Twin Paradox- a quick(ish) question

[otg]

All you've done wrong is presume that the Klingon passing by the Earth will see things at the same size, and distance as someone on Earth. They won't. Different frame, different distances and times for remote events.

Ah, but I didn't presume this. I never said that the Klingon passing the Earth would be in the same frame as me on the earth. This was only a result of the previous claim that the earth-at-rest-frame-point-of-view is different from the spaceship-at-rest-frame-point-of-view. I summarized the situations in two sentences in which I used the exact same words, except for the NOT, which made you say there is a difference between 1 and 2.

If what you write is true, that 1 will occur, but also that 2 will occur (if I remove the NOT, since it won't if I keep it), then 1 and 2 are indistinguishable. [if not, why? [my original [and so far only] question]

[I sometimes get a feeling that I think I ask something different than what you think you're answering... is this observation in my rest frame different in your rest frame? :)
You do have a good way of replying. Thank you for your patience]

 

[sylas]

Ah, but I didn't presume this. I never said that the Klingon passing the Earth would be in the same frame as me on the earth.

You effectively presumed this when you said that the distances are not changed, and when you say passing. If it is passing, then it is moving, and therefore it is NOT in the same frame. (text in blue added in edit)

If what you write is true, that 1 will occur, but also that 2 will occur (if I remove the NOT, since it won't if I keep it), then 1 and 2 are indistinguishable. [if not, why? [my original [and so far only] question]

I do not think you are defining the situation sufficiently carefully. This is why you badly need to come up with maths, or a diagram, or SOMETHING that will let you be precise. Without that, you will almost certainly make invalid assumptions without meaning to, and without even realizing it.

In your original statement for example, you say this:

1.
Spaceship-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the Earth should see the distant Earth at a different angle than I do.

2.
Earth-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the spaceship should NOT see the distant spaceship at a different angle than I do.

Note that you speak of TWO Klingons. One at Earth, one at a spaceship. And you also speak of synchronizing your watch to both of them.

That's impossible. You cannot synchronize with two events that are separated in space, because you and the Klingon don't have a common reference frame for times. I explained this previously, and you objected. This is almost certainly because you have not yet got your head around the idea that simultaneity is relative. Nearly always, the problems come down to this.

You can have an Earthbound twin, who synchronizes with a Klingon passing by the Earth, and a traveling twin, who synchronizes with another Klingon as they both pass by a distant star. You can only synchronize when you have observers at the same location in the same instant.

Try again. Spell out exactly where and when you have any synchronizations.

Here's an account of different events.

I am assuming we have Earth, and another star six light years away, which is at rest with respect to the Earth. The six light years distance is as measured from Earth, or from the star. Same thing, as they are at rest with respect to each other.

I am assuming we have a traveler, who travels from Earth to the star at 60% of the speed of light, and then returns at the same speed. It takes 20-Earth years to make the round trip. During the trip, they age 16 years.

Now. Where are the Klingons you speak of? If passing Earth, WHEN do they pass Earth. If passing the other star, WHEN do they pass the other star? At what speed, and what direction?

I'm proposing Klingon-A, who moves always in a straight line, at 60% light speed wrt Earth, and in a direction from the star towards Earth. This Klingon passes by the star in the same instant that the traveler from Earth arrives at the star.

I did this, because THIS Klingon can pick up the traveler by a short range teleportation device, and bring them back to Earth. It makes it convenient to describe where Earth is from the perspective of the inbound Klingon, or the outbound Enterprise, as the both pass by each other and by the star.

Clear enough? If you have another Klingon in mind, see how carefully you can identify where they are. You speak of them passing Earth. When? In what direction? At what speed?

Klingon-A, by the way, observes that Earth is approaching them at 60% light speed. As the star flies past the Klingon, they are observing light from Earth that was emitted twelve years previously, when Earth was at a distance of 12 light years.

The Enterprise, traveling in the other direction, observes that Earth is receding into the distance at 60% light speed. As the star flies past the Enterprise, they are observing light from Earth that was emitted three years previously, when Earth was at a distance of 3 light years.

For aliens living on the star, they are observing Earth with light that was emitted six years previously. Earth is not moving as far as these aliens are concerned, and is always 6 light years distant.

