Re: Androcles and Draper resume Einstein 1905

From: Androcles (dummy_at_dummy.net)
Date: 01/15/05 

Date: Sat, 15 Jan 2005 18:59:54 GMT

"PD" <pdraper@yahoo.com> wrote in message
news:1105806013.219512.134620@f14g2000cwb.googlegroups.com...
>
> Androcles wrote:
>> "PDraper" <pdraper@yahoo.com> wrote in message
>> news:BE0E76A0.14C4%pdraper@yahoo.com...
>> > On 1/14/05 3:26 PM, in article
>> > 8WVFd.96197$C8.28048@fe3.news.blueyonder.co.uk, "Androcles"
>> > <dummy@dummy.net> wrote:
>>
>> Ok, so it can't be done for moving frames.
>> Why is Einstein even talking about it?
>>
>> PD:
>> It can be done for a moving frame.
>> [snip]
>>
>> Androcles:
>> Does the K-observer conclude the k-rod is shorter than the K-rod,
>> or not? Yes/No.
>> [snip]
>>
>> PD:
>> Neither.
>> Shall we proceed to show that's the case?
>>
>> [snip]
>>
>> Androcles
>> No.
>>
>> Does the K-observer conclude the k-rod is shorter than the K-rod,
>> or not?
>> I want an answer, "yes" or "no".
>> [snip]
>>
>>
>> PD:
>> >>> What is the K-rod and what is the k-rod? We have one rod, moving
> in
>> >>> the
>> >>> K frame and stationary in the k frame.
>>
>>
>> Androcles:
>> >> We have three rods, two of which are called measuring-rods and
> one,
>> >> as
>> >> you
>> >> correctly point out, is the rod to be measured. Having once
> measured
>> >> the
>> >> rod with the K-frame measuring rod, or K-rod, we have already
>> >> determined
>> >> (I thought) that we cannot measure the stationary rod with the
>> >> k-rod,
>> >> aka moving
>> >> rod, because it was previously agreed the operation can only be
>> >> conducted
>> >> with both the rod to be measured and the measuring-rod were at
> rest
>> >> relative
>> >> to each other. In everyday terms, I don't know how to place chalk
>
>> >> marks
>> >> on the pavement at the wheels of the bus when the bus is moving.
>> >> Having once measured the rod to be measured, we can discard the
> rod
>> >> that
>> >> measured it, we have a duplicate. I refer to the moving measuring
> rod
>> >> as
>> >> the k-rod an the stationary measured rd (or measuring-rod or
>> >> yardstick)
>> >> as the
>> >> K-rod.
>> >> I have reduced the problem to algebraic from for you.
>> >
>>
>> PD:
>> > Not quite.
>>
>> FOR THE THIRD AND FINAL TIME, ANSWER THE QUESTION!
>> Failure to answer with yes/no terminates this discussion.
>> Does the K-observer conclude the k-rod is shorter than the K-rod,
>> or not?
>
> As you defined k-rod and K-rod, yes.

Thank you. Now proceed by all means.
You may qualify your answers, but I expect them to be unequivocal.

The length of a rod in the moving frame is less than the length of
the same rod as measured in the stationary frame.

> Ready to move on?
I have another question but I'll read on before asking it,
lest you have provided the answer below.

>>
>> Androcles
>>
>>
>>
>>
>> What we said is that the yardsticks of an observer have to be
>> > stationary in the observer's frame; for example, if we want to
>> > "synchronize"
>> > the yardsticks. Recall Einstein gives a definition for a length
>> > measurement
>> > of a moving rod. I quote from below, where we quoted Einstein:
>> >>>>>>>>> "(b)
>> >>>>>>>>> By means of stationary clocks set up in the stationary
> system
>> >>>>>>>>> and
>> >>>>>>>>> synchronizing in accordance with 1, the observer
> ascertains
>> >>>>>>>>> at
>> >>>>>>>>> what
>> >>>>>>>>> points of the stationary system the two ends of the rod to
> be
>> >>>>>>>>> measured
>> >>>>>>>>> are located at a definite time.
>> > And we also discussed earlier that "at a definite time" means
>> > simultaneously
>> > determined in that frame, simultaneity being determined by clocks
> that
>> > have
>> > been previously synchronized using an emitter/mirror/receiver
> system
>> > that is
>> > stationary in that frame. We also discussed how simultaneous
>> > measurements
>> > can be done by collaborators at rest in a given frame. It's not
>> > necessary to
>> > assume a single observer, who in practical terms can only make one
>> > measurement at a time.

See, there it is. I dispute that. "Measurement" of length is TWO
observations.
I must first look at one end of the rods and then at the other. I cannot
look at both ends at the same time. Not even with the rods at rest.
I place a ruler with the zero at one side of my *** of paper,
and then I read the number off the ruler at the mark that coincides
with the other edge of the paper, 8.5 inches.
I have less difficulty with time, I can superimpose the clocks and
make ONE observation by looking along the x-axis, like this:

        \|_

One clock is reading 2:53, the other is reading 3:00.

         |/_

One clock is reading 3:00, the other is reading 3:07

I declare the clocks are offset, but have the same gain.
I attribute this to the "slow" clock (as it is referred to in ordinary
English, but really lagging) being more distant than the nearer.

Should I observe at different times the clocks
showing 2:30 and 3:00, then 2:40 and 3:15, I declare the clocks
to be running at different gains.

We assume the clocks to be in all ways identical mechanically
 and do faithfully record time in their own frame of reference,
and it is time itself, not the measurement thereof, that differs
between frames.

I accept Sommerfield's footnote:
"We shall not here discuss the inexactitude which lurks
in the concept of simultaneity of two events at approximately
the same place, which can only be removed by an abstraction."
as it applies to time.
I do not accept it as it applies to length, I AM discussing the
inexactitude that lurks in measuring two lengths simultaneously.

