About two integrals in QCD textbook by muta

by Thor Shen
Tags: integrals, muta, textbook
 P: 9 1.How to deal with the delta functions in eq.2.3.153 to obtain the eq.2.3.154 by integrating over q'? 2.How to caculate the integral from eq.2.3.154 to eq.2.3.156, especially the theta function?
 P: 323 You already have a $\delta^4(q'-q)$ so just put q=q' and remove the integral over dq'. Then in Eq. (2.3.154) you have $\delta(k'\cdot q)$ that, in the COM, becomes just $\delta(k_0\sqrt{s})$. Therefore, taking care of the Jacobian coming out of the delta function, this just tells you that $k'_0=0$. Hence you don't need to worry about the thetas since they only give you $\theta(q_0)=\theta(\sqrt{s})=1$ since its argument is positive.
 P: 9 Actually, when I try to simplify the eq.2.3.153 for obtain the bracket in eq.2.3.154, I find we must use the two delta functions in eq.2.3.154, but I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?
P: 323
About two integrals in QCD textbook by muta

 Quote by Thor Shen I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?
No, you can't do that. You can't split one delta function in two. However, you already have two deltas:
$$\delta\left(q^2+k'^2+2q\cdot k'-4m^2\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right).$$
Now keep in mind that if you have some function $f(x)$ multiplied by a delta then $\delta(x-x_0)f(x)=\delta(x-x_0)f(x_0)$. This is true also if the function is a delta itself.

Now, the second delta tells you that $q^2+k'^2-4m^2=2q\cdot k'$. Using this is the first delta you obtain (I always omit the necessary Jacobian):
$$\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right)=\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-4m^2\right).$$
Keep also in mind that $q^2=s$.

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