I Simplify the Dirac Energy Equation?

[neilparker62]

In the above equation for Dirac energy, is it trivial to note that given:

Principal quantum number n
Orbital angular momentum quantum number l(max) = n - 1
Total angular momentum quantum number j = l + 1/2 = n - 1/2

Then n_r = n - j - 1/2 = n - (n - 1/2) - 1/2 = 0 and the energy expression simplifies considerably to:

$E=mc^2 \sqrt{1-{\left(\frac{Z\alpha}{n}\right)}^2}$

- an expression in which only the principal quantum number is in evidence ?

 

[neilparker62]

If in the above expression, we take Z=1 (Hydrogen) and n=1 (ground state), the equation becomes:

$E=mc^2\sqrt{1-{\alpha}^2}$

Noting in passing that a one term binomial expansion of the above yields the conventional Rydberg energy, we can also write the equation as:

$E=\sqrt{\left({mc^2}\right)^2-{\left(m\left({\sqrt{\alpha}c}\right)^2\right)}^2}=\sqrt{\left({mc^2}\right)^2-{\left(mv^2\right)}^2}$ where $v=\sqrt{\alpha}c$

Alternatively $mc^2=\sqrt{E^2+(mv^2)^2}$

indicating some kind of Pythagorean relationship between ground state energy E and rest mass energy $mc^2$

Does this last form of the equation have a physical interpretation ?

 

[neilparker62]

The equation written in the above form is intriguingly similar to Einstein's equation for total energy:

$E=\sqrt{(mc^2)^2+(pc)^2}$

Except that 'total energy' in the aforementioned is $mc^2$ and it appears to have two 'components':

$mc^2=\sqrt{(pc)^2+(mv^2)^2}$

The question remains as to the physical interpretation of the above ?