Do you agree?

[snip redundant debate]

>> >
>> > For your convenience, I've marked those sections below.

PD:
While we proceed, though, let me ask you this question. Suppose
you are the K observer and you have a rod that's stationary and you
follow the procedure outlined for measuring the length of a stationary
rod, and write down its length. Then you watch as that same rod is
walked some distance away from you and then sped up to go by
you at some velocity v. So then you repeat the measurement,

Androcles:
I cannot repeat the measurement. The rod is moving and
I must make two observations separated in time.

PD:
following the procedure available to you for measuring the length
of a moving rod.

Androcles:
I do not have a procedure available to me. I'll introduce one, but
it is not Einstein's. I'll count the revolutions of a wheel rolling
against the rod, or the rod rolling against the wheel.
Thus I can count the turns of the bus wheel as it passes a
distance along the pavement, making two observations at two
points in space, or I can roll a wheel against the side of the bus
(or rod) and make two observations at one point in space.
I have my own clock to time the interval between the observations,
but that is irrelevant for the length. I use that information to
compute v = x/t.
The former is not acceptable to me, I must allow for doppler
shift.
Is the second acceptable?

PD:
And suppose, suppose, lo and behold, that procedure results in a
different
 number.
 What are you going to conclude?

Androcles:
I'll conclude that counting the revolutions of the bus wheels at
two different places is an inappropriate method.
It takes finite time for light to convey the information back to me
and I'll incorrectly compute the speed of the bus
(or the trailing end of the moving rod passing the leading edge
of the stationary rod), and not agree the speed is v.

Androcles:
Does the K-observer conclude that time in the k-frame has
elapsed Less than time in the k-frame? Yes/No.

PD:
Neither.

Androcles
Ok, I require a yes/no once again.

Androcles.

PD:

