Photon-Electron Interaction

If a high energy photon interacts with an electron, the interaction can be described by the Compton scattering relationship or by the 4-vector formulation of relativistic momentum. As a specific example, consider a 10GeV photon in a head-on collision with an electron at rest.

若高能光子与电子相互作用，该相互作用可由康普顿散射关系或相对论动量的4矢量表述来描述。作为具体例子，考虑一个10GeV的光子与静止电子发生正碰。

如果我们应用康普顿公式

with λ = h/p (deBroglie relationship) for a back-scattered photon where θ = 180°, this relationship can be expressed

对于背散射光子（θ = 180°）的德布罗意关系（λ = h/p），该关系可以表示为

并重新排列得到

由于 p_i c 远大于 m_e c^2

Conservation of energy then tells us that the electron energy after the collision is 9.999744 GeV. So essentially all of the energy of the photon is transferred to the electron. This situation can be pictured conceptually as follows:

能量守恒告诉我们，碰撞后电子的能量是9.999744 GeV。因此，几乎所有的光子能量都转移到了电子上。这种情况可以概念性地描绘如下：

The conceptual description that leads you to the conclusion that the electron has most of the energy after the collision involves transforming the momenta to the zero-momentum frame, which diminishes the momentum of the photon while increasing that of the electron. Then the momenta exchange upon collision in the zero momentum frame. Then you make the reverse transformation which further reduces the magnitude of the photon momentum while increasing the momentum of the electron. For the extreme example used here, the electron gets most of the energy.

导致你得出结论，即碰撞后电子获得了大部分能量的理论描述，涉及将动量转换到零动量框架。这会减少光子的动量而增加电子的动量。然后在零动量框架中，动量发生交换。接着你进行反向转换，这将进一步减少光子的动量大小而增加电子的动量。对于这里使用的极端例子，电子获得了大部分的能量。

The transformation to the zero-momentum frame, however, is not a matter of transforming the photon down to 5 GeV and giving the electron a momentum -5GeV/c as you might expect. The transformation is much more drastic. It can be accomplished by using the energy and momentum as a 4-vector and doing the necessary Lorentz transformation.

然而，转换到零动量参考系并不是将光子降级到5 GeV并给电子赋予-5 GeV/c的动量这样的简单过程。这种转换要剧烈得多。可以通过将能量和动量作为四矢量使用，并进行必要的洛伦兹变换来实现。

The tool to be used is the invariance of the length of the 4-vector upon transformation. Evaluated in the laboratory frame, it is

所用的工具是四矢量在变换下的不变性。在实验室参考系中，它表现为

Then in the zero-momentum frame, this energy is divided between the photon and the electron in such a way that their momenta are equal in magnitude. Using the relativistic energy relationship to express the length of the 4-vector in terms of the momenta gives

然后在零动量参考系中，这种能量是被分配给光子和电子的，使得它们的动量大小相等。利用相对论能量关系将4矢量的长度用动量表示出来，得到

Solving this equation iteratively gives p*c = 50.5476 MeV. Now we can determine the frame speed of the zero-momentum frame by determining the speed required to Doppler shift the 10 GeV photon down to 50.5476 MeV.

解这个方程可得到 p*c = 50.5476 MeV。现在我们可以确定零动量参考系的速度，即确定使10 GeV光子频移至50.5476 MeV所需的速度。

这使我们能够设定能量/动量变换。

This transformation does indeed give us zero momentum and conserve energy. Reversing the sign of β should take us back to the laboratory frame.

这种变换确实给出了零动量并守恒能量。反转β的符号应能带我们回到实验室参考系。

This transformation gives us the same results that were obtained by the Compton scattering approach.