Twin Paradox: Simplifying Physics for Age Difference Explanation

[JesseM]

I see that we need to start with a much better
example, namely, my single-rod proposal.

Please tell us whether the following single-rod
proposal for absolutely synchronizing clocks
will work or not:

In any inertial coordinate system, align a rod
with the frame's x axis. While the rod is at rest
in this frame, cut it so that it fits precisely
between two arbitrary points on this x axis. Then
remove the rod, and place it at some other location
along the x axis. Finally, slide the rod inertially
toward the two points. The claim here is that since
the rod formerly fit between the points, it should
still fit, so its end points could be used to truly
or absolutely synchronize clocks at the points.

rqr

But to "move" it and then let it come to rest relative to this frame, you have to accelerate it. Changing its velocity will change its length in any frame, including this one, because of Lorentz contraction. And if in this frame both ends of the rod experience an acceleration simultaneously so the rod un-contracts and both ends touch the two points simultaneously, then in other frames the two ends began accelerating at different times, and hence touched the two points at different times too.

 

[rqr]

2 JesseM:
I am not really worried about other frames' opinions or views.

All I am worried about here is whether or not the sliding rod
will absolutely synchronize clocks in the single given frame.

What do you say?

YES _______ or NO_______

Thanks!

(And please remember that the rod is not accelerating
during the critical stage of the experiment. As I said,
it is moving merely inertially.)

rqr

 

[paw]

I am not really worried about other frames' opinions or views.

You have to be. The rod in your scenario is moving at constant velocity wrt the two marks. That is by definition a different inertial frame of reference.

You can only claim it isn't in a different inertial frame if you abandon the idea of frames of reference entirely. If you do that you abandon science in general and physics specifically.

All I am worried about here is whether or not the sliding rod
will absolutely synchronize clocks in the single given frame.

What do you say?

YES _______ or NO_______

NO.

Let's try it a slightly different way. Alice makes two marks 1 meter apart to the highest precision we can imagine, that is 1 meter +/- 1 Planck length. Bob, who is traveling toward Alice at some speed v<c, builds a meter rod to the same precision, that is 1 meter +/- 1 Planck length.

This is the same scenario as you postulated above without any explicit acceleration. Do you agree?

Now can Alice and Bob synchronize clocks with this apparatus? No! Bobs rod is shorter than the tick marks according to Alice (Bob is moving wrt Alice) and the tick marks are shorter than the rod according to Bob (Alice is moving wrt Bob).

 

[JesseM]

2 JesseM:
I am not really worried about other frames' opinions or views.

All I am worried about here is whether or not the sliding rod
will absolutely synchronize clocks in the single given frame.

What do you say?

YES _______ or NO_______

Thanks!

What do you mean by "absolutely" synchronize? Does this mean something different than just synchronizing the clocks according to that frame's definition of simultaneity? If it's supposed to be frame-independent, then you do have to worry about what happens when you repeat this experiment in different frames--an absolute method should yield the same definition of simultaneity regardless of the rest frame of the clocks or the rod (that was my point earlier about needing an experiment different physicists could perform in windowless rooms that would yield the same definition of simultaneity regardless of the different rooms' velocity relative to one another).

Anyway, this experiment doesn't even seem well-defined enough to clearly say whether it'll yield the same definition of simultaneity as the Einstein synchronization convention would in the clocks' rest frame. It really depends critically on the final acceleration which brings the rod at rest relative to the clocks.

(And please remember that the rod is not accelerating
during the critical stage of the experiment. As I said,
it is moving merely inertially.)

You mean, it's moving inertially past the two clocks? In this case it would be Lorentz-contracted relative to them. Or do you mean that it's moving inertially after being accelerated so it's at rest relative to them? In this case the moment that each end of the rod comes to rest next to a clock will depend on how the rod is accelerated.

 

[rqr]

Jeese, Jesse, youse guys are making this much too
complicated! ;-)

The rod is merely given a wee little starting shove
at the start of its wee little journey toward the
clocks, and after this wee little acceleration
period, it continues to travel inertially forever
and ever, amen!

