"Bonnie Granat" wrote in message
...

"Stephen Bint" wrote in message
.. .
David,

The formula for time dilation is a function of velocity. If I am
moving
at a
given velocity with respect to you, then you are moving at the exact
same
velocity with respect to me.

Not quite correct.

In that case, if I am moving at 10kph with respect to you, at what
velocity
will you be moving, with respect to me? Greater, or less?

Stephen


You haven't yet said what my velocity is. Am I stationary?

It doesn't matter whether you are staiuonary, or moving, with respect to
anyone or anything else. I am moving at 10kph with respect to you. It could
be that you are travelling along a road at 45kph and I am overtaking you at
55kph. Or you could be standing at the side of the road while I pass you in
a car doing 10kph.

The "v" in the time dilation equation refers only to the observed object's
speed with respect to the observer. Our speeds with respect to everything
else have no effect on the outcome of the calculation of time dilation.

--
Bonnie Granat
Granat Technical Editing and Writing
http://www.editors-writers.info

In article ,
Stephen Bint wrote:

The problem is not "deciding" which frame is the one running slow, by
clutching at an asymmetry. The problem is that time dilation fails to
prevent witnesses seeing the bouncing beam going faster than light, unless
it works both ways and the Earth-bound observer has also aged less than the
twins.

If you say either frame is the slow one, you deviate from what the formula
asserts, which is plainly that both are slow by the same factor.

In your previous example, all factors (time dilation and kinematic) are
symmetrical between the two twins, so they experience the same elapsed
time. In the classic "twin paradox", the time dilation is symmetric, but
the kinematic factors aren't. Therefore the twins experience different
elapsed times.

Here's a description of a classic "twin paradox" situation, in the same
style as my description of your previous example:

The scenario: You stay behind on Earth while your twin goes on a space
journey. From your point of view he travels away from the Earth at a
speed of 0.8c for 5 years, covering a distance of 4 light-years in the
process, then immediately turns around and returns, again at a speed of
0.8c, taking another 5 years for the return trip. The total trip duration
is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time
because of time dilation:

(5 years) * sqrt (1 - 0.8^2) = 3 years

for each leg of the trip, and from his point of view the round trip lasts
only 6 years.

The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.

By examining what both of you actually *see* by watching each other's
clocks/calendars through telescopes, we can show that both of you must
come to the same conclusion: the trip lasts 10 years for you, and 6 years
for him.

First let's look at what's happening from your point of view.

To be specific, assume that your twin starts out on his journey at the
very beginning of the year 2004. You watch his clock (calendar?) through
your telescope as he recedes. The rate at which you receive the images of
each of his "new years" depends not only on his time dilation, but also on
the fact that he is moving away from you, so we have to use the
relativistic Doppler-effect equation. You see his calendar turn over to a
new year at intervals of

(1 year) * sqrt ((1 + 0.8) / (1 - 0.8)) = 3 years.

Therefore,

At the beginning of 2007, you see his calendar turn over to 2005.

At the beginning of 2010, you see his calendar turn over to 2006.

At the beginning of 2013, you see his calendar turn over to 2007.

This is the point at which you see him turn around and begin his return
trip. In your reference frame, he actually began the return trip at the
beginning of 2009 (2004 + 5), but you don't see it until 4 years
later because it happened 4 light-years away from you.

During his return trip, we have to switch the signs in the Doppler-shift
equation because now he's approaching, not receding. You now see his
calendar turn over to a new year at intervals of

(1 year) * sqrt ((1 - 0.8) / (1 + 0.8)) = 1/3 year.

Therefore,

At the beginning of May 2013, you see his calendar turn over to 2008.

At the beginning of September 2013, you see his calendar turn over to
2009.

At the beginning of January 2014, you see his calendar turn over to 2010,
and he arrives home. Ten years have elapsed on your calendar, and 6 years
have elapsed on his.

What does this look like from your twin's point of view, as he watches
your clock/calendar through his telescope?

As he is traveling away, he sees your calendar turn over a new year at
intervals of three years, just like you do his. Therefore,

At the beginning of 2007 (2004 + 3), he sees your calendar turn over to
2005.

But three years is the length of the outbound trip, according to him,
because of time dliation. So at this point he turns around and begins his
return trip. Now he sees your calendar turn over a new year at intervals
of 1/3 year, just like you do his. Therefore,

At the beginning of May 2007, he sees your calendar turn over to 2006.

At the beginning of September 2007, he sees your calendar turn over to
2007.

