Date sön 06 januari 2019

There is a claim in Randell Mills GUTCP that because the precession of the loops, the total angular momentum is twice the spinn momentum for a single loop. Let's try to analyze this.

First of all if each loop precesses and we add up all paths we would still get a constant loop distribution. So the charge is still a constant value on the path.

Now over to the question of what we should expect when it comes to total angular momentum. Note that from wikipedia (Precession)[] we would expect the movement to be torque free and hence if \(\alpha\) is the angel between the total angular momentum and the symmetry line we have

$$ L_p = \frac{L_s}{cos(\alpha)} $$

Now take the angel \(\beta\) between the precession axis and the symmetry axis then we know from the fact that the sinus parts onto the total angular mumentum take each other out we must have

$$ sin(\alpha) = \frac{sin(\beta - \alpha)}{cos(\alpha)}, $$


$$ sin(\alpha)cos(\alpha) = sin(\beta - \alpha). $$

The magnitude of the total angular momentum is

$$ Total(\alpha) = L_s f(\alpha), $$


$$ f(\alpha) = cos(\alpha) + \frac{cos(\beta-\alpha)}{cos(\alpha)} $$

using the constraint between \(\alpha,\beta\) this can be rewritten as

$$ f(\alpha)) = cos(\alpha) + \sqrt{\frac{1}{cos(\alpha)^2}+cos(\alpha)^2 - 1} $$

or with \(f(\alpha) = g (cos(\alpha))\)

$$ g(x) = x + \sqrt{1/x^2+x^2-1}. $$

Lets find the extreme value of this, taking the derivative lead to

$$ g'(x) = 1 + \frac{-1/x^3 + x}{\sqrt{\cdots}} $$

and putting \(g'(x) = 0\) and simplify and squered we get

$$ 1 + 1/x^6-3/x^2 = 0 $$

This can be solved via a third order equations, the interesting solution, \(x_0\) is

$$ x_0 = cos(\alpha_0) = 0.80790 $$


$$ f(\alpha_0) = 1.8964 $$

These numbers are not close to the ones claimed by Mills. Not sure where the misstake is here but it is close and I find the calculation interesting. The extreme point is a minimal e.g. you will get minimal total angular momentum for a specific spinn angular momentum. Perhaps we should minimize energy in stead.


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