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Srednicki CH20

  1. Mar 15, 2010 #1
    Hi,

    Srednicki says in Ch20, we must remember that the mandelstam variable s is positive. However s is defined as [tex]s=-(k_1+k_2)^2=(E_1+E_2)^2-(\vec{k_1+k_2}).(\vec{k_1+k_2}) [/tex]. Using metric (-+++). I can't quite see why this must be always positive? or why for that matter t and u are negative?

    Also he says we get [tex] V_3(s) [/tex], from the general [tex] V_3(k_1,k_2, k_3) [/tex] by setting two of the three k's to -m^2, and the remaining one to -s. Does anyone know why these values? I would have imagined one would be [tex] sqrt(-s)=k_1+k_2 [/tex] and the other two were just left as [tex] k_1[/tex] and [tex] k_2 [/tex] for scattering in s channel kind of way.

    Thanks again
     
    Last edited: Mar 15, 2010
  2. jcsd
  3. Mar 17, 2010 #2
    just work it out,

    [tex]
    s=-(k_1+k_2)^2=(E_1+E_2)^2-(\vec{k_1+k_2}).(\vec{k_1+k_2})
    [/tex]

    write k_1 dot k_2 etc as functions of E and mass m.

    [tex]s = m_1^2 + m_2^2 + 2( E_1E_2 - cos \theta \sqrt{E_1^2-m_1^2} \sqrt{E_2^2-m_2^2} \:\: ) [/tex]

    then you'll see that s >= 0
     
  4. Mar 17, 2010 #3
    Thanks for the reply ansgar. Is the expression you posted postivie just becase [tex] \sqrt{E_1^2-m_1^2} < \sqrt{E_1^2}=E_1 [/tex] and sim for second factor. So even if [tex] cos(\theta)=1 [/tex], the [tex] cos \theta \sqrt{E_1^2-m_1^2} \sqrt{E_2^2-m_2^2} [/tex] term is less than [tex] E_1E_2[/tex]. Therefore the whole of s must be positive?

    The way I convinced myself s was postivie before seeing you post, was looking at it in the CM frame, and using the fact s is a Lorentz scalar. In the CM frame [tex] s=(E_1+E_2)^2 [/tex] which is obviously positive.

    I still have been unable to convince myself t and u are negative however, either by switching to a conveinient frame or otherwise.

    Thanks again for the help
     
  5. Mar 17, 2010 #4
    You should not struggle with this if u want to learn QFT ;)

    Yes that is also a good way to see it :)
     
  6. Mar 17, 2010 #5
    Yes, yes, I know it's trivial, just tripping myself up trying to see it exactly, but think I got there from multiple directions in the end. Having said that is there an analagous way to show t and u are negative? I've tried various frames I thought might be useful without anything that has convinced me.
     
  7. Mar 17, 2010 #6
    show us what u have tried
     
  8. Mar 17, 2010 #7
    Well for example [tex] t=-(k_1-k_3)^2[/tex]

    I naivley tried CM frame again at first:

    [tex] t=-[(E_1-E_3, 2\vec{k_1})]^2=(E_1-E_3)^2-4\vec{k^{2}_1} [/tex]

    Then subbing in the usual Einstein energy momentum relation:

    [tex] t=-[(E_1-E_3, 2\vec{k_1})]^2=(E_1-E_3)^2-4(E^{2}_1-m^{2}_1) [/tex]

    [tex] t=E^{2}_3+4m^{2}_1-2E_3E_1-3E^{2}_1 [/tex]

    I can't think of a clever way to argue why this must always be negative right now

    -----------------------------

    I also tried working in the rest frame of one of the particles with a similar result.
     
  9. Mar 18, 2010 #8
    have u used what E_3 will be in the CM frame?

    have u tried going to e.g. 3's rest frame?

    also you can try with all m -> 0 and then add a "small" mass?
     
    Last edited: Mar 18, 2010
  10. Mar 18, 2010 #9

    Avodyne

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    Science Advisor

    t and u can be positive if the mass is not zero.
     
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