Re: Androcles and Draper resume Einstein 1905
From: Androcles (dummy_at_dummy.net)
Date: 01/14/05
- Next message: stephen_at_nomail.com: "Re: too much information!"
- Previous message: Mike: "Re: Proof for E=mc2"
- In reply to: PD: "Re: Androcles and Draper resume Einstein 1905"
- Next in thread: PD: "Re: Androcles and Draper resume Einstein 1905"
- Reply: PD: "Re: Androcles and Draper resume Einstein 1905"
- Messages sorted by: [ date ] [ thread ]
Date: Fri, 14 Jan 2005 18:28:35 GMT
"PD" <pdraper@yahoo.com> wrote in message
news:1105722839.145822.145850@f14g2000cwb.googlegroups.com...
> Androcles, your news reader seems incapable of indenting properly. Can
> you get that fixed?
Dunno. Sometimes it's fine. Other times it quits halfway through.
We need to trim some of the material that is already done with, I think.
No need for a list of who said what at the beginning.
I may be undergoing some medical procedures soon, so computer
diagnostics is not a current priority for me.
That means I may vanish, too. If I looks as if I'll drop dead, I'll
try to get a message through somehow.
I'll try to leave the gist of the current sticking points evident.
>> >> Ok, so it can't be done for moving frames.
>> >> Why is Einstein even talking about it?
>> >
>> > It can be done for a moving frame. If you have a moving rod and you
>> > want to measure its length, simply go to the moving frame where it
> is
>> > at rest and repeat the same procedure.
>>
>> Oh, jump on the contracted length bus. Not easy, jumping
>> on a moving vehicle, y'know, but ok, it's a thought experiment.
>
> Not at all. The observer in k can measure the rod that is moving in K,
> because the rod is stationary in k. I didn't mean to imply that the
> observer in K becomes an observer in k, only that he goes to the
> observer in k for a consult. Not frame-jumping, just communication
> between the observers in the two frames.
It is this very commnication that is the basis for our discussion.
Does the K-observer conclude the k-rod is shorter than the K-rod,
or not? Yes/No.
Does the K-observer conclude that time in the k-frame has elapsed
less than time in the k-frame? Yes/No.
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.
>>
>> 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?
>
> 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
>
>>
>> >
>> >
>> > 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.
>> >
>> >
>> >
>> >
>> >
>
- Next message: stephen_at_nomail.com: "Re: too much information!"
- Previous message: Mike: "Re: Proof for E=mc2"
- In reply to: PD: "Re: Androcles and Draper resume Einstein 1905"
- Next in thread: PD: "Re: Androcles and Draper resume Einstein 1905"
- Reply: PD: "Re: Androcles and Draper resume Einstein 1905"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