At the beginning of January 2008, he sees your calendar turn over to 2008.

At the beginning of May 2008, he sees your calendar turn over to 2009.

At the beginning of September 2008, he sees your calendar turn over to
2010.

At the beginning of January 2009, he sees your calendar turn over to 2011.

At the beginning of May 2009, he sees your calendar turn over to 2012.

At the beginning of September 2009, he sees your calendar turn over to
2013.

At the beginning of January 2010, he sees your calendar turn over to 2014.

But now three years have elapsed (on his calendar) since he turned around,
so he has now returned home. Six years have elapsed on his calendar, and
10 years have elapsed on yours, in agreement with what you observe.

Note that although we did not use the time dilation equation directly in
predicting what your twin sees your clock/calendar doing, it is implicit
in the relativistic Doppler-shift equation. In deriving the relativistic
Doppler shift equation, one must use the time-dilation equation.

--
Jon Bell Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA

Jon,

I see that you are asserting, that the accelerations undergone by one
observer affect time dilation. Yet the velocity, v, is the only variable in
the time dilation formula, is it not? Does that not imply, that the
calculation of the degree of time dilation for either party, is irrespective
of anything but their velocity, with respect to eachother?

Will you indulge me, by considering an example of mine, that illustrates the
symmetry implicit in the assertion of time dilation?

Two skateboards glide past eachother at speed.

On each skateboard is an observer and a device which sends a light pulse
from a source which is 1m above the base, vertically down to a mirror on the
base, which reflects it directly vertically back to the source. To one of
the observers (either one) the light pulse on the other board traces a
V-shaped path, which is longer than the vertical one. If there is no time
dilation, the observer will see the pulse travel a greater distance in the
same time, and therefore, see it travelling faster than c.

Time dilation slows the event on the other skateboard, the exact amount
necessary to make the light pulse travel this longer path at exactly c. The
length of the V-shaped path and consequently, the time dilation required, is
purely a function of the relative velocity of the two skateboards.

Furthermore, the path that A sees the pulse trace on B's skateboard is
exactly the same length as the path seen by B, taken by the light pulse on
A's skateboard.

So the time dilation must be identical in both directions, regardless of how
they acheived their relative velocity.

So what if your travelling twin was passing earth and viewing such an
experiment there. Either he will see time slowed down there, or he will see
the light pulse travelling faster than c there. Is that not the case?


"Jon Bell" wrote in message
...
In article ,
Stephen Bint wrote:

The problem is not "deciding" which frame is the one running slow, by
clutching at an asymmetry. The problem is that time dilation fails to
prevent witnesses seeing the bouncing beam going faster than light,
unless
it works both ways and the Earth-bound observer has also aged less than
the
twins.

If you say either frame is the slow one, you deviate from what the
formula
asserts, which is plainly that both are slow by the same factor.

In your previous example, all factors (time dilation and kinematic) are
symmetrical between the two twins, so they experience the same elapsed
time. In the classic "twin paradox", the time dilation is symmetric, but
the kinematic factors aren't. Therefore the twins experience different
elapsed times.

Here's a description of a classic "twin paradox" situation, in the same
style as my description of your previous example:

The scenario: You stay behind on Earth while your twin goes on a space
journey. From your point of view he travels away from the Earth at a
speed of 0.8c for 5 years, covering a distance of 4 light-years in the
process, then immediately turns around and returns, again at a speed of
0.8c, taking another 5 years for the return trip. The total trip duration
is 10 years by your reckoning.

Relativity predicts that your twin experiences less elapsed time
because of time dilation:

(5 years) * sqrt (1 - 0.8^2) = 3 years

for each leg of the trip, and from his point of view the round trip lasts
only 6 years.

The relativistic time dilation equation predicts that each twin's clocks
"run slower" in the other twin's reference frame. So why can't your twin
conclude that the trip must be shorter for you, than it is for him? The
answer lies in the fact that your experiences are not symmetrical. Your
twin is at rest in two different inertial reference frames, one during the
outbound trip and another one during the inbound trip. You remain at rest
in a single inertial reference frame during the entire journey. Your twin
has to fire his spaceship's engines at the turnaround point. You do
nothing.

By examining what both of you actually *see* by watching each other's
clocks/calendars through telescopes, we can show that both of you must
come to the same conclusion: the trip lasts 10 years for you, and 6 years
for him.

First let's look at what's happening from your point of view.

