Re: Circular motion in SR

On Mar 14, 9:38 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Mar 14, 7:03 pm, rbwinn <rbwi...@xxxxxxxx> wrote:





On Mar 14, 5:29�pm, mL <mL.bey...@xxxxxxxxxxxxx> wrote:

ram.rac...@xxxxxxxxx wrote:
I'm trying to write the equations for circular motion according to SR
laws of motion. I'm doing some kind of mistakes, because I'm not
getting real solutions to the equations.

This is what's going on: You have an body of mass m0 circling around a
stationary center point that is a distance of r from the body. There
is a force F that attracts the body towards the center point, making
it move in a circle around it. What's the velocity v of the object?

OK, you have a central force field, �F_ = f(r)n_ , where the unit
vector n_ points towards a fix point.

Use the equation dE/dt = F_.v_, to show that the energy E and the speed
v = |v_| are constants when the particle moves in a circle r = R.

�> I've used these equations:
�>
�> v=sqrt(a*r)
�> a=F/(m0*gamma^3)
�> gamma=1/sqrt(1-(v/c)^2)
�>
�> I keep getting only imaginary solutions for v. What am I doing wrong?

The force-equation is wrong. For cases with constant rest mass m, you
have in general,

� �F_ �= (gamma)ma_ + (gamma)^3 m(v_.a_)v_ /c^2 ,

which for circular motion, r = R, reduces to

� �f(R) = (gamma)m a, �where �a = v^2/R,

- note that v_.a_ = 0, i.e. a_ // n_, according to comments above.

/mel

My question about all of this is, since a clock in rotation supposedly
runs slower than a clock around which the moving clock is rotating,
then since the Lorentz equations give the velocity as being the same
in both frames of reference, does the altitude of the rotating clock
differ from one frame of reference to the other?
     This is not a complicated question.  Scientists have claimed that
they put a clock in a satellite and recovered it, and the clock from
the satellite showed less time than an identical clock on earth, so
this proved that Einstein's theory was true.
    Any way you look at it, the velocity of the moving clock is the
same from either frame of reference according to Einstein's theory.
So with the distance contraction, the circumfrence of the orbit is
less from the frame of reference of the satellite than from the frame
of reference of the earth.  How does a shorter orbit relate to the
altitude of the satellite?
     I have posted this question many times, and, so far, science
seems to be silent about this, just as they are silent about many
things with regard to Dr. Einstein's famous theory.
Robert B. Winn

Science has no problem with the concepts. Your whining is misguided -
USENET seems to be silent with these things because USENET is tired of
explaining things to you just to have the explanation ignored. USENET
has figured out that decades of stupidity cannot be fixed.- Hide quoted text -

- Show quoted text -

Eric,
Here is what I think. I think that the altitude of the
satellite is the same in both frames of reference. The reason for
this is that the circumfrence of the orbit is the same in both frames
of reference. The way we calculate this is by the equation, C=2(pi)r,
where C is the circumfrence of the orbit. The secret to having the
circumfrence the same in both frames of reference is the equation
t'=t, which is from the Galillean transformation equations. So what
about the clock being slower in the satellite?
Well, the time on that clock is not t'. It is n' where n'=t(1-v/w),
and w is the velocity of light.
Notice that there is no distance contraction, the altitude of
the satellite is the same in both frames of reference, and the clock
in the satellite is running slower than the clock on earth, and the
value of pi remains 3.14 in both frames of reference. Scientists are
not going to like this.
Robert B. Winn
.

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