Most basic of thought experiments in special relativity

[Kokone]

My apologies if this should be in the homework section, but it's more just a basic conceptual wall I've come up against as I think independently (with little math---just trying to get a very basic conceptual grasp of relativity). That said:

Say there is a Clock A at Point A and a Clock B at Point B. Points A and B are not moving with respect to one another---they are in the same inertial frame. A movable Clock C, while at rest at Point A, is synchronized with Clock A. Then Clock C moves from Points A to B. While it is moving, observers at Points A and B will observe time in the Clock C frame going slower than time in Clocks A and B. An observer moving with Clock C sees time in Clocks A and B's frame going slower than time in Clock C's frame. (Time dilation is symmetric.) Up until the very moment that Clock C stops at Point B, the C observer will see time moving slower in the A/B frame. But then, when Clock C stops, both of them should agree that Clock C is behind Clock B. How is this possible?

Basically, how is the twins paradox resolved when asymmetry does not result from a return trip frame switch (if the ship twin decided to stay at his destination rather than go back to the planet)?

Thank you!

 

[cepheid]

Basically, how is the twins paradox resolved when asymmetry does not result from a return trip frame switch (if the ship twin decided to stay at his destination rather than go back to the planet)?

Thank you!

Well, for one thing, although clock C may not turn around and come back, there is still acceleration. Clock C has to slow down if it is going to stop at point B. It does not stick to one inertial frame. Maybe others can elaborate more.

 

[Kokone]

But from Clock C's perspective, where Clock C is stationary, isn't the A/B frame seen to be the one accelerating?

Actually, that's confusing for me in the twins paradox as well...doesn't the ship twin see the origin planet decelerate, switch frames, and accelerate?

Aghhhhh.

 

[sylas]

But from Clock C's perspective, where Clock C is stationary, isn't the A/B frame seen to be the one accelerating?

Actually, that's confusing for me in the twins paradox as well...doesn't the ship twin see the origin planet decelerate, switch frames, and accelerate?

Aghhhhh.

As usual... the problem is relativity of simultaneity.

You have A and B, which are separated from each other and synchronized to each other. They have clocks, which EVERY observer agrees are running at the same rate.

Furthermore, the clocks show the same time. However, this is ONLY from the perspective of someone who is in the same frame as A or B. For someone in motion towards A, clock A is running ahead of clock B.

That's the short answer to the apparent dilemma. The quality of being synchronized depends on the observer.

Note also that if your clock is synchronized with that of a co-moving remote observer, then you will actually see the remote clock showing a time in the past, back when it emitted the light with which you are seeing it.

How to tell who changes velocity

It is not actually "acceleration" that makes a difference. What makes a difference is being in a new reference frame. The same effects occur if you use a Star Trek teleporter to move into the new reference frame, with no experience of acceleration at all.

There are two observations you can use to tell the distance and velocity of a remote object... although of course you can only infer details about the remote object back when the light left that object.

The velocity of a remote object can be inferred from a redshift (receding) or a blueshift (approaching) in the light from that object. There's no difference, however, whether you are the one moving, or they are the one moving. Physically, all that matters is the relative velocity. Is the distance between you and the remove object increasing, or decreasing? You can infer this from the redshift.

The distance to a remote object can be inferred from its angular size in the sky; assuming you know how big it was to start with. Alternatively you can use the luminosity of a remote light source of known magnitude. You can account for the effect of redshift on luminosity, and then infer the distance. You'll get the same result using either luminosity or angular size; and it corresponds to the distance (in your reference frame, of course) to where the object was when the light you are seeing was emitted.

OK... with this in mind, consider two cases.

Who changes their reference frame? (At takeoff of a starship)

Suppose a star ship takes off from Earth towards another star. Assume some technology that gives almost instantaneous acceleration, with some way of protecting passengers.

From the point of view of spectators, the starship suddenly becomes redshifted, but remains the same size (though it swiftly reduces in size as it races away in the distance.)

From the point of view of passengers, they see precisely the converse. The Earth is suddenly redshifted, but remains the same size (though it swiftly reduces in size as it races away in the distance).

