$14.5.22.$ What is the maximum energy that photons with energy $\mathcal{E} = 10 \text{ eV}$ can acquire when scattered on a counter-propagating electron beam with energy $\mathcal{E}_e = 10^{10}\text{ eV}$?
### Solution
The problem can be solved simply by writing down the conservation of momentum and energy, but I want to show another method. One can notice that if we go to a reference frame where the electrons in the beam are at rest, we observe the [Compton effect](https://en.wikipedia.org/wiki/Compton_scattering). The change in energy upon scattering of a photon in this case can be obtained from the well-known formula
where $\alpha$ is the scattering angle of the photon. This formula is derived in problem [$14.5.23$](https://savchenkosolutions.com/en/14.5.23).
As a result, we need to carefully transform to such a system, calculate the energy, and then transform back to the laboratory frame.
@@ -15,6 +15,7 @@Solution
For convenience, we switch to a system of units where $c = 1$. This simplifies the notation, and it is convenient that the photon momentum equals its energy. We will denote energies in the laboratory frame (LF) by $\mathcal{E}$, and in the electron rest frame (RF) by $E$.
Thus, the gamma factor for the electron in the LF is obviously
This is the speed at which the RF moves relative to the LF. We find the photon energy in the RF using the relativistic Doppler effect:
$$
E=\mathcal{E}\gamma(1+v)
$$
Using the Compton formula, we obtain the energy change upon collision:
$$
E'=\frac{Em_e}{E(1-\cos\alpha)+m_e}
$$
The maximum energy in the LF is achieved when the photon is scattered backward in the RF ($\alpha = \pi$), because upon inverse transformation to the LF, a photon scattered backward in the RF receives the maximum Doppler shift forward (in the direction of the electron's motion):
$$
E'=\frac{Em_e}{2E+m_e}
$$
Inverse transformation (taking into account that the photon's direction has reversed):
$14.5.22.$ What is the maximum energy that photons with energy $\mathcal{E} = 10 \text{ eV}$ can acquire when scattered on a counter-propagating electron beam with energy $\mathcal{E}_e = 10^{10}\text{ eV}$?
$14.5.22.$ What is the maximum energy that photons with energy $\mathcal{E} = 10 \text{ eV}$ can acquire when scattered on a counter-propagating electron beam with energy $\mathcal{E}_e = 10^{10}\text{ eV}$?
### Solution
### Solution
The problem can be solved simply by writing down the conservation of momentum and energy, but I want to show another method. One can notice that if we go to a reference frame where the electrons in the beam are at rest, we observe the [Compton effect](https://en.wikipedia.org/wiki/Compton_scattering). The change in energy upon scattering of a photon in this case can be obtained from the well-known formula
The problem can be solved simply by writing down the conservation of momentum and energy, but I want to show another method. One can notice that if we go to a reference frame where the electrons in the beam are at rest, we observe the [Compton effect](https://en.wikipedia.org/wiki/Compton_scattering). The change in energy upon scattering of a photon in this case can be obtained from the well-known formula
where $\alpha$ is the scattering angle of the photon. This formula is derived in problem [$14.5.23$](https://savchenkosolutions.com/en/14.5.23).
where $\alpha$ is the scattering angle of the photon. This formula is derived in problem [$14.5.23$](https://savchenkosolutions.com/en/14.5.23).
As a result, we need to carefully transform to such a system, calculate the energy, and then transform back to the laboratory frame.
As a result, we need to carefully transform to such a system, calculate the energy, and then transform back to the laboratory frame.
@@ -15,6 +15,7 @@Solution
For convenience, we switch to a system of units where $c = 1$. This simplifies the notation, and it is convenient that the photon momentum equals its energy. We will denote energies in the laboratory frame (LF) by $\mathcal{E}$, and in the electron rest frame (RF) by $E$.
For convenience, we switch to a system of units where $c = 1$. This simplifies the notation, and it is convenient that the photon momentum equals its energy. We will denote energies in the laboratory frame (LF) by $\mathcal{E}$, and in the electron rest frame (RF) by $E$.
Thus, the gamma factor for the electron in the LF is obviously
Thus, the gamma factor for the electron in the LF is obviously
This is the speed at which the RF moves relative to the LF. We find the photon energy in the RF using the relativistic Doppler effect:
This is the speed at which the RF moves relative to the LF. We find the photon energy in the RF using the relativistic Doppler effect:
$$
$$
E=\mathcal{E}\gamma(1+v)
E=\mathcal{E}\gamma(1+v)
$$
$$
Using the Compton formula, we obtain the energy change upon collision:
Using the Compton formula, we obtain the energy change upon collision:
$$
$$
E'=\frac{Em_e}{E(1-\cos\alpha)+m_e}
E'=\frac{Em_e}{E(1-\cos\alpha)+m_e}
$$
$$
The maximum energy in the LF is achieved when the photon is scattered backward in the RF ($\alpha = \pi$), because upon inverse transformation to the LF, a photon scattered backward in the RF receives the maximum Doppler shift forward (in the direction of the electron's motion):
The maximum energy in the LF is achieved when the photon is scattered backward in the RF ($\alpha = \pi$), because upon inverse transformation to the LF, a photon scattered backward in the RF receives the maximum Doppler shift forward (in the direction of the electron's motion):
$$
$$
E'=\frac{Em_e}{2E+m_e}
E'=\frac{Em_e}{2E+m_e}
$$
$$
Inverse transformation (taking into account that the photon's direction has reversed):
Inverse transformation (taking into account that the photon's direction has reversed):
