compare proper distances traveled.

The difference between what? You think there's a difference between "A moved at constant velocity, B did not" and "B accelerated, A did not?" The two statements are exactly equivalent.

 

[JesseM]

One handy formula to know when thinking about the relativity of simultaneity is this: if two clocks are synchronized and a distance D apart in their own rest frame, then that means that in the frame of an observer who sees the clocks moving at speed v (parallel to the axis between them), then at a single moment in the observer's frame the clock at the rear will show a time that's ahead of the clock at the front by an amount vD/c^2. So in this example, since the Earth clock and the colony clock are synchronized and a distance of 20 light-years apart in their own frame, then for the spaceman who sees them moving at 0.5c, at any given moment in his frame the two clocks must be out-of-sync by (0.5c)(20 light-years)/c^2 = 10 years, with the rear colony clock ahead of the front Earth clock by that amount. So, at the same moment the Earth clock reads 600,000 AD, the colony clock must read 600,010 AD, according to the definition of simultaneity in the spaceman's rest frame.

I'm sorry to bother you again but I'm still confused by that 10 year head start: I mean, I sort of understand why that would happen (it's similar to the Andromeda paradox, which doesn't seem like much of a paradox to me, after all, both observers on Earth still see the same event through a telescope and that event happened in the past anyway), but I don't understand why it doesn't happen the other way around.

When I talk about the 10 year head start I'm not talking about what the ship sees visually, but rather what's true in the ship's rest frame (remember that I said the ship doesn't actually see the colony clock reading 600,010 AD until the ship's own clock reads 600,017.32 AD). The colony and the Earth share the same rest frame, so there's no difference between how the Earth defines simultaneity and how the colony defines it; if the Earth thinks the colony clock read 600,000 AD simultaneously with the ship clock reading 600,000 AD, the colony agrees. And naturally all frames must agree that the ship clock read 600,000 AD at the same time the Earth clock read 600,000 AD, since they were both at the same position in space and all frames must agree about localized facts, the relativity of simultaneity only comes in when you consider events at different locations in space.

Why doesn't the colony see the spaceship with a (0.5c)*(17.32LY)/c^2 = 8,66 year head start?

The formula only deals with two clocks at rest relative to each other. If there was a second ship at rest relative to the first one, and the ships were 17.32 LY apart in their mutual rest frame (which would make the distance between them smaller in the Earth/ship frame), then if their clocks were synchronized it would be true in the Earth/ship frame that they were 8.66 years out of sync.

And why does the D stand for the distance in the rest frame of a point, does "a point" even have a rest frame?

? I never said D stands for the distance in the rest frame of "a point", I said the formula applied in a situation where "two clocks are synchronized and a distance D apart in their own rest frame"

 

[Gulli]

When I talk about the 10 year head start I'm not talking about what the ship sees visually, but rather what's true in the ship's rest frame (remember that I said the ship doesn't actually see the colony clock reading 600,010 AD until the ship's own clock reads 600,017.32 AD). The colony and the Earth share the same rest frame, so there's no difference between how the Earth defines simultaneity and how the colony defines it; if the Earth thinks the colony clock read 600,000 AD simultaneously with the ship clock reading 600,000 AD, the colony agrees. And naturally all frames must agree that the ship clock read 600,000 AD at the same time the Earth clock read 600,000 AD, since they were both at the same position in space and all frames must agree about localized facts, the relativity of simultaneity only comes in when you consider events at different locations in space.

I know it's not what he actually sees when he leaves Earth, otherwise he could look into the future. He can only infer a head start through reasoning, like you said, and that's what I meant.

The formula only deals with two clocks at rest relative to each other. If there was a second ship at rest relative to the first one, and the ships were 17.32 LY apart in their mutual rest frame (which would make the distance between them smaller in the Earth/ship frame), then if their clocks were synchronized it would be true in the Earth/ship frame that they were 8.66 years out of sync.

I'm still trying to figure out for myself (intuitively) when I should factor in relativity of simultaneity and when not, your example with a second spaceship will probably help me out.

? I never said D stands for the distance in the rest frame of "a point", I said the formula applied in a situation where "two clocks are synchronized and a distance D apart in their own rest frame"

Sorry, my mistake.