Each perspective is equally valid, and if you were somehow protected from accelerations, you can't tell who actually changed velocity... until you look at the remote star!

When they look at the distant star, to which the star ship is bound... the spectators see no difference at all. The star ship, however, sees the distant star suddenly become blueshifted (approaching) AND it suddenly decreases[*] in size in the sky! This is because the star ship has changed its reference frame, and in the new reference frame the distance to the star is much increased[*]. Furthermore, the star ship passengers KNOW that that they are the ones who have changed perspective -- not the remote star -- because the star cannot have moved a large distance in negligible time. They must be looking at it from a new frame of reference.

*Corrected in edit. Originally I said "increases", which is wrong.​

Also, suppose there is a clock at the star, 10 light years distant, which is synchronized to a clock on Earth. It will be seen to read a time ten years previously, when the light left the star.

Now, for the star ship passengers... the clock at the star is still seen showing a time ten years previously. However, the distance to the star is less than ten light years, as measured by the star ship. Therefore the clock is NOT synchronized with the clock on Earth.

Who changes the reference frame? (A remote starship turns back to return home)

Suppose you see a distant space ship, with redshifted light, and suddenly the ship appears blueshifted. The redshift arises because it is moving away, and the blueshift because it is moving towards you. Or perhaps the other way around... you are moving away and then you are moving towards it.

Can you tell? Yes, and you don't need acceleration. Just measure the distance to the starship, using its luminosity or angular size. If the distance is unchanged in the instant of turn around, then it was the star ship that turned. However, if you find that the star ship is suddenly much much further away, then you must have altered your perspective. You are the one who has moved into a new reference frame.

Spacetime diagrams for this are given in [post=2199430]msg #50[/post] of thread "Twin Paradox- a quick(ish) question".

Cheers -- sylas

 

[Kokone]

Thank you so much, Sylas! I was already starting to get it with Minkowski diagrams, but you helped alot. It makes a lot of sense, particularly, on a basic level, the part about determining whether or not you were the one that switched frames. A few questions:

"From the point of view of spectators, the starship [...] remains the same size."
1. Wouldn't length contraction cause a sudden decrease in the starship's length from the perspective of the spectators?

"From the point of view of passengers [...] the Earth [...] remains the same size."
2. Wouldn't length contraction, in addition to reducing the distance to the star that the passengers observe, also contract the length of Earth and the star themselves?

From your diagrams, and from "However, if you find that the star ship is suddenly much much further away, then you must have altered your perspective":
3. Why would a suddenly inbound traveler (at the turnaround point) see the distance to Earth suddenly increase? Wouldn't they see it either suddenly decrease (if they were stationary at the turnaround) or stay decreased from the outbound trip? Doesn't length contraction work in all lengthwise directions?

4. I seem to have a lot of questions about length contraction. Am I misinterpreting the way length contraction works?

5. Am I correct in thinking the following: If the ship twin goes to the "turnaround point" but *does not* turn around (she goes back to the other twin's reference frame, given that the destination is not moving relative to the stay-at-home twin), then instead of jumping from an upward slash angle (forgive my terminology...) to a downward slash angle, the plane of simultaneity in a Minkwoski diagram will jump from an upward slash angle to a horizontal line. In other words, in the following diagram, if the ship twin remains stationary at the turnaround point for a moment, the missing plane of simultaneity is a horizontal line connecting the turnaround point to the stationary twin. http://upload.wikimedia.org/wikipedia/commons/6/6a/Twin_paradox_Minkowski_diagram.png And this jump in simultaneity will turn out to be enough so that the stay-at-home twin is older than the ship twin and there is no paradox.

6. Assuming that #5 is right:
Consider this diagram: http://www.askthephysicist.com/images/ask_ph12.gif
Note that the lines are not planes of simultaneity, but light pulses.
If the stationary twin does NOT return home (i.e., the upper part of the diagram is mostly irrelevent), what happens to the (red color coded) light pulses that were sent out from the stationary twin AFTER the second pulse (at the stationary twin's 2 year mark---the last to be received by the ship-while-in-transit) but BEFORE the stationary twin's 10 year mark? Does the ship ever receive those pulses? Obviously the "turnaround point" receives the pulses...does this happen after the ship arrives (from the "turnaround point"'s perspective), or before? And if it happens before: does the ship ever observe those pulses, or is it impossible for the ship to know that they happened---since there's a jump in simultaneity when the ship stops and becomes synchronized with the stationary twin's frame, did the ship just skip over events?

