Finding Approximation for X in m^2 $\gg$ $\mu^2$

[ryanwilk]

Homework Statement

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.

Homework Equations

N/A

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?:

<br /> X = m_N\&gt;\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\&gt;\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1+x)} \bigg) - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\&gt;\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1-x)} \bigg) \bigg]<br />

Expanding the logs,

<br /> \simeq m_N\&gt;\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\&gt;\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)}\&gt; \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2} \bigg) + x \bigg] \bigg]<br />

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

<br /> \simeq m_N\&gt;m^2\&gt;\mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg)\&gt; \bigg[\frac{1}{m_N^2 - (m^2+\mu^2)} - \frac{1}{m_N^2 - (m^2 - \mu^2)} \bigg]<br />

 

[ehild]

Oops... Do not ignore x . Collect the terms with the common factor ln(m_N/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(m_N/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?:

<br /> m_{\nu_L} = m_N\&gt;\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\&gt;\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1+x)} \bigg) - \frac{m^2(1-x)}{m_N^2 - m^2(1-x)}\&gt;\mathrm{ln} \bigg(\frac{m_N^2}{m^2(1-x)} \bigg) \bigg]<br />

<br /> \simeq m_N\&gt;\bigg[\frac{m^2(1+x)}{m_N^2 - m^2(1+x)}\&gt;\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)}\&gt; \bigg[ \mathrm{ln} \bigg(\frac{m_N^2}{m^2} \bigg) + x \bigg] \bigg] <br />

<br /> \simeq m_N\&gt;\bigg\{ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) \&gt; \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\}<br />

<br /> \simeq m_N\&gt;\bigg\{ \mathrm{ln} \bigg(\frac{m_N^2}{m^2}\bigg) \&gt; \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\}<br />

<br /> = 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\} \&gt; \bigg[\frac{1}{(m_N^2 - m^2(1+x))(m_N^2 - m^2(1-x))} \bigg]<br />

<br /> = 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\} \&gt; \bigg[\frac{1}{(m_N^2 - (m^2+\mu^2))(m_N^2 - (m^2-\mu^2))} \bigg]<br />

 

[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 m_N²-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 m_N²-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