Does the Twin Paradox Break Symmetry in the Zig-Zag Scenario?

[PAllen]

Look at this chapter: http://www.bartleby.com/173/11.html . Fig 2 one can see that reference frame K' also moves along with object attached to the origin of it w.r.t K with a velocity v.
.

Note that there is no reference there to changing frames, just a choice to analyze the same scenario in two different inertial frames, demonstrating how the Lorentz transform specifies how to change one set of labels (coordinates) to another. As a matter of example, there happens to be an object of interest at rest in each chosen frame. The use of two frames is not essential to any prediction, and 5 other frames could be introduced for additional comparison.

An example: when you are driving on a winding road you are not required to use a new coordinate frame every moment. If you follow your route along with GPS app, you are modeling your motion in a convenient standard frame (rather than mentally thinking of distant mountains moving superluminally whenever you make a hairpin turn). In this scenario, it is possible to construct a set of coordinates in which the car is is always at the origin, but it is rather complex to do consistently, certainly not necessary for any understanding (and the Lorentz transform would no longer be usable, since that only applies between inertial frames not to general coordinates where e.g. the time axis represents non-inertial motion). It is also possible to use a sequence of different inertial frames for such a problem, in each of which the car is momentarily at rest. But this doesn't tell you anything new about the physics of what happens in the car, and there is certain no significance to the constant relabeling of events you do each time you choose a different inertial frame.

 

[Stephanus]

Sorry if i am disturbing and if i am going off-topic. ...i was just trying to understand. Only when you discuss one knows where went wrong. It makes one better.

No, no, no. You're not disturbing. You even enlarge my question

 

[Mentz114]

I thought frames of reference are for locating position and time at which event happens. If objects are in all frames of reference,then objects measure which time?

That is a good question. Objects measure the time by looking at their clocks (which are at rest with them). Clocks show 'proper time' which is the same in all frames and is given by the formula

$\tau=\int_p \frac{d\tau}{dt} dt$ where the integral is taken along the section of the worldline between the events.

If the path segments are all straight lines this becomes the sum of the segments with each one contributing $\tau=\sqrt{(t_1-t_0)^2-(x_1-x_2)^2}$

 

[Dale]

If objects are in all frames of reference,then objects measure which time?

A correctly functioning clock measures proper time, as pointed out by Mentz114.

 

[PeterDonis]

I thought frames of reference are for locating position and time at which event happens.

It's probably worth pointing out here that, technically, a "frame of reference" and a "coordinate chart" are two different things:

A "coordinate chart" is a mapping of 4-tuples of real numbers to points in spacetime. The numbers don't have to have any physical meaning; the only requirement is that the mapping satisfy certain mathematical properties (for example, continuity--intuitively, points in spacetime which are "close together" should have 4-tuples of coordinates which are "close together"--the mathematical details make all this much more precise).

A "frame of reference" (more properly, a "frame field") is an assignment of a set of four orthonormal vectors, one timelike and three spacelike, to each point of spacetime. These vectors, intuitively, represent the 4-velocity of a clock and three mutually perpendicular "measuring rods" (to use Einstein's term) attached to the clock to define the spatial directions. These vectors, therefore, are supposed to have a direct physical meaning: we can model physical measurements made at a point of spacetime by contracting the frame field vectors at that point with other vectors (or more generally tensors of any applicable rank) to obtain scalars that represent the measurement results. For example, the "energy" of an object at a point of spacetime is the contraction of the object's 4-velocity vector with the frame field timelike vector at that point.

The reason these two terms (coordinate chart and frame of reference) are often used interchangeably, even though this is really sloppy usage, is that, for the case of flat spacetime, there is an obvious correspondence between a standard inertial coordinate chart and a frame of reference defined by the coordinate basis vectors for that chart at each point of spacetime. So the coordinates in this special case can be thought of as directly representing lengths and times as measured using the vectors of the frame of reference. However, this nice correspondence only works in that special case (an inertial coordinate chart in flat spacetime); it does not generalize. That's why it's helpful to understand the more precise definitions I gave above, since each of them does generalize (but in the general case they no longer have a nice simple correspondence with each other).

Therefore the notion that co-ordinate system having the clocks to measure time at which events happen(I think they are called reference frame) are attached to a particular object located in the origin of the reference frame is wrong?

In general, yes. See above.

 

[ash64449]

Thank you all! I think i understand now!