 

[Kokone]

Actually, maybe I know the answer to that last question...since the time at the "turnaround point" is perceived by the ship twin as being ahead of the un-synchronized home-bound-twin's time, then maybe the line of simultaneity between the "turnaround point" and the home twin is such that the ship will observe the light hit the "turnaround point," but will never itself be hit by the light?

Maybe? Erggghhhhhhh! :)

 

[Kokone]

PS...when I say "she goes back"in #5, I mean "she is in"...to prevent any confusion there. She does not return to her other twin spatially, it's just that she's now in the same inertial reference frame.

 

[ryuunoseika]

My apologies if this should be in the homework section, but it's more just a basic conceptual wall I've come up against as I think independently (with little math---just trying to get a very basic conceptual grasp of relativity). That said:

Say there is a Clock A at Point A and a Clock B at Point B. Points A and B are not moving with respect to one another---they are in the same inertial frame. A movable Clock C, while at rest at Point A, is synchronized with Clock A. Then Clock C moves from Points A to B. While it is moving, observers at Points A and B will observe time in the Clock C frame going slower than time in Clocks A and B. An observer moving with Clock C sees time in Clocks A and B's frame going slower than time in Clock C's frame. (Time dilation is symmetric.) Up until the very moment that Clock C stops at Point B, the C observer will see time moving slower in the A/B frame. But then, when Clock C stops, both of them should agree that Clock C is behind Clock B. How is this possible?

Basically, how is the twins paradox resolved when asymmetry does not result from a return trip frame switch (if the ship twin decided to stay at his destination rather than go back to the planet)?

Thank you!

By decelerating at point b, clock c is correcting itself. they will show the same time. however, if clock c was still moving as it passed by b it would be out of sync.

 

[Kokone]

No, ryuunoseika, I'm new at this but I know that's not correct.

 

[ryuunoseika]

No, ryuunoseika, I'm new at this but I know that's not correct.

Sorry, i didn't mean to say it would correct itself, but rather that the deceleration throws it off even more.

 

[granpa]

imagine a line of clocks extending through point A and point B. when C is stationary all the clocks are synchronized. when C is moving they are not synchronized and the further away they are from C the greater the difference. during acceleration some of the clocks will seem to move forwards or even backwards at different speeds depending on how far away they are from C at that time.

when C passes A it synchronizes its clock with A. if C moves at gamma=2 then every individual clock it passes will seem to be ticking at half its own rate. but the total elapsed time as told by a special clock on C that constantly synchronizes with whatever clock C is passing will be ticking at twice its own rate.

 

[sylas]

"From the point of view of spectators, the starship [...] remains the same size."
1. Wouldn't length contraction cause a sudden decrease in the starship's length from the perspective of the spectators?

Yes, in the direction of motion... which has no effect on the angular size in the sky. What you see in the sky is proportional to size that is transverse to the direction of motion (width)... and that is not Lorentz contracted.

"From the point of view of passengers [...] the Earth [...] remains the same size."
2. Wouldn't length contraction, in addition to reducing the distance to the star that the passengers observe, also contract the length of Earth and the star themselves?

Yes, in the direction of motion. Angular size depends on the "width" of the object.

It's not easy to come up with a physical observation that let's you infer the "length" of an object moving away from you. The light from the further end takes longer to reach you, so you don't actually "see" all parts of the spaceship along its length at a single instant. You do, however, see the "width" simultaneously.

You can come up with a method in principle for inferring the length of the star ship. You look at the further end of the ship at a given instant, and knowing its real transverse size, you can infer its distance at that moment. Now watch the back of the space ship, until it has an apparent size that indicates it is now at the same distance. (This is where "in principle" collides with what is practical. You have to wait microseconds, and measure size to fantastic accuracy.) Anyhow, if you do this, you can tell from the velocity of the ship, and from the time it took for the rear and the front of the ship to pass a point at some fixed distance, how long the ship must be. The result will show the Lorentz contraction. The ship, of course, sees exactly the same contraction of the Earth.

