How is twin paradox resolved in case of no/zero acceleration?

[lovetruth]

Different inertial frames are indistinguishable, so that is symmetrical also

Different inertial frames are distinguishable because they are moving at different speed. If inertial frames were indistinguishable, they must be equivalent and not relative. But we don't have Theory of Equivalency instead of Theory of Relativity.

 

[lovetruth]

That is completely symmetrical, they each observe the other to be dilated by the same amount.

If each twin observes other twin's time to be dilated, how can their time be same?

 

[jtbell]

Follow the link in post #3 of this thread.

 

[tiny-tim]

Hi San K!

no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

(if they have different velocities, two events at the same time for one will not be at the same time for the other)

hi tiny (but mighty?) tim...:)

you mean one cannot compare because they are not comparable frames of reference?

that have to be at same velocity and place? can put it more precisely?

people who are further away look smaller

if A is a long way from B, then A says "B looks smaller", and B says "A looks smaller" …

where's the paradox??​

there would only be a paradox if A was standing next to B (or if they were moving past each other, but they make their measurements at the exact moment they are passing) …

then it is paradoxical for each to regard the other as shorter

(of course, that paradox does exist … it's called the Lorentz-Fitzgerald contraction! )

similarly, there's no clock paradox unless the two clocks start at the same velocity, and finish at the same velocity (position doesn't matter)

 

[Hurkyl]

The paradox (the apparent symmetry) is resolved by noting that the situation is not symmetric after all, not even if you leave out acceleration, and that you can easily calculate the difference in proper time intervals of the clocks using simple algebraic equations.

You cannot resolve a paradox by justifying one half of the contradiction. You have to explain why the other half of the contradiction is invalid.

A resolution of the twin paradox has to explain what is wrong with the argument

Stella sees that Terra is moving with respect to her, and thus her clock running slowly for the entire trip. Therefore, when Stella returns home, she finds that Terra's clock reads less time than Stella's clock does​

Showing an asymmetry does not demonstrate a flaw with the above argument. It simply defeats the supplementary argument that Terra and Stella's points of view are indistinguishable.

One can hope that demonstrating the asymmetry, or that providing a correct way to compute things, could prompt the confused person into resolving the the paradox himself. But you also risk the rather dangerous possibility of pushing the person into accepting doublethink -- to accept the resolution despite retaining the belief that there is nothing actually wrong with the argument above.

 

[Dale]

If each twin observes other twin's time to be dilated, how can their time be same?

They aren't the same, they are symmetric. You don't seem to understand symmetry.

 

[Dale]

people who are further away look smaller

if A is a long way from B, then A says "B looks smaller", and B says "A looks smaller" …

where's the paradox??​

Excellent example of symmetry!

 

[Filip Larsen]

You cannot resolve a paradox by justifying one half of the contradiction. You have to explain why the other half of the contradiction is invalid.

A resolution of the twin paradox has to explain what is wrong with the argument

Stella sees that Terra is moving with respect to her, and thus her clock running slowly for the entire trip. Therefore, when Stella returns home, she finds that Terra's clock reads less time than Stella's clock does​

Showing an asymmetry does not demonstrate a flaw with the above argument. It simply defeats the supplementary argument that Terra and Stella's points of view are indistinguishable.

One can hope that demonstrating the asymmetry, or that providing a correct way to compute things, could prompt the confused person into resolving the the paradox himself. But you also risk the rather dangerous possibility of pushing the person into accepting doublethink -- to accept the resolution despite retaining the belief that there is nothing actually wrong with the argument above.

So, if I understand you correctly, you are saying that people who wants to resolve the paradox and are shown, quantitatively, that the Lorentz transformation of suitable events can explain and give the actual number of the difference in the twins age, these people will still in some meta-physical way believe there is a paradox unless they are presented with a much more complicated explanation that includes acceleration? If so, I don't find that very plausible. There is no doubt that changing from an outbound to an inbound trajectory is central to the resolution, but exact nature of this change is (in my book) not essential.

