Can Relativity Explain the Confusion of the Twins Paradox?

[M1keh]

No, assuming E moved inertially while T accelerated at some point (even if instantaneously), there will be no inertial reference frame where the second is true.

JesseM. Apologies. I'm getting confused now.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.

In the example, when we look at A's f.o.r, "A stays stationery for 8 years and 6.4 years have passed on B's watch" ?

Isn't that what we said earlier ?

So if we look at just A & B,

B sees himself stay stationery for 10 years and A's watch move 8 years.
A sees himself stay stationery for 8 years and B's watch move 6.4 years.

Is this right ?


Ah. Ok. I've worked through the figures from each of the three f.o.r's and answered the question for myself. I was just thinking out aloud ?

It all works perfectly well and exactly as you detailed. Thanks.

I need some thinking time to absorb this ... hurts between the ears, a bit.

 

[M1keh]

...

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.
-twin C has been moving at 0.88235c, giving a time dilation factor of 0.4706, so (8 years)*(0.4706) = 3.765 years have passed on his clock.

Between 8 years and 17 years:
-twin A has been moving at 0.88235c, so in the 17-8=9 years of this section, (9 years)*(0.4706) = 4.235 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (9 years)*(0.8) = 7.2 additional years have passed on his clock.
-twin C has been moving at 0.88235c, so 4.235 additional years have passed on his clock.

Between 17 years and 25 years:
-twin A has been moving at 0.88235c, so in the 25-17=8 years of this section, (8 years)*(0.4706) = 3.765 additional years have passed on his clock.
-twin B has been moving at 0.6c, so (8 years)*(0.8) = 6.4 additional years have passed on his clock.
-twin C has been at rest, so 8 additional years have passed on his clock.

If you add all these up, you again find that 20 years passed on twin B's clock while 16 years passed on both A and C's clocks between the beginning and end of the journey; also, twin A and C's clocks each registered 8 years before they turned around and 8 years after. This got a bit complicated, so let me know if there's any steps you don't follow.

Ok. Finally. After much head scratching and working this out from start to finish, I come up with exactly the same figures and I think I understand why.

It took a while as I couldn't believe it worked correctly. Something still didn't seem right, although the numbers proved it was.

But I've come up with another example ? This one's a slant on the original that I believe doesn't work ? I'm sure you'll prove me wrong ?

Suppose we have 3 Earths, separated by 6 ly (12 ly total). B is on E2, the Middle-Earth :-), A is on E1 and C is on E3.

Now all observers are synchronised in Earth's f.o.r, the same as the original example.

Both A & C leave at the same time and head towards E2 at 0.6c. When they get there they stop and compare watches.

How does the earlier method of explaining events work for this example ?

If you look at A's view for the 10 years, don't you get :

View From A’s Starting F.O.R.

First and only 8 years.

A remains stationery.

B[123] will travel at 0.6c and their watches will show 0.8*8years = 6.4 years. They will have traveled 0.6*8 = 4.8 ly.

C will travel towards A at (0.6+0.6)/(1+0.6^2) = 0.88235c, and his watch will show 0.4706*8years = 3.765 years. He will have traveled 0.88235*8 = 7.06 ly.


Am I messing this up somewhere ?

 

[JesseM]

No, assuming E moved inertially while T accelerated at some point (even if instantaneously), there will be no inertial reference frame where the second is true.

JesseM. Apologies. I'm getting confused now.

Between departure and 8 years:
-twin A has been at rest, so 8 years have passed on his clock
-twin B has been moving at 0.6c, giving a time dilation factor of 0.8, so (8 years)*(0.8) = 6.4 years have passed on his clock.

In the example, when we look at A's f.o.r, "A stays stationery for 8 years and 6.4 years have passed on B's watch" ?

Yes, but this was during a single section of the trip where neither A nor B accelerated, so they both had a single inertial rest frame. When you were talking about E and T before, I thought you were talking about a twin-paradox type situation where T traveled away, then turned around and returned to E so they could compare clocks at a single location...if you weren't talking about this scenario, then forget that comment.

So if we look at just A & B,

B sees himself stay stationery for 10 years and A's watch move 8 years.
A sees himself stay stationery for 8 years and B's watch move 6.4 years.

Is this right ?

Yup, that's what each one would observe in his own inertial rest frame, up until the moment A turned around.

 

[pess5]

Now by the way, do you think we "see" or only "observe" length contraction or neither?
From a twistor-space perspective we see or observe no contraction, since Lorentz transformations are shape preserving in twistor space.

The relativistic contractions are both see'able and observable. Although, it wouldn't be easy by any stretch for tiny bodies. All normally spherical planets would appear ellipsoidal after accelerating to near light speeds.

One's proper length never changes with changes in one's own state of motion, but those of relative motion >0 will see contractions of you. They are real, even though a body never sees its own proper length (or tick of tock) change.

 

[JesseM]

Ok. Finally. After much head scratching and working this out from start to finish, I come up with exactly the same figures and I think I understand why.

