Re: Another Rotating Cylinder Problem - explain from moving frame view

On 12 May 2006 07:53:33 -0700, "sal" <SpamMeHere@xxxxxxxxxx> wrote:

[I already sent this once but my news server seems to have eaten it.
So I'm sending it again, via Google....]

On Fri, 12 May 2006 13:50:29 +0000, David wrote:

On Wed, 10 May 2006 08:52:37 -0400, sal <pragmatist@xxxxxxxxxx> wrote:

On Tue, 09 May 2006 13:49:53 +0000, David wrote:

On Mon, 08 May 2006 12:20:42 -0400, sal <pragmatist@xxxxxxxxxx> wrote:

I've been away from the newsgroup, just saw this post today.

On Sun, 23 Apr 2006 15:49:46 +0000, David wrote:

On Thu, 20 Apr 2006 22:19:35 -0400, sal <pragmatist@xxxxxxxxxx>
wrote:

On Thu, 20 Apr 2006 03:23:58 +0000, David wrote:

On Wed, 19 Apr 2006 22:16:45 -0400, sal <pragmatist@xxxxxxxxxx>
wrote:

On Tue, 18 Apr 2006 14:07:07 +0000, David wrote:

On 17 Apr 2006 13:10:24 -0700, "sal" <SpamMeHere@xxxxxxxxxx>
wrote:

David wrote:
[ snip ]

The rotating cylinder problem is easy to visualize. I don't
need an extensive math explanation just some simple verbage.

<g> OK, then let me ask you a simple question about it.

Let's suppose the cylinder lies on the X axis.

So, if the wire runs straight from one disk to the other, we would
say that it makes an angle of 0 degrees with the X axis.

If, on the other hand, it is in a circle around the cylinder --
not a spiral at all, just a single loop -- then it makes an angle
of 90 degrees with the X axis.

_IF_ the wire is to appear _STRAIGHT_ in the moving frame, what's
the _MAXIMUM_ angle it can make with the X axis?

Obviously, it can't cross the X axis at 90 degrees, since in that
case it's just a circle, not a spiral. But can it cross the X
axis at angles arbitrarily close to 90 degrees? In other words,
can it form a really _tight_ spiral? Or is there some limit to
the "tightness" of the spiral it can make, and still appear to be
completely straight in the moving frame?

Please post your answer. :-)

If the wire was arbitrarily thin and the cylinder's diameter was
arbitrarily small, and the rotation rate of the cylinder was
arbitrarily high then the angles can get arbitrarily close to 90
degrees.

Then prove it.

Pick a velocity, pick a rotation rate, write down the parameters,
run the transformations, and show that there is a case where the
wire can be wrapped such that it makes and angle of, say, 89 degrees
in the center of mass frame but appears to make an angle of 0
degrees in the moving frame.

If you're right, this should be easy.

In the velocity between the moving and rest frame (the cylinder
frame) be V =0.866c. Let the length of the cylinder be L= 1
light-second/0.866 or L = 1.1547 light-seconds = 3.46 *10**8 meters.

Call the frames "M" (in which the cylinder is moving to the right at
velocity V) and frame "S" (in which the COM of the cylinder is
stationary).

OK, so if you want the time difference to be 1 second in frame "S"
then you must mean the length should be 1.1547 in frame "M".


The rod is attached to the cylinder simultaneously in the moving
frame and 1 second apart as measured in the rest frame.

If the cylinder is moving to the right in frame "M" at V, L=1.1547 in
frame "M", and the front of the cylinder is at the origin at time 0
when the rod is attached, then the coordinates of the two events are

E1 = (t=0,x=0)

E2 = (t=0, x=-(1/v)) = (t=0, x = -1.1547)

Transforming to frame "S", which is moving to the right, we get
coordinates

E1' = (t=0,x=0)

E2' = (t=1, x=-2.309)

Time difference is indeed 1 second.

But note: With your assumptions, the length of the cylinder in the
stationary frame is 2.309 LS.

Let the diameter of the cylinder be 1/3.14 millimeters.

OK, that's 1/pi mm = 1/(1000pi) meters. But we want it in
light-seconds; so we have

diam = 1.06 * 10^-12 LS

and

Circumference = 3.33 * 10^-12 LS

Let the surface speed of the rotating cylinder be close to the speed
of light say 1 * 10**8 meters/second.

This is in the "stationary" frame, right? Let's call the tangential
velocity W. Then we have

W = c/3 = 0.333 (if c=1).

But ... You seem to be assuming you can set the rotational velocity to
an arbitrary value in the stationary frame.

You CAN'T do that.

You have constrained the possible velocities by assuming that the wire
will appear straight in the moving frame; now you're going into a
different frame -- the stationary frame -- and picking arbitrary
values for its motion, and just _assuming_ that your original
constraint will still be satisfied. That's going to lead to
contradictions.

For any particular value of V there will be a _range_ of possible
rotational velocities which can be observed in the stationary frame,
and which still satisfy the constraint.

Don't believe me? Then take the parameters you've given and work out
exactly how the wire will look in the moving frame. It's not going to
appear straight. I'll show that, below.


In one second this cylinder makes about 10**10 revolutions.

