# Approximate cross section calculations

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

$$\sigma = K\frac{\Gamma_R^{AB}\Gamma_R^{CD}}{(s-m_R^2)^2+m_R^2\Gamma_R^2}$$

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

$$\Gamma_R^{CD} = \bar{\abs{M}^2}\frac{P_C}{8\pi m_R^2}$$

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.

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)

$$U=\{exp i \alpha(x)_a\lambda_a}$$

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

thanks!

Last edited:
malawi_glenn
Homework Helper
The question in your second post is not related to the first post, not so cleavr

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:
$$D_{\mu}\equiv \partial _{\mu} + i A_\mu {}^aT_a$$

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

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

Where:
$$A_{\mu} = A_{\mu}{}^aT_a$$

(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 $$\mathbf{U} = e^{i \alpha (x)^a T_a}$$

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

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

Yes alpha is function of space and time