Re: How can this work in relativity?

This is a way-too-long post. I will unfortunaely need
to stop doing this sort of thing soon; just a heads up.

bz wrote:
> russell@xxxxxxxx wrote in
> news:1123689982.619433.252370@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:
>
> > I may have missed a golden opportunity to clear everything
> > up in one stroke, so let me try to recapture it.
> >
> > russell@xxxxxxxx wrote:
> >> bz wrote:
> >
> > [snip]
> >
> >> > A = dv/dt; v = dx/dt this means that each small particle of the rod
> >> > must move the same distance at the same velocity during each instant
> >> > of time in order for the acceleration to be uniform. This keeps the
> >> > distance between particles constant [from their FoR].
> >>
> >> No, constant in the *original* FoR. Your equations are only
> >> good in an *inertial* frame.
> >
> > Oops, I was thinking here of F=ma etc., not your equations
> > which are just definitions of some coordinate velocity and
> > acceleration. Your equations are ok as far as that goes.
> >
> > However, they are useless (i.e. can't be integrated) unless
> > you specify what that coordinate system *is*.
>
> Let us pick a universe that has a wall of infinite length and width at one
> end of the universe. The thickness of the wall is sufficient to provide 1 g
> of attraction toward the wall at any point in said universe [I think 6000 or
> so km of a material with about the density of the earth should do it.]

That is a bit naive, but I'm not enough of a GR guru to
say it wouldn't work. We do have to say something about
how it got there; we could say it's been there forever.

>
> We suspend your rod [at any angle you wish] within the gravity field by
> providing a similar wall at the other end of our universe, parallel to the
> first wall.

Now you have two walls. What is holding them apart?
In GR you have to worry about such things. I could ask
other questions too, but we will soon find them moot...

>
> Everywhere within our universe, we now have zero gravity.
>
> We remove one wall.

Vanishing mass? That definitely contradicts GR. Sorry,
no can do. And, frankly, I don't think your moving giant
masses around is going to help you understand this problem.
Except perhaps in a negative sense -- that is, when you see
how difficult it is to get it right with masses, you might
come to understand that the equivalence principle will *not*
help you in your vain quest to contradict me. It's a red
herring.

>
> The rod experiences uniform acceleration along its entire length, perhaps at
> the speed of gravity.

Which btw is c, in GR. Way too slow to turn on acceleration
in the dseppala experiment, where the whole line of people is
supposed to jump onto the belt *simultaneously* in the original
frame.

But let that pass; you've made a much more serious blunder.
Your rod is in free-fall, i.e. in a GR sense, accelerating
is exactly what the rod is *not* doing. An ant on the rod,
looking at his handy-dandy accelerometer, will see a reading
of zero. The experiment analyzed previously in this thread
is completely different -- our ant sees a *nonzero* reading.
As I said last time, it matters who gets the rocket pack.

>
> What happens to the rod, if speed of gravity is c?
> What happens to the rod, if the speed of gravity is infinite?
>
> I am not sure about the c, version. In the infinite, the rod would appear to
> fall at an ever increasing rate. Observers falling with it would see not
> contraction in length.

You don't need infinite speed of gravity for that -- gravity
is *local* in GR. All you need is a uniform field (no tidal
forces) and a rod that is in freefall, with enough time to
relax any wacky initial conditions. In other words, a rod
moving inertially in flat spacetime. Fine, I'm glad you
understand that experiment, but it's not the experiment we're
supposed to be performing here.

>
> Observers in a space ship accelerating at 1 g away from the wall [thus
> maintaining a constant distance from the wall] should observe the rod
> contracting in length.

Yes, that is right. But *he* is the one accelerating, not
the rod. His accelerometer marks him indelibly as the one
accelerating. Anybody looking at that accelerometer, and
comparing its reading to the ant-on-the-rod's accelerometer,
will be able to say without a doubt who is accelerated and
who is not. (I.e. whose *proper* acceleration is nonzero.)

>
> > The easiest
> > coordinates to use are the inertial coords of the original
> > frame, because we know as a given that the acceleration is
> > uniform in that frame at least at t=0. (And as it happens,
> > the acceleration continues to be uniform in that frame, so
> > it is a particularly good choice.)
> >
> > The particles are only in that
> >> frame for an instant, so your equations can't tell you (by
> >> themselves) what the separation will be in the final frame.
> >
> > But now that our coordinates are specified, we *can* say
> > what the separation will be. It is *constant* in our
> > coordinates because, as a given, our A is everywhere the
> > same. Exactly for the reasons you say.
> >
> > Since we have our answer, we can transform it into any
> > frame we like, e.g. the final frame of the rod. Voila,
> > we see that the rod is stretched in that frame (which I
> > repeat is now the rest frame of the rod). This is what is
> > known as a physical stretch.
>
> The rod was accelerated uniformly. Was it later decelerated? If not, it
> should be in uniform motion, constitute a SR FoR, and we, from within that
> FoR see no stretch.

We don't decelerate it; it merely stops accelerating
when all parts of the rod have reached a certain speed
(3 m/s) in the original frame. As it happens, the
acceleration will stop simultaneously in this frame
(i.e. the *original*) frame, all along the length of
the rod. We can easily calculate that this will be
the case.

Since it is now in uniform motion in the original frame,
yes the rod does now define a new inertial frame in which,
after all the acceleration is finished, the rod is entirely
stationary.

(*But*, humor me and add a small caveat -- we should tell
the people holding the rod to keep their positions steady
once they are motionless in that final frame, and resist
any pull that appears in the rod. Since you think there
is no pull, that should be an easy concession to make.)

Here's the problem: the experiment ends all at once in
the original frame, but in the final frame the acceleration
ends at one end *one full second* before it ends at the
other end. Just do a Lorentz transformation and verify
this, if you doubt me. Since during that one second the
two ends of the rod are not moving at the same velocity
(one is zero, the other is -3 m/s) the rod cannot keep the
same length in that final frame. The distance from one
end to the other increases by about 3m. However, the
physical stretch works out to only 1.5m because the
original state of the rod, in this final frame, was
Lorentz-contracted by 1.5m.

Note, to say what the rod's length is in the new frame, we
really have to wait until all of the rod is in that frame,
according to the synchronized clocks of *that* frame. And
that is why I had the holders keep holding, just in case.
It turns out my caution was justified.

(Maybe your trouble is you just have not worked enough
problems with relativity of simultaneity?)

How can a stretched rod (with people holding it) be in
a state of uniform motion? Do I have to answer this?
You seem to have some weird idea that these states
are mutually incompatible. As if a pocket watch,
for example, can never be used in an inertial frame,
because it contains a spring under tension??

>
> [quote from Einstein's 1920 book on relativity]
> If the motion of the carriage is now changed into a non-uniform motion, as
> for instance by a powerful application of the brakes, then the occupant of
> the carriage experiences a correspondingly powerful jerk forwards. The
> retarded motion is manifested in the mechanical behaviour of bodies relative
> to the person in the railway carriage. The mechanical behaviour is different
> from that of the case previously considered, and for this reason it would
> appear to be impossible that the same mechanical laws hold relatively to the
> non-uniformly moving carriage, as hold with reference to the carriage when at
> rest or in uniform motion. At all events it is clear that the Galileian law
> does not hold with respect to the non-uniformly moving carriage. Because of
> this, we feel compelled at the present juncture to grant a kind of absolute
> physical reality to non-uniform motion, in opposition to the general
> principle of relatvity.
> <emphasis mine>
> But in what follows we shall soon see that this conclusion cannot be
> maintained. <end emphasis>
> [unquote]
>
> He would seem to be hinting that there are Special FoRs and General FoRs,
> just as he explicity speaks of a Special Priciple of Relativity and General
> Principle of Relativity.

He is hinting that the laws of physics can be expressed
in generally covariant form. Explicitly he is saying that
the presence of mechanical effects such as the above are
not necessarily the result of *motion*. Certainly they
are not so in any absolute sense.

But, he's not denying the *presence* of the effects; those
are real. If a real accelerometer blips from zero to nonzero
in some experiment, it must do so regardless of what frame,
coordinates, philosophy, theory, etc. you use to measure
and/or explain that experiment. And if motion is not the
cause, then your theory had better have some force (or pseudo-
force) in the chosen coordinates, to explain the observed
behavior.

>
> But I am still reading and trying to understand so I may be totally off base.

Ok, keep it up. Btw, Einstein is always good to read but
you might find more recent works to lead you more quickly
to understanding. The SR/GR teachers of today may be no
Einsteins, but they do have considerable advantages in
hindsight, longer experience with students, etc. etc.

.

Relevant Pages

  • Re: How can this work in relativity?
    ... > that as long as you do your calculations in a single frame, ... >>> uniform acceleration by applying the appropriate compensating ... >From the moving FoR that includes the rod, ...
    (sci.physics.relativity)
  • Re: TomToms stupidity (re: was always TomToms stupidity)
    ... you acknowledged yourself that this type of acceleration would ... >>>cause the rod to stretch in it's own frame. ... or accelerating the rod comes from the rocket. ...
    (sci.physics.relativity)
  • Re: TomToms stupidity (re: was always TomToms stupidity)
    ... >> acceleration is immaterial, and Daryll was right to neglect it. ... >> relative to the stationary frame. ... and this "pulling" will stretch the rod. ...
    (sci.physics.relativity)
  • Re: SR Length Contraction - how do physicists explain this
    ... >> from the pov of the rest frame. ... relative to the other endpoint on each rod. ... change length during the acceleration. ...
    (sci.physics.relativity)
  • Re: SR inconsistent?
    ... the same component of acceleration is zero. ... >> component of the magnetic force produces a nonzero corresponding ... > magnetic force in frame F' does not produce any y'-component of ... What good are equations of motion if you can't use ...
    (sci.physics)
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