I don't think I can bring anything new to this discussion at this point, only repeat what I have already said. You seem to skip or ignore the specific questions I've ask you so far, so unless you feel like answering some of them I really don't see us getting any further.

 

[Hurkyl]

So, if I understand you correctly, you are saying that people who wants to resolve the paradox and are shown, quantitatively, that the Lorentz transformation of suitable events can explain and give the actual number of the difference in the twins age, these people will still in some meta-physical way believe there is a paradox

Yes. You are explaining to them information they already have (Terra ages more than Stella), and offer absolutely no insight as to why other information they have (Stella ages more than Terra) is faulty.

unless they are presented with a much more complicated explanation that includes acceleration?

That would be even worse!

You seem to skip or ignore the specific questions I've ask you so far, so unless you feel like answering some of them I really don't see us getting any further.

I've answered everything I thought was relevant. But since you insist:

What I have been claiming, though, is that is that with regard to the calculation of proper time intervals for the two twins you can consider the three-clock scenario a limiting case of the accelerated scenario where the acceleration is arbitrarily high. Do you disagree with that?

I agree in the sense that you are considering two identical mathematical calculations. I disagree in the sense that three clocks is not a limiting case of two clocks.

it would also be nice to know if you consider the explanation using Minkowski diagram a valid resolution of the paradox?

It depends on what you do with the diagram. Wikipedia's resolution is fine. It sports a feature your argument does not -- a demonstration of where and how the rationale "Stella sees Terra's clock running slowly the whole trip" is faulty.

Was there another question youw anted answered?

 

[Filip Larsen]

It depends on what you do with the diagram. Wikipedia's resolution is fine. It sports a feature your argument does not -- a demonstration of where and how the rationale "Stella sees Terra's clock running slowly the whole trip" is faulty.

Well, I think I can see what you are getting at then. You are worried that by only showing the three-clock scenario with its associated simple calculations people would still be at loss as to the nature of the break in symmetry, that is, they would still not know the "reason" for why Stella ends up younger and not vice versa. In the words used on the Twin Paradox FAQ [1] the three-clock scenario leaves a "time gap" which I do concur seems very mysterious unless it is also explained what it "really looks like" in a scenario with acceleration.

If this was your point, then I agree. Personally I feel that the three-clock or instantaneous turn-around scenario explains a lot and that change of reference frames itself is the essential part, but I can see why some would feel something important has been left out. If I were to explain this to, say, my 15 year old kid I'd show her the three-clock version with its simple calculations and then in parallel or afterwards extend the explanation (but not the calculations) to include acceleration of material bodies so that we could make an actual twin take the trip. In fact, this would be a fun vacation exercise.

Was there another question youw anted answered?

Not unless you think I've missed your point above.


[1] http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[lovetruth]

people who are further away look smaller

if A is a long way from B, then A says "B looks smaller", and B says "A looks smaller" …

where's the paradox??​

there would only be a paradox if A was standing next to B (or if they were moving past each other, but they make their measurements at the exact moment they are passing) …

then it is paradoxical for each to regard the other as shorter

(of course, that paradox does exist … it's called the Lorentz-Fitzgerald contraction! )

similarly, there's no clock paradox unless the two clocks start at the same velocity, and finish at the same velocity (position doesn't matter)

Excellent example of symmetry!

Distant object looks smaller because they subtend smaller angle in the eye. Distant object looking small is just an optical illiusion, the object still has the same size. Do you believe that distant star comes closer when you see through telescope?

 

[lovetruth]

They aren't the same, they are symmetric. You don't seem to understand symmetry.

Symmetry is a property of geometrical figure having a mirror plane.
Time is a scalar quantity.
Saying time is symmetric is like comparing oranges to apples.
I don't get what you mean by time is symmetric. Perhaps you can express it mathematically what symmetric time means?

 

[Dale]

Distant object looks smaller because they subtend smaller angle in the eye.

