Acceleration and the twin paradox

In summary, the main problem most people seem to have is how do you get around the apparent symmetry of relative motion problem between the two twins. This becomes broken once twin A accelerates from the Earth and enters a non-inertial frame, and this is what is responsible for the slowing of twin A's aging relative to twin B.
  • #141
CKH said:
When we talk about an inertial frame, we can assign that a 4-D coordinate system (why is this now called a "chart", perhaps more general somehow?).
"Chart" is the technical term from Riemannian geometry. It is all spelled out in chapter 2 of the reference I provided ( https://www.physicsforums.com/threads/acceleration-and-the-twin-paradox.779110/page-7 ). Although the notes are from a course on general relativity, that specific chapter is only about Riemannian geometry, which is applicable to any branch of physics, including SR, QM, and even classical mechanics. If you want to learn about non-inertial frames in SR then you have to study the basics of charts and Riemannian manifolds.

CKH said:
That coordinate system covers all of space and time (in SR). So an inertial frame is not something that changes with time, as you say.
Yes.

CKH said:
The traveler experiences change over time as measured by the clock he carries with him. When he is moving inertially, he is at rest in some inertial frame and his clocks ticks at the same rate as clocks in that frame. When he moves with acceleration, at each instant of his time, he is at rest in a different inertial frame. Hyperplanes of simultaneity in each of these different frames intersect the worldline of the home clock at a different point.
All of this is correct, but ...
CKH said:
So, for the traveler, the simultaneous reading on the home clock changes because his motion becomes coincident with different inertial frames as his motion changes.
This doesn't necessarily follow. You can define a chart this way, but any other smooth one-to-one mapping which assigns a constant coordinate to the traveler is equally valid. Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

CKH said:
This seems to be the central issue here. Can you explain what you mean in more detail, in SR where spacetime is flat? ...

I need understand what we mean by the "non-inertial rest frame", if we consider it as a 4-D coordinate system. We are trying to describe 4-D rest frame for the traveler (so over the whole trip). I can understand how a traveler in uniform motion has a 4-D rest frame with coordinate axes. When his motion is not inertial, how do you define his 4-D frame? Can you create a coordinate system for this frame over all spacetime?
Any diffeomorphism from open sets in the manifold to open sets in R4 is a valid chart. See the reference above. The only requirement beyond being a valid chart is that in X's rest frame the spatial coordinates for X are constant. This leaves an immense amount of freedom for defining a non-inertial object's rest frame, and no standard convention.

Here is an example of an alternative method for constructing a non-inertial rest frame which is also commonly used: http://arxiv.org/abs/gr-qc/0104077

CKH said:
However, to me it does makes sense to say there is an instantaneous chart at each point along on the traveler's world line.
There is a MCIF. There is not an instantaneous 3D chart since that would not be a mapping from an open set of the manifold to an open set in R4.

CKH said:
Yes, the frames have to be connected using the traveler's clock, position and orientation.
More than that, you have to make sure that the resulting mapping is a diffeomorphism.

CKH said:
Your reference looks like a nice introduction to general relativity which I want to study, but that is not an issue here since we have no gravitational fields in SR.
The chapter I posted is a gentle introduction to Riemannian geometry, which is used for non-inertial frames in SR also. GR does not "own" Riemannian geometry. I agree that we should stick with only SR and not consider gravity. All of my above comments have been restricted to SR (and in fact, I have only been considering the instantaneous turn around scenario except when explicitly responding to a comment from someone else about the gradual turn around). So, there is no need to go on about GR vs SR, I am not using GR.

CKH said:
There is a diagram in Minkowski spacetime in a earlier post that assumes the simplified case with instant acceleration of the traveler. It shows the world line of the home clock and that of the traveler in the rest frame of the home clock (the base frame). The world line of the home clock is a straight vertical line segment (between these events). The world line of the traveler is a dogleg ">" segment.

The wordline for an object is the path of the object in the Minkowski spacetime. Along a world line, a tangent ray (pointing upward toward positive time) is the positive time axis of the object in its rest frame at that point. In Minkowski space, clock ticks have equal length regardless of the direction of time. So the integrated length of a world line is the elapsed time along that world line (for a clock on that world line).
Yes, that is all correct.

CKH said:
The shortest path from the event where the traveler leaves home and the event where he returns home is a straight line so such a worldline takes the least time. The dogleg path is longer in time on a clock following that path because the dogleg is longer.
Actually, a straight line is the longest timelike interval. Any timelike dogleg path is shorter. The "triangle inequality" is reversed for timelike intervals.

CKH said:
OK, now here's where I'm losing it, the equation for proper time, i.e. the time on a clock for a world line.

I don't get why squared values of spatial distance are subtracted from squared values of time to get proper time, if what I said above is true in Minkowski space about path lengths. So I do not have a firm grip on Minkowski space.
You may want to read chapter 1 of the reference I posted also: http://preposterousuniverse.com/grnotes/grnotes-one.pdf [Broken] It gives a good introduction to Minkowski space in a way that prepares you for more in the future.

CKH said:
Hill first, mountain later.
That is fine, but then you should drop the topic of non-inertial frames altogether and concentrate on understanding Minkowski spacetime and four-vectors.

CKH said:
Because spacetime is flat here, you can also use MCIFs along the traveler's worldline and determine the simultaneous readings on the home clock for the traveler.
Not necessarily. Doing so can lead to mappings that are not diffeomorphisms, even in flat spacetime.

CKH said:
A 4-D inertial frame doesn't do anything (it's fixed). As the traveler moves along his world line, he is momentarily at rest in different inertial frames. Thus over a short period of time, the traveler moves from one inertial rest frame to a different one. The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. So as you say, obviously his notion of simultaneity changes and thus the simultaneous reading on the home clock.

In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".
Please read the material provided. Until then, you are just not prepared. You may also simply want to stick to inertial frames until you have a firm grasp on four-vectors, the spacetime interval, and so forth.
 
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  • #142
CKH said:
My point was that during the main portions of the trip (in this scenario) the traveler is in inertial motion. You can analyze the trip piecewise and I think it's helpful because it highlights how the paradox is resolved.
Yes, that has been the standard treatment from the start - and your account was perfectly clear to me and several other participants. So, I hope that it was also understood by Diracpool!

