Mass from interactions

The total energy U is given by the sum of the energies of the particles, plus the potential energy stored between them. Thus, $\displaystyle U = E_1 + E_2 + \frac{\alpha e^{\beta r}}{\sqrt{r}}$ . To simplify the problem, it can be assumed that $E_1 = E_2 = E$ since the particles are identical, meaning that any asymmetry can be flipped without changing the system. Therefore, one can expect on average, identical energies. And since $E = |\vec{p}| c$ , $\displaystyle U = 2 |\vec{p}| c + \frac{\alpha e^{\beta r}}{\sqrt{r}}$

It is also known that the particles are in circular motion collinear with the center. This means that the force associated with the potential $V(r)$ must be the reason for the change in the momentum vector $|\vec{p}|$ of a particle. We therefore have $\displaystyle \frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = - \nabla V(r)$ . Taking $\vec{p} = |\vec{p}| \vec{\theta}$ and noting that $|\vec{p}|$ is constant, we have $\displaystyle - |\vec{p}| \dot{\theta} = - \alpha \frac{e^{\beta r}}{\sqrt{r}} \left( \beta - \frac{1}{2r} \right)$ . Letting $r$ be the diameter of the circle means that $\displaystyle \dot{\theta} = \frac{2 c}{r}$ . Putting everything together yields $\displaystyle 2 |\vec{p}| c = \alpha e^{\beta r} \sqrt{r} \left( \beta - \frac{1}{2r} \right)$ .

This relation can be substituted into the total energy, which can then be minimized by finding the critical points. In this case, there is only one critical point with $r$ positive, which is $\displaystyle r = \frac{\sqrt{2} - 1}{2 \beta}$ . To see this is a minimum, note that the total energy tends to $+ \infty$ as $r \rightarrow 0$ and $\rightarrow + \infty$ .

Finally, after substituting that in and calculating for the mass, one obtains

$\displaystyle m_0 = \frac{\alpha \sqrt{e}^{\sqrt{2}-1}}{c^2 \sqrt{2}} \sqrt{\frac{2 \beta}{\sqrt{2} - 1}} \approx 671.569 \mathrm{ng}$

From the interacting potential, one gets the equation for the interacting force $F(r)=-\frac{\alpha e^{\beta r}(2 \beta r - 1)}{2r^{3/2}}$

Since the particles (call them H particles) will move in a circular orbit of radius $r$ with angular velocity $\omega=2c/r$ and momentum $p$ ; from the force equation $\vec{F}=d\vec{p}/dt$ , one gets (with $E$ is the energy of each particle):

$\frac{\alpha e^{\beta r}(2 \beta r - 1)}{2r^{3/2}}=p\omega=\frac{2pc}{r}=\frac{2E}{r} \Rightarrow 2E=\frac{\alpha e^{\beta r}(2 \beta r - 1)}{2r^{1/2}}$

So the total energy of the 2-particle bound state is:

$E_{2H}=2E+V=\frac{\alpha e^{\beta r}(2 \beta r - 1)}{2r^{1/2}} + \frac{\alpha e^{\beta r}}{r^{1/2}}=\frac{\alpha e^{\beta r}(2 \beta r+1)}{2r^{1/2}}$

$E_{2H}$ is minimum when $4 \beta^2 r^2 + 4 \beta r- 1= 0$ (we take derivative of $E_{2H}$ with respect to $r$ then let it equal to $0$ ), so $r_{min}=\frac{\sqrt{2}-1}{2\beta}$ :

$E_{2Hmin}=\frac{\alpha e^{\beta r}(2 \beta r+1)}{2r^{1/2}}=\frac{\alpha e^{\frac{\sqrt{2}-1}{2}}}{\sqrt{\frac{\sqrt{2}-1}{\beta}}}$

From that, one get the mass of the 2-particle bound state:

$m_{2H}=\frac{E}{c^2}=\frac{\alpha e^{\frac{\sqrt{2}-1}{2}}}{c^2\sqrt{\frac{\sqrt{2}-1}{\beta}}}=6.71 \times 10^{-10}~\mbox{kg} = 671~\mbox{ng}$

Alternatively, one could apply the relativistic virial theorem for very relativistic particles to derive the energy and get the same result.