To be specific, assume that your twin starts out on his journey at the
very beginning of the year 2004. You watch his clock (calendar?) through
your telescope as he recedes. The rate at which you receive the images of
each of his "new years" depends not only on his time dilation, but also on
the fact that he is moving away from you, so we have to use the
relativistic Doppler-effect equation. You see his calendar turn over to a
new year at intervals of

(1 year) * sqrt ((1 + 0.8) / (1 - 0.8)) = 3 years.

Therefore,

At the beginning of 2007, you see his calendar turn over to 2005.

At the beginning of 2010, you see his calendar turn over to 2006.

At the beginning of 2013, you see his calendar turn over to 2007.

This is the point at which you see him turn around and begin his return
trip. In your reference frame, he actually began the return trip at the
beginning of 2009 (2004 + 5), but you don't see it until 4 years
later because it happened 4 light-years away from you.

During his return trip, we have to switch the signs in the Doppler-shift
equation because now he's approaching, not receding. You now see his
calendar turn over to a new year at intervals of

(1 year) * sqrt ((1 - 0.8) / (1 + 0.8)) = 1/3 year.

Therefore,

At the beginning of May 2013, you see his calendar turn over to 2008.

At the beginning of September 2013, you see his calendar turn over to
2009.

At the beginning of January 2014, you see his calendar turn over to 2010,
and he arrives home. Ten years have elapsed on your calendar, and 6 years
have elapsed on his.

What does this look like from your twin's point of view, as he watches
your clock/calendar through his telescope?

As he is traveling away, he sees your calendar turn over a new year at
intervals of three years, just like you do his. Therefore,

At the beginning of 2007 (2004 + 3), he sees your calendar turn over to
2005.

But three years is the length of the outbound trip, according to him,
because of time dliation. So at this point he turns around and begins his
return trip. Now he sees your calendar turn over a new year at intervals
of 1/3 year, just like you do his. Therefore,

At the beginning of May 2007, he sees your calendar turn over to 2006.

At the beginning of September 2007, he sees your calendar turn over to
2007.

At the beginning of January 2008, he sees your calendar turn over to 2008.

At the beginning of May 2008, he sees your calendar turn over to 2009.

At the beginning of September 2008, he sees your calendar turn over to
2010.

At the beginning of January 2009, he sees your calendar turn over to 2011.

At the beginning of May 2009, he sees your calendar turn over to 2012.

At the beginning of September 2009, he sees your calendar turn over to
2013.

At the beginning of January 2010, he sees your calendar turn over to 2014.

But now three years have elapsed (on his calendar) since he turned around,
so he has now returned home. Six years have elapsed on his calendar, and
10 years have elapsed on yours, in agreement with what you observe.

Note that although we did not use the time dilation equation directly in
predicting what your twin sees your clock/calendar doing, it is implicit
in the relativistic Doppler-shift equation. In deriving the relativistic
Doppler shift equation, one must use the time-dilation equation.

--
Jon Bell Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA

"Stephen Bint" wrote in message
.. .
Jon,

I see that you are asserting, that the accelerations undergone by one
observer affect time dilation. Yet the velocity, v, is the only variable
in
the time dilation formula, is it not? Does that not imply, that the
calculation of the degree of time dilation for either party, is
irrespective
of anything but their velocity, with respect to eachother?

Will you indulge me, by considering an example of mine, that illustrates
the
symmetry implicit in the assertion of time dilation?

Two skateboards glide past eachother at speed.

On each skateboard is an observer and a device which sends a light pulse
from a source which is 1m above the base, vertically down to a mirror on
the
base, which reflects it directly vertically back to the source. To one of
the observers (either one) the light pulse on the other board traces a
V-shaped path, which is longer than the vertical one. If there is no time
dilation, the observer will see the pulse travel a greater distance in the
same time, and therefore, see it travelling faster than c.


But if the assumption for this experiment is that c is the top speed and the
constant speed of light, the mere fact that the light path is longer tells
me that time is dilated.

Doesn't it, Bill Hobba???


--
Bonnie Granat
Granat Technical Editing and Writing
http://www.editors-writers.info

Stephen Bint wrote:

Claiming an assymetry based on the acceleration histories of the observer,
overlooks the fact that the time dilation must be real and identical in both
directions at all times or else someone will see the bouncing beam going
faster than c.
...
To say that acceleration makes time run faster, implies that if one of the
skateboarders is accelerating, the time contraction due to acceleration will
at least partially counteract the dilation caused by the velocity (predicted
by Einstein), allowing the other to see the bouncing beam move faster than
c.