From your diagrams, and from "However, if you find that the star ship is suddenly much much further away, then you must have altered your perspective":
3. Why would a suddenly inbound traveler (at the turnaround point) see the distance to Earth suddenly increase? Wouldn't they see it either suddenly decrease (if they were stationary at the turnaround) or stay decreased from the outbound trip? Doesn't length contraction work in all lengthwise directions?

From the point of view of the space ship, they see the distance increase because the distance really is larger. Distance depends on perspective, and when you have a new perspective, there's a new and larger distance.

Note that this has nothing particularly to do with Lorentz contraction. It is simply difference in the distance co-ordinate for a given point in spacetime under two different Lorentz transformations; for velocity v and velocity -v.

You can also get a different way of looking at the same thing, from the point of view of aliens on the star watching the spaceship suddenly reverse direction. The spaceship has a little pinhole camera which is capturing light from the Earth. The size of the image on the back of the camera suddenly reduces when the ship reverses direction. (And note that the size of the image is transverse to motion, and is the same for the aliens as for the space ship; no Lorentz contraction involved.)

From the point of view of the travelers, the smaller size is because of a larger distance to Earth.

From the point of view of the aliens, the smaller size is because the camera is now moving into the stream of light from the Earth. The back of the camera moves towards the point where the pin hole used to be, and so the effective length of the camera is reduced, making a much smaller image. If the camera is length d, and moving away from incoming light at velocity v/c, then the light entering the pin hole has to travel a distance (1+v/c)d before it hits the back of the camera. When it reverses, the distance is (1-v/c)d, and the image is smaller by a factor of (c+v)/(c-v).

Lorentz contraction doesn't come into this either, because although "d" is along the direction of motion, it is equally "contracted" for +v and -v.

You calculate the same number exactly by using the new distance to Earth from the point of view of the spaceship.

4. I seem to have a lot of questions about length contraction. Am I misinterpreting the way length contraction works?

I think so.

Lorentz contraction refers to the distance between two parallel world lines along a plane of simultaneity. That is, take the world line of the front, and back, of a moving object. Measure the distance between two simultaneous points on those world lines. That gives the length of the object in that instant. Lorentz contraction arises as the notion of what is simultaneous changes.

5. Am I correct in thinking the following: If the ship twin goes to the "turnaround point" but *does not* turn around (she goes back to the other twin's reference frame, given that the destination is not moving relative to the stay-at-home twin), then instead of jumping from an upward slash angle (forgive my terminology...) to a downward slash angle, the plane of simultaneity in a Minkwoski diagram will jump from an upward slash angle to a horizontal line. In other words, in the following diagram, if the ship twin remains stationary at the turnaround point for a moment, the missing plane of simultaneity is a horizontal line connecting the turnaround point to the stationary twin. http://upload.wikimedia.org/wikipedia/commons/6/6a/Twin_paradox_Minkowski_diagram.png And this jump in simultaneity will turn out to be enough so that the stay-at-home twin is older than the ship twin and there is no paradox.

You are proposing that the spaceship stops dead at the distant star, rather than reverse direction, right?

At the instant when the star ship arrives at the star, it is receiving light from an explosion set off on Earth some time previously. (Diagrams in the other thread). When the spaceship is moving away from Earth, the light is coming from a point 3 light years distant. When the spaceship reverses, the light is coming from a point 12 light years distant. When the spaceship stops dead, the light is coming from a point 6 light years distant.

From the point of view of the outward bound traveler... light from Earth has come 3 light years, and they Earth has been receding at 60% light speed, hence the light is from 5 years after the Earth started to recede. With time dilation of 1.25, the Earth was 4 years old when the bomb went off. Then there's a sudden switch to a new perspective, as the ship stops at the remote star. The Earth is seen from the same instant, so it is still seen as having aged 4 years. But the light has come from 6 light years distant, and there's no relative velocity anymore. So (if Earth remains at the same distance, and it does) the Earth must age another 6 years up to the "now" moment when the spaceship arrives to settle on the new star. Which is also what Earth calculates from their perspective.

