Re: What is the expert opinion on deriving SR with a Shubertian clock?
From: Bill Hobba (bhobba_at_rubbish.net.au)
Date: 08/27/04
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Date: Fri, 27 Aug 2004 23:49:39 GMT
"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:jwPXc.10241$D7.8914@news-server.bigpond.net.au...
>
> "Big Bird" <condor@biosys.net> wrote in message
> news:df160b8f.0408271427.34e1aec5@posting.google.com...
> > "Bill Hobba" <bhobba@rubbish.net.au> wrote in message
> news:<iRtXc.9075$D7.5030@news-server.bigpond.net.au>...
> > > "Big Bird" <condor@biosys.net> wrote in message
> > > news:df160b8f.0408261403.1a002e31@posting.google.com...
> >
> > > > THis is something of an aside. but I am not convinced that an
inertial
> > > > frame is a terribly well-defined concept.
> > > >
> > >
> > > I think you mean ill defined or not well defined. And yes would seem
it
> is
> > > not possible to completely define operationally an inertial frame (I
am
> open
> > > to be proven wrong here) - the best probably is an infinitesimal
freely
> > > falling frame. But the logical structure of SR does not depend on
such.
> >
> > Well, apparently you know what I meant :).
>
> Of course - just being pedantic :). When I reread it I laughed and
thought
> it was not really required.
>
> >
> > The problem with "freely falling" strikes me as the same as the one
> > with "free particle" -- I'd seem to need some procedure to determine
> > whether a coordinate system is freely falling if I wanted to use this
> > to define something else on top of it. But how would I test that? I
> > undertsand that I can always compare my motion to some very distant
> > fixed-star frame but that would put me squarely into the realm of a
> > global (rather than a local) concept.
>
> I said it was the best I could think of - not it was perfect. As I stated
I
> think it is operationally very difficult, perhaps even impossible, to
> operationally define an inertial frame. But one can conceptually define
it
> similar to conceptually defining lines and points in Euclidian geometry.
>
> >
> >
> >
> > > >I mean I understand what it
> > > > *is*, but spelling it out appears rather difficult. As an example
let
> > > > me use this one:
> > > >
> > > > > It usually
> > > > > is defined as one in which free particles move at constant
velocity
> > > >
> > > > which rests on the concept of a "free particle".
> > >
> > > A free particle is one that moves the same as if it was the only thing
> in
> > > the frame.
> >
> > In the coordinate system in which the moon is at rest, the moon moves
> > as if it were the only thing in the universe (namely not at all).
>
> But such a frame is not inertial.
>
> > It is thus a "free particle". Its velocity is also constant. But of
> > course the reference fram in which the moon is at rest is not an
> > inertial system.
>
> As I mentioned above before reading on.
>
> >
> > I'm posting all this, since I have found that the definition of an
> > inertial reference frame is about the hardest part of special
> > relativity. Everything else is just a bit of straightforward math.
>
> Ok here is my take.
>
> A frame of reference is a conceptual standard of rest against which
> experiments can be conducted. An inertial frame is one in which space and
> time are homogeneous and space is isotropic and stationary lines and
points
> obey Euclid's axioms. Thus we can construct a Cartesian coordinate
system.
> Sync clocks by say having two guns fire bullets from midway between the
> clocks and set them to zero when the clock is hit by the bullet. By
> homogeneity and isotropy if the guns are exactly the same the bullets have
> exactly the same transit time and hence the clocks are synced. By
> homogeneity identical clocks must remain in sync.
I want to elaborate this a bit more. First, my mistake, one requires both
homogeneity and isotropy. Suppose we have two clocks with one at the
origin. Suppose the other clock (clock 2 say) gets out of sync by delta t
ie clock 1 - clock 2 = delta t. Suppose we orientate the coordinate system
so that the other clock is as the origin and we have exactly the same
experiment with the roles of the clocks reversed (this will require rotating
the axis hence the need for isotropy) we have exactly the same experiment so
they must get out of sync by delta t ie clock 2 - clock 1 = delta t so delta
t = 0.
Thanks
Bill
> Imagine such a conceptual
> clock at each point. A space-time event is something that occurs at a
> conceptual clock thus is assigned a position (the position of the clock)
and
> a time (the time read by the clock). By homogeneity it can be shown that
> the transformation between the same space-time events recorded in
different
> inertial frames must be linear and hence the frames must be moving at
> constant velocity wrt each other. The POR now comes into play - the laws
of
> physics are the same in all inertial frames and ones moving at constant
> velocity wrt to an inertial frame. Imagine a particle as the only thing
is
> such a frame - by definition it is free. Consider an infinitesimal time
> such that it has constant velocity during that time. Go to a frame where
it
> is instantaneously at rest. By isotropy it must remain at rest. Thus
free
> particles move with constant velocity in inertial frames. From the above
> one can derive the Lorentz transformations eg
> http://arxiv.org/abs/physics/0110076.
>
> Thanks
> Bill
>
> >
> >
> > > > Which of the many dozen books that carry Mr Landau's name are you
> > > > referring to here?
> > >
> > > Page 5 Landau - Mechanics.
> > >
> >
> >
> > Ah - I don't have that at hand, but I'll look it up. Thanks.
>
>
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