The Relativistic Correction

Moving from the non-relativistic formula for the energy of an electron to the relativistic formula we make the change

\begin{displaymath}\bgroup\color{black} mc^2+{p^2_e\over {2m}}\rightarrow\left(p...
... mc^2\left( 1 + {p^2 c^2 \over {m^2 c^4}}\right)^{1/2}. \egroup\end{displaymath}

Taylor expanding the square root around \bgroup\color{black}$p^2=0$\egroup, we find

\begin{displaymath}\bgroup\color{black}\left(p^2 c^2 + m^2 c^4\right)^{1/2} = mc...
...pprox mc^2 + {p^2 \over {2m}} - {p^4 \over {8 m^3 c^2}} \egroup\end{displaymath}

So we have our next order correction term. Notice that \bgroup\color{black}${p^2\over 2m}$\egroup was just the lowest order correction to \bgroup\color{black}$mc^2$\egroup.

What about the ``reduced mass problem''? The proton is very non-relativistic so only the electron term is important and the reduced mass is very close to the electron mass. We can therefore neglect the small correction to the small correction and use

\begin{displaymath}\bgroup\color{black} H_1=-{1\over 8}{p^4_e \over {m^3 c^2}}.\egroup\end{displaymath}

Jim Branson 2013-04-22