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From: weemba@brahms.BERKELEY.EDU (Matthew P. Wiener)
Newsgroups: net.physics
Subject: Re: The death of bogus physics
Message-ID: <12675@ucbvax.BERKELEY.EDU>
Date: Wed, 26-Mar-86 02:56:03 EST
Article-I.D.: ucbvax.12675
Posted: Wed Mar 26 02:56:03 1986
Date-Received: Thu, 27-Mar-86 01:28:41 EST
References: <12603@ucbvax.BERKELEY.EDU> <875@lanl.ARPA>
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Reply-To: weemba@brahms.UUCP (Matthew P. Wiener)
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In article <875@lanl.ARPA> jlg@a.UUCP (Jim Giles) writes:
>All this is but to say that given sufficient mathematical sophistication
>you can correctly work out what's going on in either coordinate system.  I
>never doubted this.  But it still doesn't answer my original objection to
>the Russell quote: given the frames K and K' - AT MOST one of them will be
>able to spin up a gyroscope and observe NO precession.

And when did Russell say that was false?  'precession' is a frame-dependent
phenomena, as you have just shown, and is not covariant.  Russell said the
difference between the two was 'mathematical convenience'.  Perhaps you would
have been happier if he went on to say the mathematically simpler was the one
that gets the standard physics terminology.

>                                                        In addition, the
>frame which measures no precession on its gyroscopes will be fixed relative
>to the distant stars to many significant figures of measurement (in the
>neighborhood of Earth: within 0.1 arc seconds per year (MTW pp. 1119-
>1121)).  To this extent, AT LEAST, there seems to be a prefered frame with
>respect to rotation.

Preferred if you like your calculations to be simple and your terminology
to be standard.  Not preferred if you want to understand the geometry of
space-time itself.  In either K or K', the curvature will come out zero.
That is geometry.  That is where the real physics lies.

Does the above paragraph sound like it is from MTW?  It should.

>Now, I guess you are arguing that the coincident event of zero precession
>locally and being fixed relative to the distant stars is 'merely a matter
>of convenience'.  Well, I'm not buying it.  And neither did Einstein, who
>spent considerable effort to determine why this might be so.  This is why
>I brought Mach's principle into this discussion - which not only predicts
>that there would be a 'prefered' frame, but gives a relativistic
>explanation for the effect (it also predicts that any local variation
>from the 'prefered' frame would be insignificant).

But physics today rejects the notion of a preferred frame.  That is the
meaning of general covariance.  Given a frame, it is the individual framer's
responsibility to correlate his coordinate measurements with physical
reality.  There is no apriori meaning.  Thus in studying black holes, a lot
of effort must go into figuring out what 'r' and 't' refer to physically.
If you hold the earth as fixed in your frame, and then calculate that Alpha
Centauri is going 9490 times faster than light, you have abdicated your
responsibility of relating coordinate numbers to physical notions.

The only possible meaning of Mach's principle in general relativity today
is as Doug Gwyn pointed out, is in the notion of boundary conditions.

>While it is certainly convenient to do one's calculations in a non-rotating
>coordinate system, there also seems to be something physically significant
>about such systems as well.

Yes.  There is physical significance to inertial frames.

ucbvax!brahms!weemba	Matthew P Wiener/UCB Math Dept/Berkeley CA 94720