- #1
CAF123
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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!