As the rod passes the clocks, I want to know if its
wee little end points (speaking theoretically, not
Planck-length-perfect reality) will or will not hit
and start the clocks absolutely simultaneously.

[JesseM wrote:]
"You mean, it's moving inertially past the two clocks? In this
case it would be Lorentz-contracted relative to them."

Yes, yes, I mean that it's moving inertially past the
two clocks. That's the ticket!

Now we're getting somewhere, but where I don't know! ;-)

But do we really, really, really have to worry about
the rod's Lorentz contraction **relative** to the
clocks?

We do if this is a physical or intrinsic contraction,
but we do *not* if it's only an apparent or observer-
dependent "contraction."

Let me put this in "slow-motion" mode for a moment:

The rod's end points R1 and R2 exist continuously.
Therefore, whenever the left-hand end point R1 is at
the matching left-hand clock C1, we know that the
right-hand end point R2 must be somewhere, and I
would guess that it's within a few light-years of C2,
at least. ;-)

In fact, we know that R2 must be either to the
right or left of C2 or dead on C2.

Now all we have to do is to figure out where
"your" observed Lorentz contraction places R2.

CHOICE A _____To the right of C2_____

CHOICE B _____To the left of C2_____

CHOICE C _________Dead on C2_________

Vote now or forever hold your peace!
(Careful, Pee-Wee Herman!)

rqr

 

[Janus]

Let me put this in "slow-motion" mode for a moment:

The rod's end points R1 and R2 exist continuously.
Therefore, whenever the left-hand end point R1 is at
the matching left-hand clock C1, we know that the
right-hand end point R2 must be somewhere, and I
would guess that it's within a few light-years of C2,
at least. ;-)

In fact, we know that R2 must be either to the
right or left of C2 or dead on C2.

Now all we have to do is to figure out where
"your" observed Lorentz contraction places R2.

CHOICE A _____To the right of C2_____

CHOICE B _____To the left of C2_____

CHOICE C _________Dead on C2_________

Vote now or forever hold your peace!
(Careful, Pee-Wee Herman!)

rqr

The answer depends on which frame you make the determination in.


In the clocks' frame, the rod is contracted and thus R2 is to the left of C2.

In the Rod's frame, the distance between the clocks is contracted and R2 is to the right of C2.

And in a frame in which both the clocks and the rod have equal relative speeds (in opposite directions), the answer is C, dead on.

 

[JesseM]

Jeese, Jesse, youse guys are making this much too
complicated! ;-)

The rod is merely given a wee little starting shove
at the start of its wee little journey toward the
clocks, and after this wee little acceleration
period, it continues to travel inertially forever
and ever, amen!

As the rod passes the clocks, I want to know if its
wee little end points (speaking theoretically, not
Planck-length-perfect reality) will or will not hit
and start the clocks absolutely simultaneously.

OK, fair enough. If the rod is moving very slowly relative to the clocks, its Lorentz-contraction will be very small in the clocks' frame, so it will reach them very close to simultaneously in the clocks' frame. However, this does not mean it the two ends will reach the clocks very close to simultaneously in all frames, so it is not an absolute synchronization method!

Maybe it would help to give an example here. Suppose the two clocks are 1 light-second apart, and the rod is moving at 0.000001c relative to them, so that its length is around 0.9999999999995 light-seconds according to http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl . So, suppose in the rest frame of the clocks, the left clock is at coordinate x=0 light years-seconds while the right clock is at x=1 l.s., and at t=0 seconds the left end of the rod is at position x=4 l.s. and the right end is at position x=4.9999999999995 l.s. Then moving at 0.000001c, the left end will take 4/0.000001 = 4000000 s to reach the left clock, and the right end will take 3.9999999999995 / 0.000001 = 3999999.9999995 s to reach the right clock, a difference of only about 0.0000005 s...so, this method leaves them very close to synchronized in the clocks' rest frame.