There seem to be two misunderstandings here, one possibly yours and one
by those who (not having read the previous posts) I presume have been
talking about acceleration to you.

To put the record straight:

(1) An unaccelerated and an accelerated observer are NOT equivalent, but

(2) acceleration does NOT cause time dilation.

Suppose twin A stays at home and twin B travels out, accelerates towards
earth, and returns younger than twin A.

Acceleration has NOT (repeat, NOT) caused the time dilation of twin B.

In A's frame, B travelled faster than A at all times (except the small
reversal time) and so suffered a relative time dilation.

No equivalent argument can be made from twin B's point of view, because
B does not have a single reference frame. B on the outward journey will
see A suffering a dilation, but B's own return to earth later, from B's
own earlier frame, will seem like an even greater relative speed with an
even greater time dilation (twice A's) that will lead B to conclude that
B will suffer time dilation relative to A. That is, both A and B predict
that B will age less. This is because A has only one reference frame for
the whole time, but B has two different frames at different times, and
from either of these frames, B will conclude that B will age less.

But acceleration had nothing to do with it. If B were replaced by two
entities, B1, who makes the outward journey at constant speed, who
passes B2 moving in the reverse direction and who syncronises clocks
with B2 as they pass, then the time on B2's clock when he gets back will
show the same time dilation as the accelerated original B. Since neither
B1 nor B2 accelerated, this dilation is not due to the acceleration.

Any observer in a single reference frame, according to SR, will have an
interpretation of reality that is consistent with those of all other
observers, once allowance for the relative motion is made. But an
accelerated observer is not in a single reference frame, because the
acceleration changes his frame. That is the only relevance of
acceleration to the situation.

The simplest way I know to think of SR is as a geometric theory. The
formula for the length of the diagonal of a right triangle is the well
known Pythagoras' formula, but, on the time axis, bung in a minus sign.
This has the effect that, for space-like intervals, the shortest
distance between two points is a straight line, whereas for timelike
intervals, this is the longest distance. Bent timelike curves are
shorter than straight ones. Accelerated paths (by the definition of
acceleration) are bent ones, and therefore shorter.

To take this analogy a step further, suppose two points are marked on
the ground, P and Q. There is one piece of straight rope connecting P
and Q, and another curved piece doing likewise. You know just by looking
at it that the curved piece must be longer than the straight one. But it
is not that the curve in the rope has somehow stretched it. Now replace
P and Q by the timespace locations of A at the beginning and end of B's
journey, the straight rope by A's stationary (on earth) trajectory, the
curved rope by the path of the spaceship carrying B, "curve in the rope"
by "acceleration of B", and "longer" by "shorter". With these changes,
the situations are identical, and there is nothing baffling or
paradoxical about it at all.

--
Ron House
http://www.sci.usq.edu.au/staff/house

"Stephen Bint" wrote in message .. .
I started a thread about the twin paradox and cross-posted it to three
newsgroups. I am grateful to those who discussed it, even though it got a
tiny bit personal at times

Troll alert:
http://users.pandora.be/vdmoortel/di...s/AChance.html

Dirk Vdm

"Stephen Bint" wrote in message .. .
..
you are appealing to me to be open-minded enough to be able to believe it is
possible that I am wrong. I would ask the same of you.

There is a difference. Dan's view is supported by thousands
of physicists throughout the world who have studied the subject
in great detail for nearly a century.

You should start by understanding how two inertial observers
in relative motion would measure one another's clocks. It
would then be useful to understand how they would _see_
one another's clocks, which is different.

You have a choice. You can either join the list of crackpots
who frequent this group, and be treated with scorn and derision
by the physicists here, or you can make an effort to understand
the subject.

Martin Hogbin

"Martin Hogbin" wrote in message ...

"Stephen Bint" wrote in message .. .
.
you are appealing to me to be open-minded enough to be able to believe it is
possible that I am wrong. I would ask the same of you.

There is a difference. Dan's view is supported by thousands
of physicists throughout the world who have studied the subject
in great detail for nearly a century.

You should start by understanding how two inertial observers
in relative motion would measure one another's clocks. It
would then be useful to understand how they would _see_
one another's clocks, which is different.

You have a choice. You can either join the list of crackpots
who frequent this group, and be treated with scorn and derision
by the physicists here, or you can make an effort to understand
the subject.

He has been here before with another name.
Same troll, new name.

Dirk Vdm

I disagree with the subject. If you don't do the math you can never
understand the heart of the problem, which is clock synchronization.
(Well, that's _one_ heart of the problem, anyway...)