I think i haven't sufficiently understood PeterDonis's #35 post where he explains what frame of reference is. It must be because there are several terms which i haven't been exposed previously.( for example: meaning of timelike and spacelike vectors and they way vectors have physical meaning-contracting the frame field vectors at that point with other vectors)

I completely understand PeterDonis's #19 post, that B is only changing the state of motion w.r.t to A and not changing frames.

Actually changing frames do not have any meaning. All objects are in all frame of reference.( the post that i made do not have any meaning at all!)

While calculating,it is important that you work all the calculation in only one frame. We should not change reference frame for calculation mid-way.

You can describe the entire scenario using just one frame

An example of this was given by ghwellsjr in #5 post of this thread: https://www.physicsforums.com/threa...pace-time-diagram-analysis-resolution.814805/

I would like to point specifically to the part of the post where he used a space-time diagram in which he used rest frame of Stella during the first part of the scenario of Twin paradox being discussed there.

That post helped me to understand that frames are not used in general to understand one's point of view( since Stella is rest(at the origin) in frame being used for the first part). But one can find that during returning part of the journey,in the frame concerned, Stella changed position in that frame.

Reference frame are actually used for calculation purpose only. It may describe one's view if and only if object don't change the state of motion w.r.t it.
Therefore i understood that

co-ordinate system having the clocks to measure time at which events happen(I think they are called reference frame) are attached to a particular object located in the origin of the reference frame

is wrong since by this statement i was referring that reference frame is used for understanding one's point of view. I thought that this reference frame helped the object describe him when event's happen to him. This is totally wrong since this is possible so long as object don't change the reference frame.

As a result of this understanding i obtained the answer for my post:

So conclusion is if object tries to move,reference frame(which object uses to perform calculations)remains there?

 

[ash64449]

If i consider that a reference frame which helps an observer describe 'time' at which event happen(by putting clocks all over the coordinate system),coordinates at which event happen to be his frame, Can't i consider an observer an the origin of the reference frame at rest to be his frame?

If an object moves away from origin with a certain velocity, now observer has changed frames according to my consideration,right? Since now this reference frame doesn't help him describe 'time' at which event happen and coordinates at which an event happen?

 

[PeterDonis]

there are several terms which i haven't been exposed previously.( for example: meaning of timelike and spacelike vectors and they way vectors have physical meaning-contracting the frame field vectors at that point with other vectors)

These ideas are very important; if you don't understand them, it's going to be very difficult for you to make sense of any discussion of frames of reference. I suggest working through Sean Carroll's online lecture notes on relativity:

http://arxiv.org/abs/gr-qc/9712019

Chapter 1, on SR and flat spacetime, introduces the basic concepts.

Can't i consider an observer an the origin of the reference frame at rest to be his frame?

No, because, as you go on to show in your very next sentence, the observer can change his state of motion, but the frame can't. So the observer and the frame can't be identical.

If an object moves away from origin with a certain velocity, now observer has changed frames according to my consideration,right?

No, he's changed his state of motion. He was at rest relative to one frame before; now he's at rest relative to another, different frame.

this reference frame doesn't help him describe 'time' at which event happen and coordinates at which an event happen?

"Time" and "coordinates" are not absolutes; they depend on the coordinate chart you choose. Here you are implicitly choosing a coordinate chart corresponding to the first frame (the chart whose coordinate basis vectors are the frame vectors). The coordinates in this chart do give "times" and "positions" that would be measured by the first frame.

But you could just as easily choose a coordinate chart corresponding to the second frame (the one the observer is at rest relative to after he changes his state of motion). This chart would give "times" and "positions" that would be measured by the second frame, and after the observer changes his state of motion, he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

 

[CycoFin]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

At first these two rules can be seen to be in contradiction with each other but it is capturing the main idea of SR.

 

[Stephanus]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

At first these two rules can be seen to be in contradiction with each other but it is capturing the main idea of SR.

But, who moves?
The clock or us?

 

[CycoFin]

But, who moves?
The clock or us?

does not matter, result is same

 

[Stephanus]

does not matter, result is same

Okay...,
But if one of the twin comes back, why he's younger than the one who stay? Should the result be same?

 

[CycoFin]

Okay...,
But if one of the twin comes back, why he's younger than the one who stay? Should the result be same?

Because he made the turn back.

Why he knows that he made turn back and not his twin? Because all the clocks from his point of view lost sync on turn.

 

[PeterDonis]

To all these SR twin paradoxes I can give two basic rules.