 

[Eli Botkin]

Eli,

My understanding is that twin B can see the twin A clock ticking faster than his own "while B himself is non-inertial". He has to, for otherwise he could not agree with twin A on their relative aging when they are again co-located. The faster ticking is not the result or proper time speeding up, but only a change in the twin B POV due to frame transitioning, which causes the A-clock to advance faster along its worldline (per B).

GrayGhost

GrayGhost:
The phrase “twin B can see the twin A clock ticking faster than his own” can be misunderstood. I expect, though, that in this instance your SR expertise is not meaning to interpret that as “seeing through a telescope.” Rather I assume you mean the time that B assigns to events on A’s worldline based on B’s knowledge of the Lorentz transformation equations. If so, then I maintain that an observer will always assign fewer ticks to a moving clock than he experiences on his own clock. This keeps the dilation ratio < 1 even in the instantaneous inertial frames within a period of acceleration. The ratio would rise to 1 as the relative velocity approaches 0, and then fall again.

This assigned time and its resulting dilation rate are not measurable quantities. To see why there is agreement when they are again co-located, I suggest that the more usual meaning of “see” be used. If either observer views the other clock through a telescope, they will first see a slower clock than their own followed by a faster clock (Doppler). [Both of course see the same Doppler shifts.] B (who does the accelerating) will see the two Doppler-shifted periods of A’s clock-images for equal periods on his own (B’s) clock. However, A will see the slower image of B’s clock for a longer time (on A’s clock) than he sees the faster portion. Therefore, at get-together, B’s clock reads less than A’s. And they’ve both been witness to why this has occurred.

Thanks for “listening.”

 

[Mentz114]

If so, then I maintain that an observer will always assign fewer ticks to a moving clock than he experiences on his own clock.

All observers agree on the number of ticks. A tick is an event and events cannot be 'transformed away'. The interval between the ticks is observer dependent. I don't think it affects your argument if you amend the phrase in question.

 

[GrayGhost]

GrayGhost: The phrase “twin B can see the twin A clock ticking faster than his own” can be misunderstood. I expect, though, that in this instance your SR expertise is not meaning to interpret that as “seeing through a telescope.”

Yes, you're right. This is the 2nd time I've stated this poorly in this forum, as it's a bad habit when posting too quickly. Thanx. The faster ticking A-clock (per B) is ascertained from looking at the date/time, track, and navigation data after the fact. The faster ticking clock is "determined" (not seen) by analysis of the data while ignoring the relativistic effects during analysis ... which seems to be the way most folks look it unfortunately (I believe that an incorrect approach).

You're right, the light signals received from twin A will be doppler shifted, and so "the images" of the moving clock can show a clock ticking faster or slower than yor own "due to doppler effects". To know the real relative rate of 2 clocks, the spacetime dilation needs to be determined (and accounted for) from its corresponding relativistic doppler effects for twin B to know the correct A-velocity as it applies to relativity. If NOT DONE, then twin B will determine that twin A moved superluminally and the A-clock ticked (at times) faster than his own per himself. However IF DONE, then the A-velocity will always be luminal (or less) as it should be, and the A-clock will never tick faster than Bs per B.

At any rate, thanks for the correction there Eli.

GrayGhost

 

[Gulli]

@JesseM

I think I understand now.

The Earth and the colony share the same frame of reference and so agree with each other on when an event took place. The colony does not have to agree on simultaneity with the spaceman but they have confirmation from Earth (because Earth and the spaceman were at the same position during the event of the spaceship speeding by) that the spaceship passed them in 600.000. So when the signal from the spaceship speeding past Earth reaches the colony in 600.020 the signal will read "it is now 600.000 and a spaceship just passed us", so no head start. The spaceman does not have to agree on simultaneity with the Earth: the spaceman and Earth agree that the spaceship passed Earth in what they both consider to be 600.000, but the spaceman does not have to agree this event was simultaneous to the year 600.000 on the colony and can therefore infer a head start. The value of the head start depends on the relative velocity between the Earth/colony frame and that of the spaceman, and on the distance between Earth and the colony.

Thanks for this piece of enlightenment!

 

[JesseM]

@JesseM

I think I understand now.