However:
To go back to the super simple version where the traveler instantaneously reverses velocity at the turn around. His motion changes from rest in one inertial frame to rest in another inertial frame in that instant. In his new inertial frame the space of simultaneity has abruptly changed and the home clock jumps to a later time for the traveler in the new frame.
Surely you agree that simultaneity of an inertial frame is constant; thus, you likely meant something else than what you wrote (certainly no clock "jumps"!). But it's not clear to me what you meant with that, and it suggests a misunderstanding (although it was perhaps just figurative speaking). Also for Diracpool it may be useful to clarify this.

I waited to see if you would by yourself clarify this point, but this was not really the case, the physical meaning remained unclear:
CKH said:
[..] The traveler's hyperplane of simultaneity changes, so in the travel's simultaneous reading on the home clock changes over that period of time. [..]
In the case where the traveler instantly reverses his velocity, the traveler "jumps" into a new inertial frame, causing the simultaneous reading on the home clock to "jump".
I think that it was mentioned earlier in this thread that the traveler has a free choice of inertial frames. Real on-board clocks will not "jump", and consequently the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home. Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame. Thus the simultaneous reading on the home clock is adjusted by the traveler, according to his/her choice of instruments and "maps".
 
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  • #143
harrylin said:
Surely you agree that simultaneity of an inertial frame is constant; thus, you likely meant something else than what you wrote (certainly no clock "jumps"!). But it's not clear to me what you meant with that, and it suggests a misunderstanding (although it was perhaps just figurative speaking). Also for Diracpool it may be useful to clarify this.

I think that the meaning is this: At any moment, the traveling twin has an associated momentarily comoving inertial frame. According to this inertial frame, the stay-at-home twin is a certain age. Then if you define "the stay-at-home twin's age current age" (according to the traveling twin) to be his age, according to the traveling twin's current comoving inertial frame, then the stay-at-home twin's current age jumps abruptly if the traveling twin accelerates suddenly. This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.
 
  • #144
stevendaryl said:
[..]This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.
Yes, exactly that was my point. One has to be careful with fancy phrasings.
 
  • #145
The clock hypothesis states that the extent of acceleration doesn't influence the value of time dilation. In most of the former experiments mentioned above, the decaying particles were in an inertial frame, i.e. unaccelerated. However, in Bailey et al. (1977) the particles were subject to a transverse acceleration of up to ∼1018 g. Since the result was the same, it was shown that acceleration has no impact on time dilation.[27] In addition, Roos et al. (1980) measured the decay ofSigma baryons, which were subject to a longitudinal acceleration between 0.5 and 5.0 × 1015 g. Again, no deviation from ordinary time dilation was measured.[29]
 
  • #146
DaleSpam said:
"Chart" is the technical term from Riemannian geometry. It is all spelled out in chapter 2 of the reference I provided ( https://www.physicsforums.com/threads/acceleration-and-the-twin-paradox.779110/page-7 ). If you want to learn about non-inertial frames in SR then you have to study the basics of charts and Riemannian manifolds.

All of this is correct, but ...This doesn't necessarily follow. You can define a chart this way, but any other smooth one-to-one mapping which assigns a constant coordinate to the traveler is equally valid. Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

No it's not unique, but the relative time results are the same whatever free choices you make for frames, given that you synchronize the traveler's clock with the home clock at the departure event.

In the case I was elaborating, we freely pick some spatial orientation for the home frame and for simplicity choose the origin as the position of the home clock . We choose that the traveler's initial frame as identical to the home frame. Then we plot the traveler's path. The MCIFs of the traveler along the path then have no free choices. The spatial orientation remains constant for all MCIFs. The spatial origin is the traveler's position to maintain continuity with the departure frame. The time axes are tangent to the worldline, the current time in each MCIF is the traveler's proper time.

Anyhow, in this mapping all that matters to determine simultaneity are the time axes of the MCIFs (no choice there) and the departure clock synchronization of the home and traveler's clocks.

Furthermore, this method of defining a chart can violate the requirement of being smooth and one-to-one, which invalidates it in those cases.

Then let's not call it a chart if it has to satisfy such criteria.

Any diffeomorphism from open sets in the manifold to open sets in R4 is a valid chart. See the reference above. The only requirement beyond being a valid chart is that in X's rest frame the spatial coordinates for X are constant. This leaves an immense amount of freedom for defining a non-inertial object's rest frame, and no standard convention.

Just using the definitions provided in SR for inertial frames and simultaneity, I see no way of avoiding the fact the traveler's worldline has a tangent time axis at each point and the time on this axis at each point is the traveler's clock time. This time axis and proper time uniquely define a hyperplane containing all simultaneous events, including the corresponding event on the home clock's worldline.

I don't know why we need a diffeomorphism or a "valid chart" to do this analysis.

By constant spatial coordinates you mean constant orientation and constant position of the object?

Here is an example of an alternative method for constructing a non-inertial rest frame which is also commonly used: http://arxiv.org/abs/gr-qc/0104077

In that paper the first two examples use the same method I used. The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?

Then the author defines another "method". For the new method invents a new definition for the time for remote events:

[QUOTE: From the paper] ]This formula simply says that an observer can assign a time to a distant event by sending a light signal to the event and back, and averaging the (proper) times of sending and receiving.[/QUOTE]

Sure you can make such a definition and it may have some very cool properties. It is a rather arbitrary choice, since after sending a light signal, the observer may have changed position and velocity in any number of ways before he receives the reflection.

This "method" has little or nothing to do with simultaneous readings by the traveler of the time on the home clock as defined by SR. I don't think we are free to change this relationship in SR.

There is a MCIF. There is not an instantaneous 3D chart since that would not be a mapping from an open set of the manifold to an open set in R4.

More than that, you have to make sure that the resulting mapping is a diffeomorphism.

I believe what you are talking about is defining a frame for the traveler (i.e. a 4-D coordinate system along the traveler's worldline). That is an interesting subject which I asked about in the last post because I could not see how to do it with uniqueness over all of space time. But, notice that it is not really required for this analysis.

These rules for a diffeomorphism place constraints on what parts of space time can be included in a frame constructed in this way. I can visualize a variety of "tubes" surrounding the worldline of the traveler that you could include in such a chart. But no one chart can cover all of spacetime in general without overlapping itself which violates the rules. For a given event, it may be possible to include that event in some chart, but it will not be a unique chart nor will the coordinates of the event be unique in all such charts. I can't see how to join such charts to cover all of spacetime. I suspect this chart notion is more useful in GR?

On the other hand, a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame. That's not the fault of the method, it simply a consequence of SR.