1. Moving clock seems to run slower.
2. If you are looking moving clocks that are moving past you, that time seems to go faster.

The second rule doesn't make sense, and neither of them really seem helpful for understanding the twin paradox.

all the clocks from his point of view lost sync on turn.

What clocks? If the traveling twin carries a clock with him, it doesn't do anything unusual when he turns around.

If you're talking about clocks that are spatially separated from the traveling twin, then what, if anything, "happens" to them when the traveling twin turns around depends on what simultaneity convention you adopt.

 

[PeterDonis]

if one of the twin comes back, why he's younger than the one who stay?

Because his path through spacetime is shorter. The correct general rule for deciding who has less elapsed time is not "who is moving". The correct general rule is to look at the different paths through spacetime and compare their lengths.

In the case of the standard twin paradox, the comparison is simple because one of the twins, the stay-at-home twin, is in free fall the whole time; he never feels any acceleration. It's easy to show that, between any two events in flat spacetime (i.e., in the absence of gravity), the path between them with the most elapsed time is the free-fall path. One way of seeing this is to observe that the traveling twin's path through spacetime, if we idealize his turning around to take place instantaneously, forms two sides of a triangle, and the stay-at-home twin's path through spacetime is the third side of the triangle. There is a triangle inequality in spacetime similar to the one in ordinary Euclidean geometry; the only difference is that, because of the minus sign in the metric of spacetime, the third side of the triangle (the stay-at-home twin's path) is longer than the sum of the other two (the traveling twin's path).

In other words, the key thing is not "motion", it's spacetime geometry.

 

[CycoFin]

The second rule doesn't make sense, and neither of them really seem helpful for understanding the twin paradox.

If you look clocks that are moving by, each of those clocks is running slower than yours. But if you look time by those clocks that pass you, that time is running faster than yours. It seems that those moving clocks are running slower but they are out sync so that the time they show (at your point) is running faster.

What clocks? If the traveling twin carries a clock with him, it doesn't do anything unusual when he turns around.

If you're talking about clocks that are spatially separated from the traveling twin, then what, if anything, "happens" to them when the traveling twin turns around depends on what simultaneity convention you adopt.

Yes, I mean two clocks spatially separeted with traveling twin (let say front and back of the spaceship). They are not in sync after turn. Also even staying twin didn't have his clocks on sync on first place (see above, from point of view of the turning twin), they are now even different way out of sync after turn (from point of view of the turning twin).

 

[PeterDonis]

if you look time by those clocks that pass you, that time is running faster than yours.

You're going to have to give a specific scenario to illustrate this. It doesn't look right to me.

 

[Ibix]

He's imagining a stream of clocks passing in front of him at velocity v. If he always looks at the clock infront of him and ignores the fact that it isn't the same clock, it appears to be running fast since the Lorentz transform simplifies to $t'=\gamma t$ if x=0. The usual time dilation formula, of course, comes from holding x'=0, rather than x.

What relevance this has to the twin paradox I am not sure.

 

[CycoFin]

You're going to have to give a specific scenario to illustrate this. It doesn't look right to me.

Maybe it is my english that it is causing troubles, this should be very basic thing to person who knows SR.

Let say we have stationary clocks between Earth and pluto. They are on sync from our point of view.

If one person rides with spaceship from Earth to pluto with relativistic speeds, he will notice that clock on Earth and pluto and every clock between them is running slower than his own spaceship clock. But if he watch clocks from his spaceship window, the time the clocks are showing is running faster than his own spaceship clock. This is because from his point of view our clocks are NOT in sync. For him, clocks on near Pluto are advanced, not showing same time as on earth. "Why your clocks are not in sync and are running slow?" he complains to us. "Why your clocks are not in sync and are running slow", we complain to him.

I need to add two more rules

3. moving clocks are out of sync
4. stationary clocks remain on sync unless you change speed or direction

 

[PeterDonis]

If one person rides with spaceship from Earth to pluto with relativistic speeds, he will notice that clock on Earth and pluto and every clock between them is running slower than his own spaceship clock. But if he watch clocks from his spaceship window, the time the clocks are showing is running faster than his own spaceship clock.

No, that is not correct. The person on the ship will observe the clocks on Pluto running faster than his own, but he will observe the clocks on Earth running slower than his own. When he corrects both of these observations for light travel time, he computes that both sets of clocks (on Earth and Pluto) are running slower than his own.

This is because from his point of view our clocks are NOT in sync.