It took a while as I couldn't believe it worked correctly. Something still didn't seem right, although the numbers proved it was.

But I've come up with another example ? This one's a slant on the original that I believe doesn't work ? I'm sure you'll prove me wrong ?

Suppose we have 3 Earths, separated by 6 ly (12 ly total). B is on E2, the Middle-Earth :-), A is on E1 and C is on E3.

Now all observers are synchronised in Earth's f.o.r, the same as the original example.

Both A & C leave at the same time and head towards E2 at 0.6c. When they get there they stop and compare watches.

How does the earlier method of explaining events work for this example ?

If you look at A's view for the 10 years, don't you get :

View From A’s Starting F.O.R.

First and only 8 years.

A remains stationery.

B[123] will travel at 0.6c and their watches will show 0.8*8years = 6.4 years. They will have traveled 0.6*8 = 4.8 ly.

C will travel towards A at (0.6+0.6)/(1+0.6^2) = 0.88235c, and his watch will show 0.4706*8years = 3.765 years. He will have traveled 0.88235*8 = 7.06 ly.


Am I messing this up somewhere ?

Yes, in this case you need to take into account the fact that the frame where A is at rest as he approaches Earth2 defines simultaneity differently than does the rest frame of the three Earth's--in A's rest frame, Earth2's clock is ahead of the clock on Earth1 where he departed from, and Earth3's clock is further ahead still, and C departed from Earth3 well before A departed from Earth1. In general, if two clocks are at rest with respect to each other and synchronized in their rest frame, and the distance between them in their rest frame is L, then if you have an observer moving at velocity v with respect to them in the same direction as the line between the two clocks, in this observer's rest frame the back clock's time will be ahead of the front clock's time by vL/c^2. So in your example, A will observe E2's clock as being ahead of E1's by (6 l.y.)*(0.6c) = 3.6 years. So even though he only observe the clock of E2 advance by (8 years)*(0.8) = 6.4 years during the 8 years it took him to move from E1 to E2 according to his own clock, since E2's clock started out at 3.6 years at the beginning of the journey, at the end it would read 3.6 + 6.4 = 10 years.

Meanwhile, E3's clock already read 7.2 years at the moment A left E1, and since it is only ticking at 0.8 the normal rate, it would have read zero 7.2/0.8 = 9 years before A's departure. Since C left E3 when E3's clock read zero, this means that in the rest frame of A, C departed E3 9 years earlier than A left E1, which means C was traveling for 9 + 8 = 17 years before they finally met at E2. To figure out how fast C was moving in A's frame, we can use the velocity addition formula which tells us that in this frame C must have been moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, which gives a time dilation factor of 0.47059. So, in the 17 years between the time C departed and the time he met with A, his clock would have advanced forward by (17)*(0.47059) = 8 years.

If you don't want to have to make use of a bunch of specialized formulas (time dilation formula, length contraction formula, formula for how much moving clocks will appear out-of-sync, velocity addition formula), a more general way to approach any relativity problem is to use the Lorentz transformation, which transforms between the coordinates of an event in one inertial frame's coordinate system and the coordinates of the same event in other frame's system. If we have one frame which labels events with spatial coordinates x,y,z and time coordinate t, and another frame which uses spatial coordinates x',y',z' and time coordinate t', and the spatial origin of the second frame is moving along the first frame's x-axis with velocity v, with both origins coinciding at time t=t'=0, then the coordinate transform would be:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

In your problem, we can ignore the y and z directions and just concentrate on one space coordinate. Let's have the x,t coordinate system be the one in which all three Earth's are at rest, with x=0 being the position of E1 and t=0 being the time when A and C departed E1 and E3 in this frame. Then the x',t' coordinate system can be the one where A is at rest during the journey, and it will also have its origin at the event of A departing E1. So in the coordinate transform, we'd have v=0.6c, and \gamma = 1.25.

Now, it's easy to figure out the coordinates of various important events in the x,t system where the Earth's are at rest:

A leaves E1: x=0 l.y., t=0 years
C leaves E3: x=12 l.y., t=0 years
E2's clock reads 3.6 years: x=6 l.y., t=3.6 years
E3's clock reads 7.2 years: x=12 l.y., t=7.2 years

Now, use the formulas I gave above to find the corresponding x',t' coordinates of each event:

A leaves E1: x'=0 l.y., t'=0 years
C leaves E3: x'=15 l.y., t'=-9 years
E2's clock reads 3.6 years: x'=4.8 l.y., t'=0 years
E3's clock reads 7.2 years: x'=9.6 l.y., t'=0 years

So, figuring things out explicitly in terms of coordinates gives the same answers that were found earlier: in A's frame, C departed E3 nine years before A departed E1, and at the same moment that A departed E1, E2's clock read 3.6 years and E3's clock read 7.2 years. Note that I confirmed the times on E2 and E3 in a sort of backwards way, by already knowing the times they should read and then confirming that they read these times at t'=0 in A's frame, but a more logical way to do it would be to figure out the positions of E2 and E3 at t'=0 in A's frame (since they were 6 light years and 12 light years away in their own rest frame, Lorentz contraction tells us they'd be 6*0.8 = 4.8 light years and 12*0.8 = 9.6 light years away in A's frame at the moment he left E1), and then plug those coordinates into the reverse version of the coordinate transformation:

x = \gamma (x' + vt')
t = \gamma (t' + vx'/c^2)

That way, plugging in (x'=4.8 l.y., t'=0 years) and (x'=9.6 l.y., t'=0 years) would tell us that these events happened at t=3.6 years and t=7.2 years, respectively, in the Earth's' rest frame.

