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APhO 2007, Теория — Задача 3B. The Inverse Compton Scattering

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Theoretical Problem 3 · 5 т.

Условие

By collision with relativistic high energy electron, a photon can get energy from the high energy electron, i.e. the energy and frequency of the photon increases because of the collision. This is so-called inverse Compton scattering. Such kind of phenomenon is of great importance in astrophysics, for example, it provides an important mechanism for producing $X$ rays and $\gamma$ rays in space.

1. A high energy electron of total energy $E$ (its kinetic energy is higher than static energy) and a low energy photon (its energy is less than the static energy of an electron) of frequency $\nu$ move in opposite directions, and collide with each other. As shown in the figure below, the collision scatters the photon, making the scattered photon move along the direction which makes an angle $\theta$ with its original incident direction (the scattered electron is not shown in the figure). Calculate the energy of the scattered photon, expressed in terms of $E,\nu,\theta$ and static energy $E_0$ of the electron. Show the value of $\theta$ , at which the scattered photon has the maximum energy, and the value of this maximum energy. [2,4 т.]

Collision diagram: a thick arrow labelled E,p (electron) points right; a thin arrow labelled hν (photon) points left toward it; the scattered photon hν′ leaves up and to the left, making the angle θ with the electron direction.

2. Assume that the energy $E$ of the incident electron is much higher than its static energy $E_0$ , which can be shown as $E=\gamma E_0,\gamma\gg1$ , and that the energy of the incident photon is much less than $E_0/\gamma$ , show the approximate expression of the maximum energy of the scattered photon. Taking $\gamma=200$ and the wavelength of the incident visible light photon $\lambda=500\mathrm{nm}$ , calculate the approximate maximum energy and the corresponding wavelength of the scattered photon.

Parameters: Static energy of the electron $E_0=0.511\mathrm{MeV}$ , Planck constant $h=6.63\times10^{-34}\,\mathrm{J\cdot s}$ , and $hc=1.24\times10^3\,\mathrm{eV\cdot nm}$ , where $c$ is the light speed in the vacuum. [1,2 т.]

3. (1) A relativistic high energy electron of total energy $E$ and a photon move in opposite directions and collide with each other. Show the energy of the incident photon, of which the photon can gain the maximum energy from the incident electron. Calculate the energy of the scattered photon in this case. [0,7 т.]

3. (2) A relativistic high energy electron of total energy $E$ and a photon, moving in perpendicular directions respectively, collide with each other. Show the energy of the incident photon, of which the photon can gain the maximum energy from the incident electron. Calculate the energy of the scattered photon in this case. [0,7 т.]

Отговори

Решение

3B. The Inverse Compton Scattering

Let $p$ and $E$ denote the momentum and energy of the incident electron, $p'$ and $E'$ the momentum and energy of the scattered electron, and $h\nu$ and $h\nu'$ the energies of the incident and scattered photon respectively. For this scattering process (see Fig. 1) energy conservation reads

h\nu+E=h\nu'+E' \qquad \text{(1B.1)}

Scattering diagram: incident electron arrow E,p to the right, incident photon hν to the left, scattered photon hν′ up-left at angle θ, scattered electron E′,p′ down-right. — Figure 1

while the momentum conservation can be shown as (see Fig.2)

Momentum triangle with sides labelled hν′/c, p′ and p−hν/c, with the angle θ between hν′/c and the dashed extension of p−hν/c. — Figure 2

(p'c)^2=(h\nu')^2+(pc-h\nu)^2+2h\nu'(pc-h\nu)\cos\theta \qquad \text{(1B.2)}

Equations (1B.1) and (1B.2), combined with the energy-momentum relations

E^2=(pc)^2+E_0^{\ 2} \qquad \text{(1B.3)}

and

E'^2=(p'c)^2+E_0^2 \qquad \text{(1B.4)}

lead to

h\nu'=\frac{E+pc}{E+h\nu+(pc-h\nu)\cos\theta}h\nu=\frac{E+\sqrt{E^2-E_0^2}}{E+h\nu+(\sqrt{E^2-E_0^2}-h\nu)\cos\theta}h\nu. \qquad \text{(1B.5)}