I posted some of the math for the twin paradox in your earlier thread,
but I only posted it on alt.sci.physics. So, I've taken this
opportunity to spruce it up by adding the affine transformations to
rezero the clocks.

Others have addressed the issue of what each twin sees through a
telescope, so I haven't done that here. I also didn't take the extra
step of filling space with "virtual observers" and reporting what they
see. That's also informative and is easier than dealing with telescopes
since it doesn't involve Doppler shift -- you can do it just from the
transforms as given here.

With no further introduction, here it is again.

(Pick holes, if you will -- I'm always willing to learn...)

================================================== =================

The problem as stated was "two twins fly to Earth, each from a distance
of 1 ly, at a velocity of 0.75c. What happens?"

I worked this one through. There's no contradiction. It is, however,
fiendishly confusing.

Typing equations in Ascii is not a lot of fun so I'm going to try to
keep this brief. (Ha, ha.)

I'll try to describe what happens in English, with limited use of
equations. First, I made a few simplifications of the original
statement of the problem to keep the math tractable:

-- Set c=1
-- All dimensions are the same -- time and distance are both measured
as lengths (conversion factor = c, which we set equal to 1).
-- I used a velocity of 1/2 rather than 0.75
-- I assumed that A starts at location -1, B starts at location +1,
and there's an observer in the middle at location 0 who remains
stationary in the initial inertial frame (before anybody
accelerated). Call him "O".

Call the inertial frame they all started in the "base frame".

Then at time 0 in the base frame, an alarm goes off in each spaceship
and at the origin, telling A, B, and O it's time to start. A and B
immediately accelerate to +1/2 and -1/2 respectively.

Gamma and beta are as usual: beta=velocity, gamma=1/sqrt(1-beta^2).
Let's start looking at things from the point of view of 'O'.

In this frame, A is moving at beta=1/2, gamma=2/sqrt(3). The Lorentz
transform for frame(O) - frame(A) looks something like this, with a
unit width font and a 1-dimensional world (y and z components are
dropped out):

| 2/sqrt(3) -1/sqrt(3) |
| -1/sqrt(3) 2/sqrt(3) |

To get from frame(O) - frame(B) it looks like this:

| 2/sqrt(3) 1/sqrt(3) |
| 1/sqrt(3) 2/sqrt(3) |

And the transform from A -- O is the same as the one from O -- B
(just reverse the velocity, of course).

Since A starts at X=(0,-1) in the base frame, we transform that to A's
(new) frame and we see that X' = (1/sqrt(3), -2/sqrt(3)).

The distance to the origin (X'=0) went down, 'cause it was contracted,
so it's just 2/sqrt(3) in A's frame. BUT the TIME changed, too --
it's 1/sqrt(3), _not_ 0. This is where the confusing part starts, and
it's worth a few words.

No time passed for A, and his clock still reads 0. However, if, after
accelerating, he _again_ synchronizes his clock with someone located
at X=0, the result won't be the same -- as far as he's concerned, O
and B have both now got clocks that read incorrectly. In simple
terms, he needs to use light-speed signals the sync up his clocks, and
the SOL delay messes up the operation. Since A's clock hasn't
actually changed, and still reads 0, we're going to just subtract
1/sqrt(3) from all times in A's frame from now on. (This makes things
a little messier and is somewhat error prone, unfortunately.)

Anyway, now we've got A with a clock that reads 0, and an apparent
distance to the origin of 2/sqrt(3). He's coming toward O at v=1/2,
so at time t=2 in O's frame, A arrives. In O's frame, the coordinates
are (2,0). We transform that to A's frame, and get
(4/sqrt(3),-2/sqrt(3)) -- BUT we need to subtract 1/sqrt(3) to get A's
actual clock reading. So A's clock must read 3/sqrt(3) = sqrt(3).
That's ~ 1.7, which is rather less than the 2 units that elapsed for
O. A aged less than O.

The behavior of B, from the point of view of O, is identical,
including the subtraction of 1/sqrt(3) to zero his clock. So, as far
as O can see, B also ages by 3/sqrt(3). Since the math is the same
I'm not including it here.

Now, the interesting part: We work it out again from A's point of
view. In A's frame, beta(O) = -1/2, gamma(O) = 2/sqrt(3), and the
transform from A to O is the same as the one from O to B. But we also
need the transform from A -- B, and for that we need beta(B) in A's
frame -- beta'(B), if you will.

