# Kinetic Energy Relativistic Correction

When trying to add relativistic corrections to the standard Schrödinger equation, the kinetic energy correction is given as $$H'=-\frac{p^4}{8m^3c^2}$$ (e.g. page 14 https://www.tcd.ie/Physics/people/Peter.Gallagher/lectures/js_atomic/hydrogen-essay.pdf). It is usually stated that this can be found by expanding the relativistic kinetic energy formula $$KE=(\gamma-1)mc^2$$ as an infinite series.

Though when I do so (expanding $$(1-x)^{-\frac{1}{2}}$$ about $$x=0$$ as a Taylor series) I get the third term of the expansion as $$\frac{3p^4}{8m^3c^2}$$ (e.g. http://www.sjsu.edu/faculty/watkins/relamech.htm). Why is there a difference? Is a different expansion method used to generate the former of the two versions?

• Thanks for the question; for what it's worth, I think I would have given myself about a 50:50 chance of tripping up and calculating as in your second paragraph if I were having a bad hair day. – Selene Routley Jun 23 '16 at 2:36

The difference arises because you are confusing the three spatial components of the 4-momentum, which are in fact $p^j = \gamma\,m\,v^j$ with the three components of the Newtonian momentum $m\,v^j$. If you write:

$$\frac{p^2}{m^2\,\gamma^2} = v^2$$

and solve for $v$ in terms of $p$, then substitute back into the Watkin's lecture notes, you'll get your factor of $\frac{1}{8}$ in the quartic term. That's really messy. It's much easier to begin with the equation for the momentum four-vector's square norm, which I always remember as $\frac{E^2}{c^2} - p^2= m^2\,c^2$, then write $E = T + m\,c^2$ and solve for $T$ in terms of $p$.

Effectively what you've done is to use the wrong power of $\gamma$ in your expressions as a result of what seems to be your confusion.

The right (and messy) way to derive the same expression given in the Watkin's lectures and using $T = (\gamma-1)\,m\,c^2$ is most succinctly described by the following Mathematica statement:

Series[((1/Sqrt[1 - (v/c)^2]) - 1) m c^2 /. Solve[p == m v/Sqrt[1 - (v/c)^2], v], {p, 0, 4}] // Normal

in a friendly conversation with Ms Mathematica Kernel, to which she obligingly replies:

$$\left\{\frac{p^2}{2 m}-\frac{p^4}{8 c^2 m^3},\frac{p^2}{2 m}-\frac{p^4}{8 c^2 m^3}\right\}$$

The expansion is not that hard. We have $$K~=~(\gamma~-~1)mc^2~=~\left(\frac{1}{\sqrt{1~-~v^2/c^2}}~-~1\right)mc^2$$ where we do the binomial expansion on the Lorentz factor $$K~\simeq~\left(1~+~\frac{1}{2}\frac{v^2}{c^2}~+~\frac{3}{8}\frac{v^4}{c^4}~-~1\right)mc^2$$ $$=~\frac{1}{2}mv^2~+~\frac{3}{8}m\frac{v^4}{c^2}.$$

I do not know how you get the $$m^3$$ in the denominator. That looks rather odd.