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

[Stephanus]

Dear PF Forum,
Sorry if I ask (again) about twin paradox, after so many question about this topic here.

Supposed T is a star 100 ly from earth.
If B travels to T from earth
A. Is the symmetry broken?
B. If B watches A's clock at Earth and A watches B's clock at T, do they see the other clock late by 100 years, no matter how fast B travels?

Now B travels back to A, and now B brings a companion C, to travel with him, sitting beside B
C. Is the symmetry broken for A and B?
D. If the symmetry is broken for A and B, what about for B and C?
1. Does B sees C age faster along the journey?
2. Does B sees C gesture/act in fast motion?
3. Does C sees B gesture/act in slow motion?

Thanks for the answer, and pardon me for this twins paradox again.

 

[mfb]

A. Is the symmetry broken?

Which symmetry?

B. If B watches A's clock at Earth and A watches B's clock at T, do they see the other clock late by 100 years, no matter how fast B travels?

You can use the equations to calculate it. Or have a look at one of the many examples in other threads and other websites.

C. Is the symmetry broken for A and B?
D. If the symmetry is broken for A and B, what about for B and C?

Again, which symmetry?

1. Does B sees C age faster along the journey?
2. Does B sees C gesture/act in fast motion?
3. Does C sees B gesture/act in slow motion?

B and C do not move relative to each other, what do you expect? Do you see your neighbor age faster or slower than you?

 

[Dale]

When you say "see" do you mean literally what visual signal they would actually see, or do you mean what would they calculate in their reference frame.

 

[Stephanus]

Thanks Mfb for the answer.
What I meant about the symmetry, is the Twin Paradox Symmetry.
If B travels and A stays, but B doesn't come back. The clock symmetry is maintained, right?
But if B changing his inertial frame of reference, the symmetry is broken. Is that so?

 

[Stephanus]

When you say "see" do you mean literally what visual signal they would actually see, or do you mean what would they calculate in their reference frame.

They ACTUALLY see. Using a very magnificent telescope.
A can see B's clock 100 ly away. And B uses telescope can see A's clock 100 ly away.

 

[mfb]

There is no symmetry at all, that is the point of the twin "paradox".

They ACTUALLY see. Using a very magnificent telescope.

Then you have to take light travel times into account which makes the analysis a bit more complicated. Did you draw a Minkowski diagram?

 

[Nugatory]

They ACTUALLY see. Using a very magnificent telescope.
A can see B's clock 100 ly away. And B uses telescope can see A's clock 100 ly away.

For that definition "see", their respective experiences will be described by the Doppler Shift analysis section of http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[Stephanus]

There is no symmetry at all, that is the point of the twin "paradox".

Then you have to take light travel times into account which makes the analysis a bit more complicated. Did you draw a Minkowski diagram?

Sorry, I address you as "Drakith" MFB, I was reading a Drakkith answer somewhere, and the alert comes out.
No, I wasn't drawing Minkowski Diagram. I haven't studied it, yet. It's just that they say that Twin Paradox symmetry is not broken when one of the participant travels. But the symmetry is broken is the other participant, changes its inertial frame of reference. That's why I want to get a full grasp of it first, before I ask about Universe Prefered Frame of Reference (again).

 

[ash64449]

Symmetry is based on that A and B don't play the same role. One keeps changing frames while other does not.

Considering the case of original twin paradox,Here B is changing frames.

Overall, first A and B are in the same frame.But as B travels,B changes frame.

Now for the return trip,B changes the frame again(different from that of the frame which B occupies when B leaves A).

So here B keeps changing frames while A doesn't

 

[PeterDonis]

If B watches A's clock at Earth and A watches B's clock at T, do they see the other clock late by 100 years, no matter how fast B travels?

Rather than answer this question as you ask it (since it's vague), let me just describe what A and B would see as they watch each other's clocks. We assume that, when B departs from Earth, A's and B's clocks are synchronized and both read zero. When B arrives at T, he emits a light signal containing his current clock reading and the current clock reading on T. Let's suppose, for definiteness, that B travels at 87% of the speed of light relative to Earth and T, so the time dilation factor is 2. Then, when B arrives at T, his clock will read about 57.75 years, and clocks on T when he arrives will read about 115.5 years. When B's light signal, emitted when he arrives at T, reaches A on Earth, A's clock will read about 215.5 years. From then on, since B is at rest at T, the signals A receives will show clock readings for T 100 years behind A's clock reading when the signals are received, and clock readings for B about 157.75 years behind A's.

Now let's look at what B sees. When he arrives at T, he sees a light signal that was emitted at A at just the right time to arrive at T at the same instant he does. When B receives this signal, his clock reads 57.75 years, and the clock reading on A's clock that is contained in the signal is about 15.5 years. And clocks on T at this instant (assuming they were previously synchronized with Earth's using the Einstein clock synchronization convention) read about 115.5 years. From then on, since B is at rest at T, the signals B receives from A will show clock readings 100 years behind T's clock readings when the signals are received, and about 42.25 years behind B's clock readings when the signals are received.

