Difference between revisions of "Inverse Compton Scattering"

Inverse Compton Scattering

When the electron has significant kinetic energy as compared to the energy of the photon, then energy can be transferred from the electron to the photon, resulting in the scattered photon having a higher frequency.

Mathematical Derivation of Inverse Compton Scattering

To find the final energy of the photon, we must perform a couple Lorentz transformations. The standard Compton equations are valid only from the electron’s rest frame. The angles and frequencies of the photon as seen by the moving electron are different than what is measured in the lab frame.

Inverse Compton scattering geometry in two reference frames.

From the relativistic Doppler shift formula (Eqn 4.12 in R & L) we can write the initial photon energy as seen by the electron. (For this derivation, we indicate the electron rest frame as the primed reference frame and the lab frame as the unprimed frame.)

${\displaystyle E_{\gamma _{i}}^{\prime }=E_{\gamma _{i}}\gamma (1-\beta \cos \theta )\,\!}$

where ${\displaystyle \beta ={\frac {v}{c}}}$ is the beta factor from the classic Doppler shift and ${\displaystyle \theta }$ is the angle between the electron trajectory and the photon as seen in the lab frame.

Now suppose that the photon (which entered at angle ${\displaystyle \theta ^{\prime }}$ in the electron frame) rebounds at an angle ${\displaystyle \theta _{f}^{\prime }}$. To relate this to our previous derivation, we want to find the change in angle ${\displaystyle \phi }$. So note that ${\displaystyle \theta _{f}^{\prime }-\phi ^{\prime }=\theta ^{\prime }}$. Thus, in the electron’s frame, using the standard Compton energy equation, we find:

${\displaystyle E_{\gamma _{f}}^{\prime }={E_{\gamma _{i}}^{\prime }{\frac {1}{1+{E_{\gamma _{i}}^{\prime }{\frac {1}{m_{e}c^{2}}}(1-\cos \phi ^{\prime })}}}}\,\!}$

Transforming this back into the lab frame give the final Compton scattered photon energy:

${\displaystyle E_{\gamma _{f}}=E_{\gamma _{f}}^{\prime }\gamma (1+{\frac {v}{c}}\cos \theta _{f}^{\prime })\,\!}$

This a pretty messy final equation, so we will just look at the limits of the solution:

• If ${\displaystyle E_{\gamma _{i}}^{\prime }\ll m_{e}c^{2}}$, then:
{\displaystyle {\begin{aligned}E_{\gamma _{f}}&\approx E_{\gamma _{i}}^{\prime }\gamma (1+{\frac {v}{c}}\cos \theta _{f}^{\prime })\\&\approx \gamma ^{2}E_{\gamma _{i}}(1+{\frac {v}{c}}\cos \theta _{f}^{\prime })(1-{\frac {v}{c}}\cos \theta )\\\end{aligned}}\,\!}

In a “typical” collision, ${\displaystyle \theta \sim \theta _{f}^{\prime }\sim {\frac {\pi }{2}}}$, so ${\displaystyle E_{\gamma _{f}}\sim \gamma ^{2}E_{\gamma _{i}}}$. This shows that the outgoing photon can have a larger energy than the incoming photon. Even if the energy of the incoming photon as seen by the electron is low, as long as the electron has a large kinetic energy, the energy of the outgoing photon can increase.

• If ${\displaystyle E_{\gamma _{i}}^{\prime }\gg m_{e}c^{2}}$, then:
{\displaystyle {\begin{aligned}E_{\gamma _{f}}^{\prime }&={E_{\gamma _{i}}^{\prime }{\frac {1}{1+{E_{\gamma _{i}}^{\prime }{\frac {1}{m_{e}c^{2}}}}}}(1-\cos \phi ^{\prime })}\approx m_{e}c^{2}\\E_{\gamma _{f}}&=E_{\gamma _{f}}^{\prime }\gamma (1+{\frac {v}{c}}\cos \theta _{f}^{\prime })\\&\approx m_{e}c^{2}\gamma (1+{\frac {v}{c}}\cos \theta _{f}^{\prime })\approx \gamma m_{e}c^{2}\\\end{aligned}}\,\!}

This final term defines the maximum rebound of the photon.

