1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taking a limit

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    I need to find the approximation to:

    [itex] 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] [/itex]

    for [itex] m^2 \gg \mu^2 [/itex].


    2. Relevant equations

    N/A

    3. The attempt at a solution

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

    Any help would be appreciated,

    Thanks.
     
  2. jcsd
  3. Dec 12, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
  4. Dec 12, 2011 #3
    Ah ok thanks, so would this be correct?:

    [itex]
    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]
    [/itex]

    Expanding the logs,

    [itex]
    \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]
    [/itex]

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

    [itex]
    \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]
    [/itex]
     
  5. Dec 12, 2011 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  6. Dec 12, 2011 #5
    Ok, so something like?:

    [tex]
    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]
    [/tex]

    [tex]
    \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]
    [/tex]

    [tex]
    \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\}
    [/tex]

    [tex]
    \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\}
    [/tex]

    [tex]
    = 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]
    [/tex]

    [tex]
    = 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]
    [/tex]
     
  7. Dec 12, 2011 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    [tex](m_N^2-m^2)^2-\mu^4[/tex],

    and you can omit μ4 if mN2-m2>>μ2.

    ehild
     
  8. Dec 12, 2011 #7
    Awesome, thanks a lot! =D
     
  9. Dec 12, 2011 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taking a limit
  1. : Limits (Replies: 15)

  2. Limit prob (Replies: 4)

  3. Limit of a ratio (Replies: 7)

  4. Solving this limit (Replies: 11)

  5. How to limit (Replies: 1)

Loading...