Now B travels back to A, and now B brings a companion C, to travel with him, sitting beside B

How is C's clock set to start out? Is it synchronized with clocks on T, or with B's clock? I'll assume C's clock is synchronized with T's clocks to start out. And I'll assume that B and C leave T for Earth when T's clocks read 1000 years, and they travel at 87% of the speed of light, just as B did when traveling from Earth to T.

Then, when B and C start out, they emit a light signal to Earth showing both of their clock readings. Those will be 1000 years (for C) and 942.25 years (for B). When A receives this signal, his clock will read 1100 years.

When B and C arrive on Earth and meet A, B's clock will read 1000 years, C's clock will read 1057.75 years, and A's clock will read 1115.5 years.

 

[Stephanus]

For that definition "see", their respective experiences will be described by the Doppler Shift analysis section of http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

But "see" here, means that B has stopped at some point. At this example B stops at T. And A and B watches each other clock at the same frame of reference using a very powerful telescope. I think no Doppler effect there.

 

[Stephanus]

Symmetry is based on that A and B don't play the same role. One keeps changing frames while other does not.

Considering the case of original twin paradox,Here B is changing frames.

Overall, first A and B are in the same frame.But as B travels,B changes frame.

Now for the return trip,B changes the frame again(different from that of the frame which B occupies when B leaves A).

So here B keeps changing frames while A doesn't

Thanks Ash64449 for the answer. I mean A and B watch each other clock BEFORE B returns.

So B has stopped and (I think) they are at the same frame of reference.
1. Are they seeing the other clock late by 100 years?
2. Is there a doppler effect after B stops?

Thanks

 

[PeterDonis]

So B has stopped and (I think) they are at the same frame of reference.

They are at rest relative to each other. They are both "in" any frame of reference you choose; they are just not at rest relative to other frames besides the Earth-T frame.

Are they seeing the other clock late by 100 years?

No; the clock offsets they actually observe are different than that. See my post #10.

Is there a doppler effect after B stops?

No.

 

[Nugatory]

But "see" here, means that B has stopped at some point. At this example B stops at T. And A and B watches each other clock at the same frame of reference using a very powerful telescope. I think no Doppler effect there.

You can still apply the Doppler analysis. It may even be the best/simplest way of analyzing these "what do you see throiugh a telescope" thought experiments because you're tracking the light from the event that is being seen to the telescope and eyes of the person who is doing the seeing when the light reaches them.

While they're moving apart, A sees fewer flashes of light from B and vice versa (if you you don't want to count flashes of light, you can count the number of times that the sweep second hand of the clock passes a tick mark on the face of the clock, or you can count the number of times that the rightmost digit on a digital clock display changes - they'll all yield the same result as you have to wait for the light to travel from the "tick" event to your telescope before you can see it).

When B reaches T and stops, the Doppler effect will go away as you say, so A sees B's clock going back to its old undilated tick rate and vice versa after they've both come to rest relative to one another. It's a good exercise to draw a spacetime diagram showing both the light flashes that A sees as he looks a clock that is at rest relative to him at T (ticks at the same rate as his own, but he sees it consistently 100 years later because of the light travel time) and the clock that B is carrying along his trip.

 

[Stephanus]

Thanks PeterDonis for your invaluable help.

Rather than answer this question as you ask it (since it's vague), let me just describe..
...when B departs from Earth, A's and B's clocks are synchronized...
...that B travels at 87% of the speed of light relative to Earth...

Yes, yes. I supposed I forgot to mention some parameters. I tought it goes without saying about twin paradox. Clock synchronization at the beginning and travel near the speed of light.

When B and C arrive on Earth
B: 1000.00 years
C: 1057.75 years
A: 1115.50 years

But, why A's clock runs faster than C's clock. From C point of view, it's A who travels, right?
Is it the rocket acceleration? But, they say acceleration has small effect on time dilation. Only changing inertial frame of refence does.

When B and C arrive on Earth and meet A, B's clock will read...

"Arrive on Earth" suggests that B is changing it's frame of reference, right?
"Meet A" suggests that either (B and C) or (A) are moving, but then again motion is relative, right?

 

[Stephanus]

It's a good exercise to draw a spacetime diagram...

Thanks Nugatory, I'll do it.

 

[Stephanus]

Symmetry is based on that A and B don't play the same role. One keeps changing frames while other does not.
Considering the case of original twin paradox,Here B is changing frames.
Overall, first A and B are in the same frame.But as B travels,B changes frame.
Now for the return trip,B changes the frame again(different from that of the frame which B occupies when B leaves A).
So here B keeps changing frames while A doesn't

Thanks Ash64449 for the answer.
About this "changing frame".
If B travels from Earth to T, why B is changing frame? Motion is relatvie, right?
In B point of view, isn't A who changing frame?