All three observations are looking at the same photons. If a Bomb is let off on Earth at the right time, all three observers -- Enterprise, Klingon-A, and the aliens at the star, will see the bomb in the same instant as Enterprise and the Klingon pass by the star. The distance to the explosion is different for all observers. It depends not only on time and location, but on velocity as well.

The bomb will need to be detonated 4 Earth-years after the Enterprise left.

Cheers -- sylas

 

[otg]

Note that you speak of TWO Klingons. One at Earth, one at a spaceship. And you also speak of synchronizing your watch to both of them.

Still haven't read the reply thoroughly, just wanted to make clear that I did not speak of two klingons to both of which I sync my watch. I simply [think that I] had the situation described in two different ways.

Either I see it as

person being at rest in the spaceship frame in the outbound leg [spaceship moving out away from the earth] meeting a klingon going the other direction and syncing with him

OR

person being at rest in the Earth frame at the outbound leg [earth moving out away from the spaceship] meeting a klingon going the other direction and syncing with him

I still have only renamed the spaceship "earth", but there is supposed to be a difference. I'll read your last reply and hopefully never come back to this forum in this issue :)

 

[sylas]

Still haven't read the reply thoroughly, just wanted to make clear that I did not speak of two klingons to both of which I sync my watch. I simply [think that I] had the situation described in two different ways.

Either I see it as

person being at rest in the spaceship frame in the outbound leg [spaceship moving out away from the earth] meeting a klingon going the other direction and syncing with him

OR

person being at rest in the Earth frame at the outbound leg [earth moving out away from the spaceship] meeting a klingon going the other direction and syncing with him

One difference is the relative velocity of your passing Klingon. Assuming that all Klingons are moving in the direction from star towards Earth, at 60% light speed wrt to Earth. Then the Klingons are moving at over 88% light speed wrt to the outbound traveller. (Actually 15/17 of light speed.)

The Klingon passing Earth and looking back at the receding star sees light that left the star 3 years previously. The Earth sees the same light, and that light left the star six years previously. Both distances are correct, for the different observers.

The Klingon passing the star looks forward to see the approaching Earth, and sees light that left Earth 12 years previously. The Enterprise passing the star looks back to see the receding Earth, and sees light that left Earth 3 years previously. (And this IS symmetrical with the Klingon passing the Earth and leaving the star behind.)

You can calculate these numbers by using the Lorentz transformations.

I still have only renamed the spaceship "earth", but there is supposed to be a difference. I'll read your last reply and hopefully never come back to this forum in this issue :)

Don't be shy... very few people get this straight away; I certainly didn't.

Cheers -- sylas

 

[JesseM]

But I don't need the concept of acceleration to see that there is a change of frame for Betty at the turnaround. I can simply let her synchronize her clock with a third person who seems identical to Betty in Betty's outbound frame at the event of synchronization although she's moving in the opposite direction.

The fact remains that you are comparing a path made up of two segments that have different velocities as seen in every inertial frame with Alfred's clock which has a constant velocity in every inertial frame. What's more, in Betty's frame this third person's clock does not read the same time as her clock at the moment the third person passes Alfred, because in Betty's frame the third person's clock is slowed down.

Why can't I change the name "Betty" to the name "Alfred" without changing her age at the reunion?

What do you mean "change the name"? Do you mean having Alfred be the one that either accelerates to return to Betty, or just synchronizes clocks with a third person and never reunites with Betty himself?

If I remove the earth, I can place Alfred in a spaceship where the Earth "used to be". How can Betty tell that it's not the Alfredian spaceship which is moving, changing frame and returning?

Because if an accelerometer is placed on Alfred's ship it will read zero throughout, showing that it was moving at constant velocity in every inertial frame. It's implied by the definition of an inertial frame that any object at rest or moving at constant velocity in an inertial frame must experience zero G-forces as measured by accelerometers.

Why do we need the concept of acceleration to explain why her frame is non-inertial, if there is no such thing as an absolute space in relation to which one accelerates?

Because acceleration is absolute in SR--you can tell objectively whether you're accelerating or not by looking at an accelerometer.

If there are three people involved as above, how can we tell that there are two Bettys and not two Alfreds?