 K-observer says the time between two events is longer
what
>> >>>
>> >>>>> the
>> >>>>> k-observer says the time between those events is. That sounds
> like
>> >>> the
>> >>>>> same statement but it's not.
>> >>>>>
>> >>>>>>
>> >>>>>> If your answer is yes to either question, then prove it.
>> >>>>>> Otherwise we are done.
>> >>>>>> In a nutshell, we either have
>> >>>>>> Einstein) c invariant with time and length not,
>> >>>>>> Newton) time and length invariant with c just another speed.
>> >>>>>>
>> >>>>>> I say N, you say E.
>> >>>>>> I say E contains paradox and is logically inconsistent.
>> >>>>>> As you attempt your proof, I'll point out the inconsistencies.
>> >>>>>> I don't have to prove N is consistent, it is used on a daily
>> >>>>>> basis
>> >>>>>> and is self-evident and always within the limits of
> experimental
>> >>>>>> error. I will allow that E= mc^2 was not thought of by N,
>> >>>>>> but it is evident using his laws. Einstein's law is
>> >>>>>> E = mc^2/sqrt(1-v^2/c^2), and I dispute that as logically
>> >>>>> inconsistent.
>> >>>>>>
>> >>>>>>
>> >>>>>>
>> >>>>>>
>> >>>>>>>> "As regards its experimental proof, we must first of all
> note
>> >>>>> that
>> >>>>>>>> the lengthenings and shortenings in question are
>> >>> extraordinarily
>> >>>>>>>> small. We have v^2/c^2 = 1.0E-8, and thus, if epsilon = 0,
> the
>> >>>>>>>> shortenings of the one diameter of the Earth would amount to
>> >>> 6.5
>> >>>>>>>> cm." - Michelson's Interference Experiment, HA Lorentz,
> 1895.
>> >>>>>>>>
>> >>>>>>>> So now I'm moving and my tape measure also contracts, right,
>> >>>>>>>> so I still find exactly the same length of rod.
>> >>>>>>>> How does that prove the moving rod is the same length as
>> >>>>>>>> the stationary rod?
>> >>>>>>>
>> >>>>>>> It doesn't and I'm not sure why you need it to.
>> >>>>>>> All we need to know is
>> >>>>>>> that if there is a rod that is moving in K and stationary in
> k,
>> >>>>> there
>> >>>>>>> is a way for both the observer in K and the observer in k to
>> >>>>> perform
>> >>>>>>> respective length measurements, and for them to communicate
> to
>> >>>>> compare
>> >>>>>>> notes.
>> >>>>>>
>> >>>>>>
>> >>>>>> Einstein) xi = x' * gamma, where
>> >>>>>> Newton) x' = x-vt, called the Galilean Transform.
>> >>>>>>
>> >>>>>> Is the length of the k-rod changed by gamma * length of K-rod
> ?
>> >>>>>> If so, prove it. If not, we are done.
>> >>>>>> I say gamma is meaningless drivel, has the value of one,
>> >>>>>> time is invariant and so is length.
>> >>>>>> Prove otherwise, using linear algbra.
>> >>>>>> That's why we need to.
>> >>>>>> And get on with it, I haven't got forever to argue about
>> >>> definitions
>> >>>>>> or why we need to.
>> >>>>>>
>> >>>>>>
>> >>>>>>
>> >>>>>>>>
>> >>>>>>>>> As Einstein says:
>> >>>>>>>>> "We now inquire as to the length of the moving rod, and
>> >>> imagine
>> >>>>> its
>> >>>>>>>>> length to be ascertained by the following two operations:--
>> >>>>>>>>> (a)
>> >>>>>>>>> The observer moves together with the given measuring-rod
> and
>> >>> the
>> >>>>>>> rod
>> >>>>>>>> to
>> >>>>>>>>> be measured, and measures the length of the rod directly by
>> >>>>>>>> superposing
>> >>>>>>>>> the measuring-rod, in just the same way as if all three
> were
>> >>> at
>> >>>>>>> rest.
>> >>>>>>>> "
>> >>>>>>>> But he also says x' = (x-vt)/sqrt(1-v^2/c^2).
>> >>>>>>>> We have two equations to solve.
>> >>>>>>>> x' = (x-vt)/sqrt(1-v^2/c^2) is one end of the rod, and
>> >>>>>>>> X' = ((x+L) -vt)/sqrt(1-v^2/c^2) is the other.
>> >>>>>>>>
>> >>>>>>>> Show X' - x' = L
>> >>>>>>>
>> >>>>>>> I was under the impression that your quibble was with a much
>> >>>>> earlier
>> >>>>>>> equation, which we're about to get to. (See way below, where
> I
>> >>> move
>> >>>>>>> on.)
>> >>>>>>
>> >>>>>> Then you've misunderstood. My "quibble" as you call it is that
>> >>>>>> SR is pure, unadulterated excrement of the make bovine. Now,
>> >>>>>> are you going to show that X' - x' = L or not? If not, we are
>> >>> done.
>> >>>>>> Invite someone else to do so if it is too hard for you, or get
>
>> >>>>>> all
>> >>>>> the
>> >>>>>> help you need. Only quit wasting words and do the algebra,
>> >>>>>> please.
>> >>>>>> I'm giving you every opportunity to prove your case.
>> >>>>>> If you cannot, then I expect you to say
>> >>>>>> "Oh... Really?...Oh. I see I was confused. OK, I get it now."
>> >>>>>> as you expect of others that are intellectually and ethically
>> >>> honest.
>> >>>>>>
>> >>>>>> Androcles.
>> >>>>>>
>> >>>>>>
>> >>>>>>
>> >>>>>>>> I'll stop there until we have agreement.
>> >>>>>>>> I hadn't realized you were jumping frames.
>> >>>>>>>
>> >>>>>>> I'm not. Observers in the two frames can communicate - radio
>> >>> their
>> >>>>>>> results to each other after the fact, if necessary.
>> >>>>>>>
>> >>>>>>>>
>> >>>>>>>> Androcles.
>> >>>>>>>>
>> >>>>>>>>
>> >>>>>>>>> Three rods, ok. A standard rod is used to superimpose on
> the
>> >>>>> rods
>> >>>>>>> to
>> >>>>>>>> be
>> >>>>>>>>> measured. Why not three clocks, then, carrying a standard
>> >>> clock
>> >>>>>>>> between
>> >>>>>>>>> the frames?
>> >>>>>>>
>> >>>>>>> This last must be yours, Androcles?
>> >>>>>>> I'm carrying no standards between frames. The standard in k
>> >>> stays
>> >>>>> in
>> >>>>>>> k,