 

[PeterDonis]

why A's clock runs faster than C's clock

Whether or not that's true depends on which frame you use, the frame in which A, Earth, and T are always at rest, or the frame in which B and C are at rest while they are moving (but not before they start or after they arrive). Changing frames means changing simultaneity conventions; in the frame in which A, Earth, and T are at rest, B and C start out from T "at the same time" as A's clock reads 1000 years. But in the frame in which B and C are at rest while they are moving, B and C start out from T "at the same time" as A's clock reads 1086.6 years, so in this frame, A's clock only advances by 28.87 years during the journey, half of the 57.75 years that B's and C's clocks advance. This is just standard relativity of simultaneity.

"Arrive on Earth" suggests that B is changing it's frame of reference, right?

No, it says that B changes his state of motion. Changing state of motion is not the same as "changing frame of reference". Any frame of reference can be used to describe B's entire motion, before and after he arrives on Earth.

"Meet A" suggests that either (B and C) or (A) are moving

No, it says that A's, B's, and C's worldlines all coincide. There is no assumption about who is "moving" and who is not.

 

[PeterDonis]

as B travels,B changes frame.

No, B changes his state of motion relative to A, Earth, and T.

for the return trip,B changes the frame again

No, B again changes his state of motion relative to A, Earth, and T.

If B travels from Earth to T, why B is changing frame? Motion is relatvie, right?
In B point of view, isn't A who changing frame?

Neither one is "changing frame". They are moving relative to each other. There is no absolute sense in which either one is "moving" or "at rest".

As a general point: thinking of changing relative motion as "changing frames" is just going to cause confusion. It's important to keep inertial frames conceptually distinct from the states of relative motion of various objects. You can describe the entire scenario using just one frame; there is no need to "change frames" when someone changes their state of motion relative to someone else.

 

[ash64449]

So B has stopped and (I think) they are at the same frame of reference.

Yes. Both are in same reference now.

 

[PeterDonis]

Both are in same reference now.

No, both are at rest relative to each other now. Once again, changing an object's state of motion is not the same as "changing frames", and being in the same state of motion is not the same as "being in the same frame". Frames are not the same as states of motion.

 

[ash64449]

No, both are at rest relative to each other now. Once again, changing an object's state of motion is not the same as "changing frames", and being in the same state of motion is not the same as "being in the same frame". Frames are not the same as states of motion.

When do we say that a person changes reference frame?

 

[Jimmy]

When do we say that a person changes reference frame?

You don't. Objects are in all frames of reference simultaneously. If an object changes it state of motion, it still exists in whatever frame of reference you've chosen for your calculations. It just has a different velocity in that frame.

 

[ash64449]

You don't. Objects are in all frames of reference simultaneously.

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?

 

[Jimmy]

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

It just depends on which frame of reference you choose for your calculations.

 

[ash64449]

It just depends on which frame of reference you choose for your calculations.

Okay. Doesn't B use different reference frames for calculation in the whole journey? For the first trip, he used a frame of reference(for calculation) which is moving uniformly w.r.t A ,moving away from A.
And for the second trip, B uses a different frame for calculation,This time a frame of reference which moves uniformly towards A?

 

[Jimmy]

Okay. Doesn't B use different reference frames for calculation in the whole journey? For the first trip, he used a frame of reference(for calculation) which is moving uniformly w.r.t A ,moving away from A.
And for the second trip, B uses a different frame for calculation,This time a frame of reference which moves uniformly towards A?

...Frames are not the same as states of motion.

I think the point Peter was trying to make is that objects are at rest or in motion, or change their state of motion; frames of reference do not. They are simply co-ordinate systems chosen to perform calculations.

 

[PAllen]

When do we say that a person changes reference frame?

Ideally, never (despite many dubious sources that do so). A body may accelerate, decelerate etc. but this has nothing (necessarily) to do with changing frames. You can always analyze everything in one frame (any of them - that is what relativity means). Acceleration is measurable, and proper acceleration is invariant (the same in all frames). Frames are analytical constructs with no physical import whatsoever (though, you can make a frame physical by connecting clocks and rulers; but then it is better to think of these objects rather than the frame they approximate).

 

[ash64449]

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? I consider that object uses these to perform calculations. So conclusion is if object tries to move,reference frame(which object uses to perform calculations)remains there? i am confused about what you all are trying to say.

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.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.

 

[Mentz114]

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?

It is wrong. One may choose any object and use the clocks at rest with that object to work out a consistent picture.

The phrase 'changes frame of reference' does not indicate a change of location necessarily. It means 'a change in the state of motion wrt some other object'.

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.

Keep trying. One advantage of relativity theory is that coordinates are meaningless, so eventually they will be of no interest except as a ladder to reach higher understanding.

 

[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.