# Taking a limit

1. Dec 11, 2011

### ryanwilk

1. The problem statement, all variables and given/known data

I need to find the approximation to:

$X = m_N\>\bigg[\frac{m^2+\mu^2}{m_N^2 - (m^2+\mu^2)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2+\mu^2} \bigg) - \frac{m^2-\mu^2}{m_N^2 - (m^2-\mu^2)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2-\mu^2} \bigg) \bigg]$

for $m^2 \gg \mu^2$.

2. Relevant equations

N/A

3. The attempt at a solution

So obviously I can't just do $m^2 \pm \mu^2 \approx m^2$, otherwise X = 0, which makes me think that I need to expand the logs in some way...

Any help would be appreciated,

Thanks.

2. Dec 12, 2011

### ehild

Write the terms m22 in the form m2(1+x) with x=(μ/m)2. Expand the logarithm ln(1+x) with respect to x around x=0. The first power is enough: ln (1+x)=x. Ignore terms containing Edit: μ4 on higher powers.

ehild

Last edited: Dec 12, 2011
3. Dec 12, 2011

### ryanwilk

Ah ok thanks, so would this be correct?:

$X = m_N\>\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1+x)} \bigg) - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1-x)} \bigg) \bigg]$

Expanding the logs,

$\simeq m_N\>\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\>\bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) - x \bigg] - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\> \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2} \bigg) + x \bigg] \bigg]$

Then ignoring all terms involving x or x2 on the numerator,

$\simeq m_N\>m^2\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg)\> \bigg[\frac{1}{m_N^2 - (m^2+\mu^2)} - \frac{1}{m_N^2 - (m^2 - \mu^2)} \bigg]$

4. Dec 12, 2011

### ehild

Oops... Do not ignore x . Collect the terms with the common factor ln(mN/m) and those with the common factor x. Bring the terms to common denominator and try to simplify. Ignore the x^2 terms.
ehild

5. Dec 12, 2011

### ryanwilk

Ok, so something like?:

$$m_{\nu_L} = m_N\>\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1+x)} \bigg) - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\>\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1-x)} \bigg) \bigg]$$

$$\simeq m_N\>\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\>\bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) - x \bigg] - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\> \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2} \bigg) + x \bigg] \bigg]$$

$$\simeq m_N\>\bigg\{ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) \> \bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)} - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)} \bigg] - x \bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)} + \frac{m^2(1-x)}{m_N^2 - m^2(1-x)} \bigg] \bigg\}$$

$$\simeq m_N\>\bigg\{ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) \> \bigg[\frac{2 m^2 m_N^2 x}{(m_N^2 - m^2(1+x))(m_N^2 - m^2(1-x))} \bigg] - x \bigg[\frac{2m^2(m_N^2-m^2)}{(m_N^2 - m^2(1+x))(m_N^2 - m^2(1-x))} \bigg] \bigg\}$$

$$= m_N\cdot 2m^2 x \bigg\{ m_N^2 \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) - 1 \bigg] + m^2 \bigg\} \> \bigg[\frac{1}{(m_N^2 - m^2(1+x))(m_N^2 - m^2(1-x))} \bigg]$$

$$= 2 m_N \mu^2 \bigg\{ m_N^2 \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) - 1 \bigg] + m^2 \bigg\} \> \bigg[\frac{1}{(m_N^2 - (m^2+\mu^2))(m_N^2 - (m^2-\mu^2))} \bigg]$$

6. Dec 12, 2011

### ehild

It looks correct. You can further simplify the denominator. It is

$$(m_N^2-m^2)^2-\mu^4$$,

and you can omit μ4 if mN2-m2>>μ2.

ehild

7. Dec 12, 2011

### ryanwilk

Awesome, thanks a lot! =D

8. Dec 12, 2011

### ehild

You are welcome. Where did this terrible thing come from?

ehild