>> >>>>>>> the standard in K stays in K. If the rod that was stationary
> in
>> >>> K
>> >>>>> is
>> >>>>>>> sped up to become stationary in k, then I assume nothing
> about
>> >>> its
>> >>>>>>> length remaining the same in K, and in fact, I'll have to
>> >>> remeasure
>> >>>>>
>> >>>>>>> it,
>> >>>>>>> according to the procedure that has been laid out.
>> >>>>>>>
>> >>>>>>> Now, once the rod has been accelerated to be stationary in k,
> it
>> >>>>>>> satisfies the requirements for a clock-synch checker in k. So
>> >>> even
>> >>>>>>> though the clocks in k may have been recently synched by the
>> >>>>> standard
>> >>>>>>> clock in k that stays stationary in k, the new clock should
> also
>> >>>>>>> verify
>> >>>>>>> that synchronization.
>> >>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> or, if we want to measure the rod from a frame in which the
>> >>> rod
>> >>>>> is
>> >>>>>>>>> moving:
>> >>>>>>>>> "(b)
>> >>>>>>>>> By means of stationary clocks set up in the stationary
> system
>> >>>>> and
>> >>>>>>>>> synchronizing in accordance with 1, the observer
> ascertains
>> >>> at
>> >>>>>>> what
>> >>>>>>>>> points of the stationary system the two ends of the rod to
> be
>> >>>>>>>> measured
>> >>>>>>>>> are located at a definite time.
>> >>>>>>>>>
>> >>>>>>>>> What is a definite time, then?
>> >>>>>>>>
>> >>>>>>>> That means that the ends have to be measured simultaneously
> *in
>> >>>>> that
>> >>>>>>>> frame*.
>> >>>>>>>> Note that the clocks can be synchronized independently of
>> >>>>>>> synchronizing
>> >>>>>>>> the
>> >>>>>>>> yardsticks, so we don't have a chicken-and-egg problem or a
>> >>>>> hopeless
>> >>>>>>>> convolution problem.
>> >>>>>>>>
>> >>>>>>>> This, in the end, will be the nub of the issue, because the
>> >>>>>>> *procedure*
>> >>>>>>>> defines length as measuring the distance between two
>> >>> simultaneous
>> >>>>>>>> events,
>> >>>>>>>> but we'll soon see that simultaneity is not an inherent
>> >>> property
>> >>>>> of
>> >>>>>>> two
>> >>>>>>>> events.
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> The distance between these two points, measured by the
>> >>>>>>> measuring-rod
>> >>>>>>>>> already employed, which in this case is at rest, is also a
>> >>>>> length
>> >>>>>>>> which
>> >>>>>>>>> may
>> >>>>>>>>> be designated ``the length of the rod.'' "
>> >>>>>>>>>
>> >>>>>>>>> Note that in case (b), we are not using a moving rod to
>> >>>>>>> *synchronize*
>> >>>>>>>>> our yardsticks in the stationary frame. We can only do that
>> >>> with
>> >>>>> a
>> >>>>>>>>> stationary rod in the stationary frame. This doesn't mean,
>> >>>>> though,
>> >>>>>>>> that
>> >>>>>>>>> we can't define a procedure for measuring a length of a
>> >>> moving
>> >>>>> rod
>> >>>>>>> in
>> >>>>>>>>> the stationary frame.
>> >>>>>>>>>
>> >>>>>>>>> Seems to me that a third device is employed. How do you
> know
>> >>>>> where
>> >>>>>>>>> both ends are at a definite time? Looking isn't good
> enough,
>> >>>>> that
>> >>>>>>>>> requires
>> >>>>>>>>> light from one end to reach the other.
>> >>>>>>>>> Certainly that is unsatisfactory for comparing the lengths
> of
>> >>>>>>> moving
>> >>>>>>>>> rods.
>> >>>>>>>>
>> >>>>>>>> Not necessarily. All we need is a time stamp on the
> measurement
>> >>> at
>> >>>>>>> each
>> >>>>>>>> end
>> >>>>>>>> or two collaborators in the same frame agreeing ahead of
> time
>> >>> to
>> >>>>>>>> measure the
>> >>>>>>>> location of the respective ends at the same time. It's not
>> >>> crucial
>> >>>>>>> for
>> >>>>>>>> the
>> >>>>>>>> argument that the stationary frame consist of one observer
> that
>> >>>>> can
>> >>>>>>>> only
>> >>>>>>>> make one measurement at a time.
>> >>>>>>>>
>> >>>>>>>>>>>>>
>> >>>>>>>>>>>>> Moving forward a little:
>> >>>>>>>>>>>>> Take a frame k that is moving toward positive x (that's
> a
>> >>> K
>> >>>>>>>>>>>>> coordinate)
>> >>>>>>>>>>>>> with a speed v. Any point that is moving to the right
>> >>>>> (positive
>> >>>>>>>>> x)
>> >>>>>>>>>>>>> with
>> >>>>>>>>>>>>> steady speed v will be stationary in k.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> Of course. That's Galilean relativity. I've never
>> >>> disagreed
>> >>>>> with
>> >>>>>>>>>>> that.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>> This means that any fixed-length rod in K (recall
>> >>> definition
>> >>>>> of
>> >>>>>>>>>>>>> "fixed") that is moving to the right with speed v in K,
>> >>> will
>> >>>>>>>>> also
>> >>>>>>>>>>> have
>> >>>>>>>>>>>>> a fixed length in k.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> Of course. That's Galilean relativity. I've never
>> >>> disagreed
>> >>>>> with
>> >>>>>>>>>>> that.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>> This we know because either end of the rod is
>> >>>>>>>>>>>>> moving to the right with velocity v in K; therefore,
>> >>> either
>> >>>>> end
>> >>>>>>>>> of
>> >>>>>>>>>>> the
>> >>>>>>>>>>>>> rod is stationary in k; therefore, the rod has a fixed
>> >>>>> length
>> >>>>>>> in
>> >>>>>>>>> k.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> Of course. That's Galilean relativity. I've never
>> >>> disagreed
>> >>>>> with
>> >>>>>>>>>>> that.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>> What we DON'T know yet is whether the fixed length in k