The chapter I posted is a gentle introduction to Riemannian geometry, which is used for non-inertial frames in SR also. GR does not "own" Riemannian geometry. I agree that we should stick with only SR and not consider gravity. All of my above comments have been restricted to SR (and in fact, I have only been considering the instantaneous turn around scenario except when explicitly responding to a comment from someone else about the gradual turn around). So, there is no need to go on about GR vs SR, I am not using GR.

OK thanks.

Actually, a straight line is the longest timelike interval. Any timelike dogleg path is shorter. The "triangle inequality" is reversed for timelike intervals.

Aha. I guess that's a result of the subtraction I mentioned. The very strange pythagorean theorem for spacetime. So, distance along a worldline is this quantity called "s" which is not proper time, but I see formulas for proper time that look exactly like proper length, expect that the sign of the squared sum is reversed so that one of these quantities is imaginary. Is time then an imaginary number in these formulas?

Thanks for the Minkowski reference, indeed I don't understand it yet.
 
  • #147
The very strange pythagorean theorem for spacetime. So, distance along a worldline is this quantity called "s" which is not proper time, but I see formulas for proper time that look exactly like proper length, expect that the sign of the squared sum is reversed so that one of these quantities is imaginary. Is time then an imaginary number in these formulas?

The modern way of doing things doesn't use imaginary values, it uses [itex]\delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2}[/itex] for proper time (when [itex]c |\delta t| > |\delta x|[/itex]) and uses [itex]\delta L = \sqrt{ \delta x^2 - c^2 \delta t^2}[/itex] for proper distances (when [itex]c |\delta t| < |\delta x|[/itex]).

I've asked before, and there seems to be no reason ever to compute spacetime interval along a path that is sometimes spacelike and sometimes timelike. That would give a complex number. But such a path can't represent the path of an object, so it's not clear what physical meaning could be associated with such a path.
 
  • #148
CKH said:
Then let's not call it a chart if it has to satisfy such criteria.

This may or may not be an issue, depending on what you're trying to do. See below.

CKH said:
I don't know why we need a diffeomorphism or a "valid chart" to do this analysis.

You don't if all you are trying to do is "solve" the twin paradox for a particular scenario. But people use charts for a lot of things that require the chart to satisfy the criteria DaleSpam gave.

CKH said:
The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?

No.

Proper time for a particular observer is only meaningful for events on that observer's worldline. If the observer wants to assign a "time" to events not on his worldline, he has to pick a simultaneity convention, and there is no one unique way of doing that. Even the obvious way for an inertial observer, namely just using the coordinate time in a coordinate chart in which he is always at rest, is not the only possible way for that observer. Simultaneity is simply not a physically meaningful thing; it's a convention.

CKH said:
I don't think we are free to change this relationship in SR.

Sure we are. It may not make sense to do it for a particular problem, but there's nothing in the physics that forbids us from doing it. As above, simultaneity is not a physically meaningful thing; it's just a convention.

CKH said:
I believe what you are talking about is defining a frame for the traveler (i.e. a 4-D coordinate system along the traveler's worldline).

Technically, this is a "coordinate chart"; "frame" means something different. But "frame" is often used (not strictly correctly) to mean "coordinate chart".

CKH said:
I could not see how to do it with uniqueness over all of space time.

This depends on the spacetime. In flat spacetime, it is always possible to construct a valid chart (i.e., one meeting all of DaleSpam's requirements) that covers the entire spacetime, with any chosen timelike curve as its "time axis". The Dolby & Gull paper shows how to do this. Obviously, for timelike curves that are not geodesics, i.e., not inertial worldlines, the surfaces of constant time in such a chart will not match the notion of "simultaneity" that you have been using, since that notion does not lead to a valid chart on all of spacetime. That's fine because, once again, simultaneity is not a physically meaningful thing.

In curved spacetime, it may not be possible to construct a single valid chart that covers the entire spacetime, regardless of how we pick the "time axis". Even if it is possible, such a chart will probably be highly non-intuitive (for example, the Kruskal-Szekeres chart on Schwarzschild spacetime).

CKH said:
a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame.

Then this is not a "global coordinate system"; at least, not in any useful sense. How can a coordinate system be useful in a region of spacetime where it doesn't assign unique coordinates to events?
 
  • #149
stevendaryl said:
I think that the meaning is this: At any moment, the traveling twin has an associated momentarily comoving inertial frame. According to this inertial frame, the stay-at-home twin is a certain age. Then if you define "the stay-at-home twin's age current age" (according to the traveling twin) to be his age, according to the traveling twin's current comoving inertial frame, then the stay-at-home twin's current age jumps abruptly if the traveling twin accelerates suddenly. This jump can't be taken seriously as a physical change, because by this definition, the stay-at-home twin's current age can jump forward or backward, depending on how the traveling twin accelerates.

That is exactly what I was trying to do. The purpose was to find, for a given time on the traveler's clock, the corresponding simultaneous reading on the home clock as defined by SR in the traveler's frame .

As the traveler moves from the outgoing IRF to the incoming IRF (instantly), he keeps his clock reading and uses it to define current time in the inbound IRF. In switching these IRFs, there is a sudden jump in what is simultaneous in the traveler's frame. This jump is caused by a change in the "perspective" from the traveler's frame. It is not physical in the sense that no actual jump occurs on the home clock in it's own reference system, it just keeps on ticking there. Furthermore it is not something the traveler actually sees on the home clock as a local observer using a telescope at the turn around. It is just a result of the definition of simultaneity in the two different frames.

If the traveler watches the home clock through a telescope, on the outbound leg, he sees it tick very slowly and on the inbound leg he sees it ticks very fast. The apparent change in home clock's rate is abrupt because he reverses instantly.

What I haven't done are the detailed calculations, but in principle we can define what the simultaneous reading on the home clock is at all points along the traveler's journey including the reading when the traveler returns.

This may not be the best way to look at the problem, but as a novice, I was thinking specifically about inertial frames and how they affect simultaneity as a way to view the paradox.

In the home clock's inertial rest frame, there is no time discontinuity on the traveler's clock. But we can also calculate from the home frame the exact simultaneous reading on the traveler's clock at each moment on the home clock. The two mappings, traveler clock to home clock and home clock to traveler clock are different. The cause of this is non-symmetrical acceleration. The home clock does not accelerate at all.
 