B's velocity vector in B's rest frame is (1,0) (moving through time at
rate 1, stationary in space). We can transform that to O's frame, and
we get (2/sqrt(3),-1/sqrt(3)) (this is the 4-velocity with two terms
dropped out, of course). We can now transform that from O's frame to
A's frame using the transform we worked out back at the top, and we
get (5/3,-4/3). The "time" coordinate is just gamma, so gamma(B) in
A's frame is 5/3. The "x" coordinate is gamma*beta, so we divide it
by gamma and see that beta(B) in A's frame must be -4/5. If we plug
that back into the formula for gamma it works out to 5/3 again (which
is a relief): 1/sqrt(1-16/25) = 1/sqrt(9/25) = 5/3.

So the Lorentz transform to get from A's frame to B's frame is

| 5/3 4/3 |
| 4/3 5/3 |

Now, let's go back and figure out how long it takes A to get to the
origin; then we'll look at B again.

A finished accelerating at coordinates (0, -2/sqrt(3)) in his new
frame, including the clock correction. Now, we need to figure out
where the origin is in A's frame. It's _not_ 1 unit away, because
we're measuring the distance to travel in _O_'s frame (and that's the
fundamental asymmetry in this problem, by the way). Looking at O's
frame from A's frame, all distances are contracted, and the length A
needs to travel is 1/gamma = sqrt(3)/2, which is about 0.87.

Now, A sees O approaching at beta=-1/2, so the elapsed time is
(sqrt(3)/2)/(-beta) = sqrt(3). Let's look to see what's happening at
time sqrt(3) in O's frame.

We need to add back 1/sqrt(3) to A's clock and then transform it.
With the 1/sqrt(3) added in, A's coordinates at that point are
(4/sqrt(3),-2/sqrt(3)) (he's stationary in his own rest frame, of
course, so he's still sitting at -2/sqrt(3)). That transforms to
(2,0) in O's frame. Sigh of relief - A and O arrive at the same event
at the same time. Elapsed time for A = sqrt(3), just as we found
before, and elapsed time for O is 2, just as we found before.

Now, let's look at B from A's frame. In the base frame, B's starting
coordinates were (0,1). Transformed to A's frame, they're
(-1/sqrt(3),2/sqrt(3)) ... but with the clock correction of 1/sqrt(3)
that gives us -2/sqrt(3). It seems that B got a head start -- he
started his engine a EARLY from A's point of view. THIS IS IMPORTANT.
The "simultaneous" events in the base frame were NOT simultaneous in
the moving frames of A and B -- and that's where "intuition" totally
jumps the tracks.

If we transform that directly to B's frame using the matrix above
(after adding back that pesky 1/sqrt(3)), we get
(1/sqrt(3),2/sqrt(3)), which we're pleased to see agrees with earlier
results. Of course, we need to subtract 1/sqrt(3) from B's clock, too,
after which we see that B starts at time 0 by _his_ clock.

Now, at time 3/sqrt(3) in A's frame, we find
(3/sqrt(3)+1/sqrt(3),-2/sqrt(3)) maps to (4/sqrt(3),2/sqrt(3)) in B's
frame. Since 2/sqrt(3) is where B's been sitting still in his own
rest frame, A and B do indeed meet at that point. When we subtract
the clock correction of 1/sqrt(3) from B's time we see that his clock
reads 3/sqrt(3) = sqrt(3) -- the same as A's, and the same as what O
predicted. The twins aged the same amount.

Whew.

Again, the problem throughout is that once two frames are moving
relative to each other, they _cannot_ have synchronized clocks -- each
one sees that the other's clocks are skewed progressively with
distance. So, the result works out mathematically, it's supported
experimentally, but it doesn't make a whole lot of sense intuitively.

If you hate the clock corrections -- well, that's the point. The
skewed clocks are what make the math work, and they're what make the
whole thing so confusing.

=========================
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Dear Stephen Bint:

"Stephen Bint" wrote in message
.. .
David,

The formula for time dilation is a function of velocity. If I am
moving
at a
given velocity with respect to you, then you are moving at the exact
same
velocity with respect to me.

Not quite correct.

In that case, if I am moving at 10kph with respect to you, at what
velocity
will you be moving, with respect to me? Greater, or less?

I am assuming I am stationary...

Remember the title of the thread? Since I am disallowed from doing the
math, I can only answer approximately. A little less than 10kph.
Something around the 12th or 18th decimal place.

Relativity only shows up when you move *very* fast, or when you can measure
very precisely. Otherwise Newton (or another of history's geniuses) might
have spotted it.

David A. Smith