Space Forum / Astronomy / June 2008

Pentcho, be careful not to run out of question marks.
Here are some, free for you:
??????????????????????????????????????????

w.

http://www.bartleby.com/173/23.html
Albert Einstein: "An observer who is sitting eccentrically on the disc
K' is sensible of a force which acts outwards in a radial direction,
and which would be interpreted as an effect of inertia (centrifugal
force) by an observer who was at rest with respect to the original
reference-body K. But the observer on the disc may regard his disc as
a reference-body which is “at rest”; on the basis of the general
principle of relativity he is justified in doing this. The force
acting on himself, and in fact on all other bodies which are at rest
relative to the disc, he regards as the effect of a gravitational
field."

By increasing the diameter of the disk while keeping the linear speed
of the periphery constant, the "gravitational field" could be reduced
to zero and the observer who is sitting eccentrically on the disc K'
becomes virtually inertial. Then he looks at non-rotating clocks
located outside the disk but very close to the rotating periphery and
sees them, in accordance with Divine Albert's Divine Special
relativity, running:

(A) faster than his clocks (which are "at rest relative to the disc")?

(B) slower than his clocks (which are "at rest relative to the disc")?

Pentcho Valev
pvalev@yahoo.com

Intéressant c'est à creuser sérieusement ...

MA

news:e7d59c5e-a5eb-4058-9cba-3ad6199a89b9@l42g2000hsc.googlegroups.com...

With all due respect, you cannot ever reach 'zero' angular acceleration on
disk 'K' no matter how large it is.
Vince

[snip]

Hey f.cking stooopid - spin up a 78 rpm vinyl record.  Gently balance
opposite poles of a ping pong ball between two straight pins.  Spin up
the ball by touching it to the record's surface with the pins' axis
parallel to both the record's radius and the record's surface.  After
the ball spins up separate the pins so the spinning ball falls onto
the record.  It simply sits there in the lab frame, not moving.  Works
with disks, too, and wheels on axles.

This is a trivial empirical dispoof of centripetal force.  If
centipetal force existed the ball or disk would be thrown radially
outward in your stoooopid world.  You must realize that it is not the
centrifuge that impresses acceleration, it is yourself.

   There is a degree of freedom here you didn't mention. You have to
specify the relative velocity between him and whatever nonrotating
clock is located outside the disc. The clock that is outside the disc
is not rotating, but it is traveling at an arbitrary velocity relative
to the observer on the disc. There is a Lorentz time dilation factor
associated with the relative velocity, gamma, where,
gamma=sqrt(1-(v_1/c)^2)
where v_1 is the relative velocity between the two clocks and c is the
speed of light.
This time dilation factor multiplies the gravitational time dilation
seen by the observer on the disc. Thus there is time contraction
factor associated with his centripetal acceleration. In terms of
gravity, it increases with the gravitational potential relative to the
nonrotating observer. I said gravitational potential, not
gravitational field. Therefore, there will be a time contraction
factor, call it phi, that speeds up the rate he sees for clocks
outside the disc. I think in the case that both observers are on the
same radial "spoke" of the wheel,
phi=1+ax/c^2
where a is the acceleration due to the rotation and x is the distance
to the nonrotating object. As a first order approximation, i.e.,
ax<<c^2,
we can use the Newtonian formula for centripetal acceleration,
a=v_2 ^2/r
where v_2 is the linear velocity of the edge of the disc relative to
the center of the disk and r is his distance from the center of the
disc..
    Your confusion stems from the fact that v_1 and v_2 do not have
to be the same. The nonrotating observer does not have to be at rest
relative to the center of the disc.
   However, lets look at that special case. I conjecture that you may
have had that in mind, anyway. Let us look at the case where
v_1=v_2=v.
Then the total time factor is the product.
total=(1+ax/c^2)sqrt(1-(v/c)^2)
Substituting v^2/r for a,
total=(1+xv^2/rc^2)sqrt(1-(v/c)^2)
  If total>1, he sees the outside clock speed up. If total<1, he sees
the outside clock slow down. We can see that it is really determined
by v and the ratio x/r. If x/r is large enough, he sees the clock
speed up. If x/r is too small, he sees the clock slow down.