Srednicki CH20

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  • #1
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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
 
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Answers and Replies

  • #2
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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
 
  • #3
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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
 
  • #4
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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
You should not struggle with this if u want to learn QFT ;)

Yes that is also a good way to see it :)
 
  • #5
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You should not struggle with this if u want to learn QFT ;)

Yes that is also a good way to see it :)
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.
 
  • #6
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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.
show us what u have tried
 
  • #7
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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.
 
  • #8
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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?
 
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  • #9
Avodyne
Science Advisor
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t and u can be positive if the mass is not zero.
 

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