6. Assuming that #5 is right:
Consider this diagram: http://www.askthephysicist.com/images/ask_ph12.gif
Note that the lines are not planes of simultaneity, but light pulses.
If the stationary twin does NOT return home (i.e., the upper part of the diagram is mostly irrelevent), what happens to the (red color coded) light pulses that were sent out from the stationary twin AFTER the second pulse (at the stationary twin's 2 year mark---the last to be received by the ship-while-in-transit) but BEFORE the stationary twin's 10 year mark? Does the ship ever receive those pulses? Obviously the "turnaround point" receives the pulses...does this happen after the ship arrives (from the "turnaround point"'s perspective), or before? And if it happens before: does the ship ever observe those pulses, or is it impossible for the ship to know that they happened---since there's a jump in simultaneity when the ship stops and becomes synchronized with the stationary twin's frame, did the ship just skip over events?

Once the spaceship arrives at the new star, and stops, it continues to get light from Earth, with no redshift, and at the same intervals that Earth is sending them. Just extend a line vertically above the turn around point to represent the world line of new settlers at the new star.

Cheers -- sylas

 

[sylas]

imagine a line of clocks extending through point A and point B. when C is stationary all the clocks are synchronized. when C is moving they are not synchronized and the further away they are from C the greater the difference. during acceleration some of the clocks will appear to move forwards or backwards at different speeds depending on how far away they are from C at that time.

Careful! What you actually "see" is NOT simultaneous, but depends on how long ago the light was emitted.

Suppose you have a line of co-moving clocks, all synchronized in their shared frame of reference. Suppose the clocks are spaced at intervals of one light-second.

What you SEE at any point is that successive clocks show different times, by 1 second. That is invariant. It makes no difference how fast you are moving, or whether you are accelerating or not.

What changes when you are moving relative to the clocks is that the clocks are now moving in your reference frame. The distance between successive clocks is Lorentz contracted, but the clocks are all running slow. The two effects cancel exactly[* oops. no *], and you continue to see each successive clock showing values with 1 second differences. What changes is not the readings you see on the clocks, but the velocity of the clocks (their redshift) and the distance to the clocks (their apparent size in the sky).

Cheers -- sylas

PS. For simplicity, I am just considering movement and displacement in a single direction.

PPS. Oops. I should not really say that the effects "cancel". But taking into account the reduced distance between the clocks, and the slower ticking speed, AND the new notion of what is simultaneous, you end up still seeing successive clocks showing readings at 1 second differences.

 

[granpa]

nothing in my post had anything to do with what you 'see'. its what you calculate the clocks current time to be given what you see and the known distance to the clock.

 

[sylas]

nothing in my post had anything to do with what you 'see'. its what you calculate the clocks current time to be given what you see and the known distance to the clock.

OK, that's good to know. Thanks.

Consider my post a supplement, then, just in case anyone is confused by the phrase "clocks will appear to move forwards or backwards".

You are referring, correctly, to an inference, rather than what is immediately "apparent", or "seen", by an observer.

Cheers -- sylas

 

[Kokone]

Re. 1 and 2: Ah, I was misinterpreting your (correct) definition of size. Got it.

Re. 3
But now...returning to the format of the standard twin paradox...and if I take what you say to be correct:
*On the first leg of the trip, the ship twin experiences the trip to take x amount of time (I'm not sure what the number would be, but it's irrelevant for the purpose of my question) for a trip that she perceives as traversing a distance of 3 light years.
*On the return trip, the ship twin experiences the trip to take x amount of time (the same amount) for a trip that she perceives as traversing a distance of 12 light years.
So she will find that she travels faster on the first trip than the second, which just makes no sense as far as I can see.

So...maybe we're talking at cross purposes? Are you talking about SEEN distance and I'm talking about INFERRED distance? I'm not sure.

4. Yeah, I think we just had a misunderstanding there that was my fault.

5. But, forgetting for now the visual stuff, was what I said was correct about the horizontal plane of simultaneity in Minkwoski space?