The Earth and the colony share the same frame of reference and so agree with each other on when an event took place. The colony does not have to agree on simultaneity with the spaceman but they have confirmation from Earth (because Earth and the spaceman were at the same position during the event of the spaceship speeding by) that the spaceship passed them in 600.000. So when the signal from the spaceship speeding past Earth reaches the colony in 600.020 the signal will read "it is now 600.000 and a spaceship just passed us", so no head start. The spaceman does not have to agree on simultaneity with the Earth: the spaceman and Earth agree that the spaceship passed Earth in what they both consider to be 600.000, but the spaceman does not have to agree this event was simultaneous to the year 600.000 on the colony and can therefore infer a head start. The value of the head start depends on the relative velocity between the Earth/colony frame and that of the spaceman, and on the distance between Earth and the colony.

Thanks for this piece of enlightenment!

Yup, sounds like you've got it! Glad I could help.

 

[Eli Botkin]

All observers agree on the number of ticks. A tick is an event and events cannot be 'transformed away'. The interval between the ticks is observer dependent. I don't think it affects your argument if you amend the phrase in question.

Mentz114:
You are, of course, correct to say ” All observers agree on the number of ticks” on a worldline.

I should not carelessly assume and then omit clarifying phrases.
I should have written:
If so, then I maintain that an observer will always assign fewer ticks to a moving clock during any sequence of ticks on his own clock. “Fewer ticks” is what is implied by the term “time dilation.

Said another way: Time on a moving clock is “slower “ when compared to the “non-moving” clock. And again: dt/dT < 1 if t is time measured on the moving clock and T is time measured on the “non-moving” clock. Of course, the two clocks must be equivalent in every way (except for where they’re placed).

 

[Gulli]

@JesseM

So then am I correct in assuming the twin paradox has the exact same solution as my spaceman/colony problem because in the twin paradox the traveler returns and so has to turn around somewhere and thus passes a point where his velocity (well, at least the radial component) in relation to Earth is zero and that this point is analogous to the colony?

 

[DTThom]

Everyone is quick to teach and slow to learn. The tens of thousands of documents purporting to explain the Twins Paradox illustrates this as well as anything.

In the process of being so quick to teach, the purveyors of these documents have overlooked the simple and obvious truth about clock rates.

Why do people make things so difficult for themselves? Did it never occur to them that when two reunited clocks (meaning they are now once again at the same place-moment) show an ACTUAL disparity in their recorded time, that there must necessarily have been an ACTUAL difference in clock rates involved while they were in relative motion with each other?

One should never suggest (as they so often do) that there was some sort of "jump in time" involved with the change of inertial frame (meaning at the turn-around point). The simple act of starting a clock as an inbound astronaut passes an outbound astronaut cannot possibly create a "jump in time". (Remember, the outbound astronaut hands off his clock reading to the inbound astronaut.)

The time contraction formula [t' = t * sqr rt of (1 - v^2)] is not linear. That is why the party who changes frames to bring the two parties back together will register the least amount of time on his clock with the symmetry of the situation preserved.

The actual distances and speeds relative to the universe will vary depending on which party changes frames, but the parties involved cannot possibly detect that. That is in keeping with the postulates and deductions of special relativity.

Time-keeping, distance and speed are interminably bound in one equation. Therefore, actual differences in clock rates implies actual length contraction dependent on actual speed relative to the universe. Actual length contraction works in combination with actual time-keeping contraction to preserve the symmetry of measures across inertial frames.

There is clock functioning at every level, dependent on actual light speed, at even the atomic level. Our observations and measuring paradigms of every nature are constrained by the speed of light, as is our "synchronizing" of clocks.

Special relativity can be charted out in actual terms (absolute terms), where light speed is constant in an actual sense. All the results of special relativity, including the consistent measured speed of light, fall naturally into place when charting these actualities against the (experimentally undetectable) rest state of the universe.

Actual time-keeping and length contraction arise naturally from the fact that all phenomena are dependent on the speed of light, which is itself invariant in actuality, being massless.

Consider that A.P. French writes on page 150 of Special Relativity: "Note, though, that we are appealing to the reality of A's acceleration, and to the observability of the inertial forces associated with it. Would such effects as the twin paradox exist if the framework of fixed stars and distant galaxies were not there? Most physicists would say no. Our ultimate definition of an inertial frame may indeed be that it is a frame having zero acceleration with respect to the matter of the universe at large."

