Re: A challenge to non-SRians

From: sal (pragmatist_at_nospam.org)
Date: 08/09/04 

Date: Mon, 09 Aug 2004 12:45:01 -0400

On Mon, 09 Aug 2004 16:13:32 +0000, Androcles wrote:

>
> "sal" <pragmatist@nospam.org> wrote in message
> news:pan.2004.08.09.14.22.17.501312@nospam.org...
> | On Mon, 09 Aug 2004 06:40:11 -0700, Martin Miller wrote:
> |
> | > sal <pragmatist@nospam.org> wrote in message
> | > news:<pan.2004.08.07.18.11.23.475165@nospam.org>...
> | >> On Sat, 07 Aug 2004 08:32:00 -0700, Martin Miller wrote:
> | >>
> | >>
> | >> > Do you or do you not agree that when the origins of two passing
> | >> > inertial frames coincide, so do all other such corresponding points
> | >> > such as (Xa,0,0) and (Xb,0,0)?
> | >>
> | >> Oh, please. Have you ever heard of "Fitzgerald contraction"?
> | > snip
> | >> If Dirk won't answer you, try looking in a book.
> | > snip
> | >> Make a real effort to absorb some of the material, try to understand
> | >> it rather than trying to resist it, and then ask a question which is
> | >> more advanced than something I'd expect to hear on the first day of a
> | >> freshman physics class. Then, you'll be much more likely to get a
> | >> polite hearing and a serious reply.
> | >
> | > I can prove that I should have gotten "a polite hearing and a serious
> | > reply" in the first place, but do you really believe that I will ever
> get
> | > "a polite hearing and a serious reply" from any relativist?
> | >
> | > Anyway, here is my proof (actually two proofs):
> | >
> | > According to Einstein himself, whenever two passing observers
> | > coincide,
> if
> | > one is equidistant (per his own ruler) from two events, then the other
> is
> | > also (per his own ruler).
> |
> | Right. And if the frames are moving relative to each other, _ONE_ of
> | three events may share a coordinate value between the two frames. The
> | other two won't. (Three events, not two, because the observers
> | "coinciding" is also an event, of course.)
> |
> | The "left distance", for want of a better phrase, will match the "right
> | distance" in each frame. True. But the "left distance" in one frame
> | won't match the "left distance" in the other frame, which is what you
> | actually assumed was the case.
> |
> | > (See Einstein's train/embankment experiment.) (As Einstein said in
> | > that experiment, "But the events A and B **also**
> [my
> | > emphasis] correspond to positions [or frame points] A and B on the
> | > train.")
> | >
> | > Translation:
> | > According to Albert Einstein, when the origins of two passing frames
> | > coincide, all other matching frame points also coincide, such as the
> frame
> | > points (Xa,0,0,) and (Xb,0,0,).
> |
> | Wrong. You totally missed the point, as I explained out above -- but
> | then I expect you misunderstood my explanation, too, and will go on
> | claiming that Einstein's statements show Fitzgerald contraction can't
> | happen.
> |
> | Try to prove what you just said mathematically, starting with your
> | statement about equidistance in any coordinate system.
> |
> | You won't, though, because it doesn't follow.
> |
> | And that marks the end of my patience with you, and you're plonked.
> | Argue with Dirk -- he never seems to plonk anybody, as far as I can
> | tell.
> |
> |
> | --
> | I can be contacted through http://www.physicsinsights.org
>
> Harsh... he wasn't directly insulting, sal.

You're right. But oddly enough, direct insults don't bother me as much as
willful ignorance, which I felt was displayed from the start of what he
first posted, in which he just casually assumed that there was no
contraction and then went on to show that SR (which of course assumes
there _is_ contraction) didn't agree with his assumption.

> And there IS a problem with the contraction, too.

Not with what MM was talking about, tho. He quoted Einstein as saying
that (in essence) contraction is linear, and from that he concluded that
contraction doesn't exist.

I hope this diagram comes through OK... unit width font, please:

         delta 1 delta 2
..e1-----------------x -----------------e2
......e1'------------x'------------e2'
           delta 1' delta 2'

MM claimed that, because delta1 = delta2, and delta1' = delta2', we must
also have delta1 = delta1'. Of course, no such conclusion follows.

I'm not sure he even realized that that's what he was saying, by the
way, but unless I thought I had a snowball's chance of his actually
_trying_ to understand what I was saying (rather than just trying to
repudiate it, with the assumption that it _must_ be wrong) I would not be
willing to spend the time to try to convince him.

> ----------0'---------------x'---------------- k frame tau = 0
> ----------0-------------------x------------- K frame t = 0
>
> -0'---------------x'---------------- k frame tau = 0
> ------------------0-------------------x------------- K frame t = 0
>
> xi = (x-vt)/sqtr(1-v^2/c^2) given,
> but x = (x' + u.tau)/sqrt(1-u^2/c^2), u != v. (unless you can show
> dxi/dtau = dx/dt, that is). Androcles.

This wasn't MM's argument, of course.

If I understand you, you've pointed out that, viewed from the prime
coordinates, origin 0 crossed distance D' in the "same time" that point x'
crossed a smaller distance, D-eps, in the base frame. Since each sees
the other moving at the same velocity, this can't be.

Do I follow you correctly so far?

If I did, then here's the problem: Relativity of simultaneity. The above
diagram shows the prime frame contracted, which is what is observed by the
base frame. But from the PoV of the prime frame, the locations of 0' and
x' as seen by the base frame were actually evaluated at two _different_
times. When we instead use the clock of the prime frame to set the
starting points, then we see something like this:

> ----------0'------------------x'---------------- k frame tau = 0
> ----------0----------------x------------- K frame t = 0

and from the prime point of view, it's the base frame which appears
contracted. In consequence, each sees the other moving the same distance
in the same time -- but each also sees the other as timing the motion
using clocks that are "improperly" synchronized.

MM didn't get anywhere near this far in his argument, however -- he was
asserting that all points on the axis must line up "when" (no
definition given for "when") the origins line up, so no contraction at all
is allowed, symmetric or otherwise. 

-- 
I can be contacted through http://www.physicsinsights.org

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