Clearly. But why do they BOTH subtend smaller angles? How come one doesn't subtend a smaller angle and the other subtend a larger angle? The answer, unsurprisingly, is symmetry.

Symmetry is a property of geometrical figure having a mirror plane.

Symmetry is more general than that. Mathematically and physically symmetry means that something does not change under some specific transformation. For instance, an equilateral triangle is symmetric under 120º rotations about its center because it is unchanged by that transformation.

In the case of the example above with distant objects, A and B, looking at each other the angle subtended is determined by Euclidean geometry which is unchanged by rotations and translations. Since A and B are related to each other via a rotation and a translation they are symmetric, meaning that the geometry is also unchanged by transformation of swapping A with B. Therefore, if the angle subtended by A decreases for B then by symmetry we can swap A and B and state that the angle subtended by B decreases for A.

In the case of relativity (either Galileo's or Einstein's) it is a fundamental postulate of the theory that the laws of physics are unchanged by boosts. So if A and B are related to each other via a boost then they are symmetric and in any statement you can swap A and B and have an equally valid statement. So if A's clock is slow according to B then by symmetry B's clock is slow according to A.

Do you feel like you understand the important role of symmetry a little better now?

 

[tiny-tim]

… Distant object looking small is just an optical illiusion, the object still has the same size.

Yes, and a moving clock going slower is also just an optical illusion …

do you really think that a moving (non-accelerating) clock is actually going slower? ​

btw, the Lorentz-Fitzerald contraction is an optical illusion too …

do you really think that a moving train actually changes shape? ​

 

[jfy4]

Yes, and a moving clock going slower is also just an optical illusion …

do you really think that a moving (non-accelerating) clock is actually going slower? ​

btw, the Lorentz-Fitzerald contraction is an optical illusion too …

do you really think that a moving train actually changes shape? ​

I don't want to be pedantic here, since I agree with essentially everything you have said. However, I don't like your usage of the word "actually". It seems to present the idea that something is a particular way objectively. And speaking about SR, since essentially every interaction is mediated by photons, I am okay with saying that these phenomena are "optical" (the measurements/interactions/observations are all described by classical optics) however, deep down there is no pure version of the train. But, I am willing to hear out a counter argument to this idea if you think different.

 

[abbott287]

Hi San K!


no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

No, just have them take pictures of themselves they can compare later when they meet up again. :) same result will happen though. The spaceship twins picture will show him ageing at a slower rate.

 

[PAllen]

No, just have them take pictures of themselves they can compare later when they meet up again. :) same result will happen though. The spaceship twins picture will show him ageing at a slower rate.

No, this is false. While each is moving away from the other, pictures will be identical. Again, think about symmetry. Just as each sees the other getting smaller, each sees and photographs the other's clock going slower.

The symmetry goes away only if one of the twins changes direction. Then, the one that changes direction immediately sees a change in clock rate of the other; while the one that remains inertial only sees the other change clock rate after a delay. This is what would lead to a consistent correlation of the movies after both twins are together again.

 

[Dale]

No, just have them take pictures of themselves they can compare later when they meet up again. :) same result will happen though. The spaceship twins picture will show him ageing at a slower rate.

This is not ever true in any scenario, regardless of their motion. How could it possibly ever be true? If they take a photo every year according to their own clocks then they will be a year older in every photo.

In the standard twins paradox (or any other asymmetric scenario) what will differ is the number of photos, not the apparent age in the nth photo.

 

[San K]

No, this is false. While each is moving away from the other, pictures will be identical. Again, think about symmetry. Just as each sees the other getting smaller, each sees and photographs the other's clock going slower.

The symmetry goes away only if one of the twins changes direction. Then, the one that changes direction immediately sees a change in clock rate of the other; while the one that remains inertial only sees the other change clock rate after a delay. This is what would lead to a consistent correlation of the movies after both twins are together again.

Good point PAllen and DaleSpam. That's what I was thinking.

