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?

  • $\begingroup$ Note: arising from this physics discussion, in which $B_d \ll m_d c^2$ is a good limit. $\endgroup$
    – rob
    Mar 7, 2021 at 17:12

1 Answer 1


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, $-\frac18\left(\frac{2B}{mc^2}\right)^2$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .