Thompson scattering, simple integral?

In summary, the conversation discusses a plasma physics problem involving the derivation of Thompson scattering. The total cross-section is represented by the integral of (d\sigma/d\Omega) d\Omega. The integral is solved using different methods, resulting in different answers. The correct answer is obtained by changing the integral to -sin2\theta dcos\theta, which results in the required answer of 8pi*re2/3. The conversation also mentions difficulties with using latex and apologies for the long explanation.
  • #1
Kawada
2
0
In theory this is a simpel integral problem which i can't solve

So I'm doing some plasma physics, and it comes with the derivation of the Thompson scattering (please bear with the first time I've tried using latex, I am sorry some of the greek lowercase laters look like superscripts, there not supposed to)
So I'm at the point where i have the total cross-section [tex]\sigma[/tex]T = the integral of (d[tex]\sigma[/tex]/d[tex]\Omega[/tex]) d[tex]\Omega[/tex].
ok so I've been given (d[tex]\sigma[/tex]/d[tex]\Omega[/tex]) = re2sin2[tex]\theta[/tex]so the integral d[tex]\Omega[/tex] = d[tex]\theta[/tex]d[tex]\phi[/tex], with [tex]\theta[/tex] between 0 and pi, and [tex]\phi[/tex] between 0 and 2pi.

but so the [tex]\phi[/tex] integral just gives 2pi, and in my attempt the integral of sin2[tex]\theta[/tex] with respect to [tex]\theta[/tex] over 0 and pi, is just pi/2?

yet looking at my notes, and the actual thompson scattering they have [tex]\sigma[/tex]T = 8pi*re2/3

now in my notes it says that the ingtegral over sin2[tex]\theta[/tex] d[tex]\theta[/tex] can become the integral over -sin2[tex]\theta[/tex] dcos[tex]\theta[/tex] still with [tex]\theta[/tex] between 0 and pi, and this yes, gives the required answer of 8pire2/3.

but hows do they change the intergral to that?

that is my only qualm, how they change the integral, and how my method of just ingetraing sin2[tex]\theta[/tex] d[tex]\theta[/tex] is not just pi/2?

i appreciate any help! :D (sorry for the long explanation)
 
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  • #2
fail!

i've just worked it out! sorry, should i delete my original post? I am new to that is that easy to do? can an administrator easily do it?

the original integral shouldn't of been just d(theta)d(phi), but sin(theta)d(theta)d(phi), this makes the integral sin3(theta), and then when you change it to dcos(theta), you divide by -sin(theta)!

sorry about this! (also i really can't work this latex :S)
 

1. What is Thompson scattering?

Thompson scattering is a phenomenon in which a charged particle, such as an electron, scatters off of a photon, changing the direction and energy of the photon.

2. How is Thompson scattering different from Compton scattering?

Thompson scattering is a low-energy approximation of Compton scattering, which takes into account the mass and recoil of the charged particle. In Thompson scattering, the energy of the photon is assumed to be much smaller than the rest mass of the charged particle, so the mass and recoil effects can be neglected.

3. What is the simple integral in Thompson scattering?

The simple integral in Thompson scattering is the integral of the differential cross-section formula, which describes the probability of a photon being scattered at a given angle by a charged particle. This integral is used to calculate the total cross-section for Thompson scattering.

4. How is the simple integral calculated?

The simple integral in Thompson scattering can be calculated using the formula: σ = (8π/3) x (e^2/mc^2)^2 x (1-cosθ), where σ is the total cross-section, e is the elementary charge, m is the mass of the charged particle, c is the speed of light, and θ is the angle of the scattered photon.

5. What is the significance of the simple integral in Thompson scattering?

The simple integral in Thompson scattering is a fundamental calculation that helps to understand the behavior of photons and charged particles interacting with each other. It is used in various fields of science, such as astrophysics and plasma physics, to study the scattering of light in different environments.

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