  • #150
CKH said:
The purpose was to find, for a given time on the traveler's clock, the corresponding simultaneous reading on the home clock as defined by SR in the traveler's frame .

And the answer is "mu": the question is not well-posed, because it assumes that there is a unique, physically given answer to the question. There isn't. Simultaneity is not a physically meaningful thing; it's just a convention. The way you are defining simultaneity is one way to do it, but it's not the only way, it has issues (like not assigning a unique time to all events), and it's no more meaningful, physically, than other ways of doing it.
 
  • #151
harrylin said:
I think that it was mentioned earlier in this thread that the traveler has a free choice of inertial frames.

I suppose, but not if the traveler wants his position to remain fixed in his frame and his clock time to be continuous in his frame. We are not picking some random choice of the new inertial frame at turn around, we are choosing the unique inertial frame that is consistent with the traveler's local frame (which is continuous).

Real on-board clocks will not "jump", and consequently the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home.

No, the traveler's clock never jumps in the traveler's frame, but the home clock does jump in the traveler's frame. Nothing is magical here, the traveler's frame abruptly changes its plane of simultaneity from the outbound IRF to the inbound IRF. So we are forced to admit that the simultaneous reading on the home clock "jumps" in the travelers frame.

Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame. Thus the simultaneous reading on the home clock is adjusted by the traveler, according to his/her choice of instruments and "maps".

No, I intend to do it the other way around, the on-board clock defines the current time in the new rest frame. There is no free choice for time in the new rest frame because we have constrained it to be consistent with the traveler's non-inertial frame. The requirements for the traveler's frame are:

1) Orientation in space is fixed (the traveler is not rotating).
2) The traveler's position in the frame is fixed. (He is at rest in his own frame).
3) The traveler's clock is continuous (it can't jump in it's own frame, his clock is a good clock).

Finally just to be absolutely clear the "jump" in the home clock is not physical; it is rather a "jump" in the perspective in the traveler's frame. I can make the distance stars move faster than the speed of light in my frame just by rotating my body. It is not physical however.
 
  • #152
The thoughts of this whole picture becomes quite impossible the more you think about it. For instance take one single leg of this trip and look around. Bob and Alice agree what the universe looks like around them. Alice leaves and is time dilated on her trip and stops. Doesn't the universe around her look younger than what Bob experiences at this point in time? When she returns Bob looks older and so should his entire universe. How can her view of the universe, which you say doesn't deviate from her "rest frame" observations that for her Bob's universe evolves slower when the omnipotent chart of the universe around them can't possibly deviate at any point in time!
 
  • #153
jerromyjon said:
Alice leaves and is time dilated on her trip and stops. Doesn't the universe around her look younger than what Bob experiences at this point in time?

No. They are both seeing the same universe, with the same age, because they are both at the same event in spacetime. They just took different paths to get to that event, and those paths have different lengths.
 
  • #154
Then they could never reunite in reality because those realities can never coexist. If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other? When she stops nothing could have changed timewise.
 
  • #155
CKH said:
[..] the traveler's clock never jumps in the traveler's frame, but the home clock does jump in the traveler's frame. Nothing is magical here, the traveler's frame abruptly changes its plane of simultaneity from the outbound IRF to the inbound IRF. So we are forced to admit that the simultaneous reading on the home clock "jumps" in the travelers frame.
[.. + rearranging:]
Finally just to be absolutely clear the "jump" in the home clock is not physical; it is rather a "jump" in the perspective in the traveler's frame. I can make the distance stars move faster than the speed of light in my frame just by rotating my body. It is not physical however.
Such a frame of varying velocity and acceleration can be constructed in SR but does not correspond to a valid reference system for physics in SR; as a result it leads to unphysical statements. But happily you finally clarified that you did not mean it the way you made it sound - although, I'm still not totally assured:
[..+ more rearranging, and emphasis mine:]

[..] we are choosing the unique inertial frame that is consistent with the traveler's local frame (which is continuous).
[..] the on-board clock defines the current time in the new rest frame. There is no free choice for time in the new rest frame because we have constrained it to be consistent with the traveler's non-inertial frame. [..]
3) The traveler's clock is continuous (it can't jump in it's own frame, his clock is a good clock).
SR uses non-local (universal) inertial frames in which nothing can "jump". Thus I had the impression that perhaps you try to establish something that is neither SR nor GR - and possibly with an inherent mistake. But you missed the pertinent part of how I tried to check if this is just a matter of phrasing (somewhat a matter of taste), or if there is more to it. I'm still not sure!
Once more, your phrasing sounded - and it still does - as if you think that you can set up a "local" physical frame such as a rocket (note: in SR a "frame" is extended and equipped with clocks) that can be automatically consistent with both inertial frames in which the traveller is in rest most of the time. I stressed that for such a purpose you can not use a 3D "frame" that is equipped with multiple clocks at different positions. Of course you can use a single clock (a point!), but that does not define a "frame". As a reminder, I stated:
"the home clock will also not jump to a later time if the traveler passively uses an extension of his on-board clocks to determine time at home. Instead, the traveler will first have to synchronize the on-board clocks to the new rest frame."
 
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  • #156
jerromyjon said:
Then they could never reunite in reality because those realities can never coexist.

Sure they can. Since it is apparent that relativity is a new subject to you, you should avoid making such categorical statements based on your current (highly incomplete) understanding. I assure you, everything I am saying comes from a perfectly self-consistent model of a single spacetime (i.e., a single "reality"), and none of it is in any way difficult or contentious from the standpoint of relativity. This is information you can find in any good relativity textbook.

jerromyjon said:
If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other?

No. She would be younger and Bob would be older. Note that this scenario is not set in flat spacetime, so it can't really be analyzed using SR. However, there is a similar scenario which is set in flat spacetime and can be analyzed using SR: for example, suppose Bob sits at the edge of a big merry-go-round, and Alice goes around with it. Then Alice will be younger than Bob when they meet again. (The key difference between the two scenarios is that, in the latter one--the merry-go-ground--Alice is accelerated and Bob is inertial; whereas, in the Earth-orbit scenario, it's the other way around--Bob is accelerated and Alice is inertial.)
 
  • #157
PeterDonis said:
And the answer is "mu": the question is not well-posed, because it assumes that there is a unique, physically given answer to the question. There isn't. Simultaneity is not a physically meaningful thing; it's just a convention. The way you are defining simultaneity is one way to do it, but it's not the only way, it has issues (like not assigning a unique time to all events), and it's no more meaningful, physically, than other ways of doing it.

