# Reexpressing a Dirac delta

1. Jun 10, 2015

### CAF123

1. The problem statement, all variables and given/known data
Show that $$\delta(k^2) \delta[(k-q_2)^2] = \delta(k^2) \delta(k^0 - \sqrt{s}/2) \frac{1}{2\sqrt{s}},$$ where $k = (k^0, \mathbf k)$ and $s = q_2^2,$ where $q_2 = (\sqrt{s},\mathbf 0)$

2. Relevant Equations

I was going to use the fact that $$\delta(f(x)) = \sum \frac{\delta(x-a_i)}{|f'(a_i)|},$$ where the $a_i$ are such that $f(a_i)=0$.

3. The attempt at a solution

In this case let $f = (k-q_2)^2 = k^2 - 2k \cdot q_2 + q_2^2$ where $\cdot$ is the inner product of two four vectors. Then using the explicit four vector expressions I get ${k^0}^2 - |\mathbf k|^2 - 2k^0 \sqrt{s} + s = 0 \Rightarrow k^0 = \sqrt{s} \pm |\mathbf k|,$ which would imply I have a term $\delta(k^0 - (\sqrt{s} + |\mathbf k|))$ but this is not matching with what is given. The calculation is computed in the centre of momentum frame, hence the form of the four vectors. Can anyone help? Thanks!

2. Jun 10, 2015

### fzero

You can simplify the argument of $\delta[(k-q_2)^2]$ using the constraint from the factor of $\delta[k^2]=\delta[(k^0)^2 - |\mathbf k|^2]$. It is most convenient to use this before you compute the root.