Re: Circular motion in SR

On Mar 15, 9:55 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Mar 15, 7:29 pm, rbwinn <rbwi...@xxxxxxxx> wrote:





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.

We don't care.

[snip]- Hide quoted text -

- Show quoted text -

Eric,
It is just very difficult to believe that you don't care when
you go to all of the trouble to make a post that says, We don't care.
If you really didn't care, you would not respond at all. I think it
really bothers you to think that an altimeter on a weather baloon
flying at the same altitude as a supersonic jet would read the same as
an altimeter in the jet even though your mathematics shows that there
is a distance contraction in the frame of reference of the jet.
Either that, or you are bothered by pi. Probably you are bothered by
pi. Einstein certainly was.
Galileo was a scientist who was not as bothered by some of these
things. I certainly like the way his equations predict that a moving
clock would run slower than a clock that is stationary. Well, keep us
informed from time to time about what is happening in the world of
science.
Robert B. Winn
.

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