This was pretty involved, so as always, let me know if you have questions about any of the steps here.

 

[pess5]

JesseM,

Is one able to post PDFs or MS WORD docs in this forum environment offhand? Reason I ask is because a Minkowski spacetime diagram paints a picture of a 1000 verbal posts. Is this possible?

 

[jtbell]

Yes, it is possible to attach pictures etc., with restrictions on size. See the "Additional Options" section below the message-composition field when you reply or start a thread.

 

[pess5]

For the benefit of all.

Here's a portable document file (PDF) I drafted a number of years back. It's a Minkowski worldline diagram of the twins scenario. A spacetime diagram paints a 1000 words.

This scenario is called "the handoff technique". Minkowski diagrams are truly designed to represent inertial reference frames, however the effects of acceleration may be extrapolated.

This scenario has 2 figures, O stationary and A stationary ...

1. Observer O & A flyby event, where they align their clocks at flyby.
2. Observer A & C flyby, whereby C aligns his clock to A's.
3. Clock comparison at final C & O flyby.

It represents the equivalent of an observer A departing from observer O with clocks aligned, then immediately returning at some point back to O for clock comparison, where accelerations are considered instantaneous.

I also show an observer B who represents a mirrored image of observer A, although this may be ignored far as the twins scenario is concerned. I use v=0.866c outbound & inbound since gamma is conveniently = 2.

 

[JesseM]

This scenario is called "the handoff technique". Minkowski diagrams are truly designed to represent inertial only scenarios, however the effects of acceleration may be extrapolated.

I'd modify that to say Minkowski diagrams are designed to represent inertial reference frames, but within such a frame you can certainly draw in the non-straight worldline of an accelerating object.

Would it be possible for you to translate the diagrams into jpgs or gifs? I have a mac and I don't know what application to use for the doc you attached. But if it'd be much trouble don't worry about it, presumably most other people don't have this problem.

 

[pess5]

I'd modify that to say Minkowski diagrams are designed to represent inertial reference frames, but within such a frame you can certainly draw in the non-straight worldline of an accelerating object.

done.

Would it be possible for you to translate the diagrams into jpgs or gifs? I have a mac and I don't know what application to use for the doc you attached. But if it'd be much trouble don't worry about it, presumably most other people don't have this problem.

done, but I reposted it in PDF. I figure most folks have the Adobe Acrobat Reader, however the MS WORD 97 doc sure looks cleaner. But PDF isn't bad though.

 

[JesseM]

done, but I reposted it in PDF. I figure most folks have the Adobe Acrobat Reader, however the MS WORD 97 doc sure looks cleaner. But PDF isn't bad though.

Thanks! Very nice (and detailed) diagrams there.

 

[M1keh]

Yes, in this case you need to take into account the fact that the frame where A is at rest as he approaches Earth2 defines simultaneity differently than does the rest frame of the three Earth's--in A's rest frame, Earth2's clock is ahead of the clock on Earth1 where he departed from, and Earth3's clock is further ahead still, and C departed from Earth3 well before A departed from Earth1.

Ouch. Didn't see that one coming. How can that be ?

If A starts on E1 and C starts on E3 and E1-3 are 'stationery', A & C start at exactly the same f.o.r as E1-3 and all of their clocks are sync'd ?

When A accelerates (instantly) to 0.6c. His time will dilate by a factor of 0.8 compared with all 3 ? The distance will also compress by a factor of 1.25 compared with all 3 ?

In general, if two clocks are at rest with respect to each other and synchronized in their rest frame, and the distance between them in their rest frame is L, then if you have an observer moving at velocity v with respect to them in the same direction as the line between the two clocks, in this observer's rest frame the back clock's time will be ahead of the front clock's time by vL/c^2. So in your example, A will observe E2's clock as being ahead of E1's by (6 l.y.)*(0.6c) = 3.6 years. So even though he only observe the clock of E2 advance by (8 years)*(0.8) = 6.4 years during the 8 years it took him to move from E1 to E2 according to his own clock, since E2's clock started out at 3.6 years at the beginning of the journey, at the end it would read 3.6 + 6.4 = 10 years.

Ok. Now I'm struggling to keep up.

Before the journey starts, tA = tE1 = tE2 = tE3 = tC ?

At what point does A see E2's time jump ahead by 3.6 years ? and E3's time jump ahead by 7.2 years ?

The only 'rules' I've seen so far state that time slows down, it doesn't jump ?