That is true, but it has nothing to do with the different rates of the clocks. Clock synchronization and clock rate are two different things.

 

[ash64449]

Because his path through spacetime is shorter

Clocks of both the twins measure proper time.
Clock with the traveling twin traverses lesser distance in space-time,therefore lesser proper time is for the clock with the traveling twin.

But the home twin traverses bigger distance in the space-time and therefore has more proper time than traveling twin.

Is that the reason why you consider that traveling twin ages less than home-twin?

 

[ash64449]

No, because, as you go on to show in your very next sentence, the observer can change his state of motion, but the frame can't. So the observer and the frame can't be identical.

No, he's changed his state of motion. He was at rest relative to one frame before; now he's at rest relative to another, different frame.

The only reason why you did not agree with me is that you have ignored this consideration or assumption i made:

If i consider that a reference frame which helps an observer describe 'time' at which event happen(by putting clocks all over the coordinate system),coordinates at which event happen to be his frame

You said:

he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

I assumed that when such a correspondence is produced i would "call" it observer's frame.

Since you have ignored it, i as well will ignore it. I will talk in such a manner that 'changes frame' and 'his frame' would not come next time.

"Time" and "coordinates" are not absolutes; they depend on the coordinate chart you choose. Here you are implicitly choosing a coordinate chart corresponding to the first frame (the chart whose coordinate basis vectors are the frame vectors). The coordinates in this chart do give "times" and "positions" that would be measured by the first frame.

But you could just as easily choose a coordinate chart corresponding to the second frame (the one the observer is at rest relative to after he changes his state of motion). This chart would give "times" and "positions" that would be measured by the second frame, and after the observer changes his state of motion, he is at rest relative to the second frame, so its measurements of time and position would correspond to his.

I completely agree with you.

 

[PeterDonis]

I assumed that when such a correspondence is produced i would "call" it observer's frame.

And I was telling you that this way of speaking leads to confusion, because it doesn't keep distinct observers and frames. It looks like you are not going to use that way of speaking any longer, which was my objective in pointing all this out.

 

[PeterDonis]

Clock with the traveling twin traverses lesser distance in space-time,therefore lesser proper time is for the clock with the traveling twin.

But the home twin traverses bigger distance in the space-time and therefore has more proper time than traveling twin.

Is that the reason why you consider that traveling twin ages less than home-twin?

Yes.

 

[ash64449]

It is right that twin paradox can be solved by using any frame of reference-since in any frame of reference concerned traveling twin always ages less than the home twin.

But the original reason why twin paradox arises is due to one's point of view-because each twin considers other's clock to run slower-which correspond's to one's point of view.

But it is to be understood that any frame of reference in general doesn't express one's point of view.

For example, we can describe home-twin's point of view by using a frame of reference in which the home-twin is at rest. Since home-twin is always at rest and consider him to be at the origin,frame's measurement of 'time' and position would correspond to home-twin's.

But not so for traveling twin.

For example,if we use the frame of reference in which the traveling twin is rest in the first part of the scenario, traveling twin moves in this frame of reference in the second part of the scenario.
Similarly,we cannot use a frame of reference in which traveling twin is rest in the second part of the scenario.(i.e either frame unable to express traveling twin's point of view)

To express what i am saying, i am going to state what 'paradox' does twin paradox presents.

To keep the calculations simple,

Let the distance between the Earth and planet be 8 light years.(rest length between them)
Let speed of the traveling twin(TT) be 0.8c in both part's of the scenario.

Now the factor 'gamma' is 1/0.6

Now let the Home Twin (HT) calculate the time TT take to come back according to his point of view.Since the frame he uses has distance between the planet and Earth as 8 light years and speed is 0.8c,
Time taken by the TT to reach the planet is 10 years according to HT.
Consequently, Total time to reach the Earth is 20 years. So according to HT, TT takes 20 years to reach earth.
Now the HT makes calculations how much time TT would take for the same.

In the frame HT uses, TT is moving and hence time runs slower for TT. So HT predicts TT would measure 20*0.6=12 years to reach earth. And hence concludes that in between the events 'TT leaves the earth' and 'TT reaches the earth', proper time elapsed for TT is 12 years and proper time elapsed for HT is 20 years and hence TT is 8 years younger than HT.

Now let us go for the calculation of TT

In TT's point of view, Earth and the planet are moving and hence represent moving length. the length is 8*0.6=4.8 light years.