We have assumed that the kinetic energy of the incident electron is higher than its static energy, and the energy of the incident photon $h\nu$ is less than $E_0$ , so that $\sqrt{E^2-E_0^2}>h\nu$ . Therefore from Eq. (1B.5), it can be easily seen that $\theta=\pi$ results in the maximum of $h\nu'$ , and the maximum $h\nu'$ is

(h\nu')_{\max}=\frac{E+\sqrt{E^2-E_0^2}}{E+2h\nu-\sqrt{E^2-E_0^2}}h\nu. \qquad \text{(1B.6)}

Substitution of $E=\gamma E_0$ into Eq. (1B.6) yields

(h\nu')_{\max}=\frac{\gamma E_0+\sqrt{\gamma^2-1}E_0}{\gamma E_0-\sqrt{\gamma^2-1}E_0+2h\nu}h\nu=\frac{\gamma+\sqrt{\gamma^2-1}}{\gamma-\sqrt{\gamma^2-1}+2h\nu/E_0}h\nu\ . \qquad \text{(1B.7)}

Due to $\gamma\gg1$ , $\sqrt{\gamma^2-1}\approx\gamma(1-\frac{1}{2\gamma^2})=\gamma-\frac{1}{2\gamma}$ , and $h\nu/E_0\ll1/\gamma$ , then we have

(h\nu')_{\max}\approx\frac{\gamma+\gamma-1/2\gamma}{\gamma-\gamma+1/2\gamma+2h\nu/E_0}h\nu\approx4\gamma^2h\nu. \qquad \text{(1B.8)}

In the case of $\gamma=200$ and the wavelength of the incident photon $\lambda=500\mathrm{nm}$

h\nu=\frac{hc}{\lambda}=\frac{1.24\times10^3}{500}=2.48\ \mathrm{eV},

\frac{h\nu}{E_0}=\frac{2.48}{0.511\times10^6}=4.85\times10^{-6}\ll\frac{1}{\gamma}=\frac{1}{200}=5.0\times10^{-3},

satisfying expression (1B.8). Therefore the maximum energy of the scattered photon

(h\nu')_{\max}\approx4\times200^2h\nu=1.6\times10^5\times2.48=3.97\times10^5\mathrm{eV}\approx4.0\times10^5\mathrm{eV}=0.40\mathrm{MeV}

corresponding to a wavelength $\lambda'=\frac{hc}{h\nu'}=\frac{1.24\times10^3}{4.0\times10^5}=3.1\times10^{-3}\mathrm{nm}.$

(1) It is obvious that if the incident electron gives its total kinetic energy to the photon, the photon gains the maximum energy from the incident electron through the scattering process, namely the electron should become at rest after the collision. In this case, we have (see Fig. 3)

Before the collision the electron E,p moves right and the photon hν moves left; after it the electron E₀ is at rest (small circle) and the photon hν′ moves right. — Figure 3

h\nu+E=h\nu'+E_0 \quad \text{(Conservation of energy)} \quad \text{(1B.9)}

p-h\nu/c=h\nu'/c \quad \text{(Conservation of momentum)}

pc-h\nu=h\nu' \qquad \text{(1B.10)}

Subtracting (1B.10) from (1B.9) leads to the energy of the incident photon

h\nu=\frac{1}{2}(E_0-E+pc)=\frac{1}{2}(E_0-E+\sqrt{E^2-E_0^2}). \qquad \text{(1B.11)}

In above equation the energy- momentum relation

(pc)^2=E^2-E_0^{\ 2} \qquad \text{(1B.12)}

has been taken into account. Therefore from Eq. (1B.9) we obtain the energy of the scattered photon

Right-triangle momentum diagram: horizontal arrow E,p, vertical arrow hν/c downward, hypotenuse hν′/c. — Figure 4

h\nu'=h\nu+E-E_0=\frac{1}{2}(E-E_0+\sqrt{E^2-E_0^2}). \qquad \text{(1B.13)}

(2) Similar to question 3. (1), now we have (see Fig. 4)

h\nu+E=h\nu'+E_0 \quad \text{(Conservation of energy)} \quad \text{(1B.9)}

p^2+(h\nu/c)^2=(h\nu'/c)^2 \quad \text{(Conservation of momentum)}

That is,

(pc)^2+(h\nu)^2=(h\nu')^2\ . \qquad \text{(1B.14)}

Substitution of Eq. (1B.12) into Eq. (1B.14) yields

E^2-E_0^{\ 2}+(h\nu)^2=(h\nu')^2.