>> >>> is
>> >>>>> the
>> >>>>>>>>> same
>> >>>>>>>>>>> as
>> >>>>>>>>>>>>> the fixed length in K.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> Ah, well, we need to synchronize them before moving,
> don't
>> >>>>> we?
>> >>>>>>>>>>>
>> >>>>>>>>>>> No, why do we need to do that?
>> >>>>>>>>>>>
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>> That's what we aim to find out.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> No problem. Synchronize the lengths when stationary and
>> >>>>> employ
>> >>>>>>>>>>>> the definition of "fixed" in each frame.
>> >>>>>>>>>>>
>> >>>>>>>>>>> What does "when stationary" mean? You mean when the two
>> >>> frames
>> >>>>>>> are
>> >>>>>>>>> at
>> >>>>>>>>>>> rest with respect to each other, and then you speed one
> up?
>> >>>>>>>>> Einstein
>> >>>>>>>>>>> assumes no such synchronization procedure and either do
> I.
>> >>>>>>>>>>>
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>> All we know is
>> >>>>>>>>>>>>> that the length of the rod stationary in k has a fixed
>> >>>>> length
>> >>>>>>> in
>> >>>>>>>>> k
>> >>>>>>>>>>> and
>> >>>>>>>>>>>>> a fixed length in K.
>> >>>>>>>>>>>>
>> >>>>>>>>>>>> Of course it does. If one rod is 1 unit stationary and
> the
>> >>>>> other
>> >>>>>>>>> rod
>> >>>>>>>>>>> is
>> >>>>>>>>>>>> 1 unit
>> >>>>>>>>>>>> moving, then the 1 unit stationary isn't going to grow
>> >>> past
>> >>>>> one
>> >>>>>>>>> unit
>> >>>>>>>>>>>> just
>> >>>>>>>>>>>> because someone decided to move another 1 unit rod past
>> >>> it.
>> >>>>>>> That's
>> >>>>>>>>>>>> absurd.
>> >>>>>>>>>>>> What if I moved a third rod in the opposite direction?
>> >>>>>>>>>>>
>> >>>>>>>>>>> That's not what I said. I did not say that the fixed
> length
>> >>> in
>> >>>>> K
>> >>>>>>>>>>> changes.
>> >>>>>>>>>>
>> >>>>>>>>>> Two rods. Same length when together, no relative motion.
>> >>>>>>>>>> Move relatively, and A is shorter than B.
>> >>>>>>>>>> Therefore B is longer than A.
>> >>>>>>>>>>
>> >>>>>>>>>> How am I to know which changed?
>> >>>>>>>>>
>> >>>>>>>>>> Does it matter?
>> >>>>>>>>>
>> >>>>>>>>> Yes. It certainly does.
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>> It only does if you assume that length is an inherent
>> >>>>>>>>>> property of an object, which it turns out is not -- that's
>> >>> one
>> >>>>> of
>> >>>>>>>> the
>> >>>>>>>>>> things that Einstein insists is a preconception that we
> must
>> >>>>>>>> abandon.
>> >>>>>>>>>> Length is something that is derived from a *procedure*,
>> >>> nothing
>> >>>>>>>> more.
>> >>>>>>>>>> That procedure is what's outlined above.
>> >>>>>>>>>
>> >>>>>>>>> The procedure doesn't preclude the stationary rod having
>> >>>>> increased
>> >>>>>>>>> its length when compared to the moving rod. That it is the
>> >>>>> moving
>> >>>>>>>>> rod that shrinks is a preconception. Why isn't it the
>> >>> stationary
>> >>>>>>> rod
>> >>>>>>>>> that grows?
>> >>>>>>>>
>> >>>>>>>> I presume that the stationary observer would detect that by
>> >>> doing
>> >>>>> an
>> >>>>>>>> immediate synchronization of his rods. Moreover, we said
> that
>> >>> the
>> >>>>> rod
>> >>>>>>>> at
>> >>>>>>>> rest in the observing frame is a *fixed*-length rod, that
> is,
>> >>> it
>> >>>>> does
>> >>>>>>>> not
>> >>>>>>>> change with time in that frame. Moreover, the moving rod is
> a
>> >>>>>>>> *fixed*-length
>> >>>>>>>> rod in the moving frame (where it is stationary to the
> moving
>> >>>>>>>> observer).
>> >>>>>>>>
>> >>>>>>>> Moreover, we said that even the moving rod as viewed in the
>> >>>>>>> stationary
>> >>>>>>>> frame
>> >>>>>>>> is of fixed length -- that is, it does not change its length
>> >>> with
>> >>>>>>> time
>> >>>>>>>> in
>> >>>>>>>> the observing frame. This we know because the leading end
> flies
>> >>> by
>> >>>>>>> with
>> >>>>>>>> velocity v, as will be confirmed by any of a long chain of
>> >>>>>>>> collaborating
>> >>>>>>>> observers in the stationary frame, and the trailing end also
>> >>> flies
>> >>>>> by
>> >>>>>>>> with
>> >>>>>>>> velocity v, as will be confirmed by any of a long chain of
>> >>>>>>>> collaborating
>> >>>>>>>> observers in the stationary frame.
>> >>>>>>>>
>> >>>>>>>> Finally, as a check, the conclusion of the previous
> paragraph
>> >>> is
>> >>>>>>>> consistent
>> >>>>>>>> with the fixedness of the length of the moving rod as seen
> in
>> >>> the
>> >>>>>>>> moving
>> >>>>>>>> frame. The reason is that any point that moves to the right
>> >>> with
>> >>>>>>>> constant
>> >>>>>>>> speed v in the stationary frame is stationary in the moving
>> >>> frame.
>> >>>>>>>> Since the
>> >>>>>>>> leading end and the trailing end are both traveling at speed
> v
>> >>> in
>> >>>>> the
>> >>>>>>>> stationary frame, they will both be stationary points in the
>> >>>>> moving
>> >>>>>>>> frame.
>> >>>>>>>> Since they are both stationary points in the moving frame,
> the
>> >>>>> length
>> >>>>>>>> of the
>> >>>>>>>> rod in the moving frame is fixed; that is, it doesn't change
>> >>> with
>> >>>>>>> time.
>> >>>>>>>>
>> >>>>>>>> The confusion perhaps stems from the terms "shrink" and
> "grow"
>> >>>>> which
>> >>>>>>>> imply
>> >>>>>>>> that there is a compression or expansion process that is
>> >>> happening
>> >>>>> to
>> >>>>>>>> the
>> >>>>>>>> moving rod. That's not the case. What's true of a rod that
> is
>> >>>>> moving
>> >>>>>>>> with
>> >>>>>>>> respect to a stationary frame and stationary with respect to
> a
>> >>>>> moving
>> >>>>>>>> frame
>> >>>>>>>> is this: the fixed length of that rod as measured in the
>> >>>>> stationary
>> >>>>>>>> frame is
>> >>>>>>>> *different* than the fixed length of the rod as measured in
> the
>> >>>>>>> moving
>> >>>>>>>> frame. That does not imply a transition from one length to
>> >>> another
>> >>>>>>>> length.
>> >>>>>>>> It does say something about the procedure used to measure
>> >>> length.
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> My indentation appears to have failed.
>> >>>>>>>>> Androcles:
>> >>>>>>>>>> It is known that Einstein's equations --as usually
>> >>> understood
>> >>>>> at
>> >>>>>>> the
>> >>>>>>>>>> present time--when applied to moving bodies, leads to
>> >>>>> asymmetries
>> >>>>>>>>> which
>> >>>>>>>>>> do not appear to be inherent in the phenomena. Take, for
>> >>>>> example,
>> >>>>>>>> the
>> >>>>>>>>>
>> >>>>>>>>>> reciprocal lengths of two rods. The observable phenomenon
>> >>> here
>> >>>>>>>>> depends
>> >>>>>>>>>> only on the relative motion of the rod A and the rod B,
>> >>> whereas
>> >>>>>>>>>> Einstein's view draws a sharp distinction between the two
>> >>> cases
>> >>>>> in
>> >>>>>>>>> which
>> >>>>>>>>>> either the one or the other of these bodies is in motion.
>> >>> For
>> >>>>> if
>> >>>>>>> the
>> >>>>>>>>> rod
>> >>>>>>>>>> A is in motion and the Rod B at rest, there arises in the
>> >>>>>>>>> neighbourhood
>> >>>>>>>>>> of the Rod A a change in length. But if the rod A is
>> >>> stationary
>> >>>>>>> and
>> >>>>>>>>> the
>> >>>>>>>>>> rod B in motion, a change in length arises in the
>> >>> neighbourhood
>> >>>>> of
>> >>>>>>>>> the
>> >>>>>>>>>> Rod B. In the Rod A, however, we find a length change to
>> >>> which
>> >>>>> in
>> >>>>>>>>> itself
>> >>>>>>>>>> there is no corresponding energy, but which gives
>> >>>>> rise--assuming
>> >>>>>>>>>> equality of relative motion in the two cases discussed--to
> a
>> >>>>>>> change
>> >>>>>>>>> in
>> >>>>>>>>>> length of the same intensity as those produced by the
> motion
>> >>>>> the
>> >>>>>>>>> former
>> >>>>>>>>>> case.
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> No response?
>> >>>>>>>>
>> >>>>>>>> Not here. I had hoped that I addressed it above.
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>>
>> >>>>>>>>>>
>> >>>>>>>>>>
>> >>>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>>>> What I said was if a rod of fixed length is moving with
>> >>> speed
>> >>>>>>>>>>> v in K, it will have a fixed length where it is
> stationary
>> >>> in
>> >>>>> k,
>> >>>>>>>>> not
>> >>>>>>>>>>> necessarily the same fixed length as measured in K. The
>> >>> only
>> >>>>>>> thing
>> >>>>>>>>> I'm
>> >>>>>>>>>>> establishing is that a rod that isn't changing in length
>> >>> in
>> >>>>> one
>> >>>>>>>>> frame
>> >>>>>>>>>>> is also not changing in length in the other frame. A rod
>> >>> with
>> >>>>>>>>> length
>> >>>>>>>>>>> 1.2 (and always 1.2 -- that is, "fixed") in K may have
>> >>> length
>> >>>>>>> 0.80
>> >>>>>>>>>>> (and
>> >>>>>>>>>>> always 0.80 -- that is, "fixed") in k. Or it may have
>> >>> length
>> >>>>> 1.2
>> >>>>>>>>> (and
>> >>>>>>>>>>> always 1.2 -- that is, "fixed") in k. Of the two latter
>> >>>>> choices,
>> >>>>>>> we
>> >>>>>>>>>>> haven't figured it out, but it's fixed -- not changing in
>> >>> time
>> >>>>> --
>> >>>>>>>>> in
>> >>>>>>>>>>> either case.
>> >>>>>>>>>>>
>> >>>>>>>>>>>>
>> >>>>>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>>>>>> OK so far?
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>>>>> No. You haven't agreed on a method of synchronizing the
>> >>> rods
>> >>>>>>> yet.
>> >>>>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>>>> Once you explain what you mean by "synchronizing the
> rods,"
>> >>> we
>> >>>>>>> can
>> >>>>>>>>>>> sort
>> >>>>>>>>>>> that out.
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>>> I mean the same as synchronizing the clocks. I cut one rod
>> >>> to
>> >>>>> the
>> >>>>>>>>> same
>> >>>>>>>>>> length as the other.
>> >>>>>>>>> PD:
>> >>>>>>>>> This I can only do if both the template and the
> synchronized
>> >>>>> rods
>> >>>>>>> are
>> >>>>>>>>> at rest with respect to the cutter/observer.
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> Template? What template? I need no template. If I want to
>> >>> cut a
>> >>>>>>> rod
>> >>>>>>>> to
>> >>>>>>>>> the same length as another, the master rod is the template.
>> >>>>>>>>> I'll agree the two rods are relatively at rest when the
>> >>>>> operation
>> >>>>>>> is
>> >>>>>>>>> conducted.
>> >>>>>>>>> I can then carry the master rod to the moving frame and cut
> a
>> >>>>> third
>> >>>>>>>> rod.
>> >>>>>>>>> I can do the same with a clock.
>> >>>>>>>
>> >>>>>>> Honestly, I can't tell if this is new from you or not. Does
> this
>> >>>>> need
>> >>>>>>> a
>> >>>>>>> response?
>> >>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> Androcles.
>> >>>>>>>>>> I set one clock to the same time as the other.
>> >>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>> How do you do that if the clocks are separated by a
> distance?
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> Each clock is an oscillator and a counter. I'm not
> concerned
>> >>>>> about
>> >>>>>>>> the
>> >>>>>>>>> offset, as I've previously explained. The counter part of
> the
>> >>>>> clock
>> >>>>>>>> can
>> >>>>>>>>> be anywhere. Changing the distance will change the number
> of
>> >>>>>>>>> oscillations
>> >>>>>>>>> "in flight" between the oscillator and the counter, and
> hence
>> >>>>> the
>> >>>>>>>>> offset.
>> >>>>>>>>> The issue to be addressed is any change in the oscillator,
>> >>> the
>> >>>>>>>> "gain".
>> >>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>> Einstein
>> >>>>>>>>> proposes a solution with the emitter/mirror/receiver
> scheme.
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> Sure. When the clocks are at rest they can be synchronized.
>> >>>>>>>>> No argument from me on that score. My concern isn't the
>> >>> offset
>> >>>>>>>>> nor do I need to send a signal to the distant clock. I
> simply
>> >>>>>>>>> count its oscillations as they arrive, or read its dial.
> I'm
>> >>>>>>> waiting
>> >>>>>>>> for
>> >>>>>>>>> you to show the gain of the oscillator changes.
>> >>>>>>>>
>> >>>>>>>> I don't intend to show you that. I think you *do* need more
>> >>> than
>> >>>>> what
>> >>>>>>>> you
>> >>>>>>>> propose, though, because your test is *merely* a gain-gauge,
>> >>>>>>> something
>> >>>>>>>> that
>> >>>>>>>> checks rate. Einstein's scheme sets *simultaneity* (warns
> about
>> >>>>>>> offset)
>> >>>>>>>> each
>> >>>>>>>> time the synchronization is done. Repetition of the
>> >>>>> synchronization
>> >>>>>>>> tells
>> >>>>>>>> you about gain.
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>>> In other words, I take out any offset. Offset can be quite
>> >>>>> useful,
>> >>>>>>>>>> New York time is offset from London time by -5 hours. It
>> >>> would
>> >>>>>>>>>> become a problem if the clocks (or the rods) had different
>> >>>>> gains,
>> >>>>>>>>>> though.
>> >>>>>>>>>> Why, a New York clock would eventually record a different
>> >>>>> offset
>> >>>>>>>>>> to the London clock if it had a greater gain, and the two
>> >>> rods
>> >>>>>>> would
>> >>>>>>>>> no
>> >>>>>>>>>> longer be of the same length if I heated one of them and
> not
>> >>>>> the
>> >>>>>>>>> other.
>> >>>>>>>>>> The difference in time or the difference in length is the
>> >>>>> offset.
>> >>>>>>>>>> Synchronizing rods is no different to synchronizing
> clocks.
>> >>> All
>> >>>>> we
>> >>>>>>>>>> do is remove the offset.
>> >>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>> OK, I can work with that. That is, if we apply a
>> >>> synchronization
>> >>>>>>>>> procedure to two clocks with no difference in gain, then we
>> >>>>> would
>> >>>>>>>> find
>> >>>>>>>>> that the synchronization equation that we use to define it
> is
>> >>>>>>>> satisfied
>> >>>>>>>>> -- that is, the clocks are still in synchronization.
> However,
>> >>> if
>> >>>>>>>> there
>> >>>>>>>>> is a difference in gain, then the synchronization equation
>> >>> would
>> >>>>>>> not
>> >>>>>>>> be
>> >>>>>>>>> satisfied and we would have to make an adjustment to one
>> >>> clock
>> >>>>> or
>> >>>>>>>>> another to correct that. The greater the time since the
> last
>> >>>>>>>>> synchronization, the less confident we would be that the
>> >>> clocks
>> >>>>>>> would
>> >>>>>>>>> still be synchronized. Indeed, the difference in the gains
> in
>> >>>>> the
>> >>>>>>>>> clocks would determine how frequently we would have to
>> >>> perform
>> >>>>> the
>> >>>>>>>>> synchronization so that the agreement would be within an
>> >>>>> agreed-on
>> >>>>>>>>> tolerance.
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> For the purpose of this debate, let us stipulate to there
>> >>> being
>> >>>>> no
>> >>>>>>>>> mechanical, electrical or otherwise difference in the
> clocks,
>> >>>>> both
>> >>>>>>>>> of which are perfect. Likewise, no heat expansion or
> similar
>> >>>>> change
>> >>>>>>>>> in the rods.
>> >>>>>>>>> We do not live in an ideal universe, but this is a
>> >>> theoretical
>> >>>>>>>> exercise
>> >>>>>>>>> and we allow no tolerance whatsoever. Any difference in the
>> >>>>> clocks
>> >>>>>>>>> that may be found is to be solely a result of calculation,
>> >>> not
>> >>>>> of
>> >>>>>>>>> some practical consideration. Thus we can imagine the clock
>> >>> to
>> >>>>>>>>> be many light years distant, although we cannot place it
>> >>> there
>> >>>>>>>>> in practice.
>> >>>>>>>>
>> >>>>>>>> OK, that's fine too. And what that means is that, once the
>> >>> clocks
>> >>>>> are
>> >>>>>>>> synchronized using Einstein's scheme, every repetition of
> the
>> >>>>>>>> synchronization will verify that the clocks are still
>> >>>>> synchronized.
>> >>>>>>>> (Again,
>> >>>>>>>> as long as the clocks are stationary in that frame.)
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>> A similar procedure would apply to rods with different
> gains.
>> >>>>> Note,
>> >>>>>>>>> however that the two rods I'm synchronizing and a master
> rod
>> >>> all
>> >>>>>>> have
>> >>>>>>>>> to be at rest with respect to each other to do this. I
> cannot
>> >>>>> apply
>> >>>>>>>>> this procedure to a rod at rest and one flying by.
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> Agreed. All synchronization to be done at rest. Let us now