Am I missing a 'rule' somewhere ? Or misunderstanding what you're saying ?

Meanwhile, E3's clock already read 7.2 years at the moment A left E1, and since it is only ticking at 0.8 the normal rate, it would have read zero 7.2/0.8 = 9 years before A's departure. Since C left E3 when E3's clock read zero, this means that in the rest frame of A, C departed E3 9 years earlier than A left E1, which means C was traveling for 9 + 8 = 17 years before they finally met at E2. To figure out how fast C was moving in A's frame, we can use the velocity addition formula which tells us that in this frame C must have been moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, which gives a time dilation factor of 0.47059. So, in the 17 years between the time C departed and the time he met with A, his clock would have advanced forward by (17)*(0.47059) = 8 years.

If you don't want to have to make use of a bunch of specialized formulas (time dilation formula, length contraction formula, formula for how much moving clocks will appear out-of-sync, velocity addition formula), a more general way to approach any relativity problem is to use the Lorentz transformation, which transforms between the coordinates of an event in one inertial frame's coordinate system and the coordinates of the same event in other frame's system. If we have one frame which labels events with spatial coordinates x,y,z and time coordinate t, and another frame which uses spatial coordinates x',y',z' and time coordinate t', and the spatial origin of the second frame is moving along the first frame's x-axis with velocity v, with both origins coinciding at time t=t'=0, then the coordinate transform would be:

x' = \gamma (x - vt)
y' = y
z' = z
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}

In your problem, we can ignore the y and z directions and just concentrate on one space coordinate. Let's have the x,t coordinate system be the one in which all three Earth's are at rest, with x=0 being the position of E1 and t=0 being the time when A and C departed E1 and E3 in this frame. Then the x',t' coordinate system can be the one where A is at rest during the journey, and it will also have its origin at the event of A departing E1. So in the coordinate transform, we'd have v=0.6c, and \gamma = 1.25.

Now, it's easy to figure out the coordinates of various important events in the x,t system where the Earth's are at rest:

A leaves E1: x=0 l.y., t=0 years
C leaves E3: x=12 l.y., t=0 years
E2's clock reads 3.6 years: x=6 l.y., t=3.6 years
E3's clock reads 7.2 years: x=12 l.y., t=7.2 years

Now, use the formulas I gave above to find the corresponding x',t' coordinates of each event:

A leaves E1: x'=0 l.y., t'=0 years
C leaves E3: x'=15 l.y., t'=-9 years
E2's clock reads 3.6 years: x'=4.8 l.y., t'=0 years
E3's clock reads 7.2 years: x'=9.6 l.y., t'=0 years

So, figuring things out explicitly in terms of coordinates gives the same answers that were found earlier: in A's frame, C departed E3 nine years before A departed E1, and at the same moment that A departed E1, E2's clock read 3.6 years and E3's clock read 7.2 years. Note that I confirmed the times on E2 and E3 in a sort of backwards way, by already knowing the times they should read and then confirming that they read these times at t'=0 in A's frame, but a more logical way to do it would be to figure out the positions of E2 and E3 at t'=0 in A's frame (since they were 6 light years and 12 light years away in their own rest frame, Lorentz contraction tells us they'd be 6*0.8 = 4.8 light years and 12*0.8 = 9.6 light years away in A's frame at the moment he left E1), and then plug those coordinates into the reverse version of the coordinate transformation:

x = \gamma (x' + vt')
t = \gamma (t' + vx'/c^2)

That way, plugging in (x'=4.8 l.y., t'=0 years) and (x'=9.6 l.y., t'=0 years) would tell us that these events happened at t=3.6 years and t=7.2 years, respectively, in the Earth's' rest frame.

This was pretty involved, so as always, let me know if you have questions about any of the steps here.

That'll take a while to sink in !

 

[pess5]

JesseM,

I've reposted the Handoff Scenario rev c (PDF attachment) with a few cleanup tweeks at the original post ...

https://www.physicsforums.com/showpost.php?p=1104227&postcount=68"

pess

 

[pess5]

Where discussing the twins scenario gets real interesting, is when one addresses the what the non-inertial twin sees of the inertial twin, especially over periods of active acceleration. Also, how the spacetime grid of one twin frame maps into the spacetime grid of the other.

 

[M1keh]

JesseM,

I've reposted the Handoff Scenario rev b (PDF attachment) with a few cleanup tweeks at the original post ...

https://www.physicsforums.com/showpost.php?p=1104227&postcount=68"

pess

Cool. Thanks. It's take some time to decypher & digest. Gives me something to consider over the weekend.

 

[M1keh]

JesseM,

I've reposted the Handoff Scenario rev b (PDF attachment) with a few cleanup tweeks at the original post ...

https://www.physicsforums.com/showpost.php?p=1104227&postcount=68"

pess

pess. What about the scenario I mentioned earlier ?

If the twins start from opposite sides of the Earth, 6ly away (Earth distance) and stationary compared with the Earth. They then accelerate to 0.6c and travel to Earth. What are the time lines ?

