Given the following equation:
$$ p_{\gamma}^2+(-2m_dc)p_{\gamma} +2m_dB_d=0 \tag 1$$
Solving it for $p_{\gamma}$, using the quadratic formula, gives me:
$$ p_{\gamma} = \frac{2m_dc \pm \sqrt{4m_d^2c^2-4(2m_dB_d)}}{2}$$
$$ p_{\gamma} = m_dc \pm \sqrt{m_d^2c^2-(2m_dB_d)}$$
$$ p_{\gamma} = m_dc \pm m_dc \sqrt{1-\frac{2B_d}{m_dc^2}}$$
$$ p_{\gamma} \approx m_dc \pm m_dc\left(1 -\frac{B_d}{m_dc^2} \right)$$
But Im supposed to get the following solution for $p_{\gamma}$:
$$ p_{\gamma} = \frac{B_d}{c} \left( 1+ \frac{B_d}{2m_dc^2} \right) \tag 2$$
and I am aware that I have to use the quadratic formula and Binomial expansion to get to equation (2) but I can get there. Any help?
Best Answer
Let's simplify the notation by dropping subscripts, and turn this into an arithmetic problem by making everything dimensionless:
$$ \frac{p}{mc} = 1 \pm \sqrt{1 - \frac{2B}{mc^2}} $$
We can treat the square root as a binomial and expand. The binomial expansion is
$$ (1 + \epsilon)^n = 1 + n\epsilon + \frac{n(n-1)}{2!}\epsilon^2 + \cdots + {n\choose m} \epsilon^m + \cdots $$
and the trick is to neglect the higher-order terms; the trick works better when $n\epsilon \ll 1$. Taking the negative discriminant in your solution, the first-order approximation gives
\begin{align} \frac{p}{mc} &\approx 1 - \left( 1 - \frac12 \frac{2B}{mc^2} \right) = \frac{B}{mc^2} \end{align}
To get the correction you're looking for, retain the second-order term, $-\frac18\left(\frac{2B}{mc^2}\right)^2$.