It is well-posed if you consider all the details of how simultaneity is defined for the traveler. It is definitely not the same for all observers (but neither is the length of an object which is "physical"). I believe simultaneity is well-defined for an inertial frame. MCIFs along the path of a non-inertial object can be well-defined also by imposing time continuity and fixed spatial coordinates (to not do so yields meaningless results). I'll call this construction the "rest frame of that object". Given all that, you can consistently define simultaneity for non-inertial travelers in SR.

Maybe I forgot to note this; it's obvious but important. If you want to ask "in my rest frame what is the simultaneous reading on a clock at rest in a different frame", you must first synchronize clocks in those frames at some common event (such as departure in the example).

Concerning clock synchronizing conventions in SR. Bear in mind that if you pick one you must use it consistently. By convention I'm assuming the definition in Einstein's 1906 paper. Will the simultaneous readings on the home clock in the traveler's frame be different with a different choice of clock synchronization convention? If you don't already know the answer we can pick a different convention and try it out. I suspect it doesn't matter, because I don't see where that convention matters in the scenario of the twins.

Any clock synchronization scheme has to obey some rules to be consistent, so they are not arbitrary.

Einstein said "simultaneity is relative [to the observer's frame]". Measurements in general are relative but have meaning so long as you declare how the measurement is made.

Can you elaborate on "Simultaneity is not a physically meaningful thing; it's just a convention"? In particular, what criteria apply to "physically meaningful things".

Simultaneity is simply fixed time in an inertial coordinate system, a hyperplane. You have to pick such a system to talk about it. Would you go further and say that coordinates in general are not physically meaningful? They are a choice, but we have decided how we make that choice in an given situation, and thus we can make well-defined measurements of "physical things" in a coordinate system.
 
  • #158
CKH said:
No it's not unique, but the relative time results are the same whatever free choices you make for frames, given that you synchronize the traveler's clock with the home clock at the departure event.
I am not sure what you mean by "relative time results". If you mean the final numbers at the end, then yes, I agree. Those are invariant quantities so it doesn't matter what chart you use to calculate them (inertial or not), you will always get the same answer.

CKH said:
The MCIFs of the traveler along the path then have no free choices.
Yes, but the MCIF's are not the same as the traveler's rest frame.

CKH said:
Then let's not call it a chart if it has to satisfy such criteria.
The criteria are not optional if you want to use it to do physics. Chapter 2 explains why.

CKH said:
Just using the definitions provided in SR for inertial frames and simultaneity, I see no way of avoiding the fact the traveler's worldline has a tangent time axis at each point and the time on this axis at each point is the traveler's clock time.
If you use the SR definition for simultaneity on a non-inertial worldline then you get the radar coordinates of Dolby and Gull, not the kind of coordinates you have been describing so far.
CKH said:
The author complains that there are events to which multiple values of time can be assigned by the traveler and that the home clock can jump backward and what not. Isn't this exactly what the definition of simultaneity and proper time in SR require?
No. Each inertial chart assigns one and only one time to every event.

CKH said:
This "method" has little or nothing to do with simultaneous readings by the traveler of the time on the home clock as defined by SR.
It has everything to do with it. That is the way that Einstein defines simultaneity in his 1905 paper, but he used only inertial observers. The only thing that they did was to simply say "let's apply the standard SR method to a non-inertial observer".

CKH said:
On the other hand, a global coordinate system for the travelers frame can be defined using the MCIFs on the world line. But in that case you have to accept that events do not a have unique coordinates in this frame. That's not the fault of the method, it simply a consequence of SR.
It is not even a coordinate system (aka chart) let alone a global one, the problem is not a consequence of SR, it is the fault of the method, and there exist other methods without that problem.

I may not have much time for replying the rest of this week, so I would recommend that you really focus on reading and learning the material suggested.
 
  • #159
They can both be outside the vicinity of gravity, in deep interstellar space between two stars, I'll diagram it:
revolving dilation.png

Well Bob sits still (blue circle) and Alice plays the hadron which definitely has an experimentally proven time dilation. They both see the same stars, and SR says Alice's stars should evolve slower, as her existence is slowed. If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".
 
  • #160
PeterDonis said:
Proper time for a particular observer is only meaningful for events on that observer's worldline. If the observer wants to assign a "time" to events not on his worldline, he has to pick a simultaneity convention, and there is no one unique way of doing that. Even the obvious way for an inertial observer, namely just using the coordinate time in a coordinate chart in which he is always at rest, is not the only possible way for that observer. Simultaneity is simply not a physically meaningful thing; it's a convention.

OK. Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time? Hopefully I'll finally get what you mean by "convention" here.
 
  • #161
PeterDonis said:
This may or may not be an issue, depending on what you're trying to do. See below.

This depends on the spacetime. In flat spacetime, it is always possible to construct a valid chart (i.e., one meeting all of DaleSpam's requirements) that covers the entire spacetime, with any chosen timelike curve as its "time axis". The Dolby & Gull paper shows how to do this.

Are you sure? I was just quoting this paper in another thread (or earlier in this one, I forget where exactly), and my recollection is that Dolby & Gull only claimed that their coordinate system would cover the region of space-time that the observer on the timelike curve could send and receive signals from. Since the method is radar based, I don't see how the method could assign coordinates to regions of space-time that the observer could not both send and receive signals from. It's good - but not THAT good.

Unfortunately, when we consider the specific example of a uniformly accelerated observer, said observer can't send and receive signals from all of space-time :(.
 
  • #162
CKH said:
Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time?

The Dolby & Gull paper gives one, as DaleSpam has repeatedly pointed out to you.

Another one would be to just pick a particular inertial frame (say Bob's in the twin paradox scenario under discussion), and use that frame's simultaneity convention for everybody, regardless of their state of motion. For the scenario we've been discussing, this won't seem very natural, because only linear motion is involved. But, for example, if we have a bunch of observers at various points on a rotating disk, it can be very useful for everybody to adopt the simultaneity convention of the observer at the center of the disk (who is the only one who is at rest in an inertial frame), even though that convention does not match up with your definition for any other observer on the disk.
 
  • #163
jerromyjon said:
SR says Alice's stars should evolve slower, as her existence is slowed.

SR says no such thing. Alice's state of motion does not affect the evolution of the stars, except as that evolution is perceived by Alice. And the effect of Alice's motion will be that she perceives the stars to be evolving faster, compared to the rate of her clock, than Bob sees them evolving compared to the rate of his clock.

