Simplifying Argument of Dirac Delta for Reexpressing a Dirac Delta

[CAF123]

Homework Statement

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. Homework 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!

 

[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.