Actually, that's W/circumference , or

0.333 / 3.33 * 10^-12

or about 10^11 revolutions.


So when a steady-state conditiion is reached there will be about
10**10 wrappings of the attached rod for every 3.46*10**8 meters

Say what? Length is 2.3 meters, there are 10^11 turns, which comes to
about 4.3 * 10^10 turns per meter.


or about 29 wrappings per meter which approaches the values you
wanted demonstrated.

I asked you to determine the _angle_ the wire made with a line
parallel to the X axis. You didn't do that. But you also have
contradictory assumptions here.

Let's just unwind one turn. Its ends are separated by an X distance
of

dX = 1/(4.3*10^10) meters

= 7.75 * 10^-20 LS

and its perpendicular length is

dY = 1/1000 meters

= 3.33 * 10^-12 LS

So the _angle_ it makes with the X axis is

atan(dY/dX) = atan((4.3*10^10)/1000)

= 1.57 radians

That's nice and close to 90 degrees, of course. BUT...

Let's check the constraint by looking at the motion of our
straightened out "turn" of wire, and transform it back to the moving
frame. At time tau=0 in frame "M", if E1' is event at the "front" end
of the piece of wire and E2' is the event at the "back" end,

E1' = tau=0, x'=0, y'=0

E2' = tau=0, x'=-dX, y'=dY
I guess I don't understand these equations. Help me one small step at
a time. The "M" frame is the moving frame. The prime coordinates
are the coordinates in the moving frame. The front end of the wire as
measured in the moving frame is E1' and the back end of the wire is
E2'. Now in the problem, I stated that the moving observer at time
tau places the wire on the surface of the cylinder on a line parallel
to the x-axis. That's the scenario, the given information. In the
E1' equation you have y'=0 at time tau and in the E2' equation you
have y'=dY at time tau. I interpret this to mean your equations are
saying the moving frame observer did not put the two ends of the wire
onto the surface of the cylinder along a line that is parallel to the
x-axis. If he did both y' coordinates would have to be the same value
at time tau (otherwise the wire would not be parallel the x-axis). If
the cylinder wasn't rotating, when he put the wire onto the surface of
the cylinder, the y coordinates would have to be the same at time tau
for the wire to be parallel to the x-axis. I don't follow what you
are doing.
David

And a little later, at the moment when the "back" end arrives at the
same Y coordinate that the "front" end started at (and so lies on a
"straight line" lying along the cylinder),

E3' = tau=dY/W, x'=-dX, y'=0

In frame "M" we are interested in the coordinates of events E1 and E3.
They must occur at the _same_ _time_ in the moving frame, if the wire
is to appear straight. In that frame, we have

E1 = t=0, x=0

E3 = t=gamma*(dY/W + V*dX), x=(we don't care about x)

If the wire is straight in frame "M" then we must have

time(E3) = time(E1) = 0.

Then we must have

dY/W = -V*dX

dY = -V*W*dX

3.33*10^-12 = 0.866 * 0.333 * 7.75*10^-20

3.33*10^-12 = 2.23 * 10^-20

and it doesn't add up. The constraint was violated; you picked an
impossible rotational velocity for the cylinder as viewed from the
stationary frame.

Forget about the moving frame for a moment. If I have a cylinder
aligned along the x-axis, why can't the surface rotate at a speed
close to c. What do you think is the maximum rotation rate allowed?

Go back to the beginning and read what I said, slower this time.

You can have the cylinder rotate at any speed you want, obviously.
HOWEVER, if you do that, the wire attached to it won't necessarily
appear to be STRAIGHT in the moving frame.

I do not follow what you are saying.

Of course you don't, you skipped all the math.

Go back and read what I wrote, slowly, line by line, and ask about the
specific points where you didn't follow it, and I'll be happy to try
to explain the details in more detail, so to speak. I didn't snip it;
it's all still there, up above, in this message.

Just waving at the whole thing and saying "I do not follow..." doesn't
do anyone any good.


In the stationary frame, let the cylinder aligned on the x-axis rotate
with a surface velocity close to c, and let the length of the cylinder
be very long. Instead of a wire, let's just say the moving frame
observers paint a line on the cylinder. At time t0 as measured in their
frame, they paint a straight line parallel to the x-axis by painting
each point of the surface of they cylinder along the x-axis that is at
the 12 o'clock position. I don't understand why you think these moving
observers say that this is not a straight line.

You missed the point; you apparently skipped everything I wrote that
contained equals signs.

Here's the point again, read it slowly, then go back and look at what I
wrote before:

Starting with the moving-frame picture, as you have just done, figure
out
HOW TIGHT the spiral appears to be in the STATIONARY frame.

Got that?

Then, figure out the very TIGHTEST the spiral can appear to the
stationary-frame observers, given that it appears straight to the
moving-frame observers, with any combination of interframe velocity
and rotation rate in the moving frame you like.

Once you've done that, you can claim to have some understanding of the
statement of the problem. Right now, you're still in the position of
claiming the problem is obvious without the use of math, which is
patently false, since you apparently still don't understand the
relationship between the stationary and moving frames in this
case. Your every response makes that more clear. You need to work
through the transformations to understand even the most rudimentary
aspects of this problem and you still haven't done that.

[ snip ]
.

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