- #1

Pnin

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Can someone please explain what the author does here in 15.59? I do not understand both steps. Neither the rewriting of the derivative, nor the integral.

thank you

thank you

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- Thread starter Pnin
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- #1

Pnin

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thank you

Last edited:

- #2

mathman

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Attachment seems hard to open.

- #3

Pnin

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I pasted an image.

- #4

strangerep

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I'm guessing the bit that's confusing you about the derivative rewrite involves the following:$$s ~=~ p^2 ~=~ p^\alpha p_\alpha ~,~~~~ \Rightarrow~~ p^\alpha = \hat p^\alpha \, s^ {1/2} ~,$$where the overhat denotes a unit vector. (This assumes ##p^\alpha## is not lightlike.) Then,$$\frac{\partial p^\alpha}{\partial s} ~=~ \frac{\partial s^ {1/2}}{\partial s} \; \hat p^\alpha ~=~ \frac{1}{ 2 s^ {1/2}}\; \hat p^\alpha ~=~ \frac{s^{1/2}}{ 2 s}\; \hat p^\alpha ~=~ \frac{p^\alpha}{2s} ~.$$Can someone please explain what the author does here in 15.59? I do not understand both steps. Neither the rewriting of the derivative, nor the integral.

View attachment 256512

thank you

So, using the chain rule,$$\frac{\partial M}{\partial s} ~=~ \frac{\partial p^\mu}{\partial s} \; \frac{\partial M}{\partial p^\mu} ~=~ \frac{p^\alpha}{2s} \; \frac{\partial M}{\partial p^\mu} ~.$$

Next, the ##\partial/\partial p^\mu## operator passes through the integral sign (since the variables ##k## and ##p## are independent). Then it's just a matter of performing $$\frac{\partial}{\partial p^\mu} \; \frac{1}{(p - k)^2} $$which is merely an application of the usual quotient rule for derivatives. (Can you do that bit?)

- #5

Pnin

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Fantastic! Crystal-clear now. Thanks strangerep, much appreciated!

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