This way they have the perfect hand-off as they all meet at Earth at the same time, or not, or they do, or they don't ?

 

[jtbell]

Ouch. Didn't see that one coming. How can that be ?

Don't feel too bad about it. You're yet another victim of incomplete presentations of basic relativistic phenomena in popular-level science books, and even in many general-physics textbooks at the college/university level.

They all talk about length contraction and time dilation, but most of them either don't mention relativity of simultaneity at all, or present it in a way that doesn't make clear from the beginning that it is at the same level of importance as length contraction and time dilation.

Start with a line of clocks, all stationary in inertial reference frame S, and all synchronized in S. In reference frame S' which is moving along that line with respect to S, three things about these clocks are different:

1. The clocks are closer together than in S: length contraction.

2. The clocks tick more slowly than an identical clock which is at rest in S': time dilation.

3. The clocks are not synchronized in S': relativity of simultaneity.

In general, you need to take all three of these phenomena into account, not just the first two.

 

[pess5]

Mikeh,

I just retweeked it again, so please reprint the updated (newer) rev b version I just posted. You got the 1st rev b. See ...

https://www.physicsforums.com/showpost.php?p=1104227&postcount=68"​

In this 2nd rev c version, I added a line-of-simultaneity in the bottom figure. This in relation to JesseM's point regarding synchronized clocks appearing to lag each other in clock readouts per a moving observer vantage. You might want to print this post out when looking at the figure printout...

The top figure shows the O stationary perspectve. O's line-of-simultaneity is horizontal, and 12:34am-C, 1:09am-O, 12:34am-B all ly on the horizontal black line. A horizontal line-of-simultaneity is a line representing NOW across space for the stationary observer perspective at some moment. So in this case, O always sees B & C clocks aligned in time readout at any moment, ie they appear synchronized per O.

The bottom figure shows the A stationary perspective. O's line-of-simultaneity is slanted because O is in motion with slanted worldline, and 12:34am-C, 1:09am-O, 12:34am-B all ly on the slanted gray line. This represents O's sense of NOW at 12:09am-O. A slanted line-of-simultaneity is a line representing NOW across space for the moving observer perspective. Note that the observers C,O,B times which intersect this slanted line-of-simultaneity match that of the upper figure!

To determine a moving observer's line-of-simultaneity, draw his x'-axis such that the angle it makes with the 45 deg light beam is identical to the angle his slanted worldline makes with the 45 deg light beam, but on the other side of the light beam. This is to say ...

the faster an observer moves, the more slanted his worldline becomes. Light paths are always 45 deg on Minkowski worldline diagrams, since they travel one unit space per 1 unit time, and units of space & time are set equal (normalized), and so speed c=1. The more a moving observer's worldline tilts away from your own stationary vertical worldline (ie your time axis), the more his x'-axis tilts upward toward the 45 deg ray path from your x-axis. That x'-axis is his line-of-simultaneity, his sense of NOW!

Note that in the top diagram, O stationary, the observer A & B clocks read the same in that O instant of 1:09am-O. In the bottom A stationary figure, observers B & C never possesses the same time readouts in any observer A moment, because B & C are in motion. The most aft clock (C's) reads well ahead of the foremost clock (B's). A's line-of-simultaneity is of course horizontal in the lower figure, since that's an A stationary vantage. Lines-of-simultaneity are your spatial axis, so x-axis for one guy, x'-axis for the other guy. Minkowski spacetime diagrams portray 1-space vs time.

pess

 

[pess5]

pess. What about the scenario I mentioned earlier ?

If the twins start from opposite sides of the Earth, 6ly away (Earth distance) and stationary compared with the Earth. They then accelerate to 0.6c and travel to Earth. What are the time lines ?

This way they have the perfect hand-off as they all meet at Earth at the same time, or not, or they do, or they don't ?

Mikeh,

Ahhh. Assuming all clocks are in sync from the start, and accelerate/decelerate identically, the twins will arrive on Earth with identical clock readings. However their clocks will lag behind the Earth clock. If we assume the magical instant accelrations, each twin traverses 0.8(6ly)=4.8ly, since a v=0.6 produces a contraction of 0.8. So 4.8ly/0.6c=8yr, thus the twins age 8 yrs in their trek to earth.

Earthlings see it differently of course. They have to await the twins to transcend 6ly at 0.6c, so 6ly/0.6c=10 yr. So although the twins only age 8 yr over the trek, the earthlings age 10 yr. Personally, I'd rather age 10 yr on Earth with my friends than age 8 yr alone in a tin can alone.

I know you been thru this with JesseM already with all the math, but I'm just shooting from the cuff here.

PS ... my twins handoff scenario PDF is now at a rev c. That's the best one you'll want to look at. thanx.

pess

 

[pess5]

Don't feel too bad about it. You're yet another victim of incomplete presentations of basic relativistic phenomena in popular-level science books, and even in many general-physics textbooks at the college/university level.

They all talk about length contraction and time dilation, but most of them either don't mention relativity of simultaneity at all, or present it in a way that doesn't make clear from the beginning that it is at the same level of importance as length contraction and time dilation.

