- Rest Mass ? -

[Only registered users see links. ] \(formerly\) (dlzc1.cox@net), in article nntp:/<D1zQa.7165$u51.2965@fed1read05> , wrote:
quoted Dear Jeff Relf:


Because, on average, neither the billiard ball nor its contents
are moving relative to your frame of reference. The same would be true if
the thing were spinning, and these wave-particles you speak of don't
escape certain "orbits" of brownian motion and nuclear spin.

A question on a tangent to yours is whether a spinning billiard
ball would gain mass according to its rate of spin or if it would gain
mass according to its temperature.

"''" <brewhaha@ecn.ab.ca> wrote in message
news:f4PQa.3007$Fy1.136496@localhost...
nntp:/<D1zQa.7165$u51.2965@fed1read05> , wrote:

A spinning ball has more kinetic energy than a ball at rest. So yes. It
would weight more i.e. it's mass increases

Pmb

Dear Pmb:

"Pmb" <[Only registered users see links. ]> wrote in message
news:IkSQa.2985$[Only registered users see links. ]...
....

Its RELATIVISTIC mass increases. Its RELATIVISTIC mass increases for both
increasing spin and increasing temperature.

David A. Smith

"dlzc@aol.com (formerly)" <dlzc1.cox@net> wrote in message
newsjTQa.8471$u51.3725@fed1read05...

In this case the rest mass increases as well. Think of what "rest mass"
means. It's the mass as measured in the frame of referance in which the
total momentum is zero. Think of a top which is at rest and not spinning in
frame S. The mass is m_o. Now start spinning the top in a way such that the
total momentum is zero. Then in S the total energy has increased and
therefore the total mass in this frame - i.e. the total rest mass has
increase. The mass went from m_o to m'_o > m_o.

Pmb

Pmb wrote:

And the formulation is?

"Sam Wormley" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
in
the

Let E_o = total energy in the rest frame. Then the rest mass M_o will be

M_o = E_o/c^2

The additional mass comes from the kinetic energy of all the parts of the
body

Pmb

Pmb <[Only registered users see links. ]> wrote in message
news:UpWQa.3767$[Only registered users see links. ]...
mass"
the
spinning
that

For once, Pmb is correct, but what is the formula? For mall
rotation rate, Omega, and angular moment of inertia, I0, it might be

M0*c^2 = m0*c^2 + (1 / 2) I0*Omega^2

But what is the relativistic formula? [Old Man]

"Old Man" <[Only registered users see links. ]> wrote in message news:3f13035f_1@newsfeed...
has
the

I'm always correct. :-) Especially since I never take a stand here on
anything that I'm not 100% certain of.



Very complex. Not sure if there is a simple formula.

If it was as easy as just adding up the kinetic energies of all the pieces
then that would be it. However to keep the pieces attached to the body (i.e.
hold particles at a fixed distance from the center of rotatio) one has to
force them to follow a circular path. That means that a rotating body is
under stress - and that stress contributes to the energy in the zero
momentum frame.

Pmb

"Pmb" <[Only registered users see links. ]> wrote in message news:<yVYQa.106$[Only registered users see links. ]>...

Hilarious. What is 100% dertain is the brown probability amplitude for
being wrong is closer to unity than zero.

Pmb wrote :
" Let E_o = total energy in the rest frame .
Then the rest mass M_o will be M_o = E_o / c^2
The additional mass comes from the kinetic energy
of all the parts of the body .


Cool !

This is easy to understand ,
add spin or heat and they rest mass goes up by E / c^2 .