Speed of both planet and Earth is 0.8c, Therefore time taken for TT to reach planet is 4.8/0.8=6 years and therefore 12 years to reach the Earth for TT-same as predicted by HT.

Now TT calculates how much time HT measures for the same from his point of view.

According to TT, HT is moving. So time is running slower for HT. In all parts of the journey, it is the HT that moves, therefore time measured by HT should be 12*0.6=7.2 years for HT. So TT calculates elapsed proper time for him is 12 years but for HT is 7.2 years Therefore HT is 4.8 years younger that TT.This is the manner how both get contradicting result. Each predicts the other one should be younger.So the solution to the paradox should be given which includes one's point of view. Since the calculation error comes when one predicts the elapsed proper time of the other from his point.

Error is in the analysis on TT's and not HT. In fact HT's point of view is same as that of calculations made in a frame of reference where HT is rest.

But not for TT. TT's point of view can't be explained by any single frame since he is always moving in one part of the scenario or the other in whatever frame of reference one takes.

So original question of twin paradox should be: 'What is the error in the analysis of the age of the HT done by TT'?

 

[PeterDonis]

if we use the frame of reference in which the traveling twin is rest in the first part of the scenario, traveling twin moves in this frame of reference in the second part of the scenario.

This is only true if we use inertial frames. But that is not required. We can use a non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time. But the metric in this chart will be different than it would be for an inertial chart, so when we use this non-inertial chart to calculate the elapsed proper time for the traveling twin and the home twin, we find that the home twin still experiences more elapsed time.

This is the manner how both get contradicting result. Each predicts the other one should be younger.

The HT's calculation is correct, because he is calculating the length of the entire path through spacetime followed by the TT. But the TT's calculation is not correct, because he is not calculating the length of the entire path through spacetime of the HT. There is a section of the HT's path that is not covered by the TT's calculation because of the change in simultaneity convention when the TT turns around.

I highly recommend reading the Usenet Physics FAQ article on the twin paradox:

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

Pay particular attention to the Spacetime Diagram Analysis; it clearly shows how everything fits together, and what is missing in the TT's analysis that you have described.

 

[ash64449]

This is only true if we use inertial frames. But that is not required. We can use a non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time. But the metric in this chart will be different than it would be for an inertial chart, so when we use this non-inertial chart to calculate the elapsed proper time for the traveling twin and the home twin, we find that the home twin still experiences more elapsed time.

This method of treatment is new to me.

No, they don't. The HT's calculation is correct

Yes. I did say HT's calculation is correct. As i said here :

Error is in the analysis on TT's and not HT

There is a section of the HT's path that is not covered by the TT's calculation because of the change in simultaneity convention when the TT turns around.

I actually know the answer! i was trying to say that This paradox's error should be pointed out from one's point of view.

But definitely the answer i thought was also due to change in simultaneity convention but method is not in terms of space-time diagram analysis.

 

[PeterDonis]

This paradox's error should be pointed out from one's point of view.

I'm not sure exactly what "one's point of view" means, but it seems to me that pointing out to the TT that his calculation leaves out a portion of the HT's worldline should count as pointing out his error "from his point of view".

method is not in terms of space-time diagram analysis.

You can use any method you like, as long as you get the right answer. If the TT does an analysis that leaves out a portion of the HT's worldline, he is not going to get the right answer, so whatever method he is using is wrong.

 

[ash64449]

non-inertial frame (more precisely, a non-inertial coordinate chart) in which the traveling twin is at rest the whole time.

I don't think this is absolutely necessary. I mean what is the reason why we want TT rest the whole time?

Because we want to describe his point of view. The technique you think is not familiar to me. But i do know that switching frames helps to describe one's point of view.( here to describe TT's point of view, for the first part of the scenario,frame of reference in which Earth and the planet are moving towards the left. Second part of the scenario: frame of reference in which Earth and the planet move towards the right)

This helps us to understand the change in simultaneity convention satisfactorily to solve twin paradox.

 

[PeterDonis]

switching frames helps to describe one's point of view

If you're willing to accept that the TT's "point of view" can switch simultaneity conventions instantly, sure.

You also have to ignore the portions of the TT's trip where he is accelerating, not inertial--for any real scenario, the TT's turnaround will take a finite time, during which he is accelerating and so is not at rest in any inertial frame. During these times, the only way to describe the TT as being at rest is to use non-inertial coordinates. And once you've admitted that you need to do that, why not just construct a non-inertial coordinate chart in which the TT is at rest the whole time?