Scattering of Single Electron off of a Sea of Photons

Recall the following rules for photons bouncing off of relativistic electrons:

{\displaystyle {\begin{aligned}E_{f}&\sim \gamma ^{2}E\\max(E_{f})&\sim \gamma m_{e}c^{2}\\\end{aligned}}\,\!}

We’d like next to discuss the behavior of a single ${\displaystyle e^{-}}$ swimming through a sea of photons. We’d like to know how much power this ${\displaystyle e^{-}}$ is going to scatter by Compton-upscattering these photons. To order of magnitude, the power scattered by a single relativistic (${\displaystyle \gamma \gg 1}$) electron should depend on the cross-section for scattering, the speed of the electron (${\displaystyle c}$), the # density of photons having various energies, and the energy these photons take from scattering off of the electron:

{\displaystyle {\begin{aligned}P&\sim \sigma _{T}c\int {\eta (E)dE\,E_{f}(E)}\\&\sim \sigma _{T}c\int {\eta (E)dE\,\gamma ^{2}E}\\&\sim \gamma ^{2}c\sigma _{T}U_{ph}\\\end{aligned}}\,\!}

where ${\displaystyle U_{ph}}$ is the energy density of the photon field. This looks very similar to the power radiated by the synchrotron magnetic field, which is ${\displaystyle P_{synch}\sim U_{B}c\sigma _{T}\gamma ^{2}}$.

This was an order of magnitude derivation, but we can do better–we just need to be more careful about how much the photons end up with, versus how much the electron actually gave to the photons (they had energy to begin with). We made the assumption that the initial energy of the photons was negligible because the electron was relativistic. However, what follows will be true for any ${\displaystyle \gamma }$.

First, note that the energy bequeathed to the photon bath is given by ${\displaystyle P_{net}=P_{scat}-P_{incident}}$:

{\displaystyle {\begin{aligned}P_{inc}&=c\sigma _{T}\int {\eta (E)E\,dE}\\&=c\sigma _{T}U_{ph}\\\end{aligned}}\,\!}

The power scattered out by the electron should be equal to the power radiated by the ${\displaystyle e^{-}}$ in its own rest frame, due to accelerations caused by ${\displaystyle {\vec {E}}^{\prime }}$-fields of photons seen in its rest frame (recall that power is Lorentz-invariant quantity). In the ${\displaystyle e^{-}}$’s rest frame, the ${\displaystyle {\vec {B}}}$’s of the photons don’t produce accelerations (there’s no velocity), so all we need to consider are the Lorentz-transformed ${\displaystyle {\vec {E}}}$’s of the photons:

${\displaystyle P^{\prime }={2 \over 3}{e^{2}(a^{\prime })^{2} \over c^{3}}={2 \over 3}{e^{2} \over c^{3}}\left({e{\vec {E}}^{\prime } \over m_{e}}\right)^{2}={2 \over 3}{e^{4} \over m_{e}^{2}c^{3}}({\vec {E}}^{\prime })^{2}\,\!}$

Recall that the Lorentz-transformed ${\displaystyle {\vec {E}}}$’s look like:

{\displaystyle {\begin{aligned}E_{x}^{\prime }&=E_{x}\\E_{y}^{\prime }&=\gamma E_{y}-{\gamma v \over c}B_{z}\\E_{z}^{\prime }&=\gamma E_{z}+{\gamma v \over c}B_{y}\\\end{aligned}}\,\!}

for an electron moving in the ${\displaystyle {\hat {x}}}$ direction. Substituting these values into our equation for ${\displaystyle P^{\prime }}$:

{\displaystyle {\begin{aligned}\left\langle {P^{\prime }}\right\rangle &={2 \over 3}{e^{4} \over m_{e}^{2}c^{3}}\left(\left\langle {E_{x}^{2}}\right\rangle +\gamma ^{2}\left\langle {E_{y}^{2}}\right\rangle +\gamma ^{2}\beta ^{2}\left\langle {B_{z}^{2}}\right\rangle -\underbrace {2\gamma \beta \left\langle {E_{y}B_{z}}\right\rangle } _{=0}+\gamma ^{2}\left\langle {E_{z}^{2}}\right\rangle +\gamma ^{2}\beta ^{2}\left\langle {B_{y}^{2}}\right\rangle +\underbrace {2\gamma ^{2}\beta \left\langle {E_{z}B_{y}}\right\rangle } _{=0}\right)\\&={2 \over 3}{e^{4}\left\langle {E_{x}^{2}}\right\rangle \over m_{e}^{2}c^{3}}\left(1+2\gamma ^{2}+2\gamma ^{2}\beta ^{2}\right)\\&=\underbrace {{2 \over 3}{e^{4} \over m_{e}^{2}c^{4}}} _{=\sigma _{T} \over 4\pi }{c \over 3}\underbrace {3\left\langle {E_{x}^{2}}\right\rangle \over 4\pi } _{=U_{ph}}4\pi (1+2\gamma ^{2}+2\gamma ^{2}\beta ^{2})\\&=\sigma _{T}cU_{ph}{1 \over 3}(2\gamma ^{2}+2(\gamma ^{2}-1)+1)\\\end{aligned}}\,\!}

