Approximate cross section calculations

  • Thread starter kaksmet
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  • #1
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Hello
Ive come across some approximate cross section formulas during a course of particle physics. And in a reaction like
A + B --> R --> C + D
the cross section should be something like

[tex]
\sigma = K\frac{\Gamma_R^{AB}\Gamma_R^{CD}}{(s-m_R^2)^2+m_R^2\Gamma_R^2}
[/tex]

Where K is a constant (depending on spin and color multiplicity), s is the center of mass energy sqared, m is the mass, and Gamma is the width calculated from

[tex]
\Gamma_R^{CD} = \bar{\abs{M}^2}\frac{P_C}{8\pi m_R^2}
[/tex]

My question is now how to handle a situation when the intermediate particle is mass less. Say a photon or a gluon. Can I take them as being virtual and thus having a mass which should be equal to the CM energy?

Greatfull if anyone can shed some light on this.
 

Answers and Replies

  • #2
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I also wonder if anyone know where I can find some info about how the gluon fields transforms under a local gauge transformation in SU(3)

[tex]
U=\{exp i \alpha(x)_a\lambda_a}
[/tex]

where alpha is a function of space (and time?) and lambda are the eight generators of the group.

thanks!
 
Last edited:
  • #3
malawi_glenn
Science Advisor
Homework Helper
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The question in your second post is not related to the first post, not so cleavr :-p

But anyway, you are asking in general how the Gauge fields of a non-abelian Gauge theory transforms. Information about this can be found in e.g. Peskin and Shroeder Quantum Field Theory.

Covariant derivative is:
[tex] D_{\mu}\equiv \partial _{\mu} + i A_\mu {}^aT_a [/tex]

Now we use the fact that the covariant derivative makes our lagrangian gauge invariant, thus we have that [tex] D_{\mu}\phi [/tex] transforms as: [tex] D_{\mu}'\phi' = \mathbf{U} D_{\mu}\phi [/tex].

we must have for the lagrangian to be invariant under gauge transformation:
[tex] \left( \partial _\mu + iA'_{\mu} \right) \mathbf{U} \phi = \mathbf{U}\left( \partial _\mu + iA_{\mu} \right)\phi [/tex]

Where:
[tex] A_{\mu} = A_{\mu}{}^aT_a [/tex]

(So T = lambda, this is something from a LaTeX doc- I did as project work a time ago, and I did for SU(2) but the arguments are exactly the same, but you will have index a going from 0 to 7 and not the Levi-Chivita but the structure constants of SU(3))

And [tex] \mathbf{U} = e^{i \alpha (x)^a T_a}[/tex]

It is easy to see that [tex] A'_{\mu} = \mathbf{U} A_{\mu} \mathbf{U}^{\dagger} + \xi [/tex], where [tex] \xi [/tex] now is to be determined. Using the product rule for derivatives, we find that [tex] \xi = i(\partial _\mu\mathbf{U})\mathbf{U}^{\dagger} [/tex] is suitable:
[tex] \left( \partial _\mu + iA'_{\mu} \right) \mathbf{U} \phi = \left( \partial _\mu + i\mathbf{U} A_{\mu} \mathbf{U}^{\dagger} - (\partial _\mu\mathbf{U})\mathbf{U}^{\dagger} \right) \mathbf{U} \phi = [/tex]
[tex] \partial _\mu (\mathbf{U} \phi ) - (\partial _\mu \mathbf{U}) \phi + i\mathbf{U}A_{\mu} \phi = \mathbf{U}\left( \partial _\mu + iA_{\mu} \right)\phi [/tex]

So the gauge field transforms as: [tex] A'_{\mu} = \mathbf{U} A_{\mu} \mathbf{U}^{\dagger} +i (\partial _\mu\mathbf{U})\mathbf{U}^{\dagger} [/tex]. Using an infinitesimal transformation: [tex] \mathbf{U} = 1 + i \alpha ^a(x)T_a \equiv 1 + i \alpha [/tex], we obtain to first order in [tex] \alpha [/tex]:
[tex] A'_{\mu} = (1 + i \alpha) A_{\mu} (1 - i \alpha) - \partial _\mu \alpha \Rightarrow [/tex]
[tex] A'_{\mu} = A_{\mu} - \partial _\mu \alpha - i\left[ A_{\mu} , \alpha \right]. [/tex]
Now we use that [tex] A_{\mu} = A_{\mu}{}^aT_a [/tex] :
[tex] A'_{\mu}{}^a = A_{\mu}{}^a -\partial _\mu \alpha ^a(x) + 2 \epsilon^{abc}A_{\mu}{}^b \alpha ^c(x) [/tex]

Yes alpha is function of space and time
 

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