week10 - 1 Monday November 21 Inverse Compton Scattering...

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1 Monday, November 21: Inverse Compton Scattering When I did the calculations for the scattering of photons from electrons, I chose (for the sake of simplicity) the inertial frame of reference in which the electron was initially at rest. However, as we look at the universe around us, we see electrons in motion. The electron velocities, in our frame of reference, range from the highly non-relativistic motions of free electrons in partially ionized warm gas to the highly relativistic electrons that emit synchrotron radiation. When an electron is initially moving relative to a photon, we can redo the analysis of the electron-photon collision, with the usual assumptions of the conservation of momentum and energy. 1 If the electron has an arbitrary initial velocity ~v i , the calculations become a bit tedious, and require a few envelope backs to compute. However, the special case in which the initial velocities of photon and electron are in opposite directions requires only a single envelope back. 2 Suppose that the photon is initially moving in the positive x direction, and the electron is initially moving in the negative x direction with velocity - v i ˆ e x . The photon is scattered by an angle θ from its initial direction, and the electron is scattered by an angle ϕ from its initial direction. The initial momentum of the system will be ~ p i = - γ i m e v i ˆ e x + i c ˆ e x (1) and its initial energy will be ² i = γ i m e c 2 + i . (2) The final momentum of the system will be ~ p f = - γ f m e v f (cos ϕ ˆ e x + sin ϕ ˆ e y ) + f c (cos θ ˆ e x + sin θ ˆ e y ) (3) 1 Equivalently, we can take our calculations with an initially stationary electron, and transform them to another frame of reference in which the electron initially has a constant velocity. 2 I’ll refer to this case as a “head-on” collision, in analogy with a collision between cars going in opposite directions; a “head-on” collision, however, doesn’t imply that the photon and electron are necessarily reflected straight back the way they came. 1
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and its final energy will be ² f = γ f m e c 2 + f . (4) Aside from the addition of the electron’s initial momentum, and an acknowl- edgment that the electron’s initial energy is greater than the rest energy, this is the same as the problem we solved last week. I will therefore leave it as an “exercise for the reader” to demonstrate that the final energy, in its dimensionless form, X f f / ( m e c 2 ), is X f = X i γ i (1 + β i ) γ i (1 + β i cos θ ) + X i (1 - cos θ ) , (5) where θ is the angle through which the photon is scattered, β i = v i /c is the initial speed of the oncoming electron, in units of the speed of light, and γ i is the initial Lorentz factor for the electron. 3 The fractional amount of energy lost or gained by a photon in a “head-on” collision with an electron is X f - X i X i = ( γ i β i - X i )(1 - cos θ ) γ i (1 + β i cos θ ) + X i (1 - cos θ ) . (6) Thus, there is no energy transfer when X i = γ i β i . If the electron is initially highly relativistic ( γ i 1), photons with energy i < γ i m e c 2
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