But now let's look at how things look in a frame which is moving at 0.6c relative to the clocks' rest frame. In this frame the clocks are moving at 0.6c, and we can use velocity addition to show the rod is moving at (0.6c + 0.000001c)/(1 + 0.6*0.000001) = 0.600000639999616c, so its Lorentz-contracted length must be 0.799999519999888 l.s. Meanwhile, the Lorentz-contracted distance between the clocks is 0.8 l.s. in this frame. So, you'd think the rod would reach both ends almost simultaneously in this frame too...but you'd be wrong. At t'=0 seconds, the left clock is at x'=0 l.s., the right clock is at x'=0.8 l.s. Using the Lorentz transformation, the event of the left end of the rod being at x = 4 l.s. at t = 0 s in the first frame translates to it being at position x' = 5 l.s. at t' = -3 s in this frame, so 3 seconds later it will have moved a distance of 3*0.600000639999616 = 1.80000191999885 l.s. closer to the origin, putting it at position x' = 5 - 1.80000191999885 = 3.19999808000115 l.s. at time t'=0 s in this frame. Since its length is 0.799999519999888 l.s. in this frame, at time t'=0 s the right end must be at position x' = 3.19999808000115 + 0.799999519999888 = 3.99999760000104 l.s. in this frame.

So, now we have enough information to calculate when the left end of the rod will reach the left clock, and when the right end of the rod will reach the right clock. Since the left clock starts at x'=0 and moves to the left at 0.6c, its position as a function of time is given by the function x'(t') = -0.6t'. Since the left end of the rod starts at x'= 3.19999808000115 and moves to the left at 0.600000639999616c, its position as a function of time is given by x'(t') = 3.19999808000115 - 0.600000639999616t'. So to figure out when the left clock meets the left end of the rod, we need to set the two functions equal to one another and solve for t':

-0.6t' = 3.19999808000115 - 0.600000639999616t'

which simplifies to:

0.000000639999616t' = 3.19999808000115

So, solving for t' gives:

t' = 5000000.0000018

Now we can similarly figure out when the right end of the rod catches up with the right clock. The right clock starts at x'=0.8 l.s. and is moving to the left at 0.6c, so its position as a function of time is given by x'(t') = 0.8 - 0.6t'. The right end of the rod starts at x' = 3.99999760000104 l.s. l.s. and moves to the left at 0.600000639999616c, so it's position as a function of time is given by x'(t') = 3.99999760000104 - 0.600000639999616t'. So, once again let's set them equal:

0.8 - 0.6t' = 3.99999760000104 - 0.600000639999616t'

which simplifies to:

0.000000639999616t' = 3.19999760000104

Solving for t':

t' = 4999999.25000118

So, that's the time the right end of the rod catches up with the right clock. Now we can figure out the difference in time between each end of the rod catching up with its respective clock:

5000000.0000018 - 4999999.25000118 = 0.750000620260835

At around 0.75 seconds, this is a pretty substantial difference. And, not-so-coincidentally, it turns out that if you use the coordinate systems defined by the Lorentz transformation, then if you have two events which are simultaneous and 1 light-second apart on the x-axis in one frame, then in another frame moving at 0.6c relative to the first frame's x-axis, then in this frame the two events happen exactly 0.75 seconds apart.

So you see, your method is of absolutely no help at all in finding an absolute synchronization method, although it can get arbitrarily close to properly synchronizing two clocks in their own rest frame, in the same way they'd be synchronized by the Einstein synchronization convention. But imagine that Lorentzian relativity is correct, and the frame where the two clocks are moving at 0.6c also happens to be the "absolute" frame whose definition of simultaneity is the "absolutely" correct one, and with rulers shrinking in absolute terms and clocks slowing down in absolute terms when they move relative to this frame. In this case, if you have two clocks moving with an absolute speed of 0.6c and an absolute distance of 0.8 light-seconds, and you use the method of moving a rod with a rest length of 1 light-second at an absolute velocity which is almost identical to 0.6c but differs from it by some tiny amount delta (so that the rod's absolute length is contracted to almost exactly 0.8 light-seconds), then it is not true that the two ends of the rod will pass the two clocks almost simultaneously in absolute terms--in fact, in the limit as delta approaches zero, the two ends will pass the two clocks arbitrarily close to 0.75 seconds apart. So, if you don't happen to know which frame is the absolute frame in the first place, so you can't guarantee in advance that your clocks are at absolute rest and your rod is moving at a very small absolute velocity, then this method will not help you to get your clocks synchronized in absolute terms, it will only help you to get them synchronized according to the definition of simultaneity of their rest frame in SR.