6. Got it. Thanks.

 

[granpa]

when C passes A it synchronizes its clock with A. if C moves at gamma=2 then every individual clock it passes will seem to be ticking at half its own rate. but the total elapsed time as told by a special clock on C that constantly synchronizes with whatever clock C is passing will be ticking at twice its own rate. when C stops at B the line of clocks will again be synchronized (from C's perspective).

the line of clocks can be used as 'mile markers' (to measure distance) too. one can include a line of clocks moving with C so that C can use them to measure distances from his perspective too. but don't forget that the length of an object (from your perspective) is the distance between the front and back at one simultaneous (from your perspective) moment.

 

[Kokone]

sylas:
Also, if, when the ship leaves Earth the alien star suddenly appears closer and bigger, wouldn't the Earth similarly appear bigger and closer when departing the alien star, instead of smaller and farther?

I don't really understand why there's any size or distant shift of that sort at all, but if there's going to be one, shouldn't it be consistent given that there's no practical difference between leaving an "alien star" and leaving a "native star"?

 

[sylas]

Re. 1 and 2: Ah, I was misinterpreting your (correct) definition of size. Got it.

You are too kind... I was ambiguous, but we're all on the same page now.

But now...returning to the format of the standard twin paradox...and if I take what you say to be correct:
*On the first leg of the trip, the ship twin experiences the trip to take x amount of time (I'm not sure what the number would be, but it's irrelevant for the purpose of my question) for a trip that she perceives as traversing a distance of 3 light years.
*On the return trip, the ship twin experiences the trip to take x amount of time (the same amount) for a trip that she perceives as traversing a distance of 12 light years.
So she will find that she travels faster on the first trip than the second, which just makes no sense as far as I can see.

The example I've given is at 60% light speed, because it works out to nice even numbers. The gamma factor is 1.25; so the stay-at-home twin is 1.25 times older than the spaceship captain. I consider a trip to a star 6 light years away (measured from Earth). So at 60% light speed, the ship takes 10 Earth-years to get there, but the twin on board only ages 10/1.25 = 8 years.

When you speak of the twin on board traversing a certain distance, that's a problem... because in her time frame, she doesn't move at all. It is the Earth, and star, that is moving.

As far as an observer on Earth is concerned, she is moving to the star. In her frame, the star is moving towards her vehicle; and the Earth is moving off into the distance; and then they reverse and move back again.

Now... how does she interpret events? She is in several different frames, so there are going to be some sudden jumps when she moves from one frame of reference to another, but she is a seasoned space farer, and now finds this perfectly natural, just as we find nothing odd about something that was on our left suddenly appearing on our right when we turn around.

Frame 1. At the spaceport on Earth

Initially, at the briefing room in the spaceport, she checks the flight plan, which is to visit a star that is 6 light years distant. Most recent laser communications from the star confirm all is well to receive her... although of course that message left the star six years previously.

Frame 2. En route to the stars

She engages warp drive, and instantaneously shifts into a condition of outward velocity 60% light speed. The warp drive, which I don't have time to explain in detail, gets around all the difficulties of accelerations. As the ship exits the launch chute, she's already at 60% light speed. But from her perspective, the spaceport has suddenly moved off to the rear at 60% light speed. The star is suddenly blueshifted, and the Earth redshifted, by a factor of 2.

$$2 = \sqrt{\frac{1 + 0.6}{1 - 0.6}}$$​

The star is also smaller in the sky than it was a second previously, because it is further away. The signals from the star which are wishing her well for the trip, which a second ago were coming from 6 light years distant, are now coming from 12 light years distant, having left the star 12 years previously (in her new frame of reference). But... on the other hand the star is approaching at 60% light speed. Assuming all is going to plan, over the 12 years since the light left that star, it must have moved 0.6 * 12 = 7.2 light years closer, so "now" it must be only 4.8 light years away. So in her new frame of reference, the star is "now" closer, but the light she sees at present is coming from a point that is much further away. At 60% light speed, the star should be here in another 8 years time. Sure enough, 8 years later the star arrives at the space ship.