This is not ever true in any scenario, regardless of their motion. How could it possibly ever be true? If they take a photo every year according to their own clocks then they will be a year older in every photo.

In the standard twins paradox (or any other asymmetric scenario) what will differ is the number of photos, not the apparent age in the nth photo.

The time-space for both, Earth and ship, are different.

Now if we take photos according to a third stationary (or for that matter the stationary "earth") clock say at times...noon, 1 pm, 2 pm...etc i.e. per the Earth's clock time and for now ignore the clock time on the "moving" ship.

a) what would we notice about the photos?

if we take photos per the "moving" ships clock and ignore the "stationary" Earth's clock time

b) what would we notice about the photos?

Note: I know that that comparing photos this way might not be "complete" (for the purposes of showing the consistency of relativity) since we are not accounting for other things.

 

[Hurkyl]

Now if we take photos according to a third stationary (or for that matter the stationary "earth") clock say at times...noon, 1 pm, 2 pm...etc i.e. per the Earth's clock time and for now ignore the clock time on the "moving" ship.

A clock doesn't make a reference frame -- you need a way for observers to decide what the clock says "now".

e.g. if Stella flies past the sun and uses her telescope to look at the clock on Earth, when does she take the "1 PM" photograph"? When she sees "1:00" on the clock through the telescope? When she sees "12:52" (minus a few seconds)? Some other scheme?

(note: The sun is just over 8 light-minutes away from the Earth)

 

[San K]

A clock doesn't make a reference frame -- you need a way for observers to decide what the clock says "now".

e.g. if Stella flies past the sun and uses her telescope to look at the clock on Earth, when does she take the "1 PM" photograph"? When she sees "1:00" on the clock through the telescope? When she sees "12:52" (minus a few seconds)? Some other scheme?

(note: The sun is just over 8 light-minutes away from the Earth)

- by "now", do you mean to say both the frames of references (terra's Earth and stella's ship) have to agree on what is "now"?
i.e. they have to be at same speed/velocity?

- the photons that hit Stella's telescope at 1:08 pm would show 1 pm?, however there is some point you are trying to make, what is it? Stella's clock would say something like 12:30 pm?

note: the 1:08 pm is per the clock's time on earth/sun/Terra (or any other object that is "stationary" with respect to earth/sun) and not on Stella's ship

 

[tiny-tim]

hi jfy4!

… your usage of the word "actually". It seems to present the idea that something is a particular way objectively.

i think we're basically in agreement

as a solid life-form, i have a prejudice in favour of rigidity (or form) …

i consider that for something to be real, it must at least be rigid, it must have a particular shape …

my prejudice tells me that a square rigid object must still be rigid when it is moving (even if it is no longer square) …

but the Lorentz-Fitzgerald contraction makes it a rectangle, and the shorter side changes if you rotate it 90° …

that's not rigid!​

…

so i say that the contraction is an illusion, and the original square is the objective reality

 

[San K]

people who are further away look smaller

if A is a long way from B, then A says "B looks smaller", and B says "A looks smaller" …

where's the paradox??​

there would only be a paradox if A was standing next to B (or if they were moving past each other, but they make their measurements at the exact moment they are passing) …

then it is paradoxical for each to regard the other as shorter

(of course, that paradox does exist … it's called the Lorentz-Fitzgerald contraction! )

similarly, there's no clock paradox unless the two clocks start at the same velocity, and finish at the same velocity (position doesn't matter)

don't they have to simply finish at the same velocity? ...start at same velocity not required?

 

[tiny-tim]

start at same velocity not required?

required

 

[San K]

required

you mean...same time duration and same velocity at start and stop?...got it

because this would mean same time dilation (degree and duration combo) etc

a requirement for things to be symmetric (a term that I newly discovered here)?

 

[Dale]

as a solid life-form, i have a prejudice in favour of rigidity (or form) …

i consider that for something to be real, it must at least be rigid, it must have a particular shape …

so i say that the contraction is an illusion, and the original square is the objective reality

This is a terrible argument. By this argument liquids and gasses are not real. In fact, a person who loses a limb or even goes on a diet is not real. Argument by prejudice is never valid, but this is a particularly bad one.

Also, illusions are not measurable. They are when your senses, particularly vision, disagree with what is measurable. So length contraction is not an illusion.

 

[Dale]

Now if we take photos according to a third stationary (or for that matter the stationary "earth") clock say at times...noon, 1 pm, 2 pm...etc i.e. per the Earth's clock time and for now ignore the clock time on the "moving" ship.

a) what would we notice about the photos?

if we take photos per the "moving" ships clock and ignore the "stationary" Earth's clock time

b) what would we notice about the photos?

There is too much left unspecified to answer this directly. Basically, to determine the age of something, or the reading on a clock, you simply calculate the proper time.

http://en.wikipedia.org/wiki/Proper_time

The proper time is invariant, so you can calculate it in any frame and all frames will agree.

 

[San K]

Let's take a new case (please see the analysis below and feel free to suggest modification):

A rocket moving at a constant velocity of, say, 80% speed of light between Earth and sun, towards the sun.

on Earth Mr. E is sitting with his clock
on Sun Mr S is sitting with his clock
on rocket Mr R is sitting with his clock

R says to both S and E (separately) that you are the ones that are moving with .8c

R says that E is moving away, from it, at .8c and S is moving towards it at .8c.

Thus from R's perspective/Frame-of-reference the clocks on E and S are moving slower.

E and S agree that the distance between them is constant and neither is moving.

Correction: E and S agree that the distance between them is constant, however they cannot say, for sure, that neither is moving. All they can say is either both are moving at same velocity or both are stationary. i.e. the relative velocity between them is zero.

Now all three (E, R and S) are correct because they are in different frames of references that are not comparable.

If apple says I am more redder than the orange is oranger ...how do you compare?

To compare you have to bring them to the same speed/velocity (plus you have to know the earlier conditions speed etc to make exact calculations on time dilation).

When we bring them to the same speed and do the time dilation calculations all three E, R and S will agree and there would not be any contradictions/paradox.

 

[Dale]

how do you compare?

Using the Lorentz transform:
http://en.wikipedia.org/wiki/Lorentz_transformation

At this point, you should probably take some time to learn the basics of the Lorentz thransformation and also invariants such as the spacetime interval. Your recent questions seem more like random flailing around than a directed and purposeful line of questioning.

 

[Hurkyl]

the photons that hit Stella's telescope at 1:08 pm would show 1 pm?

If Stella didn't accelerate much, so she was moving "slowly" relative to Terra, then yes, if Stella is near the sun when her clock reads 1:08 PM, her telescope will see photons from Earth that read 1:00 PM, and she would measure the Earth-Sun distance as being 8 light-minutes.

"Slowly" means that all numbers are small enough that Newton mechanics still holds approximately. So you can still pretend there is an absolute notion of time and of simultaneity.


However, if Stella took off at a somewhat faster velocity, things would be different. At a certain rapidity, if she takes off at noon, she would find her clock reads 12:02:00 when she passes the Sun. At this instant, her inertial reference frame reckons that Earth's clock reads 12:00:28. Through her telescope, she sees a reading of 12:00:14 on Earth's clock.

However, the Terra's inertial reference frame reckons that Stella passes the sun when Earth's clock reads 12:08:33. (Of course, it agrees that Stella's clock reads 12:02:00 when it happens)


If Stella were to take a picture when her clock reads 12:02:00, how does she label the picture?

• This is a picture of me at 12:02:00
• This is a picture of me at 12:00:14
• This is a picture of me at 12:00:28
• This is a picture of me at 12:08:33
• Something else?

, however there is some point you are trying to make, what is it?

That it makes no sense to say things like "do something when Earth's clock says noon" unless you are actually on Earth. You need to include some way of deciding precisely when Earth's clock says noon.