For example, if Bob sees each star emit a flare once per hour by his clock, Alice will see the same flares happening less than an hour apart by her clock, i.e., she will see them happening at a faster rate.

jerromyjon said:
If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".

No, that's not what would happen. Suppose Alice and Bob use some particular flare to synchronize their clocks; they pick a flare whose image (the light signal emitted from the flare) just happens to reach both of them at the same instant they are passing each other. They will both see this flare at the same instant, and at that instant, they both set their clocks to time ##t = 0##.

Now suppose that the interval between the flares is just right so that the next flare's image reaches Alice and Bob the next time they happen to be passing each other (i.e., Alice has completed half of an orbit while Bob sat still). Then they will see the next flare, once again, at the same instant; but at that instant, Bob's clock will read a later value than Alice's clock does. (For example, if Bob's clock reads exactly 1 hour, Alice's will read less than 1 hour.) That's how they know that Alice's clock is the one that is running slower: they have the same common pair of events (seeing the two flares), at which they are both co-located, to compare to, and Alice's clock shows less elapsed time between those two events.
 
  • #164
pervect said:
my recollection is that Dolby & Gull only claimed that their coordinate system would cover the region of space-time that the observer on the timelike curve could send and receive signals from. Since the method is radar based, I don't see how the method could assign coordinates to regions of space-time that the observer could not both send and receive signals from.

Hm, you're right, their basic method is radar-based. I thought I remembered that there was a way of extending it into the rest of spacetime, but re-reading the paper there doesn't seem to be.

One can still construct a coordinate chart with the properties I described, without using the Dolby & Gull method, if one is willing to drop orthogonality. Such a chart would still be one-to-one and diffeomorphic to other charts; it just would not have orthogonal axes at all events. (Also, of course, the physical interpretation of such a chart in a region which could not exchange light signals with one's chosen worldline would be problematic.)
 
  • #165
jerromyjon said:
They can both be outside the vicinity of gravity, in deep interstellar space between two stars, I'll diagram it:
https://www.physicsforums.com/attachments/75385
Well Bob sits still (blue circle) and Alice plays the hadron which definitely has an experimentally proven time dilation. They both see the same stars, and SR says Alice's stars should evolve slower, as her existence is slowed. If Bob really is older he could say " a solar flare happened at my "t=?" time and Alice could say, at some point in her slower time line, "Oh, yeah! Now I see it!".
Not necessarily. Alice might be closer the the flare and see it first.

Think about another scenario. Alice is moving very fast relative to Bob and as Alice passes Bob, the light from a solar flare hits their common area and both see it at the same time. Why do you think Alice's speed alone dictates when she sees things?

"Alice's existence is slowed" is meaningless. It only has meaning relative to Bob, not to Alice.
 
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  • #166
CKH said:
OK. Can you give me an example of another way (in SR) to define what is simultaneous with an observer's proper time? Hopefully I'll finally get what you mean by "convention" here.
Dalespam gave you a link to a very well known alternative back in post #141 (http://arxiv.org/abs/gr-qc/0104077). Did you look at it? It directly extends the Einstein simultaneity convention to non-inertial frames, and the result is quite different from MCIF simultaneity. Both are just definitions, with no physical content. The one in the referenced paper, though, is actually measurable, while the MCIF simultaneity is solely a computed quantity that cannot be measured.
 
  • #167
I imagine we could have 2 way video links between A and B which would allow Bob to see Alice moving slower while Alice would see Bob move faster... just like the pulses of a 5 second pulsar for Bob would be every 4 seconds for Alice spinning circles around Bob at .6c. Would the 60Hz video sent by Bob be received as 72hz by Alice and her 60hz be received by Bob at 48hz (frames per second)?

I've been thinking about this so much I'm confusing myself, lol. Good night.
 
  • #168
jerromyjon said:
Then they could never reunite in reality because those realities can never coexist. If she traveled that speed in orbit around Earth and Bob sat still where she could fly by him periodically very close to Bob and see him age and he see her they would both appear younger to each other? When she stops nothing could have changed timewise.

I really do think it's worth while to explore how the questions about mutual time dilation and the twin paradox have very close analogies in Euclidean geometry.

As for mutual time dilation: You have two roads that run between points A and B. One road, R, goes straight from A to B, and the other road, R', takes a round-about path. The length of R between [itex]A[/itex] and [itex]B[/itex] is [itex]L[/itex]. How do you compute the length of R'?

Well, let [itex]x[/itex] measure the distance along R, with A being at [itex]x=0[/itex]. At every point along R, you can look perpendicular to the road to see a corresponding point on R'. Let [itex]x'[/itex] be the distance along R' from A to this point. Then Euclidean geometry tells us that

[itex]\delta x' = \sqrt{1+m^2} \delta x[/itex]

where [itex]m[/itex] is the relative slope between the two roads (slope = the tangent of the angle between them). So to compute the length of a path along R', just integrate: [itex]L' = \int \sqrt{1+m^2} dx[/itex]. As long as [itex]m^2 > 0[/itex], [itex]L' > L[/itex]. So the nonstraight road is longer.

At this point, you could point out a "road paradox": Someone traveling along road R' could just as well look perpendicular to the road to see a corresponding point on R. The distance [itex]x[/itex] along R is related to the distance [itex]x'[/itex] along R' via the formula:

[itex]\delta x = \sqrt{1+m^2} \delta x'[/itex]

So the traveler along road R' would see that [itex]\delta x > \delta x'[/itex]. So he should conclude that the distance along R is longer than the distance along R'. He should conclude [itex]L > L'[/itex]. Paradox!

The resolution of the paradox is to realize that the recipe of finding a point [itex]x'[/itex] on road R' that corresponds to a point [itex]x[/itex] on road R gives different answers for R and for R'. If the line between [itex]x'[/itex] and [itex]x[/itex] is perpendicular to road R, then it WON'T be perpendicular to road R'. The two roads disagree about which points correspond. This is the Euclidean equivalent of the "relativity of simultaneity" in SR. R uses one convention for setting up corresponding points, and R' uses a different convention. The two conventions have different notions of which points correspond. They both correctly compute the relative lengths of segments of the other road, but they compare DIFFERENT segments.

