why no change of variable to polar coordinates inside multi-loop integral ??


by zetafunction
Tags: coordinates, inside, integral, multiloop, polar, variable
zetafunction
zetafunction is offline
#1
Jun12-10, 05:48 AM
P: 399
given a mul,ti-loop integral

[tex] \int d^{4}k_{1} \int d^{4}k_{2}.................\int d^{4}k_{n}f(k_{1} , k_{2},.....,k_{n}) [/tex]

which can be considered a 4n integral for integer n , my question is why can just this be evaluated by using a change of variable to 4n- polar coordinates ?

one we have made a change of variable and calculated the Jacobian, and integrated over ALL the angular variables we just have to make an integral

[tex] \int_{0}^{\infty}drg(r)r^{4n-1} [/tex] which is just easier to handle
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mathman
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#2
Jun12-10, 05:38 PM
Sci Advisor
P: 5,935
I don't what specific integral you have in mind, but it depends very much on the form of f as it depends on the k's. You seem to imply that it can be represented as a function g of one variable. This may be true for some particular f, but it certainly is not true in general.
zetafunction
zetafunction is offline
#3
Jun13-10, 03:43 AM
P: 399
for example

[tex] \iint dx dy \frac{x^{3}}{1+xy} [/tex] its divergent if taking the limits (0,oo)

making a change of variable to polar coordinates one gets

[tex] \int du \int_{0}^{\infty}dr\frac{r^{4}cos^{3}(u)}{1+(1/2)r^{2}sin(2u)} [/tex]

integrating over the angular variable 'u' you have now a simple one dimensional integral

mathman
mathman is offline
#4
Jun13-10, 04:10 PM
Sci Advisor
P: 5,935

why no change of variable to polar coordinates inside multi-loop integral ??


In general if you have an m dimensional integral and integrate over m-1 dimensions, you will have a one dimensional integral. In your general case (4n) I am not sure what you mean by polar coordinates.

This question belongs in the mathematics forum. There isn't apparent connection with Beyond the Standard Model (physics).


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