And I feel very sorry for any physicist who doesn't understand that.

Michio Kaku states on page 80 of Einstein's Cosmos that bringing the twins together "determines which twin was "really" moving."

Martin Gardner writes on page 114 of Relativity Simply Explained: "There is one all important difference between the relative motion of the astronaut and the relative motion of the stay-at-home. The stay-at-home does not move relative to the universe."

Both Kaku and Gardner were using the simplest of twins paradox scenarios, in which one party is assumed to be at rest with the cosmos. But that need not be the case. There can be any number of "in between" situations, leading to a lesser time differential. It is also not necessary for the twins to reunite to determine which one was "really moving". The noted asymmetry (noted by both parties) in the time-keeping difference builds incrementally, beginning at the moment of inertial change for one party, when radio or light signals are regularly sent forth and back to check on current clock status.

One should do a search on Einstein's clock synchronization, and its bearing on spacetime diagrams. He or she will find that the notorious "jump in time" is built into that clock synchronization, because it is a one way synchronization, which gets instantly replaced with a different synchronization when a new inertial frame is adopted.

There is all the difference of night and day between predicting and explaining. We can use Einstein's clock synchronization and spacetime to predict a time differential, but we must look at relativity in the universal frame of reference to explain not only that time differential, but also all the mutually symmetrical measures made across inertial frames.

The preceding remarks were copied off my copyrighted web document.

By the way, "observations" and "measures" are strictly synonomous, and constrained by light speed. You might think that a "visual observation" is something different than measuring, but the combination of eye to brain processing is precisely a form of measuring. Identically, biological aging is precisely synonomous with clock functioning of all types, right down to the atomic level and things such as the Doppler effect.

Also, the following phrase is a meaningless muddling of terminology: ".. the distance between them really is shorter in the spaceship's own rest frame.

"Really" means "really", a reality independent of anyone's inertial frame.

You may call it the "God's eye view" or the "view from a higher dimension". Such view is the instantaneous view, free from the constraint of light speed, which is finite.


---------------------------------------------------------------
"First they tell you you're wrong and they can prove it;
then they tell you you're right but it isn't important;
then they tell you it's important but they knew it all along."

Charles Kettering, former head of General Motors
---------------------------------------------------------------

 

[DTThom]

I need to clarify one paragraph in the mid portion of my previous post. Here is the first half of that paragraph, better stated:

Both Kaku and Gardner were using the simplest of twins paradox scenarios, in which one party is assumed to be at rest with the cosmos, and the other party both has motion relative to the universe and changes frames. But that need not be the case. There can be any number of "in between" situations -- such as both parties having motion relative to the universe, with still only one party changing frames (same time differential); or both parties changing frames, leading to a lesser time differential.

 

[Dale]

Hi DTThom, welcome to PF!

Why do people make things so difficult for themselves? Did it never occur to them that when two reunited clocks (meaning they are now once again at the same place-moment) show an ACTUAL disparity in their recorded time, that there must necessarily have been an ACTUAL difference in clock rates involved while they were in relative motion with each other?

Consider two cars that travel from Miami to New York, one travels via Washington DC and the other travels via Denver. When the two cars are reunited in New York they show an ACTUAL disparity in their recorded mileage. Does that necessarily imply that there must necessarily have been an ACTUAL difference in their odometer rates? No, both odometers could still have accurately measured one mile per mile and yet obtained different odometer readings.

 

[JesseM]

@JesseM

So then am I correct in assuming the twin paradox has the exact same solution as my spaceman/colony problem because in the twin paradox the traveler returns and so has to turn around somewhere and thus passes a point where his velocity (well, at least the radial component) in relation to Earth is zero and that this point is analogous to the colony?

If you choose to analyze the twin paradox from the perspective of the Earth frame this could be one way of looking at it, but the answer to which twin ages less is frame-indpendent, you could equally well analyze everything in the frame of an inertial ship which was traveling at the same velocity as the traveling twin during his outward journey, but which kept traveling inertially in the same direction when the traveling twin turned around.

 

[bobc2]

Hi DTThom, welcome to PF!Consider two cars that travel from Miami to New York, one travels via Washington DC and the other travels via Denver. When the two cars are reunited in New York they show an ACTUAL disparity in their recorded mileage. Does that necessarily imply that there must necessarily have been an ACTUAL difference in their odometer rates? No, both odometers could still have accurately measured one mile per mile and yet obtained different odometer readings.

