# Power Series Expansion in Inverse Powers

In a book I am reading at the moment it says that the expression $$H = \sqrt{P^2 c^2 + m^2 c^4}$$ can be "formally expanded in inverse powers of $c$" to obtain $$H = mc^2 + \frac{1}{2m} P^2 + \dots$$ I can get this result in Wolfram Mathematica using the 'Series' function and expanding about $c = \infty$ as in this answer. However, I can not arrive at this result my self, as when I take the Taylor expansion I just get something that goes off to infinity. If someone could explain how to get this result without software I would be very thankful.

• @DavideMorgante Fair enough, I've removed it. Seeing as I thought Laurent series and other objects in complex analysis might have something to do with this I added it. Aug 17 '18 at 8:16
• Ahah nothing to do with Laurent series! I'll post an answer Aug 17 '18 at 8:16
• Rewrite as $mc^2\sqrt{1+x}$ and expand with Taylor or the generalized binomial theorem.
– user65203
Aug 17 '18 at 8:32

With physical concepts of energy and mass for both positive and negative mean, you need to know binomial expansion $$(1+x)^r=1+rx+\dfrac{r(r-1)}{2}x^2+\dfrac{r(r-1)(r-2)}{2\times3}x^3+\dfrac{r(r-1)(r-2)(r-3)}{2\times3\times4}x^4+\cdots$$ then $$H=\sqrt{p^2c^2+m^2c^4}=\sqrt{m^2c^4}\sqrt{1+\dfrac{p^2}{m^2c^2}}=\sqrt{m^2c^4}\left(1+\dfrac{p^2}{m^2c^2}\right)^\frac12$$ now use binomial expansion and find what you want.

• Thanks for the answer Aug 17 '18 at 8:49
• You are welcome Thomas. Aug 17 '18 at 8:51

First of all we want to expand around zero, so we have to expand with some factors that goes to zero (considering that the speed of light is huge and we are in non relativistic case, so $v\lt\lt c$). So we rearrange the equation as follows $$E = \sqrt{p^2c^2+m^2c^4} = mc^2\sqrt{1+\left(\frac{p}{mc}\right)^2}$$Now we expand in the variable $u=\frac{p}{mc}$ (I'll leve it to you to understand the physical meaning of this quantity). Basically we have to expand the function $$f(u) = \sqrt{1+u^2}$$ in Taylor series. We have $$f(u)=f(0)+f'(0)u+\frac{f''(0)}{2}u^2\cdots = 1+\left.\frac{u}{\sqrt{1+u^2}}\right|_{u=0}u+\left.\frac{1}{(1+u^2)^{3\over2}}\right|_{u=0}u^2\cdots$$ Then we take up to the second order and substitute back for $u$ and we get $$E\sim mc^2\left(1+\left(\frac{p}{mc}\right)^2\right) = mc^2+\frac{p^2}{2m}$$

• Ahh it's that easy! Thanks. Aug 17 '18 at 8:48
• You're welcome! Aug 17 '18 at 8:58

Note that $H=mc^2\sqrt{1+\frac{P^2}{m^2c^2}}=mc^2\sqrt{1+\left(\frac P{mc}\right)^2}$. Now you use that Taylor series of $\sqrt{1+x^2}$ about $0$:$$1+\frac{x^2}2-\frac{x^4}8-\frac{x^6}{16}-\cdots$$getting$$mc^2+\frac{P^2}{2m}-\frac{P^4}{8mc^2}-\cdots$$

• Thanks for the answer Aug 17 '18 at 8:49

$$H = \sqrt{P^2 c^2 + m^2 c^4}=mc^2\sqrt{\frac{P^2 c^2 + m^2 c^4}{m^2c^4}}=mc^2\sqrt{1+\frac{P^2 }{m^2c^2}}$$

Now, since $c$ is large, consider $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ Replace $\epsilon=\frac{P^2 }{m^2c^2}$ making $$H=mc^2\left( 1+\frac{P^2}{2 c^2 m^2}+O\left(\frac{1}{c^4}\right)\right)=mc^2 +\frac{P^2}{2 m}+O\left(\frac{1}{c^2}\right)$$

• Thanks for the answer Aug 17 '18 at 8:49