Taking a limit

[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.

[ehild]

Write the terms m2+μ2 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

[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]

[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

[ryanwilk]

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

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]

[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

[ryanwilk]

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

Awesome, thanks a lot! =D

[ehild]

Awesome, thanks a lot! =D

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

ehild