# Approximate cross section calculations

1. Jan 11, 2009

### kaksmet

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.

2. Jan 11, 2009

### kaksmet

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: Jan 11, 2009
3. Jan 11, 2009

### malawi_glenn

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