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Breo

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- Thread starter Breo
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In summary, the conversation is about changing the momentum integrals to spherical coordinates in order to solve loop integrals using dimensional regularization. This involves interpreting the integral in 4-d as a limit of an integral in d dimensions and using the volume element of a hypersphere to solve the integral.

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Breo

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I suspect they mean spherical coordinates in momentum space ...

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Breo

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Orodruin said:I suspect they mean spherical coordinates in momentum space ...

"[...]Change the momentum integrals to spherical coordinates[...]"

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ChrisVer

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[itex] d^3 p \rightarrow p^2 dp d \Omega[/itex]

is quite useful ...

is quite useful ...

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Breo

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Did you perhaps use the standard forms of the integrals that you end up with in dim-reg which you can find, eg, in the end of Peskin? How do you think these integrals are evaluated?Breo said:But why? because I solved loop integrals using dim regularization and other tricks and never used the change of variables.

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Breo

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oh!

On the unit sphere area?

On the unit sphere area?

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ChrisVer

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why "unit" sphere area?

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Breo

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ChrisVer said:why "unit" sphere area?

I am still wondering haha

Do not know for now... tip for clairvoyance?

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ChrisVer

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[itex] \int \frac{d^3 p}{(2 \pi)^3} \frac{a }{|p|^2+m^2} e^{-i\vec{p} \cdot \vec{x}}[/itex]

?

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ChrisVer

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Orodruin said:area of a d-1-dimensional unit sphere

Why saying a unit sphere? This doesn't make sense, exactly because the ##d^d p## is a volume. The [itex]d^d p = |p|^{d-1} dp d\Omega_{d-1}[/itex] is a volume element of an "hypersphere" between radius ##p## and ##p+dp##.

edit: I think I figured it out

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Yes, exactly. I took one step extra and integrated the ##d\Omega##. This gives you the area of the d-1-dimensional unit sphere.ChrisVer said:Why saying a unit sphere? This doesn't make sense, exactly because the ##d^d p## is a volume. The [itex]d^d p = |p|^{d-1} dp d\Omega_{d-1}[/itex] is a volume element of an "hypersphere" between radius ##p## and ##p+dp##.

Spherical coordinates are often used in physics and mathematical calculations because they simplify complex integrals and make them easier to solve. In momentum space, integrals can be quite complicated and difficult to solve, whereas in spherical coordinates, the integrals can often be separated into simpler components.

Spherical coordinates use a different coordinate system than Cartesian coordinates. In spherical coordinates, a point is defined by its distance from the origin, its polar angle (measured from the z-axis), and its azimuthal angle (measured from the x-axis). This is different from Cartesian coordinates, which use the familiar x, y, and z axes to define a point.

No, spherical coordinates are not suitable for all types of integrals. They are most commonly used for integrals involving spherical objects or symmetrical systems, such as the gravitational potential of a sphere or the electric field of a charged sphere. For other types of integrals, different coordinate systems may be more appropriate.

One limitation of using spherical coordinates is that they are not as intuitive as Cartesian coordinates, which can make it difficult to visualize the problem at hand. Additionally, spherical coordinates are only applicable in three-dimensional space, so they cannot be used for problems in higher dimensions.

To convert from momentum space integrals to spherical coordinate ones, you will need to use the appropriate transformation equations for each variable. These equations can be found in most mathematical or physics textbooks. Alternatively, there are also online calculators and conversion tools available that can help with this process.

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