Re: Androcles and Draper resume Einstein 1905

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

Date: Sat, 15 Jan 2005 13:54:22 GMT

"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?

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.
>
> But it is true that "simultaneous" is defined by the clock
> synchronization
> procedure, and that what is simultaneous in the K frame is not
> simultaneous
> in the k frame. That is precisely what leads to the disagreement about
> whether the observer in a given frame has followed the procedure for
> measuring the length. Because that is an unresolvable disagreement --
> there
> is no "true" simultaneity -- the definition of length is also
> frame-dependent. There is no "true" length. Length is something that
> is
> defined by procedure. The result of that procedure depends on the
> observer.
> Length is not an inherent property of the rod.
>
>
>>>
>>> There is at least one yardstick at rest in the K frame, and at least
>>> one yardstick at rest in the k frame, but none of these are the rod
>>> in
>>> question.
>>
>> The rod in question has become redundant. It was measured by the
>> K-measuring rod. It has yet to be measured by the k-rod (moving) and
>> the comparison of measurement can be between the rods k and K,
>> when brought to rest.
>>
>>> I don't understand the question.
>>>
>>> PD
>> How to measure. Place the ruler on the rod,
>> the mark labelled zero at one end of the rod and read the number
>> on the ruler at the other end of the rod to be measured. Do not
>> let the ruler move relative to the rod as you do so, or you will not
>> correctly measure the rod.
>> Children understand this instinctively, but sometimes fail to make
>> a correct measurement because they let the ruler slip. The trained
>> and skilled craftsman looks again at the zero mark to verify the rod
>> and the ruler have not slipped.
>> That I should have to explain this simple procedure to one reputed
>> to be of intelligence and capable of supposedly understanding
>> the nonsense of Einstein is regrettable in the extreme.
>>
>> I am not sure we can continue if you are confused as to the meaning
>> of "measuring".
>
> See the above. You are obviously placing priority on one of the two
> methods
> that Einstein gives for measuring a length of a moving rod. Why? What
> is
> inherently suspect about one that is not suspect in the other?
>
>>
>> "Never underestimate yourself, let others do it for you"- (Inspector
>> Morse)
>> Colin Dexter.
>>
>>
>
> Androcles, please note that I have moved on a bit in recent postings,
> trying
> to get at some of the algebra in terms of the coordinates that
> Einstein
> defines. You haven't responded to any of that yet.
>
> For your convenience, I've marked those sections below.
>
>>
>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>>
>>>>> 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, following the procedure
>>>>> available to you for measuring the length of a moving rod. And
>>>>> suppose,
>>>>> suppose, lo and behold, that procedure results in a different
>>> number.
>>>>> What are you going to conclude?
>>>>>
>>>>>> Does the K-observer conclude that time in the k-frame has elapsed
>>>>>> less than time in the k-frame? Yes/No.
>>>>>
>>>>> Neither. 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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