On the other hand, square of Eq. (1B.9) results in

(h\nu')^2=E^2+E_0^{\ 2}+(h\nu)^2+2Eh\nu-2EE_0-2E_0h\nu

Combination of the above two equations leads to

E^2+E_0^{\ 2}+(h\nu)^2+2Eh\nu-2EE_0-2E_0h\nu=E^2-E_0^{\ 2}+(h\nu)^2

That is,

2(E-E_0)h\nu=2E_0(E-E_0)\ .

Then, we obtain the energy of the incident photon

h\nu=E_0. \qquad \text{(1B.15)}

Substitution of (1B.15) into Eq. (1B.9) gives the energy of the scattered photon

h\nu'=h\nu+E-E_0=E \qquad \text{(1B.16)}

Explanatory notes about the solution of Question 3:

Question 3. (1) can also be solved as follows. According to Eq. (1B.6), the maximum energy $\Delta$ that the photon of energy $h\nu$ gains from the electron is

\Delta=h\nu'_{\max}-h\nu=2\frac{pch\nu-h^2\nu^2}{E+2h\nu-pc},

where $pc=\sqrt{E^2-E_0^2}$ . To obtain the maximum $\Delta$ , we use the extreme condition

\frac{d(\Delta/2)}{d(h\nu)}=\frac{(pc-2h\nu)(E+2h\nu-pc)-2(pch\nu-h^2\nu^2)}{(E+2h\nu-pc)^2}=0.

Let the numerator equal to zero, a quadratic equation results:

2(h\nu)^2+2(E-pc)h\nu-(Epc-p^2c^2)=0.

Its two roots can be shown as

h\nu=\frac{1}{2}\left[-(E-pc)\pm\sqrt{E^2-p^2c^2}\right]=\frac{1}{2}(-E+pc\pm E_0).

Since the negative sign leads to a meaningless negative $h\nu$ , we have

h\nu=\frac{1}{2}(\sqrt{E^2-E_0^2}-E+E_0).

where $pc=\sqrt{E^2-E_0^2}$ has been taken into account. This result is just the same as Eq. (1B.11). The expression for $h\nu'$ is then the same as Eq. (1B.13).

Question 3. (2) can also be solved as follows

For the sake of simplification, it is assumed that the scattered photon and electron move in the same plane which the incident photon and electron moved in. Meanwhile the angles which the directions of the scattered photon and electron make with the direction of the incident electron are denoted by $\psi$ and $\varphi$ respectively (see the figure). Then, we have

Perpendicular collision: electron E,p moving right, photon hν moving down; after the collision the photon hν′ leaves at angle ψ and the electron E′,p′ at angle φ below the dashed horizontal line.

E+h\nu=E'+h\nu',\quad \text{(Conservation of energy)}\quad \text{(1B.1')}

p=\frac{h\nu'}{c}\cos\psi+p'\cos\varphi,

pc-h\nu'\cos\psi=p'c\cos\varphi,\quad \text{(Conservation of horizontal momentum)}\quad \text{(1B.2')}

and

\frac{h\nu}{c}=\frac{h\nu'}{c}\sin\psi+p'\sin\varphi,

h\nu-h\nu'\sin\psi=p'c\sin\varphi.\quad \text{(Conservation of vertical momentum)}\quad \text{(1B.3')}

(1B.2') $^2$ +(1B.3') $^2$ leads to

p'^2c^2=p^2c^2+(h\nu')^2\cos^2\psi-2pch\nu'\cos\psi+(h\nu)^2+(h\nu')^2\sin^2\psi-2h\nu h\nu'\sin\psi \qquad \text{(1B.4')}

Square of Eq.(1B.1') results in

E'^2=E^2+(h\nu)^2+(h\nu')^2+2Eh\nu-2Eh\nu'-2h\nu h\nu'.

