Re: Caltech Football Team
- From: "Ken S. Tucker" <dynamics@xxxxxxxxxxxx>
- Date: 8 May 2005 13:39:04 -0700
Randy M. Dumse wrote:
> "Ken S. Tucker" <dynamics@xxxxxxxxxxxx> wrote in message
> news:1115565012.419673.83130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > But
> falling back on the Principle of
> > Equivalence (PoE) I think the beam would
> > not be deflected relative to SUE as she
> > orbits.
>
> I didn't get enough details about satellite SUE to really say. We'd
> really have to talk about if the perfect mirrors were "closed" so we
had
> a light box. Also you're talking about SUE being in freefall. So PoE
> says you have (the equivalent of) a gravity free condition.
Yes. I read your post, and I guess you can see a bit
of GP-b is rendering, because an initial orientation
at aphelion, if varied would change the beam path in
SUE.
> So given the 2x difference between the geodesics of mass compared to
> light, I''m pretty sure the light would not stay trapped between two
> initially "parallel" mirrors in free fall. If we had a closed box,
the
> light would occassionally graze the bottom, and gain an upward moment
> back up the mirror faces.
Ok, but using the Schwarz. sol, how can that be justified,
that's where I get into trouble.
> But the problem is really difficult, because you have an orbit which
> isn't circular, and the satelite that would need some angular
momentum
> to keep the parallel mirrors always with the same orientation to the
> direction of the gravitational field, while it was changing angular
> positions in the orbit, etc. etc.
Yes, but suppose we orientated SUE at a distant quasar,
then as she orbited (along the lines of GP-b) would her
axis of gyroscopic orientation vary as she approached
perehelion? (I'm recognizing that problem).
> As you rightly point out, even the concept of parallel is exceedingly
> difficult to figure, given a gravitational field. We know there are
no
> global frames to count on.
:-)
> The experiement I was thinking of was something that could be done in
a
> lab on Earth. Of course the problem with measuring the gravitational
> deflection of the light is light is so fast. If we just watched a
beam
> pass the earth, it would go from side to so in .04s. How far does
> something fall in .04s? about .4m and across 13,000,000m Worse, the
> g-field is only 1g as it grazes the surface, and averages less, so
even
> that deflection on the order of 1 part in 30 million.
:-)
> So the problem is how do we keep the light in a fairly constant field
> long enough to measure the effects of gravity? So my thought for the
> experiment is to take an omniguide (tube made out of these perfect
> mirror material), and shine the light in the bottom, let it reflect
back
> and forth inside the vertical tube. Between reflections it will be in
> free fall in a 1-g field. We should be able to "watch" it fall if we
can
> start it at just the right angle, and we can tell if we're watching
that
> angle by finding when the light falls out.
:-)
> Note this tube is not a closed light box, but only a (nearly)
vertical
> tunnel with the ends open to light traps. The only light of interest
is
> the light that "hangs" longest in the tube.
>
> Notice if the tube isn't vertical, the problem doesn't change. At
worst
> minor variations in the tube might cause it to get an unnatural kick
> upwards or downwards if not cancelled by the opposite angle on the
other
> side. These minor variations will shorten the total time in the tube.
So
> the last light out followed the most perfect path. Injecting the
light
> with a slight dispersion spread should allow some portion of the
light
> to have the perfect entry angle for an extremal flight.
Listening...
> The mass equivalent experiment would be like throwing racket balls
(glad you changed the balls :-)...
> up a
> round chimney. (It doesn't really matter who's racket balls these
are,
> btw.) Given an initial velocity and angle of entry, the ball that
stays
> longest in the chimney would be the one that bounces, zig zagging up
the
> sides, without clearing the top, and continues to bounce down the
sides,
> side to side, on the way back down. For a given entry speed, there is
a
> perfect angle that allows a trajectory which peaks near the top
without
> exiting.
Understood
> Randy M. Dumse
Did you just avoid the problem or is it just me?
With a pinch blame SUE.
I'm concerned we're not directly addressing the
problem.
Ken S. Tucker
.
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