Exactly. Right on the money, DaleSpam. (Could it be that each twin travels his own world line at light speed--but their world lines have different path lengths, just like your two different paths from Miami to New York?)

 

[DTThom]

Hi DTThom, welcome to PF!Consider two cars that travel from Miami to New York, one travels via Washington DC and the other travels via Denver. When the two cars are reunited in New York they show an ACTUAL disparity in their recorded mileage. Does that necessarily imply that there must necessarily have been an ACTUAL difference in their odometer rates? No, both odometers could still have accurately measured one mile per mile and yet obtained different odometer readings.

I can easily deduce that the two odometers in your example traveled actual different distances, due to the actual difference showing up in the number of odometer ticks at the same place-moment in New York. I can identically deduce that the two reunited clocks I spoke of ticked an actual different number of times while they were apart. I don't care what distance the clocks traveled. You provided a distance problem. An odometer is bound to read true to the distance of the road. I provided a clock ticking problem, and the number of ticks are dependent on the combination of speed and distance covered relative to the universe.

* I care only about the number of clock ticks involved, compared to the number of clock ticks registered on the clock at rest with the universe. *

You can do relativity the easy way, or you can make it hard for yourself, and never explain the missing time of the twins paradox.

Also, you cannot travel on a world line. A world line is a spacetime construct, not something physical.

 

[JesseM]

I can easily deduce that the two odometers in your example traveled actual different distances, due to the actual difference showing up in the number of odometer ticks at the same place-moment in New York. I can identically deduce that the two reunited clocks I spoke of ticked an actual different number of times while they were apart. I don't care what distance the clocks traveled. You provided a distance problem. An odometer is bound to read true to the distance of the road. I provided a clock ticking problem, and the number of ticks are dependent on the combination of speed and distance covered relative to the universe.

There is a rather precise analogy between problems involving paths through space and odometers, and problems involving paths through spacetime and clocks--see [post=2972720]this post[/post].

* I care only about the number of clock ticks involved, compared to the number of clock ticks registered on the clock at rest with the universe. *

The phrase "at rest with the universe" has no meaning in relativity. You can choose any inertial frame you like, each of which with a different definition of which objects are "at rest", and they will all make the same prediction about how much time elapses on a given clock between two events on its world line (like the event of leaving a space station and the event of returning to it).

Also, you cannot travel on a world line. A world line is a spacetime construct, not something physical.

"Travel on a world line" is just a common shorthand for taking a path through space and time that matches the world line (i.e the coordinates are the same). And how are you defining "something physical"? Would you say a path through space is also not "something physical" and that therefore you cannot travel on a path through space?

 

[DTThom]

Exactly. Right on the money, DaleSpam. (Could it be that each twin travels his own world line at light speed--but their world lines have different path lengths, just like your two different paths from Miami to New York?)

You can travel through space (a distance dimension) at light speed (a speed dimension).

You cannot travel through spacetime at any speed. No meaning can be attached to such a statement.

Distance = speed * time (relativity or not)

"At rest with the universe" has a clear meaning, relativity or not. Relativity can be fully developed in absolute (universal) terms, and in doing so, it is seen just why it is that we cannot determine our true state of motion relative to the universe.

For the benefit of the recent posters, I'll present these few more comments:


One can hold two reunited clocks in ones hand and see that one registered more clicks than the other WHILE they were apart. "While" sounds like a "time" word to me. We must say that the two clocks ticked at different rates, i.e., ticks per unit "time". That "time" can only be some "time" by which to distinguish the "time" recorded by the two clocks. The only way to avoid circular reasoning is to acknowledge a "time" as kept by a clock at rest with the universe.

Light has a finite and constant speed relative to the universe. It is the speed by which we define all lesser speeds. That is your clue as to the functioning of a photon clock, which must necessarily complete fewer cycles as it (the clock itself) increases its translatory speed through the universe.

Without absolutes (actualities) we are left with only circular definitions (strictly relative definitions). But the time differential showing up at the same place-moment does not fit with circular definitions. The disparity in the clock readings is real. Therefore the two clocks did actually tick a different number of times while they were apart.

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I have better things to do. Goodbye.