Substitution of the energy-momentum relation $E'^2=E_0^2+p'^2c^2,E^2=E_0^2+p^2c^2$ into the above equation of energy conservation leads to

p'^2c^2=p^2c^2+(h\nu)^2+(h\nu')^2+2Eh\nu-2Eh\nu'-2h\nu h\nu' \qquad \text{(1B.5')}

Comparison between Eq. (1B.4') and Eq. (1B.5') yields the energy of the scattered photon

h\nu'=\frac{Eh\nu}{E+h\nu-(pc\cos\psi+h\nu\sin\psi)}. \qquad \text{(1B.6')}

From (1B.6') it can be seen that if $\psi=\cos^{-1}\frac{pc}{\sqrt{p^2c^2+h^2\nu^2}}$ , the energy of the scattered photon reaches the maximum,

h\nu'_{\max}=\frac{Eh\nu}{E+h\nu-\sqrt{p^2c^2+h^2\nu^2}}. \qquad \text{(1B.7')}

The energy that the photon gets from the electron is

\Delta=h\nu'_{\max}-h\nu=\frac{\sqrt{p^2c^2+h^2\nu^2}h\nu-(h\nu)^2}{E+h\nu-\sqrt{p^2c^2+h^2\nu^2}}. \qquad \text{(1B.8')}

The extreme condition for $\Delta$ is

\frac{d\Delta}{d(h\nu)}=\frac{A}{(E+h\nu-a)^2}=0, \qquad \text{(1B.9')}

where $\sqrt{p^2c^2+h^2\nu^2}=a$

and $A=(a+\frac{(h\nu)^2}{a}-2h\nu)(E+h\nu-a)-(h\nu a-h^2\nu^2)(1-\frac{h\nu}{a})$ .

$A=0$ results in

Ea+h\nu a-p^2c^2-(h\nu)^2+\frac{E(h\nu)^2}{a}+\frac{(h\nu)^3}{a}-(h\nu)^2-2h\nu E-2(h\nu)^2+2h\nu a

=h\nu a-(h\nu)^2-(h\nu)^2+\frac{(h\nu)^3}{a}

Simplifying this equation leads to

Ea-p^2c^2+\frac{E(h\nu)^2}{a}-2h\nu E-2(h\nu)^2+2h\nu a=0

i.e.,

(E+2h\nu)a+\frac{E(h\nu)^2}{a}=p^2c^2+2h\nu E+2(h\nu)^2.

Squaring both of the two sides of this equation yields

(E+2h\nu)^2a^2+\frac{E^2(h\nu)^4}{a^2}+2(E+2h\nu)E(h\nu)^2

=(pc)^4+4(h\nu)^2E^2+4(h\nu)^4+4(pc)^2h\nu E+4(pc)^2(h\nu)^2+8(h\nu)^3E

Substitution of $\sqrt{p^2c^2+h^2\nu^2}=a$ into the above equation and making some simplifications yield

E^2p^2c^2+\frac{E^2(h\nu)^4}{p^2c^2+h^2\nu^2}=p^4c^4+h^2\nu^2E^2,

that is,

E^2(pc)^4+E^2(pc)^2(h\nu)^2+E^2(h\nu)^4=(pc)^6+(pc)^4(h\nu)^2+E^2(pc)^2(h\nu)^2+(h\nu)^4E^2.

After some simplifications we obtain

(pc)^4(h\nu)^2=(pc)^4(E^2-p^2c^2)=(pc)^4E_0^2,

which yields

h\nu=E_0. \qquad \text{(1B.10')}

Substitution of (1B.10') into Eq. (1B.7') leads to

h\nu'_{\max}=\frac{EE_0}{E+E_0-E}=E. \qquad \text{(1B. 11')}

The results (1B.10') and (1B.11') are just the same as Eqs. (1B.15) and (1B.16) in the former solution.

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf