Comparing Time in Different Galaxies: A Dilemma?

[nucleartear]

Sorry if this doubt has been posted anytime before, but in a little view I have not seen it! (well, I have not make an exhaustive search, it could take long time!). I think this doubt must be easy to resolve:

So, we have two galaxies, A and B, and they are moving... If we are at A, we can say that B is moving with respect to us with a determinate velocity, so, the time at B goes slower than at A (let me know if I'm wrong!)

But the people at B can say the same: the time at A goes slower than at B!

And a C galaxy could be in a way that A and B have the same velocity, and say that the time at A and at B goes equal...

If there are not a "better" reference, how can we determinate the relation between "velocity" of time on each galaxy??

Thanx to all!

 

[Janus]

The upshot is that you can't. All you can say is that from A, B's time is measured as moving slow, and that from B, A's time is measured as slow, and that from C, both A's and B's times are measured as moving slow at the same rate.

Here's an analogy.

You have two men standing some distance from each other.
From man A's position, man B looks smaller.
From man B's position, man A looks smaller.

And from the position of a third man an equal distance from A and B, both A and B look the same size.

It is meaningless to ask which man "really" has a smaller apparent size than the other in this context.

 

[nucleartear]

I understand what u say! but seems like we still have a problem:

If those three people were born at the same moment at the same place (call it D), equidistant from A, B and C, and each one goes to each galaxy at the same moment (moment at D), and some time after, they come back to D at the same moment (at D), they will have differents ages, because the time goes different, isn't it?...

Surely there is an error there, but I can't find it...

 

[C.Kelly]

No error I am afraid. The thing is, your gedanken experiment has a flaw: An object can never move from one place to another instaneously, as that would involve moving faster than c.

Your paradox is the same as the classic twin paradox. If one twin heads off to some distant place and the other twin stays at home, when the rocket twin returns they will be different ages. IF the rocket twin could instantaneously assume its traveling velocity, and when it gets there, instantaneously gain a velocity back in the other direction, there would be no problem; the twins would return the same age and the simpler time dilation formula holds true. It remains true because of the symmetry of the argument. However in real life, the rocket twin has to undergo acceleration, both to get up to its initial speed, and in turning round. If the Earth twin looks at the rocket twin, the acceleration it sees the rocket undergo will, by relativity, not be the mirror of the acceleration that the rocket twin observes for the Earth, and the symmetry is broken. The rocket twin will then return to Earth some years younger than their twin.

This has actually been proven. They put a highly accurate atomic clock on a round the world trip, and the centripetal acceleration caused the clocks to show different times when they were again in the same reference frame.

I hope this sorts the problem out. Relativity seems weird, but all you have to remember is that it is always internally consistent, and the weirdness only steps in when you try to look at the events in a different reference frame.

Chris

 

[Hurkyl]

IF the rocket twin could instantaneously assume its traveling velocity, and when it gets there, instantaneously gain a velocity back in the other direction, there would be no problem; the twins would return the same age and the simpler time dilation formula holds true. It remains true because of the symmetry of the argument.

It's still not symmetric; one twin underwent an instantaneous acceleration, and the other twin did not. The rocket twin will again be younger.

If you insist on looking at things from the rocket twin's POV, the instantaneous acceleration is equivalent to changing one's reference frame... which includes changing the time coordinate of distant places. (see the lorentz transformations!)

 

[nucleartear]

Originally posted by C.Kelly
No error I am afraid. The thing is, your gedanken experiment has a flaw: An object can never move from one place to another instaneously, as that would involve moving faster than c. [...]

I'm agree with you! But I don't trying to say they move instantaneously! Thanx anyway!

We have theese three galaxies, A, B and C, where A and B is moving at the same velocity at the C reference...

So:

- If we are at A, B is moving, so the time at B goes slower than at A.
- If we are at B, A is moving, so the time at A goes slower than at B.
- If we are at C, A and B are moving with the same velocity, so the time at A and at B goes equal.

If there is not a "privileged" reference, how can we know what's the real relation between time-velocities??

So, if we have three people (a-man, b-man and c-man) who were born at the same time at the same point (call it D) and if they start a trip towards the universe at the same moment (D-moment), where the a-man goes to A and spends the same time that b-man spends to go to B and the same time that c-man spends to go to C... If they stay some time at their galaxies, and go back and spend the same time to arrive to D, What will be the relation between ages?

- a-age > c-age > b-age ?
- b-age > c-age > a-age ? or
- c-age > a-age = b-age ?

I hope to have explained it better this time!

So, if there are someone who can tell me an explanation, or find an error, I'll be hightly gratefull!

 

[Hurkyl]

In terms of understanding why the paradigm shift must be made, and to solve some simple problems, length contraction and time dilation are great tools... but actually learning Minowski geometry is, IMHO, the best way to understand special relativity.

Proper time τ and proper distance s are the physical relevant quantities. One computes the proper time elapsed over a period of constant velocity motion by the formula:

(Δτ)² = (Δt)² - (Δx / c)²

Where Δτ is the interval of proper time experienced during the trip, Δt is the change in coordinate time as measured in a particular inertial reference frame, and Δx is the change in coordinate position as measured in the same reference frame.


For example, suppose we choose an inertial reference frame where point D is always at the position x = 0. (i.e. D's frame of reference). A traveller leaves D at time t = 0, traveling at 0.5c eastward and arrives at galaxy A one year later, meaning he is at position x = 0.5 light-years and t = 1 year. The amount of proper time experienced by the traveller is then

Δτ² = (1 year)² - (0.5 light-years / c)² = 1 year² - 0.25 year² = 0.75 year²

so Δτ = [squ](0.75 year²)= 0.866 years

If you have the coordinates (both time and position) of all the events in your scenario, as measured in a fixed frame of reference, you can use this formula to compute the proper time experienced by each traveller, and thus their age.

If they stay some time at their galaxies, and go back and spend the same time to arrive to D, What will be the relation between ages?

- a-age > c-age > b-age ?
- c-age > c-age > a-age ? or
- c-age > a-age = b-age ?

There isn't enough information to answer this problem; The first and the third could both be solutions depending on the details. (The second can't as typed, but fix the typo and it could be as well)

 

[nucleartear]

Originally posted by Hurkyl
In terms of understanding why the paradigm shift must be made, and to solve some simple problems, length contraction and time dilation are great tools... [...] ...and thus their age.

quote:
If they stay some time at their galaxies, and go back and spend the same time to arrive to D, What will be the relation between ages?

- a-age > c-age > b-age ?
- c-age > c-age > a-age ? or
- c-age > a-age = b-age ?

There isn't enough information to answer this problem; The first and the third could both be solutions depending on the details. (The second can't as typed, but fix the typo and it could be as well)

Sorry!, it must be:

- a-age > c-age > b-age ?
- b-age > c-age > a-age ? or
- c-age > a-age = b-age ?

I have edit it in the other one!

So, What's the information we need? What are the details?

If we can't know this, that means that we can't know the kinetik energy, either... not even the total energy... Do u really think so? We can't have much more information in this way if we see at the sky... is that mean that we can't know all those things of the universe?

(Thanx for the other information!)

 

[Hurkyl]

One way to specify everything is to list the trajectories of the galaxies and the times when the travellers leave D, arrive at their galaxy, leave their galaxy, and arrive at D (all measured in D's reference frame).



And yes, kinetic energy depends on your choice of coordinates. If we restrict our attention to any particular reference frame, we have conservation of energy, but the actual total quantity of energy will generally change if we change reference frames.


And if we delve into General Relativity, measuring anything is tricky business... it's astonishing how deeply our procedures for measuring things depend on the peculiar "niceness" of Euclidean geometry and thus fail in a more general context.

 

[zoobyshoe]

Janus,

My limited understanding of Rela-
tivity is that Einstein was mere-
ly pointing out that "Moving clocks appear to run slow."Here at the Forums many people
seem to be asserting that moving
clocks actually run slow.Your analogy of the three men
standing at a distance from each
other leads me to believe that
your understanding of Einstein
is more in line with my own.

It seems to me that in order for
A to view B's clock as running
slow at the same time B views A's
clock as running slow when they
are moving relative to each other
it is absolutely necessary that
their clocks actually be
running the same.

If A views B as about as tall as
his thumb, and B actually is as tall as A's thumb
then it will not be possible for
B to mistake A as being as tall as
his (B's) thumb. If B is as tall
as A's thumb A will look very
large to B.

Any thoughts?

-Zooby

 

[Sonty]

Originally posted by zoobyshoe

If A views B as about as tall as
his thumb, and B actually is as tall as A's thumb
then it will not be possible for
B to mistake A as being as tall as
his (B's) thumb. If B is as tall
as A's thumb A will look very
large to B.

For B to be exactly as tall as A measures him, B would have to be exactly in front of A, meaning no "speed" between them so B would measure A to be very large indeed, but doesn't this beat the purpose of a relativity problem?

I myself find only one answer to this problem (and many others where you try to put classical logic into relativity): "There is NO spoon".

 

[zoobyshoe]

Sonty,

No, you missed the point.

 

[nucleartear]

Originally posted by Hurkyl
One way to specify everything is to list the trajectories of the galaxies and the times when the travellers leave D, arrive at their galaxy, leave their galaxy, and arrive at D (all measured in D's reference frame).



And yes, kinetic energy depends on your choice of coordinates. If we restrict our attention to any particular reference frame, we have conservation of energy, but the actual total quantity of energy will generally change if we change reference frames.


And if we delve into General Relativity, measuring anything is tricky business... it's astonishing how deeply our procedures for measuring things depend on the peculiar "niceness" of Euclidean geometry and thus fail in a more general context.

So, I was thinking in it, and, if we see at the experiments we have:

- The satellites orbit around the Earth with a big velocity respect to the Earth, and their time goes slower than at the Earth (the GPS system needs to coordinate the clocks of the satellites every time!)

- The experiment of the two planes crossing the Earth, one from West to East and the another one from East to West. The clocks on the two planes had different time, because one of them had more velocity respect to the Sun (even they had the same velocity respect to the Earth).

That minds that:
- respect to the Earth, the satellites of the Earth have all the kinetik energy.
- respect to the Sun, the satellites of the Earth and the Earth have all the kinetik energy, where "satellites Ek" > "Earth Ek"
- and (that's a bit speculative) the Solar System has all the kinetik energy respect to the center of the Milky Way.

So, that minds that the best reference is who has more inertia?? (or more mass, or whatever).

Because we can say that the satellites have the K energy... that minds that it doesn't depend on the reference!(let me know if I'm wrong)... If we try to make the same orbit-problem with the satellite as the central-force and the Earth with the Kinetik and the Potencial energy, it could be a bit dificult...

So, back to the problem... Could the right answer be knowed if we know the relation of masses? (It's not a rethoric question, I really don't know it!)

... the times when the travellers leave D, arrive at their galaxy, leave their galaxy, and arrive at D (all measured in D's reference frame).

Oh, don't worry about that! we don't need it to know the relation between ages, we only meed the relation between the time-velocity at the galaxies! the story of the three men is only to understand it better, but I can say: "If I am God, and I can know everything in my universe, what would the relation between time be in the three galaxies?"

And, about the trayectory, if we are at A, we can have them in the same line in the order A-C-B where C has a velocity of V and B a velocity of 2V...

 

[Sonty]

Originally posted by zoobyshoe

It seems to me that in order for
A to view B's clock as running
slow at the same time B views A's
clock as running slow when they
are moving relative to each other
it is absolutely necessary that
their clocks actually be
running the same.

There is supposed to only one way to measure time for both A and B. If you were a god and could leap from one system to another you could measure the same amount of time for the coffee to boil. Is this the point?

 

[nucleartear]

To Janus, zoobyshoe and Sonty:

That point of view is very interesting and makes me thinking...

You actually say that the time flows at the same velocity, but if we make a mesure of the time in the moving-object while we are at the static-reference we will detect it flows slower at the moving-object... (let me know if I'm wrong)

...but, if it's thru, the rocket twin of the famous paradox would have the same age as his brother!

And the clocks of GPS satellites wouldn't need to be checked, and the celsius-clocks of the two-planes-experiment would have the same time, but one of them was delayed...

 

[nucleartear]

Originally posted by nucleartear
[...]

So, that minds that the best reference is who has more inertia?? (or more mass, or whatever).

[...]

So, back to the problem... Could the right answer be knowed if we know the relation of masses? (It's not a rethoric question, I really don't know it!)

Well, I'm still thinking on it... and maybe the "natural" reference is the center of masses (I don't know if that's the word in english, u know: Rmc = [R1·m1 + R2·m2 +...+ Rn·mn] / [m1 + m2 +...+mn], where R is de vector-position and m the mass)

Can anyone tell me if that could be right? Oh, please, I try to understand all these puzzles but it's so hard!

 

[Janus]

Originally posted by nucleartear
To Janus, zoobyshoe and Sonty:

That point of view is very interesting and makes me thinking...

You actually say that the time flows at the same velocity, but if we make a measure of the time in the moving-object while we are at the static-reference we will detect it flows slower at the moving-object... (let me know if I'm wrong)

No, because that would require some external point of view by which time rate can be measured independent of any frame of reference. There isn't one.

If A see's B's clock moving more slowly, then from A's frame B's clock really does run slowly. Conversely, the same is true for B.

To go back to the men separated by distance: Each man measures the other as having a smaller angular measure. This is a real measurement.

Time rate and velocity bear a simular relationship as distance and angular measure.

...but, if it's thru, the rocket twin of the famous paradox would have the same age as his brother!

With the Twin paradox you have to consider more than just the time dilation due to constant motion.

For instance, you have to consider length dilation. The distance traveled will be measured differently by each twin. Thus the Earth twin mght say that the other twin traveled out to a distance of 10 ly at .5c and took 40 yrs to return, but since the other twin's clock was running slow it would only measure 34.641 yrs upon return.

The other twin would say that the maximum separation between the two was only 8.66 ly , and thus at .5c it was only 34.641 years until they were reunited.

Thus both twins agree that 34.641 years was measured by the "traveling" clock.

The only argument that could arise is over how much time passes on the Earth clock. The Earth twin says 40 yrs, and at first glance, it appears that the other twin would say that only 30 yrs should have passed on Earth.

But this doesn't take into account the effects of acceleration has on the "traveling" twin.

When he takes off from Earth, he will feel himself pushed back towards the rear of the his ship. Under these conditions SR says that he will measure an additional, time dilation effect On the Earth clock. The degree of this effect is determined by the rate of acceleration and the distance to the Earth. since the distance to the Earth is small at this time is effect is small.

At this point of the trip this dialtion willl make the Earth clock run even slower.

Once the ship cuts its engines, only the regular constrant velocity dilation is measured.

The ship nears the turn around point, and begins to brake. It is now feeling an acceleration in the opposite direction from before. This causes a time dilation on Earth's clock which causes it to speed up its time rate. And since the distance to the Earth is now large, the Earth clock will run really fast.

This continues until the Relative velocity between the Earth and Ship is such that the two are approaching each other again.

The engines cut off and we're back to the SR dilation with the Earth clock running slow again.

Once back near Earth, the engines are used to brake the ship and you get hte same time dilation as you did when you left.

So to recap. The Earth twin will see his brother's clock as running slow the whole duration of the trip, resulting is less time expiring on his brother's clock.

The ship twin will see the Earth clock as running slow, then running really fast, and then running slow, with the really fast part over shadowing the slow part, so he will measure more time expiring on the Earth clock.

Both twins will agree as to who aged more and by how much, they will just disagree as to how it came about. both twins have an equally valid claim in this respect.

This has nothing to do with which twin was 'really' moving. The same thing will happen even if you consider both twins as moving at .5c (say with respect to the Earth) to start with, and one twin decelerates to a stop (wrt Earth) and then accelerates again, catches up with his brother, and matches speeds again. While his brother just continues on at .5c wrt Earth.

The twin who does all the manuveuring will be the one that ages less; And this would be true as far as someone watching from the Earth is concerned as well.

All observers moving at any speed will agree on the end results, they just won't agree as to how the results came to be.

 

[zoobyshoe]

Originally posted by Janus
No, because that would require some external point of view by which time rate can be measured independent of any frame of reference. There isn't one.

If A see's B's clock moving more slowly, then from A's frame B's clock really does run slowly. Conversely, the same is true for B.

It is demonstrated, then, that
each will see the other's clock
as running slow. Something that
hasn't been brought up is the
fact that they are checking each
others time by sending light sig-
nals back and forth.

It seems to me that ths reciprocal
perception of the other's clock
as being slow can only arise when
both clocks are, in fact, running
at the same rate.

How would the perceptions change
if we deliberately gave one a
clock that ran at a slower rate
to begin with?

 

[Janus]

Originally posted by zoobyshoe

Originally posted by Janus
No, because that would require some external point of view by which time rate can be measured independent of any frame of reference. There isn't one.

If A see's B's clock moving more slowly, then from A's frame B's clock really does run slowly. Conversely, the same is true for B.

It is demonstrated, then, that
each will see the other's clock
as running slow. Something that
hasn't been brought up is the
fact that they are checking each
others time by sending light sig-
nals back and forth.

It seems to me that ths reciprocal
perception of the other's clock
as being slow can only arise when
both clocks are, in fact, running
at the same rate.

How would the perceptions change
if we deliberately gave one a
clock that ran at a slower rate
to begin with?

Whenever one talks about time rates in Relativity, one means as measured by clocks of identical construction or by indentical physical processes. IOW, clocks that, if set side by side, in the same frame of reference, would keep indentical time.

Throwing in clocks that run at different rates under these conditions doesn't add anything to the discussion.

Having said that, the only time the clocks can be said to be running at the same speed is when they are in the same frame of reference.

 

[nucleartear]

Originally posted by Janus
No, because that would require some external point of view by which time rate can be measured [...]

Thanx for the detailed explanation of the twin-paradox! I have saved a copy of your text in my computer!

The twin who does all the manuveuring will be the one that ages less; And this would be true as far as someone watching from the Earth is concerned as well.

So, one of them will be younger and it doesn't depend on the reference. It only depends on who has received the forces. (let me know if I have understood it badly)

So, in the three-galaxies-problem, we need to know which forces are in scene to know who is the younger...

but... if is there no forces? Let me explain the same problem with a different way:

There are three galaxies (A, B and C) in a way that A and B have the same velocity if we use the C-reference. We can put them at the same line in the order A-C-B where C has velocity of v and B has velocity of 2v if we use the A-reference. The three galaxies are so far enought to don't receive a significant gravity force.

We are at a point called D at a perpendicular line, so far enought to say the three galaxies are at the same distance:

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

We send three twin brothers (a, b and c) with the same velocity, so, with our super-telescope, we will see they arrives to each galaxy (a to A, b to B and c to C)at the same moment.

Some time after, we use our super-telescope to see the galaxies and we see that the three brothers start the trip back to D at the same moment, call it a casuality (the galaxies are almost at the same distance, so the light arrives to D at the same moment). They spend the same time to arrive to D (they trip at the same velocity).

The unic forces involved are used to make the trip, so the three brothers have been put under the same forces, so we can't use them to try to find the solution.

So, who is the younger?? We have three possibilities:

A is not moving respect to D, so: a-age > c-age > b-age
B is not moving respect to D, so: b-age > c-age > a-age
C is not moving respect to D, so: c-age > a-age = b-age

But the solution must be unic! If we make this experiment in the real life, the solution is unic! but, which of them? How can we know it?

Well, this problem has three-level-liberty, so we need something else to know the answer...

What is what we need?

Thanx!

 

[Janus]

Originally posted by nucleartear
Thanx for the detailed explanation of the twin-paradox! I have saved a copy of your text in my computer!



So, one of them will be younger and it doesn't depend on the reference. It only depends on who has received the forces. (let me know if I have understood it badly)

So, in the three-galaxies-problem, we need to know which forces are in scene to know who is the younger...

but... if is there no forces? Let me explain the same problem with a different way:

There are three galaxies (A, B and C) in a way that A and B have the same velocity if we use the C-reference. We can put them at the same line in the order A-C-B where C has velocity of v and B has velocity of 2v if we use the A-reference. The three galaxies are so far enought to don't receive a significant gravity force.

We are at a point called D at a perpendicular line, so far enought to say the three galaxies are at the same distance:

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

We send three twin brothers (a, b and c) with the same velocity, so, with our super-telescope, we will see they arrives to each galaxy (a to A, b to B and c to C)at the same moment.

Some time after, we use our super-telescope to see the galaxies and we see that the three brothers start the trip back to D at the same moment, call it a casuality (the galaxies are almost at the same distance, so the light arrives to D at the same moment). They spend the same time to arrive to D (they trip at the same velocity).

The unic forces involved are used to make the trip, so the three brothers have been put under the same forces, so we can't use them to try to find the solution.

So, who is the younger?? We have three possibilities:

A is not moving respect to D, so: a-age > c-age > b-age
B is not moving respect to D, so: b-age > c-age > a-age
C is not moving respect to D, so: c-age > a-age = b-age

But the solution must be unic! If we make this experiment in the real life, the solution is unic! but, which of them? How can we know it?

Well, this problem has three-level-liberty, so we need something else to know the answer...

What is what we need?

Thanx!

There will always be forces involved if you are changing sometihing's velocity(which you are, by having the twins travel to the galaxies and back. )

If D is at rest with respect to C(or has a velcocity along the line joining D and C), then under this situation, twin's A and B will remain the same age while aging less than C.

 

[nucleartear]

There will always be forces involved if you are changing sometihing's velocity(which you are, by having the twins travel to the galaxies and back. )

Yes, for this reason I say: "The unic forces involved are used to make the trip, so the three brothers have been put under the same forces, so we can't use them to try to find the solution."

If D is at rest with respect to C(or has a velcocity along the line joining D and C), then under this situation, twin's A and B will remain the same age while aging less than C.

Yes, I have considered this situation and the other two possible situations:

A is not moving respect to D, so: a-age > c-age > b-age
B is not moving respect to D, so: b-age > c-age > a-age
C is not moving respect to D, so: c-age > a-age = b-age

If there is not a privileged reference, how can we know the real age-relation??

 

[Janus]

Originally posted by nucleartear
Yes, for this reason I say: "The unic forces involved are used to make the trip, so the three brothers have been put under the same forces, so we can't use them to try to find the solution."

The three brothers can't be "put under the same forces". If A and B are both moving away from C, then twins A and B have to take paths that will cause them to intercept their respective galaxies.

Even if you start from a position such that A B &C are for all intents and purposes the same distance from D, They won't be by the time the twins arrive. Since the twins must travel at a finite velocity, it will take time for them to travel the distance they must, and this allow galaxies A & B to move farther from.

No matter how you cut it, twins A and B have to travel longer distances, and in order for them to arrive at their respective galaxies at the same time as twin C arrives at his, they have to travel faster and experience greater forces during acceleration.

Yes, I have considered this situation and the other two possible situations:

A is not moving respect to D, so: a-age > c-age > b-age
B is not moving respect to D, so: b-age > c-age > a-age
C is not moving respect to D, so: c-age > a-age = b-age

If there is not a privileged reference, how can we know the real age-relation??

If D is motionless with respect to different galaxies, then the forces felt by each twin will be different during their trips, and you will get different age relations.

Different circumstances give different results. But no matter what the initial conditions, everyone involved will agree as to the end result (observers at D, A, B, C and all three twins) though they may not agree as how the end result came to be.

There is no need to invoke a preferred frame of reference.

 

[nucleartear]

Originally posted by Janus
The three brothers can't be "put under the same forces". If A and B are both moving away from C, then twins A and B have to take paths that will cause them to intercept their respective galaxies.

Even if you start from a position such that A B &C are for all intents and purposes the same distance from D, They won't be by the time the twins arrive. Since the twins must travel at a finite velocity, it will take time for them to travel the distance they must, and this allow galaxies A & B to move farther from.

No matter how you cut it, twins A and B have to travel longer distances, and in order for them to arrive at their respective galaxies at the same time as twin C arrives at his, they have to travel faster and experience greater forces during acceleration.




If D is motionless with respect to different galaxies, then the forces felt by each twin will be different during their trips, and you will get different age relations.

Different circumstances give different results. But no matter what the initial conditions, everyone involved will agree as to the end result (observers at D, A, B, C and all three twins) though they may not agree as how the end result came to be.

There is no need to invoke a preferred frame of reference.

Thanx for your reply!

But... (please, sorry if I'm bothering to you!) ...I'm trying to find the relation between A, B and C, so I'm trying to make a mental-experiment where the trip towards the universe doesn't create any problem...

You are analyzing the differences between time-rate created in the travel, but I just know how it is... what I don't know is the differences between time-rate they can experiment when they are at A, B and C, and back to D.

So, ok, there are differences at the distance. We'll see with our super-telescope that they don't arrive at the same moment...

Imagine that they have a chronometer (a-chronometer, b-chronometer and c-chronometer) and they enable it *only* when they arrive to their galaxy, and disable it when they leave the galaxy.

We can imagine that (for example) the first one to arrive at the galaxy is the a-brother, and we enable our chronometer (d-chronometer). After a time t_1 b-brother arrives. After a time t_2 c-brother arrives. We disable the chronometer and resert it.

Some time later, we see with our super-telescope that the a-brother start the trip back to D, and we enable the chronometer. Exactly after a time t_1, we see the b_brother start his trip, and after a time t_2, the c-brother do the same. So, for us at D, they have rested at the galaxies the same time.

When they arrive to D (in a different order, it doesn't mind now) and we see the chronometers, which will mark more time, if the velocity-relations of the galaxies are the same exposed in the other post?

A is not moving respect to D,
so time in: a-chronometer > c-chronometer > b-chronometer
B is not moving respect to D,
so time in: b-chronometer > c-chronometer > a-chronometer
C is not moving respect to D,
so time in: c-chronometer > a-chronometer = b-chronometer

Thanx again for your time!

 

[Janus]

Originally posted by nucleartear
Thanx for your reply!

But... (please, sorry if I'm bothering to you!) ...I'm trying to find the relation between A, B and C, so I'm trying to make a mental-experiment where the trip towards the universe doesn't create any problem...

You are analyzing the differences between time-rate created in the travel, but I just know how it is... what I don't know is the differences between time-rate they can experiment when they are at A, B and C, and back to D.

So, ok, there are differences at the distance. We'll see with our super-telescope that they don't arrive at the same moment...

Imagine that they have a chronometer (a-chronometer, b-chronometer and c-chronometer) and they enable it *only* when they arrive to their galaxy, and disable it when they leave the galaxy.

We can imagine that (for example) the first one to arrive at the galaxy is the a-brother, and we enable our chronometer (d-chronometer). After a time t_1 b-brother arrives. After a time t_2 c-brother arrives. We disable the chronometer and resert it.

Some time later, we see with our super-telescope that the a-brother start the trip back to D, and we enable the chronometer. Exactly after a time t_1, we see the b_brother start his trip, and after a time t_2, the c-brother do the same. So, for us at D, they have rested at the galaxies the same time.

When they arrive to D (in a different order, it doesn't mind now) and we see the chronometers, which will mark more time, if the velocity-relations of the galaxies are the same exposed in the other post?

A is not moving respect to D,
so time in: a-chronometer > c-chronometer > b-chronometer
B is not moving respect to D,
so time in: b-chronometer > c-chronometer > a-chronometer
C is not moving respect to D,
so time in: c-chronometer > a-chronometer = b-chronometer

Thanx again for your time!

I don't really see what this example is intended to show, other than that the time rates of each galaxy depend upon which frame they are measured from, which is exactly what Relativity says.

For each circumstance given (D being at rest with different galaxies) you will get different end results, but also in each case, all observers will agree with those end results.

You seem to be searching for some absolute way to determine which galaxy's time rate is "really" moving slower. There isn't any. In fact, it is a meaningless question.

 

[nucleartear]

Oh! I see where is the problem with the example...

Thanx! I thought that we could make a kind of time-rate-map if we know all the objects and their velocity at the scene... and it would depend on the reference as we designated the zero-point in a metric... I see now that was not right...

And, in the example, now I see the motion-relation of D with the other galaxies must be knowed because it's unic... I have missed that point...

Lots of thanks for your help! I think I understand it now a little better!

 

[wakeball32]

To Nucleartear

Hi,
I am not a physics genius who has a degree specializing in relativity or anything, I just love the subject. Anyway, dealing with special relativity, just remember that nothing is absolute (i.e. time, velocity, etc,) because as your 1st post shows, all measurements are different with respects to different galaxies, hence none of them is absolute. Also, there is no such thing as simultaneity. Thats just some basics.

