Fasten your belts, part II

According to Einstein

$\frac{\Delta t'}{\Delta t_{0}} = \frac{1}{\sqrt{1- \frac{v^{2}}{c^{2}}}} \approx 1 + \frac{v^{2}}{2c^{2}}$

if $v << c$ . As $v = \omega r << c$ , we'll be able to use this approximation. Now, of course $\Delta t' = \Delta t(r)$ must be a function of $r$ , and $\Delta t_{0} = \Delta t(0)$ . Then the above equation becomes

$\frac{\Delta t'}{\Delta t_{0}} \approx 1 + \frac{(\omega r)^{2}}{2c^{2}}$

Now

$\left | \overrightarrow{\mathrm{F}} \right | = \frac{mv^{2}}{r} = m \omega^{2} r = \frac{\partial }{\partial r} V(r) = m \frac{\partial }{\partial r} \varphi(r)$

where $V(r)$ is the potential energy associated with the force (and all forces have a "gravitational potential field" associated with it, according to Einstein). Solving for $\varphi(r)$ will give us

$\varphi(r) = \frac{(\omega r)^{2}}{2}$

And, of course,

$\frac{\Delta t'}{\Delta t_{0}} \approx 1 + \frac{\varphi(r)}{c^{2}}$

Therefore $A + B = 2$ .