The rod's end points R1 and R2 exist continuously.
Therefore, whenever the left-hand end point R1 is at
the matching left-hand clock C1, we know that the
right-hand end point R2 must be somewhere, and I
would guess that it's within a few light-years of C2,
at least. ;-)

In fact, we know that R2 must be either to the
right or left of C2 or dead on C2.

Now all we have to do is to figure out where
"your" observed Lorentz contraction places R2.

CHOICE A _____To the right of C2_____

CHOICE B _____To the left of C2_____

CHOICE C _________Dead on C2_________

Vote now or forever hold your peace!
(Careful, Pee-Wee Herman!)

The answer would depend on what frame we are using. Or, if we imagine that Lorentzian relativity is correct and there is a single absolute frame, then in absolute terms, whether R2 was to the right or left of C2 would depend on whether the rod's absolutely velocity was slightly higher or slightly lower than the absolute velocity of the clocks (in my example above it was slightly higher, so the rod was slightly shorter than the distance between the clocks) and also whether R2 is approaching C2 from the left or from the right (in my example above it was approaching from the right). In my example, R1 passed C1 about 0.75 seconds after R2 passed C2 (in spite of the fact that the rod was only moving at 0.000001c in the clocks' rest frame), so by the time R1 reached C1, R2 was already to the left of C2 (choice B).

 

[RandallB]

rqr, rqr, rqr

Jeese, Jesse, youse guy are making this much too
complicated!
The rod is merely given a wee little starting shove ...
...
Now all we have to do is to figure out where
"your" observed Lorentz contraction places R2.

CHOICE A _____To the right of C2_____

CHOICE B _____To the left of C2_____

CHOICE C _________Dead on C2_________

Vote now or forever hold your peace!

I agree simple is better - but you need to be complete and detailed.

Yes to all three
r2’ to the left of C2
r2 to the right of C2’
r2’ dead on with C2’
r2 dead on with C2

At least cut the rod in half for two identical equal length rods one in each reference frame as shown below. Everything with a prime mark C1’ etc moves together in a fixed reference frame. Every thing with no prime in the “non-moving” reference frame.

 r1’       --> -->          r2’
 |--------------------------|
 C1’                        C2’                       Cx’


r1                                                    r2
 |----------------------------------------------------|
C1                         Cx                         C2

Note: the rods are long and heavy so the workers are needed in both frames at both ends of the two rods to pull it just “wee little” enough so that the clocks all start at t=0 & t’=0. It is total up to the workers to synchronize there watches “only within there own reference frames!” such that they all pull when C1’ reaches C1.

Note:
I’ve added Cx which at t’=0 will be at the same place as C2’ one rod’ from C1’.
For Cx t≠0 you should be able to figure t & how many rods it is from C1.
Also:
added Cx’ which at t=0 will be at the same place as C2 one rod from C1.
For Cx’ t’≠0 you should be able to figure t’ & how many rods it is from C1’.

Only after you carefully get these numbers, change you view to using the primes ie. rod’ as the fixed frame of reference. You should get the same results – a tip when you change frames C1 & r1 will be at the front of the moving rod, not at the back like C1’ & r1’ are. For real numbers a recommend a speed of 0.6c

If this doesn’t do it for you, PM me

 

[rqr]

[JesseM wrote:]
"However, this does not mean it the two ends will reach
the clocks very close to simultaneously in all frames,
so it is not an absolute synchronization method!"

Thanks!

I fully agree. Can we consider this rod case closed?

By closed, I mean that we seem to agree that a rod's
intrinsic length is change by its absolute motion
through space, so its endpoints cannot be used to
absolutely synchronize two clocks it the rod is
moving at a speed that differs from that of the
clocks.

But before I go on to the next case, I have a few
things to say about something else you wrote.
Something's fishy.

[JesseM wrote:]
"The answer would depend on what frame we are using."

How a clock-synchronization experiment's result can
possibly vary depending on the frame in use?

I am speaking of the *general* result, i.e., either we
have a resulting absolute synchronization or we have
a resulting absolute asynchronization.

Yes, you gave various examples, but the only
examples that count (or are valid) are those which
acknowledge the facts that there is only one rod
with only one intrinsic length which can set the
clocks in only one way.

(Remember your own warning that this must be a
closed-lab experiment; since it is, we do not have to
worry about any outside-observer viewpoints; all that
matters is that which occurs *within* the closed lab.)

(Note: I am not really concerned about how fast the
rod is moving either absolutely or wrt the given clock
frame; any speed(s) will do. All I want to know is if
the experiment will work to absolutely synchronize
the clocks or not. (If it won't work, then it simply
*won't work* at any speed(s), generally speaking.))

Here, now, is the next case:

It is basically just the clock transport case.

In a given frame A, clocks C1 and C2 are at rest
and separated. Let a 3rd clock C3 that is moving
inertially toward C1 be set to match C1 as they
meet. As this 3rd clock moves on, it eventually
reaches clock C2. As they meet, C2 is made to match
C3. (This rigmarole eliminates all accelerations.)

matches C1
C3--->
C1------------------------------------C2

...........C3--->
C1------------------------------------C2
.........matches C3

Clocks C1 and C2 will be absolutely synchronized by
this simple, closed-lab procedure (given that the
physical internal atomic rhythms of all three clocks
are identical during the experiment).

rqr

 

[RandallB]

As this 3rd clock moves on, it eventually
reaches clock C2. As they meet, C2 is made to match
C3. (This rigmarole eliminates all accelerations.)

matches C1
C3--->
C1------------------------------------C2

...........C3--->
C1------------------------------------C2
.........matches C3

Clocks C1 and C2 will be absolutely synchronized by
this simple, closed-lab procedure (given that the
physical internal atomic rhythms of all three clocks
are identical during the experiment).

Nope
Working the problem in post #38 to complete and understand the rod case, will explain the issues to you.

The only tool useful synchronized C1 & C2 in this version are light signals sent back and forth between those areas of your “closed-lab”. C3 needs to be marked as C3’ as part of a different moving reference frame and t’ used by moving C3’ will not be the same as t just because you observe that clock within the imaginary boundaries of a “closed-lab” as it passes by.
Time t=t’ can only be true at one and only one place & time in both reference frames.
In this case you have defined that as when C1 & C3’ are together for the one and only one time they will be so using these moving frames.
When C3’ gets to C2 those two clocks will not read the same and if you change C2 it will not read the same time t C1 any longer.

Note:
when editing your posts with quoted text from other posts be sure to retain the (/quote) command in the correct position to close the quote.
Also making simple diagrams is sometimes easier when using the fixed spacing from wrapping that part with (/ code) (the # tool at the top of the edit screen)

Just use “preview text” to see if it looks OK

 

[JesseM]

[JesseM wrote:]
"However, this does not mean it the two ends will reach
the clocks very close to simultaneously in all frames,
so it is not an absolute synchronization method!"

Thanks!

I fully agree. Can we consider this rod case closed?

By closed, I mean that we seem to agree that a rod's
intrinsic length is change by its absolute motion
through space, so its endpoints cannot be used to
absolutely synchronize two clocks it the rod is
moving at a speed that differs from that of the
clocks.

I don't personally agree there is such a thing as "absolute motion through space", I just described things in terms of Lorentzian relativity since this is how you apparently prefer to think about things. Of course the difference between the usual interpretation of SR and Lorentzian relatiity is just a philosophical one, they predict the same thing about all experimental results.

But before I go on to the next case, I have a few
things to say about something else you wrote.
Something's fishy.

[JesseM wrote:]
"The answer would depend on what frame we are using."

How a clock-synchronization experiment's result can
possibly vary depending on the frame in use?

I am speaking of the *general* result, i.e., either we
have a resulting absolute synchronization or we have
a resulting absolute asynchronization.

This sentence doesn't really make clear what you mean by "general result", the two alternatives you offer are identical...maybe you mistyped? Anyway, all I was talking about was the question you asked about whether the right end of the rod would be to the left or the right of the right clock at the "same moment" that the left end of the rod reached the left clock. Of course different frames define simultaneity differently, so they give different answers. If you prefer to imagine an absolute frame with absolute simultaneity, then the answer depends on whether the clocks or the rod have the higher absolute velocity, which we mere mortals would have no way of knowing, since there's no experimental method that can determine the absolute rest frame (see below).

