0

When going over an article Sterile Neutrinos as Dark Matter the following expression is given:

$$\sin ^2 2\theta = \frac{\mu ^2}{\mu^2 + (\frac{c \Gamma E}{M} + \frac{M}{2})^2}$$

I have been told by colleagues that I can Taylor expanding this up to the power of 2:

$$\sin^2 2\theta \approx \frac{\mu^2}{(\frac{c \Gamma E}{M} + \frac{M}{2})^2}$$

I don't understand how this expansion was reached.

Note: don't think this matters for the expansion, but to contextualise, $\mu$ is the Dirac Mass, $M$ is the Majorana mass, $E$ is the energy of a particle and $\Gamma$ represents the total interaction rate of neutrinos with plasma

user7077252
  • 173

1 Answers1

1

One quick way is to use geometric series ideas. Recall that $\frac{1}{1-r} = 1 + r + r^2 + \cdots$ when $-1 < r < 1.$ By various algebraic methods (manipulations such as I'll do shortly, partial fractions for certain rational functions, etc.) this can be exploited quite a bit. For example, if $a$ is a nonzero constant, then $\frac{1}{a-r} = \frac{1/a}{1 - (r/a)}$ (divide numerator and denominator by $a),$ and thus $\frac{1}{a-r}$ can be expanded as $(1/a)[1 + (r/a) + (r/a)^2 + (r/a)^3 + \cdots].$

In your case, putting $ c^2 = \left(\frac{c \Gamma E}{M} + \frac{M}{2}\right)^2$ we get

$$ \frac{\mu^2}{\mu^2 \; + \; c^2} \;\; = \;\; \mu^2 \left[ \frac{1}{c^2 \; - \; (-\mu^2)}\right] \;\; = \;\; \mu^2 \left[ \frac{c^{-2}}{1 \; - \; \left(-\mu^2/c^2\right)} \right] \;\; = \;\; \frac{\mu^2}{c^2} \left[ \frac{1}{1 \; - \; \left(-\mu^2/c^2\right)} \right] $$

$$ = \;\; \frac{\mu^2}{c^2}\left[1 \; + \; \left(-\mu^2/c^2\right) \; + \; \left(-\mu^2/c^2\right)^2 \; + \; \left(-\mu^2/c^2\right)^3 \; + \; \cdots \right] $$

$$ = \;\; \frac{\mu^2}{c^2} \; - \; \frac{\mu^4}{c^4} \; + \; \frac{\mu^6}{c^6} \; - \; \frac{\mu^8}{c^8} \; + \; \cdots $$

Dave L. Renfro
  • 36,843