Well, I'd have to say you are correct on this matter. I therefore will post an attachment here for the benefit of all regarding the ...

the Relativity of Simultaniety​

Please see the following PDF attachment. Goes back a number of years. It's a scenario depicting vulcano eruptions on Mt Hood & Mt Ranier and investigates how a seismologist & passing spaceship capt witness the events.

jtBell ... does this suffice?

 

[robphy]

Please see the following PDF attachment. Goes back a number of years. It's a scenario depicting vulcano eruptions on Mt Hood & Mt Ranier and investigates how a seismologist & passing spaceship capt witness the events.

Hmmm... sounds like something from the Physics Education Group at U. Washington.

 

[JesseM]

Ouch. Didn't see that one coming. How can that be ?

If A starts on E1 and C starts on E3 and E1-3 are 'stationery', A & C start at exactly the same f.o.r as E1-3 and all of their clocks are sync'd ?

The difference in synchronizations shouldn't exactly be thought of as a physical effect--if you accelerate and then begin moving inertially again, your rulers and clocks can generally assumed to now measure length and time the same as rulers and clocks that have always been at rest in your current inertial rest frame, but clocks on different parts of your ship will not automatically by synchronized the same way as your current rest frame's definition of what it means for different clocks to be "synchronized".

Ultimately, all the special equations we've used--lorentz contraction equation, time dilation equation, velocity addition equation, and that formula for how clocks will be out-of-sync in frames where they're moving if they're in-sync in their own rest frame--are derived from the general coordinate transformation I gave you, which is known as the "Lorentz transformation". So it might help if I gave you an idea of what assumptions are needed to derive the Lorentz transformation in the first place. Basically, Einstein started out by assuming each inertial observer should define the coordinates of any event in terms of local measurements made on a system of rulers and synchronized clocks which are at rest with respect to that observer. So, for example, if I look through my telescope and see a firecracker going off in the distance, then if I see that it happened right next to the 300-meter mark on a ruler laid out parallel to my x-axis, I'll assign the event an x-coordinate of 300 meters, and if I see that as the explosion was happening the clock at the 300-meter mark was reading 25 seconds, I'll assign the event a t-coordinate of 25 seconds. The fact that I'm relying only on local measurements made in the immediate area of the event means I don't have to worry about the delay between the time an event happens and the time I actually see it, due to the finite speed of light.

But this brings up a new problem--what does it mean for clocks at different positions in my system of rulers and clocks to be "synchronized"? The most obvious method of synchronizing distant clocks is to move them to a common position, synchronize them right next to each other, and then move them apart again. But by the time Einstein was working on SR, it had already been suggested by Lorentz that moving clocks might slow down, in order to account for the results of the Michelson-Morley experiment. The story behind this is that physicists had always imagined that light only traveled at exactly c in the rest frame of the "ether", which was supposed to be the medium that light waves were vibrations in just like sound waves are vibrations in the air, and that if you were moving at velocity v relative to the ether, you'd measure light to move at c+v in one direction and c-v in the other, just as would be true if you measured sound waves while in motion relative to the rest frame of the air around you. But the Michelson-Morley experiment had shown that light was measured to move at c in all directions at different points in the Earth's orbit, when its velocity relative to any inertial frame like the frame of the ether should have been noticeably different, and Lorentz found that this could be accounted for if you imagine that rulers shrunk when moving relative to the ether, and clocks slowed down.

So back to clock synchronization: if clocks change their rate of ticking depending on how they move, this causes problems for the obvious method of synchronizing clocks at a common location and then moving them apart, because they could get out of sync during the process of moving them. Einstein suggested a different method: what if each observer synchronizes different clocks in their system using light-signals, making the assumption that light travels at the same speed in all directions in their own rest frame? (This assumption wouldn't make any sense if you believed in the ether, but the results of making this assumption led to some interesting results which made Einstein think the ether should be abandoned altogether.) If each inertial observer makes this assumption, they could each synchronize their own clocks by setting off a light flash at the midpoint of two clocks, then setting both clocks to read the same time at the moment the light from the flash reaches them.

But a little thought shows that this method must lead different inertial observer to disagree about simultaneity. Suppose I am in a ship which is moving forward in your rest frame, and I decide to synchronize two clocks at the front and back of my ship using this method of setting off a flash at the center and setting them to read the same time when the light reaches them. In your frame, both clocks were still the same distance from the point where the flash was set off, but the clock at the front of the ship is moving away from that point as the ship moves forward, while the clock at the back is moving towards that point, so naturally if you assume that both light beams move at the same speed relative to yourself, you must conclude that the light beam moving towards the back clock caught up with that clock at an earlier time than the other light beam caught up with the front clock. But if I set both clocks to read the same time when the light reached them, naturally this leads you to conclude that my front clock is running behind my back clock! As long as each observer assumes light moves at the same speed in all directions in their own rest frame, there is no way to avoid this sort of disagreement about simultaneity.

Note that this definition of what it means for two clocks at rest in a given frame to be "synchronized" in that frame is not really forced on you by nature--you'd be free to use a different method of synchronization if you wanted, like having one special observer synchronize his clocks using this light-signal method, but then having every other observer synchronize their clocks so that they'd agree with the special observer's definition of simultaneity. For this reason, the standard synchronization method in special relativity is known as the "Einstein synchronization convention." But it's not arbitrary either, there is in fact a good reason to prefer this method of synchronization to any other. If two inertial observers both define their coordinate systems in terms of measurements made on the sort of ruler-clock system described by Einstein, then if the first observer figures out the correct equations to describe the laws of physics in terms of his coordinates x,y,z,t, and the second observer figures out the correct equations for the laws of physics in terms of his own coordinates x',y',z',t', they will find that they are both using identical equations, save for the replacement of x with x' and y with y' and z with z' and t with t'. This would not be true if they had used a different clock synchronization convention in defining their coordinate systems. It's a particular feature of the laws of physics in our universe that they all have the property of looking the same in the different coordinate systems constructed in this way, known as "Lorentz-invariance" (and you can show that these coordinate systems will be related by the Lorentz transformation which I mentioned before). It's this Lorentz-invariance of all the known fundamental laws which makes this the most "natural" way to define the coordinate systems of inertial observers, including each system's differing definition of simultaneity.

As always, let me know if you have questions about any of these ideas. Conceptually, it may also help you to learn more about "Minkowski diagrams", which show the x and t axes of different reference frames in a single diagram (lines of simultaneity, or constant t-coordinate, for different frames will appear slanted with respect to one another, for example), along with the worldlines of different objects. The ones pess5 provided are good examples, and you might want to find some general tutorial as well.

When A accelerates (instantly) to 0.6c. His time will dilate by a factor of 0.8 compared with all 3 ? The distance will also compress by a factor of 1.25 compared with all 3 ?

By "all 3" you mean the 3 planets, not including C whose velocity relative to A is different from that of the planets, right? If so, yes, in the rest frame of the planets A's clock will be compressed to 0.8 the length they were previously, and the length of his rulers will be divided by 1.25 (or shrunk by a factor of 0.8, equivalently).

Ok. Now I'm struggling to keep up.

Before the journey starts, tA = tE1 = tE2 = tE3 = tC ?

At what point does A see E2's time jump ahead by 3.6 years ? and E3's time jump ahead by 7.2 years ?

Again, it's best not to think of the differences in simultaneity as this sort of physical effect. If you imagine two networks of rulers and synchronized clocks moving past each other inertially, one at rest in the Earth's' frame and one which will be at rest with respect to A after his acceleration, then you can just imagine A switching which network of rulers/clocks he uses to measure distances and times before and after he accelerates, since he always wants to make measurements on a system that is currently at rest with respect to him.

As I've said in earlier posts, I think the best course is to avoid trying to think about the "point of view" of a non-inertial observer altogether, and just analyze the problem from start to finish in a single inertial frame (after which you can of course go back and analyze it from start to finish in a different frame, the point is not to try to switch frames in the middle).

 

[pess5]

Hmmm... sounds like something from the Physics Education Group at U. Washington.

Sounds like you didn't like the scenario much? I thought it was pretty decent.

I just looked up the website for Physics Education Group at U. Washington here, and you were right of course. I debated this scenario online long ago, but couldn't remember what university had it posted. At the time, we debated about 8 or 9 presumed paradoxes of relativity theory. As it turned out, none were paradoxes after all. The error always arises from injecting absolute simultaneity midstream into a relativistic scenario, whereby the debates always began assuming the postulates true. Anyways, thanks much. It's an excellent scenario for grasping relative simultaneity. Here's the link...

http://arxiv.org/ftp/physics/papers/0207/0207081.pdf"​

See Section VI ... Assessing Student Understanding of Simultaniety

pess

 

[robphy]

Sounds like you didn't like the scenario much? I thought it was pretty decent.

I just made the comment to note a similarity.
It wasn't intended to complain about the scenario.
(I haven't had a chance to really look at it in detail...been a little busy.)

 

[M1keh]

Don't feel too bad about it. You're yet another victim of incomplete presentations of basic relativistic phenomena in popular-level science books, and even in many general-physics textbooks at the college/university level.

They all talk about length contraction and time dilation, but most of them either don't mention relativity of simultaneity at all, or present it in a way that doesn't make clear from the beginning that it is at the same level of importance as length contraction and time dilation.

Start with a line of clocks, all stationary in inertial reference frame S, and all synchronized in S. In reference frame S' which is moving along that line with respect to S, three things about these clocks are different:

1. The clocks are closer together than in S: length contraction.

2. The clocks tick more slowly than an identical clock which is at rest in S': time dilation.

3. The clocks are not synchronized in S': relativity of simultaneity.

In general, you need to take all three of these phenomena into account, not just the first two.

Thanks for the reply.

The sqrt( 1 - (v**2/c**2) ) accounts for 1 & 2. I think JesseM gave me the formula for 3. I'll check.

