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I was wondering if anyone could explain how Srednicki gets to his eqn 26.7:

[tex] \tilde{dk_1}\tilde{dk_2} \sim (\omega^{d-3}_{1}d\omega_1) (\omega^{d-3}_{2}d\omega_2)(sin^{d-3}\theta d\theta) [/tex]

I thought this would be to do with transforming into some kind of d-dimensional polar coords so I start as:

[tex] \tilde{dk_1}\tilde{dk_2}=\frac{d^{d-1}k_1}{(2\pi)^{d-1}2\omega_{1}}\frac{d^{d-1}k_2}{(2\pi)^{d-1}2\omega_{2}}=\frac{\vec{k_1}^{d-2}d\vec{k_1}d\Omega_{d-2}\vec{k_2}^{d-2}d\vec{k_2}d\Omega_{d-2}}{(2\pi)^{d-1}2\omega_{1}(2\pi)^{d-1}2\omega_{2} } [/tex]

Now since he's working in the massless limit [tex] \omega_{1,2}=\vec{k}_{1,2} [/tex]

[tex] \tilde{dk_1}\tilde{dk_2}=\frac{\omega^{d-3}_{1}d\omega_{1}d\Omega_{d-2}\omega^{d-3}_{2}d\omega_{2}\Omega_{d-2}}{4(2\pi)^{d-1}(2\pi)^{d-1} } [/tex]

[tex] \tilde{dk_1}\tilde{dk_2}=(\omega^{d-3}_{1}d\omega_{1})(\omega^{d-3}_{2}d\omega_{2}) \frac{d\Omega_{d-2}d\Omega_{d-2}}{4(2\pi)^{d-1}(2\pi)^{d-1}} [/tex]

Which looks quite similar to what he has, but not there yet. I'm guessing that the solid angle must go something like

[tex] d\Omega_{d-2}=sin^{d-3}d\theta \times d\phi_{1}d\phi_{2}... [/tex]

Which probably cancels out a few [tex]\pi[/tex]'s but then why doesnt he have two lot's of the sin term?

Thanks for any help on this

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# Srednicki CH26

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