So why does the calculation of relative length by R turn out to be right, and the calculation of relative length by R' turn out to be wrong? It's because R' is not straight. The convention that R' uses to figure out a corresponding point on R has sudden jumps whenever R' makes a turn. Because of these jumps, either a segment of R is skipped (when R' turns away from R), or a segment of R is counted twice (when R' turns toward the road). Because of these jumps, the length calculation by R' does not give the right answer. You can't use a nonstraight road to compute the length of a straight road (at least not in a straight-forward way). This is exactly analogous to the twin paradox: you can't use a noninertial path to compute the proper time for an inertial path without a lot of extra work.

The above is all about roads on a flat Earth, which is analogous to paths through spacetime in flat spacetime. Your question about orbits brings up curved spacetime, so the analogy would be roads on a curved Earth. Consider two roads that both run straight from the North Pole to the South Pole. Near the North Pole, the length of one road can be approximately computed in the same way as for a flat Earth, because the curvature is only noticeable when you've traveled a long distance. So right near the North Pole, you can calculate the length of one road relative to the other, and use the formula [itex]\delta x' = \sqrt{1+m^2} \delta x[/itex], and each road will conclude that the other road must be longer. But the resolution here is that the formula [itex]\delta x' = \sqrt{1+m^2} \delta x[/itex] is only good for small distances, where you can ignore the curvature of the Earth. It can't be used to compute the length of a road running all the way from the North Pole to the South Pole.
 
  • #169
jerromyjon said:
I imagine we could have 2 way video links between A and B which would allow Bob to see Alice moving slower while Alice would see Bob move faster... just like the pulses of a 5 second pulsar for Bob would be every 4 seconds for Alice spinning circles around Bob at .6c. Would the 60Hz video sent by Bob be received as 72hz by Alice and her 60hz be received by Bob at 48hz (frames per second)?

I've been thinking about this so much I'm confusing myself, lol. Good night.
No, that doesn't make sense. WHATEVER the speed of Bob relative to Alice it is exactly the same as the speed of Alice relative to Bob.
 
  • #170
I did a search on this forum for "Brian Greene", together with the key words "accelerating viewpoint", "chart" and "co-moving", and I found an old thread that has a lot of discussion about some of the issues that have been raised on this thread.

Here's a link to one of the posts on that thread that talks about charts (it is post #14):

https://www.physicsforums.com/threa...he-accelerating-traveler.671804/#post-4278370Here's a link to one of the posts that talks about how to do the co-moving method (it is post #34):

https://www.physicsforums.com/threa...elerating-traveler.671804/page-2#post-4293784 Here's a link to a post that describes how to get the result that Brian Greene talked about on that NOVA show (it is post #289 of a different thread):

https://www.physicsforums.com/threa...-dilation-of-time.605080/page-15#post-4050221
 
  • #171
Philosophy reference removed
PAllen said:
Dalespam gave you a link to a very well known alternative back in post #141 (http://arxiv.org/abs/gr-qc/0104077). Did you look at it? It directly extends the Einstein simultaneity convention to non-inertial frames, and the result is quite different from MCIF simultaneity. Both are just definitions, with no physical content. The one in the referenced paper, though, is actually measurable, while the MCIF simultaneity is solely a computed quantity that cannot be measured.
I did go back and read that article about the "radar conventions" some more, but the full significance of how this (radar time) convention is "different" has not sunk in yet. In all honesty the light hasn't dawned yet on what the "conventionality" you refer to is about, how it affects our analysis of motion (I obviously need to study it more). You say radar time directly extends Einstein's convention (meaning it matches in any inertial frame and so it is the same convention in that context) but then you say the result is different from MCIF simultaneity. That sounds paradoxical because any MCIF already defines planes of simultaneity. Can two different definitions coexist in a consistent system?

There is also a "radar distance" defined in that paper. It's interesting that Einstein didn't explicitly establish a convention (AFAIK) for spatial measurement (he refers only to normal Euclidean measurement of space using rulers) but he did define a convention for time measurements in different locations. It hardly a convention at all in Einstein's description of SR, since he takes the one-way constancy of c as a given; no different convention that makes sense with those givens.

Simultaneity of two events means that they have the same time coordinate value in some "valid" 4-D coordinate system, agreed? So time coordinates must also be conventional. Is the Lorentz transformation just "conventional" whatever that means? The transformation (as I've seen it written) transforms time and space coordinates in different inertial frames. If the transformations are actually independent of these "conventions" that implies that all simultaneity conventions must be consistent in some way for all inertial frames.

What about Minkowski space, does that define simultaneity as Einstein's convention? Do you break Minkowski space by adopting some other convention?

By the way, I don't believe that simultaneity cannot be measured in a MCIF. It is nothing more that a certain inertial frame. In principle, a physical frame can be pre-equipped with synchronized clocks linked in a rigid framework at any required points in space. At the spatial position of an event, the clock can be read, when the event arrives. Hence we can measure when events in an MCIF occur and directly determine whether two of them were simultaneous by experiment. So doing this mathematically could be confirmed by experiment.

I believe that in the simplified twin paradox scenario, that simultaneous readings on the home clock and traver's clock can be made in an experiment. I'll elaborate if necessary, but to summarize, comoving (3D) frames for the outbound and inbound journeys can be pre-equipped with instruments that record both the home clock and traveler clocks readings simultaneously in each frame near the turn around. One recording is made just before turnaround in the outbound frame and one just after in the inbound frame. In the travel's 3D frame these two readings are made as close as you like to simultaneously.

The resulting records will show a jump in the home clock reading before and after the turn around while the traveler's time has barely changed. These frames move inertially with the traveler on each leg of the journey, thus simultaneity for the travel is defined by each in each leg. Their recording for the traveler's clock and the home clock all occur simultaneously with the turn-round event. Simultaneity in these frames is as described by Einstein's 1905 SR. Thus it seems that in SR (at least as described by Einstein), we are stuck with the result that home clock is discontinuous with the traveler's clock time at the turn around, and in principle it can be tested.

Does this jump mean anything non-physical occurred? No, the observation can be explained without any magic. Blame relativity.
 
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  • #172
CKH said:
Simultaneity of two events means that they have the same time coordinate value in some "valid" 4-D coordinate system, agreed? So time coordinates must also be conventional. Is the Lorentz transformation just "conventional" whatever that means?

The Lorentz transforms assumes that one is using Einstein clock synchronization in an inertial frame. It requires conventions to define, conventions which Einstein gave. Whether or not you want therefore to call it "conventional" is a semantic issue.