I don't understand this question--obviously there must be an objective fact of the matter as to whether the third person passes Alfred or Betty first (and synchronizes clocks with whoever he passes first at the moment they pass). Relativity doesn't involve uncertainty about the identity of different worldlines, after all.

 

[matheinste]

Yes, it does change immediately at turn around. It changes immediately at turn around in precisely the same way as the direction to the other twin changes immediately at turn around.

If you switch from 60% light speed outbound, to 60% light speed inbound, the angular size of the objects in your direction of motion change instantly by a factor of 4, reflecting their new distance in the new reference frame. Recall that the size of an object in the sky depends not on the "current distance", but the distance that has been traveled by the photons now reaching you.

Cheers -- sylas

Addendum: Oops. Actually the solid angle is 16 times smaller... each of the two transverse angles is 4 times smaller.

One thing I cannot grasp is that if we consider a position just before turnaround and a position just after turnaround where the relative speeds are equal and opposite in direction
how can the "new" distance between traveller and Earth be different on these two occasions. Surely the distance depends only on relative speed and not on direction. As nobody else has any objection to this description I must assume that there is something lacking in my understanding of the situation. Perhaps you could throw in some mathematics to make it clearer to me.

I have never had a problem with the resolution of the twin paradox before but have never come across this aspect of it. I know that we can also resolve the paradox using the Lorentz contracted distance for the traveller which the earthbound twin does not experience, but as far as the traveller is concerned his inbound and outbound legs are symmetrical with regards to distance, although of course his line of simultaneity is different in and outbound.

This point is really bothering me and I would be glad of some help in its resolution.

To recap, my query is, how is the distance between the traveller and Earth different immediately before turnaround from immediately after turnaround.

Matheinste

 

[sylas]

One thing I cannot grasp is that if we consider a position just before turnaround and a position just after turnaround where the relative speeds are equal and opposite in direction
how can the "new" distance between traveller and Earth be different on these two occasions. Surely the distance depends only on relative speed and not on direction. As nobody else has any objection to this description I must assume that there is something lacking in my understanding of the situation. Perhaps you could throw in some mathematics to make it clearer to me.

Use the Lorentz transformations. Remember… velocity is signed. The reference frame for a ship passing the star with v = 0.6 is different from the one in which the ship passes the star with v = -0.6. I am using units where c=1.

Here is the general form of the Lorentz transform. Let (t,x) be a spacetime event in the frame of observer A. Assume that observer B has a velocity of v with respect to A.

To map (t,x) into the frame of B, use the following:

$$\begin{equation*}\begin{split} t' & = \gamma ( t - vx ) \\ x' & = \gamma ( x - vt ) \\ \gamma & = \frac{1}{\sqrt{1 - v^2}} \end{split}\end{equation*}$$​

To map back into the frame of A, note that A has velocity of -v with respect to B. Hence we use

$$\begin{equation*}\begin{split} t'' & = \gamma ( t' + vx' ) \\ & = \gamma^2 ( t - vx + v(x- vt) ) \\ & = \gamma^2 ( 1 - v^2 ) t \\ & = t \end{split}\end{equation*}$$​

A similar proof applies for x'' = x. In other words, it's all consistent. You map from the frame of A to the frame of B and back again with the same transformations, but remember that the sign of velocity is reversed.

OK. Make it concrete. Let (0,0) be the turn around event. In the frame of the star, the Earth is 6 light years away. Hence, in the frame of the star, the light that they are now seeing from Earth was emitted six years ago, and at a distance of six light years. Assume that the positive direction is from Earth towards the star. Thus we have the light from Earth being emitted at the spacetime event (-6,-6), in the frame of the star.

We can transform that event at (-6,-6) for a passing spaceship moving at speed 0.6. We get different results depending as the velocity is positive, or negative. (We are keeping to one dimension for simplicity throughout.)

Here's what you have for v=0.6, which is a spaceship outbound from Earth:

$$\begin{equation*}\begin{split} t & = x = -6 \\ v & = 0.6 \\ \gamma & = 1.25 \\ t' & = \gamma ( t - vx ) \\ & = 1.25 \times -6 \times (1-v) \\ & = 1.25 \times -6 \times 0.4 \\ & = -3 \\ x' & = \gamma ( x - vt ) \\ & = 1.25 \times -6 \times (1-v) \\ & = -3 \end{split}\end{equation*}$$​

They are seeing Earth from a distance of 3 light years, with light that left 3 years previously.

