Integrating the Yukawa Potential: Understanding the 3rd Step

In summary, the steps in the resolution of this integral involve using spherical coordinates and integrating out the \phi variable, then integrating with respect to the variable u, which results in a difference involving an exponential term. The third step involves changing the integration variable from k to -k and factorizing the denominator before solving the complex integral. If you require more details on a specific step, you can ask for it.
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  • #2
1st step: go to "spherical coordinates" and integrating out the [itex]\phi[/itex]... [itex] u = \cos \theta [/itex]
2nd step: Integrating wrt to [itex]u[/itex], will give a difference of [itex]e^{i~something} - e^{-i~something} \propto \sin (something) [/itex]. In fact I wouldn't ever write it in terms of sin...
3rd step: uses that he has two integrals one with the exponential with + and the other with the exponential with - ... then changes the integration variable of the one from k to -k, and gets this result.
4th step: factorizes the denominator.
5th step and then final: solves the (complex) integral

If you want to see a step in more details, you can ask for a specific one.
 
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Likes vanhees71
  • #3
I actually get it with your points! Thank you very much!
 
  • #4
ChrisVer said:
1st step: go to "spherical coordinates" and integrating out the [itex]\phi[/itex]... [itex] u = \cos \theta [/itex]
2nd step: Integrating wrt to [itex]u[/itex], will give a difference of [itex]e^{i~something} - e^{-i~something} \propto \sin (something) [/itex]. In fact I wouldn't ever write it in terms of sin...
3rd step: uses that he has two integrals one with the exponential with + and the other with the exponential with - ... then changes the integration variable of the one from k to -k, and gets this result.
4th step: factorizes the denominator.
5th step and then final: solves the (complex) integral

If you want to see a step in more details, you can ask for a specific one.
hello sir,
i can't get the 3rd step. is change the integration variable, change the value from k to -k in that term, including the dk -> d(-k)?
 
  • #5
Muh. Fauzi M. said:
i can't get the 3rd step. is change the integration variable, change the value from k to -k in that term, including the dk -> d(-k)?

like everytime you change your integration variable from [itex]x[/itex] to [itex]y(x)[/itex] what has to change is of course the differential, the integrand and the limits of the integral.
 

What is the Yukawa Potential?

The Yukawa Potential, also known as the screened Coulomb potential, is a mathematical expression that describes the interaction between two charged particles in quantum field theory.

Who first derived the Yukawa Potential?

The Yukawa Potential was first derived by Japanese physicist Hideki Yukawa in 1935 as a modification to the traditional Coulomb potential to account for the effects of quantum field theory.

How is the Yukawa Potential derived?

The Yukawa Potential is derived by considering the exchange of virtual particles between two charged particles. This exchange creates an attractive or repulsive force, depending on the charges and distances between the particles.

What is the significance of the Yukawa Potential?

The Yukawa Potential is significant because it provides a more accurate description of the interaction between particles at short distances, such as within atomic nuclei, where the traditional Coulomb potential breaks down.

What are the applications of the Yukawa Potential?

The Yukawa Potential has applications in various fields, such as nuclear physics, particle physics, and astrophysics. It is used to explain the forces between particles and the structure of atomic nuclei. It also plays a role in the study of dark matter and the behavior of neutron stars.

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