(Remember your own warning that this must be a
closed-lab experiment; since it is, we do not have to
worry about any outside-observer viewpoints; all that
matters is that which occurs *within* the closed lab.)

Not true, the whole point of a physical definition of "absolute simultaneity" (as opposed to a purely philosophical one) is that different closed-lab experiments must give the same definition of simultaneity. Closed-lab experiment means that as the experiment is performed the experimenter must have no access to information about anything outside the lab, but once the experiment is done and the clocks have been synchronized, you must open the different labs up and compare the clocks synchronized in different labs to see if they have arrived at the same definition of simultaneity or different ones. As long as the physical aspects of the theory of relativity are correct--and these are entirely compatible with the interpretation known as Lorentzian relativity, in which there is such a thing as absolute simultaneity--then it will be impossible to come up with a closed-lab method that is guaranteed to give the same definition of simultaneity for different labs.

Here, now, is the next case:

It is basically just the clock transport case.

In a given frame A, clocks C1 and C2 are at rest
and separated. Let a 3rd clock C3 that is moving
inertially toward C1 be set to match C1 as they
meet. As this 3rd clock moves on, it eventually
reaches clock C2. As they meet, C2 is made to match
C3. (This rigmarole eliminates all accelerations.)

matches C1
C3--->
C1------------------------------------C2

...........C3--->
C1------------------------------------C2
.........matches C3

Clocks C1 and C2 will be absolutely synchronized by
this simple, closed-lab procedure (given that the
physical internal atomic rhythms of all three clocks
are identical during the experiment).

They won't be absolutely synchronized, no. Presumably you want C3 to move very slowly relative to C1/C2 so there is little difference in the time dilation factor between C3 and C1/C2, but the difference in time dilation factors cannot be eliminated entirely. Assume the distance between C1 and C2 in their rest frame is 10 light-second, and the velocity of C3 in their rest frame is v. In this case, in their frame it takes a time of 10/v seconds for C3 to move between the two clocks. But C3 is slowed down by a factor of \sqrt{1 - v^2/c^2}, so it advances forward by

\frac{10 \sqrt{1 - v^2/c^2}}{v}

in that time. So, what's the difference between this, the time on C3 as it reaches C2, and the time that C1 reads at the same moment in this frame, namely 10/v? Well, it'd just be:

\frac{10(1 - \sqrt{1 - v^2/c^2})}{v}

If we take the limit of this quantity as v approaches zero, it will work out to zero (use L'Hospital's rule along with the chain rule of calculus to prove this), which means C2 will get arbitrary close to being synchronized with C1 in this frame as the velocity of C3 becomes very slow compared to c.

But that's just for the C1/C2 rest frame; now consider what happens in another frame where C1 and C2 move at some nonzero velocity, say 0.6c, and C3 moves with some slightly different velocity (0.6c + v)--again, we're going to be taking the limit as v approaches zero. In this frame the distance between C1 and C2 is shrunk to 8 light-seconds.

So if C3 and C1 start at x=0 light-seconds at time t=0 seconds, C3's position as a function of time x(t) is:

x(t) = (0.6c + v)*t

And if C2 starts at x=8 l.s. at time t=0 s, C2's position as a function of time is:

x(t) = 0.6c*t + 8 l.s.

So to figure out when C3 catches up with C2, set them equal:

0.6c*t + vt = 0.6c*t + 8 l.s.