Here's what you have for v=-0.6, which is a spaceship inbound for Earth:

$$\begin{equation*}\begin{split} t & = x = -6 \\ v & = -0.6 \\ \gamma & = 1.25 \\ t' & = \gamma ( t - vx ) \\ & = 1.25 \times -6 \times (1-v) \\ & = 1.25 \times -6 \times 1.6 \\ & = -12 \\ x' & = \gamma ( x - vt ) \\ & = 1.25 \times -6 \times (1-v) \\ & = -12 \end{split}\end{equation*}$$​

They are seeing Earth from a distance of 12 light years, with light that left 12 years previously.

You can also use these transformations for a pinhole camera on board the spaceships, and use that to show that the angle subtended by light moving in the direction of motion has an angle that scales with the redshift.

Cheers -- sylas

 

[matheinste]

Hello sylas.

Thanks for a quick reply. I see what you are saying but I look at it as distance being the same but the line of simultaneity being different. That is, we are seeing the Earth as it is/was at different times, not from different distances.

I wil take a more detailed look later. Thanks.

Matheinste.

 

[sylas]

Hello sylas.

Thanks for a quick reply. I see what you are saying but I look at it as distance being the same but the line of simultaneity being different. That is, we are seeing the Earth as it is/was at different times, not from different distances.

I wil take a more detailed look later. Thanks.

Matheinste.

If you are seeing it at different times, it MUST also be seen at different distances. You are seeing it with light, that travels at the same speed for everyone. And note here that "different time" means the SAME instant, the SAME spacetime event -- which occurred a different amount of time previously depending on your perspective.

Just remember that what you see is not simultaneous with the observation. Hence simultaneity is not what you want for the angular size of something in the sky. What you see in the sky is events that have already occurred in the past.

For both the inbound and outbound ship, the distance to the Earth "now" is 4.8 light years. The problem is that they don't have the same idea of "now", and they aren't seeing the Earth "now".

They CAN identify a common event along a photon light path. They are both seeing light from the Earth that was emitted at an unambiguously identified event in the past. They just locate it at a different time and distance. The sudden change in angular size is another way to recognize that you have turned around, even if you fail to notice an acceleration for some reason.

Cheers -- sylas

 

[matheinste]

Hello sylvas.

In a standard spacetime diagram using the inertial frame of Earth there is no distance change immediately before/after the traveller's turnaround. Is this change represented in the spacetime diagram using the successive frames of the traveller?

Matheinste

 

[sylas]

Hello sylvas.

In a standard spacetime diagram using the inertial frame of Earth there is no distance change immediately before/after the traveller's turnaround. Is this change represented in the spacetime diagram using the successive frames of the traveller?

Matheinste

Yes, there is a distance change.

Recall, we are talking of the distance traveled by light from Earth, as seen at the turn around; not the distance between two events along anyone's plane of simultaneity.

Also, your standard spacetime diagram shows the perspective of one reference frame; and so you would need TWO diagrams to show what is seen before, and after, the turn around. Or you can use a non-standard diagram.

Here I show FOUR spacetime diagrams, all on the one grid. Scale is years (vertical) and lightyears (horizontal). There are four world lines shown.

The diagrams show four ways to look at the same history of events. This involves Capt. Kirk traveling outbound from Earth to another star, 6 light years from Earth as measured on Earth, at which point Kirk teleports across to a passing inbound Klingon freighter. A bomb is set off on Earth just at the right time to be seen by Kirk and the Klingons as teleportation occurs.

World lines:

• Green is the Earth, and a light green circle for the bomb event.
• Red is light from the bomb, received at the turn around; and also light from the turnaround going back to Earth.
• Brown is Capt. Kirk, outbound.
• Purple is the Klingon freighter, headed inbound for Earth at constant velocity.