Frame 3. In the alien bar

When the star has reached her ship, the pilot engages the warp drive once more, and the star stops. Of course, according to local aliens, she was moving into the docking port; but I'm sticking with the pilot's viewpoint.

Some time previously, Earth sent her a message of congratulations, timed to arrive at the same time she did. She was listening to the first part of that message when she docked.

Just prior to docking, the message from Earth was coming from a point 3 light years distant and was therefore sent 3 years previously. Hence, it was sent 5 years after she and Earth parted company 8 years ago. Since Earth is moving at 60% light speed, Earth is time dilated, and has aged 4 years (= 5/1.25) from the departure to when the message was sent.

A second later, however, after docking, the message is coming from 6 light years distant. The Earth is no longer redshifted, and the angle subtended in the sky by Earth reduces by half, since it is now 6 light years distant, and motionless, rather than being seen from a distance of 3 light years.

Frame 4. Coming home

After a quick drink at the alien spaceport bar, and an unfortunate disagreement with a belligerent bounty hunter, our heroine decides now is good time to return. Wasting no time, she's in the ship again, and Earth is approaching at 60% light speed, while the star (and the bounty hunter) recedes at 60% light speed. The long winded congratulations from Earth is still coming over comms... but now it is coming from 12 light years distant. And, of course, Earth was still 4 years older at the time of sending those congratulations.

At 60% light speed, Earth takes 20 years to travel the 12 light years to the spaceship. And, being time dilated, Earth will age 16 years during the trip. Now of course, that message was transmitted (in her current frame of reference) 12 years ago... so the Earth should arrive in another 8 years... and so it does.

Frame 5. Debriefing

When Earth finally arrives at the spaceship, 8 years later, the spaceport commander is indeed, just as expected, 20 years older than when the spaceport first departed... 16 years ago by the spaceship clock.

Earth has traveled a total of 3 light years (from departure to transmission of the congratulations message) and 12 light years (from transmission of the congratulations messages to arrival again). It may seem odd that Earth moved away a smaller distance than the distance it moved to return... but that's normal for the spaceship captain; not worth another thought. At 60% light speed, those 15 light years took 25 years total travel time, before you consider time dilation effects. It may seem odd to think of 25 years total time when it was only 16 years on the ship... but the captain is far too much of an old hand at this game to find it at all unusual. In any case... give the time dilation with Earth traveling at 60% light speed, the spaceport is now 25/1.25 = 20 years older.

The spaceport has a different perspective on the matter... the spaceship captain is the one that was time dilated, and she left 20 years ago, and so she is now 20/1.25 = 16 years older. They are quite right, of course. Either perspective works.

5. But, forgetting for now the visual stuff, was what I said was correct about the horizontal plane of simultaneity in Minkwoski space?

I'm not completely sure... I've never calculated it like that.

The plane of simultaneity for the outbound twin through the turn around points intersects with the Earth world line at a point where the Earth clock reads 8/1.25 = 6.4 years. At rest in the alien bar, the Earth clock is "now" at 10 years. Inbound, it intersects a point where the Earth clock reads 13.6 years, by symmetry. That's a gain of 7.2 years. Now 16 years travel time (by ship clock) gives 16/1.25 = 12.8 years elapsed time, add the 7.2 and you do get 20 years. But I have not proved to myself mathematically how you could apply this safely as a general principle. I'm sure it can be made to work in general, but I don't feel competent to describe how to do it and be sure of getting a correct answer, in more complex situations.

Cheers -- sylas

 

[sylas]

sylas:
Also, if, when the ship leaves Earth the alien star suddenly appears closer and bigger, wouldn't the Earth similarly appear bigger and closer when departing the alien star, instead of smaller and farther?

Yes. I screwed up, and have edited my previous post to include a correction.

When you are moving towards something, it appears smaller. When you are moving away, it appears larger. But just to confuse the matter, when you are moving towards something OR away from it, the distance to that object "now" is reduced.

For example. Consider a spaceship 6 light years from earth.

If it is at rest, then it receives light from 6 years in the past, from a point 6 light years distant, and Earth is "now" 6 light years away.