Recall that ${\displaystyle \gamma ^{2}\beta ^{2}=\gamma ^{2}-1}$, so ${\displaystyle \left\langle {P^{\prime }}\right\rangle =\sigma _{T}cU_{ph}{1 \over 3}(4\gamma ^{2}-1)}$, and our net power scattered is:

${\displaystyle {P_{net}={4 \over 3}\sigma _{T}U_{ph}c\beta ^{2}\gamma ^{2}}\,\!}$

This is an exact expression. Note that if ${\displaystyle \gamma \gg 1}$ then the average photon energy of Compton-upscattered radiation is:

{\displaystyle {\begin{aligned}\left\langle {h\nu }\right\rangle _{1}&={P \over {U_{ph} \over \left\langle {h\nu }\right\rangle }\sigma _{T}c}\\&=\left\langle {h\nu }\right\rangle {4 \over 3}\gamma ^{2}\beta ^{2}\\\end{aligned}}\,\!}

Another note: Compare our expression for the Compton-upscattering radiation to the power radiated by a single electron undergoing synchrotron radiation:

${\displaystyle P_{sync}=2\sigma _{T}cU_{B}\beta ^{2}\gamma ^{2}\left\langle {\sin ^{2}\alpha }\right\rangle \,\!}$

The only difference (ignoring the ${\displaystyle \sin ^{2}\alpha }$), is the exchange of ${\displaystyle U_{B}}$ for ${\displaystyle U_{ph}}$.

Synchrotron Self-Compton (SSC)

Electrons undergoing synchrotron radiation create a photon bath which other electrons will then interact with via inverse Compton scattering. Recall that for original (unprocessed) synchrotron radiation, that ${\displaystyle F_{\nu }}$, between some minimum and maximum frequency cut-off, goes as ${\displaystyle K\nu ^{\alpha }}$, and that the number of photons per ${\displaystyle \gamma }$ is ${\displaystyle {dN \over d\gamma }=N_{0}\gamma ^{s}}$, where ${\displaystyle \alpha ={1+s \over 2}}$. These frequency cut-offs were set by ${\displaystyle \gamma _{min}^{2}\nu _{cyc}}$ and ${\displaystyle \gamma _{max}^{2}\nu _{cyc}}$. After this radiation is processed by SSC, approximately every photon is upscattered to a new energy ${\displaystyle {4 \over 3}\gamma ^{2}\nu }$. We are assuming that the relationship between an incoming photon frequency and it’s final frequency are related via a delta function. Thus:

${\displaystyle F_{\nu ,SSC}(\nu )=\tau \int _{\tilde {\nu }}{K{\tilde {\nu }}^{\alpha }d{\tilde {\nu }}\delta \left({\tilde {\nu }}-{\nu \over \gamma ^{2}}\right)\int _{\gamma }{N_{0}\gamma ^{s}d\gamma }}\,\!}$

Keep in mind that ${\displaystyle N_{0}}$ is normalized to so the integral comes out to 1 (it just accounts for the “shape” of the energy distribution function). ${\displaystyle \tau }$ is what contains the actual # density of ${\displaystyle e^{-}}$’s. It is the fraction scattered, and is generally ${\displaystyle \ll 1}$. ${\displaystyle \nu \sim {\tilde {\nu }}\gamma ^{2}}$.

For a fixed ${\displaystyle \nu \sim {\tilde {\nu }}\gamma ^{2}}$, we find that ${\displaystyle \gamma \sim \left({\nu \over {\tilde {\nu }}}\right)^{\frac {1}{2}}\propto {\tilde {\nu }}^{-{\frac {1}{2}}}}$.