 

[yogi]

SR problems always lead to much debate and further questions - there are several explanations - Einstein took the position in his original publication that the turn around twin experinced acceleration and therefore the problem was not one that could be solved by SR per se - but there are other views and many proposed tests - the many tests that have verified SR also confirm Lorentz Ether Theory (actual time dilation and actual length contraction). These tests also cannot distinguish between one of the latest explanations of MMx experiment that proposes that the one way speed of light is constant in the Earth centered system because the Earth's gravity modifies the local space (this is not an either drag hypothesis) and is distinquishable therefrom because distant sources such as the CBR and star light aberration will be detectable even though local over and back and one way light transmission will be isotropic. Finally, the notion that the twin problem gets explained because the turn around twin experiences forces - fails in the triplet thought experiment where there is no turn around - the outbound twin simply transfers his clock reading to an inbound brother who left years earlier - and this guy continues on back to Earth with the info from the outbound brother - but the two clocks back on Earth do not read the same. arrives back at the Earth in th

 

[Brad_Ad23]

Yogi, please read what you are saying. Any version of the twin situation, or triplet, or quadruplet situation matches up well. Why? Two situations.

1) Transfer of a physical clock. The outbound twin transfers the clock to the inbound twin. Seems nice and clean there for you doesn't it? But if I recall, is not a change in direction (remember velocity is a vector) an acceleration? Why yes it is. And wound not the clock experience quite a large acceleration when being transferred at such high velocities? Why yes, yes it would.

2)A signal is sent. This one is easier because the signal will be automatically shifted one way or the other depending on when it is sent. Remember even though c is constant, it will take some time for it to catch up with an object traveling a good fraction of c.

 

[Janus]

Originally posted by yogi
Finally, the notion that the twin problem gets explained because the turn around twin experiences forces - fails in the triplet thought experiment where there is no turn around - the outbound twin simply transfers his clock reading to an inbound brother who left years earlier - and this guy continues on back to Earth with the info from the outbound brother - but the two clocks back on Earth do not read the same. arrives back at the Earth in th

This scenerio changes nothing, As it doesn't address the supposed "paradox" of the "Twin paradox"

If twin 1 stays on Earth and watches twin 2 travel at .866c for a distance of .866 ly, where he meets twin 3 inbound, he will say that 1 yr will have passed on his clock, while only .5 yr will have passed on twin 2's clock, due to time dilation.

Twin 2, upon meeting twin 3, will also say that .5 yr has passed on his clock. Because of length contraction, the distance between him and twin 1 at the time of meeting twin 3 is only .433 ly.

Thus all three twins agree as to the time shown on twin 2's clock when he transfers his clock reading to twin 3 and there is no paradox.

The "paradox" arises from the fact that Twin 2 should also see only .25 yr as having passed on Earth when he meets twin Three.

But this doesn't lead to a problem, either, as the only problem would be if twin 2 disagreed as to how much time had passed for twin 1 when twin 3 arrives with the time reading.

Assuming that Twin 3 is moving inward at .886 wrt to Twin 1.

That means according to Twin 1, 2 yrs will pass from the departure of Twin 1 to the Arrival of Twin 3.

Now the velocity of twin 3 to Twin 2 as measured by twin two is

(.866c+.866c)/(1+(.866c)²) = .9897c

This means, that measured by twin 2, twin 3's velocity wrt Twin 1 is

.9897c - .9=866c = .1237 c

Twin 3 was .433 ly from twin 1 when they met, so by twin 2's clock, it would take 3.499 years. Due to time dialtion, twin 2 would measure 1/2 this time as passing for twin 1 or 1.75 yr. add this to the .25 yrs, at the time of meeting, and twin 2 will say that 2 years will have passed for Twin 1, the same time that twin 1 measures. Again, no paradox.

The only other way to force a paradox is to have twin 1 watching the Twin 1 the whole time, and bring him back to the same frame as Twin 1 so that he can physically compare the time passed for twin 1 to the Time he Saw passing for twin1. On the outbound and Inbound trips, he would see less time pass for twin 1 than on his own clock. But the only way for there to be a paradox is if the two are brought back into the same frame and compared, (thus twin's 1 and 2 not moving relative to each other, are looking at the same clocks and both are seeing different readings, or each sees the other twin as being younger.)

But the only way for this to happen is for one twin to experience an acceleration, and this acceleration alters how he sees time pass on the the other twin's clock. It will cause him to see time as passing faster for his brother during this time of acceleration, to the extent that once the brother's are brought back together, he will agree that he aged less than his brother over the entire time period of separation.