Diagrams:

• Left most diagram is from Earth's perspective. Turn around is six light years away. The bomb is set off four years after Kirk's departurn. The turn around is seen 16 years after Kirk's departure. The Klingon freighter drops Kirk back home 20 years after departure.
• Next diagram across is from Kirk's outbound perspective. Note the distance to the bomb.
• Right most diagram is from the perspective of the Klingon freighter. Note also that they see Kirk approaching at 15/17 of the speed of light.
• The unconventional diagram is in the upper right. This is what Kirk sees during his round trip. All events are located according to Kirk's frame at the time he receives photons from that event. It is obtained by combining views of Kirk outbound with the view of the freighter.

All locations in these spacetime diagrams can be calculated using Lorentz transformations.

Cheers -- sylas

 

[otg]

I get a feeling that my original question, which I think was quite straightforward, got lost somewhere in all the lengthy discussions about Klingons... I have no issue with simultaneity being relative, and no problem accepting that different frames give different results, nor the relativity theory itself.
The issue here is just that it's not always easy to get the head right on what happens in which frame when one tries to see it in different ways, and where those ways are different even though they seem not to be.

My original question is this attached pdf. Why is the away-traveller older than the other in one view, and younger in the other? I only changed symbols and letters.
Correction: I think I only changed symbols and letters.

 

[sylas]

I get a feeling that my original question, which I think was quite straightforward, got lost somewhere in all the lengthy discussions about Klingons... I have no issue with simultaneity being relative, and no problem accepting that different frames give different results, nor the relativity theory itself.
The issue here is just that it's not always easy to get the head right on what happens in which frame when one tries to see it in different ways, and where those ways are different even though they seem not to be.

My original question is this attached pdf. Why is the away-traveller older than the other in one view, and younger in the other? I only changed symbols and letters.
Correction: I think I only changed symbols and letters.

Here is your diagram:

You don't say "older" with respect to which events.

Let me see if I understand correctly. In the upper part of the diagram, you have "A" being the Earth, at rest. "B" travels outbound from Earth, and passes "C", who travels back to Earth again. No problem there.

What is the next diagram trying to show? Is it the same events from the perspective of "C"? If so, you have "B" passing by "A" on the right, then synchronizing with "C", and then "C" ariving at "A"... except that this time, "A" is moving towards "C". So you really need to indicate "A" moving towards "C", but more slowly than "B".

When you say "older", what are you comparing? You have only one reference point where ages of "A" and "C" can be compared. If you try to get their ages from some other point, then you have the same problem that ALWAYS shows up in these discussions. You are (effectively) assuming simultaneity.

We can relate your diagrams to my diagrams as follows. "A" is the Earth (green world line in my diagrams) "B" is Capt. Kirk, who travels outbound from Earth (brown world line). "C" is the Klingon (purple world line) who travels back towards Earth again, having passed "B" (Capt. Kirk) and synchronized, at the turn around point.

Hence the upper part of your diagram corresponds to my left-most spacetime diagram, and the lower part of your diagram is... what? "C" being at rest? If so, this is the rightmost of my spacetime diagrams, for the Klingon's reference frame, and there's no basis to compare ages of "A" and "C", because the Klingon only has one point at which they can synchronize with Earth.

You have three events. "B" leaves from "A". Then "B" synchronizes with "C" as they pass by each other. Then "C" arrives at "A".

The age of "A" in this series of events is unambiguous, because "A" is there at the start, and at the finish. But what age of "C" can you compare? You have to identify a point in time when you start counting "C"'s age. If you make it simultaneous with "B" leaving "A", then you are not actually identifying a starting point at all, because you cannot synchronize "C" with the event of "B" leaving "A". They weren't there. I pointed this out previously that you can't synchronize with remote events.

If "C" and "B" are the same individual, however, (with a turn around) then you have what I have described for Capt. Kirk, who teleports into the Klingon ship for the return part of his journey.

Cheers -- sylas

 

[matheinste]

Hello sylvas.

I make no mention of Klingons. I think we are at cross purposes here. It's my fault for interupting someone elses thread so causing confusion. I will back out now and perhaps start another thread on this changeing distance point.

Thanks for your replies.

Matheinste.

 

[sylas]

Hello sylvas.

I make no mention of Klingons. I think we are at cross purposes here. It's my fault for interupting someone elses thread so causing confusion. I will back out now and perhaps start another thread on this changeing distance point.

Thanks for your replies.

Matheinste.