>> >>>>> declare
>> >>>>>>>>> the clocks and the rod to be synchronized, and you may
> begin
>> >>>>>>>>> your experiment.
>> >>>>>>>>
>> >
>> > Section 1 moving forward starts here...
>> >
>> >>>>>>>> OK, so now what we're going to do is we're going to take a
> rod
>> >>>>> that
>> >>>>>>> was
>> >>>>>>>> stationary in frame K, and we'll walk it backwards a few
> miles,
>> >>>>> and
>> >>>>>>>> we're
>> >>>>>>>> going to speed it up to speed v toward the right in frame K.
>> >>> It's
>> >>>>>>>> reached v
>> >>>>>>>> by the time it passes the origin of K. (It actually doesn't
>> >>> matter
>> >>>>>>> that
>> >>>>>>>> it
>> >>>>>>>> was ever stationary in the frame K, but Einstein says we'll
> do
>> >>> it
>> >>>>>>> that
>> >>>>>>>> way,
>> >>>>>>>> so we'll go along). After the rod has been sped up, this rod
> is
>> >>>>>>>> stationary
>> >>>>>>>> in the frame k. That is, the location of the emitter and
>> >>> receiver
>> >>>>> at
>> >>>>>>>> one end
>> >>>>>>>> of the rod are at a fixed location in k, and the location of
>> >>> the
>> >>>>>>> mirror
>> >>>>>>>> at
>> >>>>>>>> the other end of the rod is at another fixed location in k.
>> >>>>>>>>
>> >>>>>>>> Now, the observer in K knows that this rod can't be used to
> to
>> >>>>>>>> synchronize
>> >>>>>>>> any yardsticks in K, because it's not at rest. Moreover, the
>> >>>>>>>> emitter/mirror/receiver can't be used to synchronize the
> clocks
>> >>> in
>> >>>>> K,
>> >>>>>>>> because the system is not at rest. That doesn't really
> matter.
>> >>> The
>> >>>>> K
>> >>>>>>>> observer can recheck the synchronization of K's yardsticks
> and
>> >>>>> clocks
>> >>>>>>>> with
>> >>>>>>>> another rod and emitter/mirror/receiver stationary in K. As
> you
>> >>>>> say,
>> >>>>>>>> we'll
>> >>>>>>>> assume there is no drift, so the resynching would just
> verify
>> >>> that
>> >>>>>>>> clocks
>> >>>>>>>> are still synched and yardsticks are still the same length.
>> >>>>>>>>
>> >>>>>>>> However, we *can* measure a length of the moving rod in K.
> The
>> >>>>>>>> prescription
>> >>>>>>>> above will do it.
>> >>>>>>>>
>> >>>>>>>> Moreover, the moving rod *can* be used as a synching system
> in
>> >>> k.
>> >>>>>>>>
>> >>>>>>>> Agree so far?
>> >
>> > Section 2 moving forward starts here...
>> >
>> >>>>>>>
>> >>>>>>> OK, now there are three events:
>> >>>>>>> The "0" event, where a flash of light is emitted from one
> end,
>> >>> the
>> >>>>>>> "left end", of the rod that is moving in K and stationary in
> k.
>> >>>>>>> The "1" event, where the light bounces off the mirror at the
>> >>> other
>> >>>>>>> end, the "right end", of the same rod.
>> >>>>>>> The "2" event, where the light returns to the receiver at the
>> >>> left
>> >>>>> end
>> >>>>>>> of the same rod.
>> >>>>>>>
>> >>>>>>> The observer in k labels the coordinates of these events as
>> >>>>>>> (xi0,eta0,zeta0,tau0), (xi1,eta1,zeta1,tau1),
>> >>>>> (xi2,eta2,zeta2,tau2).
>> >>>>>>>
>> >>>>>>> Now because the rod is stationary in k, we know that xi0 =
> xi2.
>> >>>>>>> We have also set it up so that
>> >>> eta0=eta1=eta2=zeta0=zeta1=zeta2=0.
>> >>>>>>> We know that the length of the rod as measured in k is (xi1 -
>> >>> xi0),
>> >>>>>>> because the rod is stationary in k and the rod has a fixed
>> >>> length
>> >>>>> in
>> >>>>>>> k.
>> >>>>>>> That is, the right-end of the rod is always at xi1 in this
>> >>> frame,
>> >>>>>>> which
>> >>>>>>> means that the right end of the rod is at xi1 at tau0 as well
> as
>> >>> at
>> >>>>>>> tau1. We've satisfied the procedure for measuring the length
> of
>> >>> a
>> >>>>>>> stationary object.
>> >>>>>>>
>> >>>>>>> Finally, because the rod is stationary in k, we can use it as
> a
>> >>>>> clock
>> >>>>>>> synchronization rod in k. Since we previously synched the
> clocks
>> >>> in
>> >>>>> k,
>> >>>>>>> and since the clocks are ideal (don't drift), then this synch
>> >>> check
>> >>>>>>> will simply confirm the synchronization condition in this
> frame.
>> >>>>>>> Namely:
>> >>>>>>> tau1 = (1/2)(tau0 + tau2).
>> >>>>>>>
>> >>>>>>> Before we move on to how the observer in K sees these events,
>> >>> are
>> >>>>> we
>> >>>>>>> in
>> >>>>>>> agreement that this is a good picture of what Einstein
> intends
>> >>> in
>> >>>>> his
>> >>>>>>> set-up so far?
>> >>>>>>>
>> >>>>>>> PD
>> >
>> > And this ends the sections added recently.
>> >
>> >>>>>>>
>> >>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>>> You are going to try to tell me that relative motion
> changes
>> >>>>> the
>> >>>>>>>> gain
>> >>>>>>>>>> of the rod and the gain of the clock, are you not?
>> >>>>>>>>>
>> >>>>>>>>> PD:
>> >>>>>>>>> Not at all! Far from it. That would be true if length were
> an
>> >>>>>>> innate
>> >>>>>>>>> property of the rod, because then the only thing that could
>> >>>>> change
>> >>>>>>>> the
>> >>>>>>>>> length of the rod would be a process that physically
> affected
>> >>>>> the
>> >>>>>>>> rod.
>> >>>>>>>>> However, that's not what's going on here.
>> >>>>>>>>>
>> >>>>>>>>> Androcles:
>> >>>>>>>>> I'm pleased to hear it. Do continue.
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>> Androcles.
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>>>
>> >>>>>>>
>> >>>>>
>> >>>
>> >>
>> >>
>> >
>

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