...and solving this for t gives t=(8 l.s.)/v. So, this is the time it takes for C3 to go from C1 to C2 in this frame. Now, since C1 is slowed down by a factor of 0.8 in this frame, it will have elapsed a time of (0.8)*(8 l.s.)/v in this time. And C3 is slowed down by a factor of \sqrt{1 - (0.6c + v)^2 /c^2} in this frame, so it will have elapsed a time of:

\frac{\sqrt{1 - (0.6c + v)^2 /c^2} * 8 \, l.s.}{v}

So, the difference between the reading of C1 and the reading of C3 at the moment C3 reaches C2 will be:

\frac{(0.8 - \sqrt{1 - (0.6c + v)^2 /c^2}) * 8 \, l.s.}{v}

Now we want to know what this difference will approach in the limit as v approaches zero. Since both the numerator and the denominator of this fraction individually approach zero in the limit as v approaches zero, we can again use L'Hospital's rule, taking the derivative of both the top and bottom and seeing what the new fraction approaches in the limit as v approaches zero...to take the derivative of the numerator we must also use the chain rule again. This gives us the following complicated-looking fraction:

\frac{8 \, l.s. * [-(1/2) * (1 - (0.6c + v)^2 / c^2)^{-1/2} * (\frac{-1.2c - 2v}{c^2})]}{1}

Which simplifies to:

\frac{8 \, l.s. * (\frac{0.6c + v}{c^2})}{\sqrt{1 - (0.6c + v)^2 / c^2}}

And if we take the limit of this as v approaches zero, it just turns out to be:

\frac{8 \, l.s. * (\frac{0.6c}{c^2})}{0.8}

Which is 6 seconds. So, even in the limit as the velocity of C3 relative to C1 and C2 gets arbitrarily small, this method will still leave C1 and C2 6 seconds out-of-sync in this frame (and you're free to imagine that this frame in which C1 and C2 move at 0.6c also happens to be the absolute rest frame of Lorentzian relativity). And not-so-coincidentally, it turns out that if you had synchronized C1 and C2 using the Einstein synchronization convention, they would also be 6 seconds out-of-sync in this frame. So again, this method is useless for absolute synchronization, all it does is to replicate the same type of synchronization as the Einstein convention, which will cause clocks that are in-sync in one frame to be out-of-sync in another (or, if you prefer, will cause clocks that are moving relative to the absolute rest frame to be absolutely out-of-sync).

Incidentally, if you don't trust my calculus, feel free to take the equation I gave earlier for the difference between the reading of C1 and C3 at the moment C3 reaches C2:

\frac{(0.8 - \sqrt{1 - (0.6c + v)^2 /c^2}) * 8 l.s.}{v}

...and instead of taking limits, just plug in some very small v like v=0.000001c, you should end up with an answer very close to 6 seconds.

 

[rqr]

JeeseM,

One problem is that your math analysis was
based entirely on Einstein's definition of
clock synchronization, a definition which
did not yet exist in my given proposal for
absolute synchronization.

Additionally, all of Einstein's values are
incorrect; e.g., in his time dilation equation,
the "v" was found by using SR's asynchronous
clocks, so it cannot be correct.

Another problem is your assumption that
I wanted the clock to move very slowly
when transported; I couldn't care less how
fast it moves because clock motion is not
supposed to affect a clock's atomic rate.

Indeed, if we are speaking of SR's merely
relative motion, then we all know that such
motion cannot affect anything physical.
(I am of course speaking of the mere
passing of frames in the night, and NOT
any collision of them!)

In fact, if SR's mere relative motions
could somehow effect a clock's internal
atomic, then each passing frame would
of course affect this rate differently,
so a single clock moving at a constant
velocity would at once have to have an
infinite number of physically different
atomic rates, and we all know that this
is not possible.

Since you are not allowed to apply any
Einstein-definition-based formula to my
clock-synchronization proposals, what are
you going to do?

I suggest that you must look at the physics
involved.

For example, in my rod experiment, either
it still fits precisely between the two
points, or it doesn't. And if it doesn't,
then its intrinsic length has changed.
And that has to be a function of its
speed through space because there is
simply no other explanation.

Let me put it this way:
I challenge you to show on paper two
or more inertial frames getting a
null result for the Michelson-Morley
experiment. To assure that the only
real difference is the use of two
frames, we need to let them share
the light source. But this is fine
because light is source-independent.
To keep it simple, we shall use only
the x-axis of each frame, and we shall
use only one clock - an origin clock -
for each frame.

I hereby claim that you cannot do this
unless you physically contract at least
one frame's x axis, and also physically
slow at least one frame's clock.

rqr