No problem. I don't think you caused any confusion. Here again are the three stock standard space time diagrams for the traveling twins, one for each perspective, but with all mention of Klingons removed. There is also a mixed diagram, in which the traveler is always at rest, and all other events are at spacetime co-ordinates for the frame of the traveler at the time the photons arrive to let the traveler see that event.

The traveler goes to another star at 60% light speed and then returns. At the point when they turn around, the light coming from Earth is coming from a point at 3 years previously (just before the jump) and from a point at 12 years previously (just after the jump). The Earth is being seen at the same instant in spacetime, but the distance and time to that event changes with the new inbound perspective of the traveler.

The only reason for introducing Klingons was to emphasize that acceleration is not what matters... only the new perspective. The same diagrams apply whether the traveler turns using an almost infinite acceleration, or a spacetime warp, or by teleporting into a passing spaceship headed back the way they have just come.

Cheers -- sylas

 

[otg]

Let me see if I understand correctly. In the upper part of the diagram, you have "A" being the Earth, at rest. "B" travels outbound from Earth, and passes "C", who travels back to Earth again. No problem there.

Correct

What is the next diagram trying to show? Is it the same events from the perspective of "C"? If so, you have "B" passing by "A" on the right, then synchronizing with "C", and then "C" ariving at "A"... except that this time, "A" is moving towards "C". So you really need to indicate "A" moving towards "C", but more slowly than "B".

No, it is the same events from the perspective of "A" [who I named "B" in the previous pic]. In your words:
In the lower part of the diagram, you have "A" being the spaceship, at rest. "B" travels outbound from spaceship, [that is, Earth moves away from "A" in "A":s perspective] and passes "C", who travels back to spaceship again.

This is [or should be] identical to the above situation. Why is it not? Why must I say that it is the spaceship that shifts frames?

The age of "A" in this series of events is unambiguous, because "A" is there at the start, and at the finish. But what age of "C" can you compare? You have to identify a point in time when you start counting "C"'s age. If you make it simultaneous with "B" leaving "A", then you are not actually identifying a starting point at all, because you cannot synchronize "C" with the event of "B" leaving "A". They weren't there. I pointed this out previously that you can't synchronize with remote events.

But in all explanations, we compare "C" [the home coming twin] with "A" [the staying twin] even though "C" "used to be "B"".
"A" is there at the start, and at the finish, yes. But why can't "A" be the spaceship? What if we replace the Earth with a spaceship? Where is "there", the place where "A" is all the time, if there is no absolute space?

 

[JesseM]

But in all explanations, we compare "C" [the home coming twin] with "A" [the staying twin] even though "C" "used to be "B"".
"A" is there at the start, and at the finish, yes. But why can't "A" be the spaceship? What if we replace the Earth with a spaceship? Where is "there", the place where "A" is all the time, if there is no absolute space?

You are certainly free to modify the physical situation so that A was the one who changed velocities while B did not. But then all frames will agree that A was the one who changed velocities and B did not. The point is that in the same physical situation you can't have one frame that thinks A changed velocities while B moved inertially, and a different frame which thinks B changed velocities and A moved inertially; however you choose to set up the physical situation, all inertial frames agree about which one moved inertially and which changed velocities, and they all will predict that the one who changed velocities will be younger when he reunites with the inertial twin. Velocity is relative but acceleration (change in velocity) is absolute.

 

[sylas]

This is [or should be] identical to the above situation. Why is it not? Why must I say that it is the spaceship that shifts frames?

It is not identical because you are changing who shifts frames. That MEANS it is no longer identical, of course!

Once you decide what situation to consider, involving moves and accelerations and velocities (but not gravity) SR will let you calculate all the results of all observations of any observers involved.

 

[otg]

All frames will agree that A was the one who changed velocities and B did not.

Ah this was what I needed to hear :) I think all I really needed was for someone to tell me [again].
Been quite busy lately and my mind is not really following so this struck me as really annoying since I had long accepted the twin paradox and all of relativity, especially the mathematical part. It is all too easy to fall back to pure mathematics and "leave reality behind" [to forget that the $\Gamma^{\mu}_{\nu\sigma\tau}$ actually stands for something and just "mechanically" move indices back and forth]. But arguing that "if you calculate, you'll see" is not useful when trying to explain for non-mathematicians.

Thank you so very much for all your time and effort