If the spaceship is inbound at 60% light speed, then in the spaceship frame, the signals from Earth are coming from a point 12 light years distant, but Earth is "now" 4.8 light years away, having moved inwards 7.2 light years since then.

If the spaceship is is outbound at 60% light speed, then in the spaceship frame the signals from Earth are coming from a point 3 light years distant, and Earth is "now" 4.8 light years distant (having moved 1.8 more light years in those 3 years).

I don't really understand why there's any size or distant shift of that sort at all, but if there's going to be one, shouldn't it be consistent given that there's no practical difference between leaving an "alien star" and leaving a "native star"?

I have fixed my error in the previous post. Alien stars and native stars are the same, as you say.

For the rest, you can just use the Lorentz transformations.

In the frame of the alien star, and taking the positive direction as outbound AWAY from Earth, signals from Earth come from a point at (-d,-d), meaning "d" years ago and "d" light years away. In these units, c=1.

Applying Lorentz transformations for conversion into the frame of spaceship headed outbound from Earth at velocity v, that point becomes

$$(-\gamma(1-v)d, -\gamma(1-v)d)$$​

That factor is

$$\gamma(1-v) = \frac{1-v}{\sqrt{1-v^2}} = \sqrt{\frac{1-v}{1+v}}$$​

which is the factor for Doppler shift. A positive v represents increasing separating from Earth, and negative v represents reducing distance to Earth.

Cheers -- sylas

 

[sylas]

Excuse some self-indulgence...

The notion that the angular size of an object in the sky depends on your velocity is not one that is often raised in traveling twin discussions, but it is one I have found useful, so I thought it worth while to explore a bit further.

You can use the Lorenz transformations as shown in the previous post to show that the light you see from a distant remote object is actually emitted at distances that depend on your velocity, according to the factor

$$\sqrt{\frac{1-v}{1+v}}$$​

where v is your velocity away from the object. Hence you should expect the apparent size in the sky to vary by this factor as well, though inversely. (I am using units where c=1 throughout.)

You can also get the same result by reasoning from the point of view of an observer stationary with respect to a distant object, and thinking about the size of the image obtained within a pinhole camera.

Suppose you have a pinhole camera of length d. It will project an image of a certain size at the back of the camera, which is proportional to the length of your camera, and to the size of the object, and inversely proportional to the distance to the object.

Now take your camera, and move it rapidly towards, or away, from your light source. The image in the camera changes size. Here's a diagram of how it works:

At the top is a simple pin hole camera, length d. The distance to an object is inversely proportional to the size of its image. The images were constructed assuming velocity of 0.6 light speed, in which case the gamma factor is 1.25, and the Doppler scale factor is 2. The image on moving towards the object is half size; and on moving away it is double size.

Suppose you move the camera rapidly away from the object. The camera is Lorenz contracted to be length d/γ. If the light moves a distance D inside the camera, then the back of the camera has moved a distance Dv away from the pinhole over the same time. Hence the total distance moved by the light from entering the pinhole until it hits the back of the camera is

$$\begin{equation*}\begin{split} D &=d/\gamma + Dv \\ D &= \frac{d}{\gamma(1-v)} \\ &= d\frac{\sqrt{1-v^2}}{1-v} \\ &= d\sqrt{\frac{1+v}{1-v}} \end{split}\end{equation*}$$​

Moving towards the source is the same, except that v is negative.

Since special relativity is self consistent, we obtain the same answer as was given just by working in the frame where the camera is at rest and the source is moving, in which case the size of the image is given by the distance to the object, which is obtained from Lorenz transformations.

Cheers -- sylas

 

[Kokone]

Hm. It's all getting a bit complex for me, having zero physics background yet and barely even remembering trigonometry at the moment. And then I get obsessed with perceived paradoxes and misunderstood concepts that are difficult to resolve when I don't have the necessary foundations and it drives me crazy and makes me want to avoid physics at all cost. So for the sake of my sanity, I am going to try to put relativity out of my mind until I have real physics classes in Newtonian and basic stuff. :)

Thank you for all your help! You made several things make a lot more sense. You are awesome for putting this much time. Thank you.