Now, I'm probably wrong, but my understanding was that two observers in different f.o.r's can synchronise their watches as they pass each other, using light singals ? So they can confirm that their watches show the same times ?

This presumably also confirms that they ARE at the same place ?

 

[jtbell]

Yes, two observers (each carrying a clock with her) that pass each other closely (ideally two pointlike observers passing infinitesimally close to each other) can unambiguously synchronize their clocks at the instant that they pass. Of course, the clocks go out of sync again immediately, because they run at different rates in any inertial reference frame.

Similarly, even if the two observers don't synchronize their clocks as described above, they always agree on the readings of each others' clocks at the instant that they pass.

For that matter, if any two moving clocks pass each other infinitesimally closely, all observers agree on the readings of those two clocks at the instant that they pass.

 

[M1keh]

Yes, two observers (each carrying a clock with her) that pass each other closely (ideally two pointlike observers passing infinitesimally close to each other) can unambiguously synchronize their clocks at the instant that they pass. Of course, the clocks go out of sync again immediately, because they run at different rates in any inertial reference frame.

Similarly, even if the two observers don't synchronize their clocks as described above, they always agree on the readings of each others' clocks at the instant that they pass.

For that matter, if any two moving clocks pass each other infinitesimally closely, all observers agree on the readings of those two clocks at the instant that they pass.

Ok.

And all observers in a common f.o.r, no matter how far apart, agree on the time in their f.o.r ?

ie. If I'm one side of the Universe, stationery compared with another observer on the other side of the Universe, we will agree on the current time ?

 

[jtbell]

Yes, all observers who are at rest in the same inertial reference frame, and therefore at rest with respect to each other, can synchronize their clocks so that they all agree on what time it is, in that inertial reference frame.

In any other inertial reference frame, those clocks will not be synchronized.

 

[M1keh]

Yes, all observers who are at rest in the same inertial reference frame, and therefore at rest with respect to each other, can synchronize their clocks so that they all agree on what time it is, in that inertial reference frame.

In any other inertial reference frame, those clocks will not be synchronized.

Ok. Here goes.

If A1, A2 & A3 are all in the same f.o.r, but 6ly apart, and B1, B2, B3 are all traveling at 0.6c relative to the A's but, locally, 4.8ly apart, they could reach B1, B2 & B3 respectively, simultaneously. B1, B2 & B3 could then synchronise their watches with A1, A2 & A3 respectively.

Now A1, A2 & A3 must agree on their local time & B1, B2 & B3 must agree on their local time. If A3 agree's with B3's time (now synchronised) A1 must agree with B3's time ?

What am I missing ?

 

[JesseM]

Ok. Here goes.

If A1, A2 & A3 are all in the same f.o.r, but 6ly apart, and B1, B2, B3 are all traveling at 0.6c relative to the A's but, locally, 4.8ly apart, they could reach B1, B2 & B3 respectively, simultaneously.

Only in the B frame will they pass the A clocks simultaneously. In the A frame, the B clocks are 4.8*0.8 = 3.84 ly apart while the A clocks are 6 ly apart, so each B clock will pass its respective A clock at a different time.

B1, B2 & B3 could then synchronise their watches with A1, A2 & A3 respectively.

If they do that, then their watches will no longer be synchronized in their own rest frame. Your definition of simultaneity is not based on what your clocks actually happen to read, since your clocks could very well be out of sync, it's based on what an ideal network of clocks at rest in your frame and synchronized according to Einstein's clock synchronization procedure would define as simultaneous.

Now A1, A2 & A3 must agree on their local time & B1, B2 & B3 must agree on their local time. If A3 agree's with B3's time (now synchronised) A1 must agree with B3's time ?

In the A frame, the B clocks pass their corresponding A clocks at different times, as I mentioned above, so even if each B clock sets itself to the same time as the A clock at the moment it passes it, since the B clocks are still ticking slow in A's frame, the B clocks won't be synchronized (first B1 passes A1 and sets itself to the same time as all the A clocks, but then 3.6 years later in the A frame, B2 passes A2 and sets itself to the same time as the current reading on all the A clocks, but B1 has only advanced by 3.6*0.8 = 2.88 years in this time so it's 3.6-2.88 = 0.72 years behind B2, and likewise B2 will be 0.72 years behind B3 when it sets itself to the same time as B3). Meanwhile, in the B frame all three B clocks pass their A clocks simultaneously, but the three A clocks are out of sync in this frame--as I mentioned before, if two clocks are x distance apart and synchronized in their own rest frame, then in a frame where they're moving at v along the axis joining them, the back clock's time will be ahead of the front ones by vx/c^2. So since the A clocks are 6 ly apart and moving at 0.6c in the B frame, in this frame each one will be out of sync with the one ahead of it by (0.6 ly/year)*(6 ly)/(1 ly/year)^2 = 3.6 years. So when the B clocks synchronize with the A clocks as they pass, they'll now each be 3.6 years out of sync from their nearest neighbor in their own rest frame.