The transformation (as I've seen it written) transforms time and space coordinates in different inertial frames. If the transformations are actually independent of these "conventions" that implies that all simultaneity conventions must be consistent in some way for all inertial frames.

See above.

What about Minkowski space, does that define simultaneity as Einstein's convention? Do you break Minkowski space by adopting some other convention?

You would change the metric (usually diagonal -1, 1, 1, 1) if you changed the coordinates or the conventions on which they were based, but I would say that you wouldn't change the space itself. It'd still be Minkowskii space, but with unusual coordinates. the different coordinates would imply a different metric. Exactly what the metric would be would depend on the coordinate choice.

By the way, I don't believe that simultaneity cannot be measured in a MCIF. It is nothing more that a certain inertial frame. In principle, a physical frame can be pre-equipped with synchronized clocks linked in a rigid framework at any required points in space. At the spatial position of an event, the clock can be read, when the event arrives. Hence we can measure when events in an MCIF occur and directly determine whether two of them were simultaneous by experiment. So doing this mathematically could be confirmed by experiment.

Certainly simultaneity can be measured in a MCIF. Every different inertial frame has a different notion of simultaneity, though. I am suspecting from some of your later comments you might not be aware of this. Are you familiar with Einstein's train experiment?

See for instance http://www.bartleby.com/173/9.html

I'll omit quoting the whole page, and just stress the important part of the conclusion. I would recommend reading the original , my summary is a bit choppy. Though if your time is limited, I'd just skim over the reference by Einstein, and study the Scherr et al paper below instead.

Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result: 3
Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity).

The following paper may be useful as well, it is oriented towards the teacher but it has some expereince behind it on the best way to present the issue.

If you only have time to study one paper in depth, I'd suggest studying this one, assuming you have the time to study any at all.

"The challenge of changing deeply held student beliefs about the relativity of simultaneity", Scherr et al, http://arxiv.org/abs/physics/0207081

The original version of the train paradox used in tutorial differs slightly from the one
described above. Students are told that two sparks occur at either end of a train that
moves with relativistic speed relative to the ground. The sparks leave char marks on
the ground and on the train.

The ground-based observer, Alan, who is at rest midway between the marks on the
ground, receives the wavefronts from the sparks at the same time. (See Fig. 2(a).)

The question is what order Beth, who is on the train, and in the middle of the train, receives the events. Without quoting the paper in full, I'll state that Beth, in the center of the train, does not receive the signals at the same time. Due to student resistance in drawing the correct conclusion from the above facts, the following reformulation was used by the authors to try to drive the point home.

We decided to modify the tutorial to help students recognize the ‘paradox’ in the
train paradox. The approach we took was to shift the focus from the time order of two
events (the reception of each wavefront) to whether or not a single event occurs.

In the modified tutorial, students are told that Beth has a tape player that operates as
follows. When wavefront F reaches the tape player, it starts to play music at top volume.
When wavefront R reaches it, the tape player is silenced. If both wavefronts reach
the tape player at the same instant, it remains silent. Students are asked whether
the tape player plays (i) in Alan’s frame and (ii) in Beth’s frame.
 
  • #173
At this point there is no reason to continue this thread. Plenty of references have been provided. There is no value in continuing to address the same basic misconceptions that have already been addressed multiple times.

If any of the participants feel they have questions remaining then they should read the provided references carefully prior to posting a specific question in a new thread.
 
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<h2>1. What is acceleration and how does it relate to the twin paradox?</h2><p>Acceleration is the rate of change of an object's velocity over time. In the context of the twin paradox, it refers to the change in velocity of one twin as they travel through space, while the other twin stays on Earth. This acceleration is a key factor in understanding the time dilation effect of the twin paradox.</p><h2>2. Can you explain the twin paradox in simple terms?</h2><p>The twin paradox is a thought experiment that explores the effects of time dilation on two individuals, one of whom travels through space at high speeds while the other stays on Earth. Due to the effects of special relativity, the traveling twin experiences time at a slower rate, resulting in a younger age when they return to Earth compared to their twin who stayed on Earth.</p><h2>3. Is the twin paradox a real phenomenon or just a thought experiment?</h2><p>The twin paradox is a thought experiment used to explain the concept of time dilation in special relativity. While it is not possible for humans to travel at the speeds necessary for the twin paradox to occur, the effects of time dilation have been observed in experiments with particles traveling at high speeds.</p><h2>4. What are some real-life examples of the twin paradox?</h2><p>One example often used to illustrate the twin paradox is the scenario of one twin traveling in a spaceship near the speed of light while the other twin stays on Earth. Another example is the time dilation experienced by astronauts on the International Space Station due to their high speeds in orbit.</p><h2>5. Can the twin paradox be resolved?</h2><p>The twin paradox can be resolved by considering the effects of acceleration and the fact that the traveling twin must turn around and return to Earth, which results in a different experience of time for each twin. This resolution is in line with the principles of special relativity and does not contradict any known laws of physics.</p>

1. What is acceleration and how does it relate to the twin paradox?

Acceleration is the rate of change of an object's velocity over time. In the context of the twin paradox, it refers to the change in velocity of one twin as they travel through space, while the other twin stays on Earth. This acceleration is a key factor in understanding the time dilation effect of the twin paradox.

2. Can you explain the twin paradox in simple terms?

The twin paradox is a thought experiment that explores the effects of time dilation on two individuals, one of whom travels through space at high speeds while the other stays on Earth. Due to the effects of special relativity, the traveling twin experiences time at a slower rate, resulting in a younger age when they return to Earth compared to their twin who stayed on Earth.

3. Is the twin paradox a real phenomenon or just a thought experiment?

The twin paradox is a thought experiment used to explain the concept of time dilation in special relativity. While it is not possible for humans to travel at the speeds necessary for the twin paradox to occur, the effects of time dilation have been observed in experiments with particles traveling at high speeds.

4. What are some real-life examples of the twin paradox?

One example often used to illustrate the twin paradox is the scenario of one twin traveling in a spaceship near the speed of light while the other twin stays on Earth. Another example is the time dilation experienced by astronauts on the International Space Station due to their high speeds in orbit.

5. Can the twin paradox be resolved?

The twin paradox can be resolved by considering the effects of acceleration and the fact that the traveling twin must turn around and return to Earth, which results in a different experience of time for each twin. This resolution is in line with the principles of special relativity and does not contradict any known laws of physics.

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