DaleSpam said:
OK, I am not working today so I decided to start this. Here is an equation for the worldline of a particle on a rotating hoop in the reference frame where the axis of rotation is stationary.
s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)
in units where c=1 and r=1
Which can be differentiated to obtain the four-velocity
u=\left(\frac{1}{\sqrt{1-\omega ^2}},-\frac{\omega \sin (\phi +t<br />
\omega )}{\sqrt{1-\omega ^2}},\frac{\omega \cos (\phi +t \omega<br />
)}{\sqrt{1-\omega ^2}},0\right)
Which can be differentiated to obtain the four-acceleration
a=\left(0,-\frac{\omega ^2 \cos (\phi +t \omega )}{1-\omega<br />
^2},-\frac{\omega ^2 \sin (\phi +t \omega )}{1-\omega ^2},0\right)
A quick check shows u.u=1 and u.a=0
I then transformed the hoop into the frame where it rolls without friction
s'=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br />
^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega<br />
^2}},\sin (\phi +t \omega ),0\right)
and then differentiated to obtain u' and a'. I verified that u'=L.u and a'=L.a and that u'.u'=1 and u'.a'=0. So, kinematically everything works out just fine. I didn't actually work out the four-forces nor the tension on a differential element.
So then, to answer the question about the shape of the hoop in the primed frame I need to look at all of the points for a single value of t'.
t'=\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br />
^2}}
I could not solve this equation for t as described above, but I could solve it for phi to obtain
\phi =-t \omega \pm \cos ^{-1}\left(\frac{t' \sqrt{-(\omega -1)<br />
(\omega +1)}-t}{\omega }\right)
Substituting this back into the expression for s' gives an expression for the shape of the hoop at a given t' in the frame where it is rolling
\left(t',\frac{t'-t \sqrt{1-\omega ^2}}{\omega<br />
},\pm \sqrt{1-\frac{\left(t-t' \sqrt{1-\omega<br />
^2}\right)^2}{\omega ^2}},0\right)
I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.
It is not clear to me how Dalespam was able to plot the last equation that contains both t' and t when he has stated it is not possible to solve for t in terms of t'.
Let's take the equation he derived that equation from and insert some numerical values.
Say time = 0 in both frames and and R=1 and w=0.8c in the rest frame.
We take the the positions of 4 elements a,b,c and d at the 4 points of the compass where phi = 0, 90, 180 and 270 degrees. Now insert these values into the equation for the rest frame (s):
s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)
and we get (x,y) coordinates a=(1,0), b=(0,1), c=(-1,0) and d=(0-1) in frame s.
Now if we insert the same values into the transformed equation:
s \single-quote=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\sin (\phi +t \omega ),0\right)
we get (x',y') coordinates a'=(1.67,0), b'=(0,1), c'=(-1.67,0) and d'=(0,-1) in frame s'.
It can be seen that the transformed ball is wider than it is high by this equation which is a false result. This is due to the "catch 22" situation of not being able to calculate phi without knowing how the proper time of each element transforms and it is not possible to calculate how the the proper time of each element transforms without knowing how phi transforms. On the face of it, the equation is not taking the simultanity issues into account between the transformed frames.
This can be seen when we do the calculations for the transformed time and space coordinates (t',x',y') to get a'=(1.33,1.67,0), b'=(0,0,1), c'=(-1.33,-1.67,0) and d'=(0,0,-1) in frame s'. The time t' is not simultaneous in the s' frame. I do not think Dalespams equation are wrong, but it does not seem that they resolved the issues of how the dimensions of the ball would be measured simultaneously in the transformed frame without doing numerical aproximations.
