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Chapter 5: Problem 12

A fast electron of rest mass \(m\) decelerates in a collision with a heavy nucleus and emits a (bremsstrahlung) photon. Prove that the energy of the photon can range all the way up to \((\gamma-1) m c^{2}\), the kinetic energy of the electron. [Hint: use a four-vector argument.]

Short Answer

Expert verified
Question: Prove that the energy of the emitted bremsstrahlung photon in a collision between a fast electron and a heavy nucleus can range up to the kinetic energy of the electron. Answer: Using four-momentum conservation and the energy-momentum four-vector, we showed that the maximum energy of the emitted photon is given by \((\gamma - 1)mc^2 - mc^2 = (\gamma - 2)mc^2\), where \(\gamma\) is the Lorentz factor of the electron and \(m\) is its rest mass. This means that the energy of the photon can range all the way up to \((\gamma-1)mc^{2}\), which is the kinetic energy of the electron.

Step by step solution

01

Write the four-vector conservation equation for the collision

We'll start by writing the conservation of four-momentum for the collision of the electron and the nucleus. Before the collision, the four-momentum of the electron is \((E/c, p_x, p_y, p_z)\), and the four-momentum of the nucleus is \((E_N/c, P_{Nx}, P_{Ny}, P_{Nz})\). After the collision, the four-momentum of the electron is \((E'/c, p'_x, p'_y, p'_z)\) and the four-momentum of the photon is \((\frac{E_\text{photon}}{c}, k_x, k_y, k_z)\). The conservation equation, therefore, is: \[(E + E_N, p_x + P_{Nx}, p_y + P_{Ny}, p_z + P_{Nz}) = \left(\frac{E' + E_\text{photon}}{c}, p'_x + k_x, p'_y + k_y, p'_z + k_z\right).\]
02

Calculate \(\gamma\) for the electron

The Lorentz factor \(\gamma\) is given by the following equation: \[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}},\] where \(v\) is the speed of the electron. As we are not given the speed, we leave \(\gamma\) as is and find that the energy of the electron can be written in terms of \(\gamma\) as \(E = (\gamma - 1)mc^2\).
03

Use the conservation equation to find the maximum energy of the photon

Now we have the four-momentum conservation equation and the energy of the electron in terms of \(\gamma\). We can use these to find the maximum energy of the photon. Since the mass of the nucleus is much larger than the mass of the electron, we can assume that the nucleus doesn't move significantly during the collision. And since the photon is emitted, the collision is elastic, so the following equation holds: \[\frac{E + E_N}{c} = \frac{E' + E_\text{photon}}{c}.\] Rearranging this equation and substituting in the electron's energy expression, we have: \[E_\text{photon} = E - E' = (\gamma - 1)mc^2 - E'.\] To find the maximum energy of the photon \(E_\text{photon}\), we need to minimize the final energy of the electron \(E'\), which occurs when the electron comes to rest. In this case, \(E' = mc^2\) and, therefore, the maximum energy of the photon is given by: \[E_\text{photon\,max} = (\gamma - 1)mc^2 - mc^2 = (\gamma - 2)mc^2.\] This proves that the energy of the photon can range all the way up to \((\gamma-1) m c^{2}\), which is the kinetic energy of the electron.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Four-Vector Conservation
Understanding the principle of four-vector conservation is paramount when dealing with collisions in particle physics, especially within the framework of special relativity. In such scenarios, we consider not only the three spatial components of momentum but also time—or energy, when considered in units consistent with momentum. This gives rise to the concept of the four-momentum vector.

In the case of a collision, just like the one involving the bremsstrahlung photon emission by an electron, the total four-momentum before and after the interaction must be conserved. This means that the sum of the four-momenta of the electron and the nucleus before the collision equals the sum of the four-momenta of the resulting particles post-collision. This adherence to conservation is a cornerstone in calculating the possible outcomes of particle interactions and is crucial in determining the energy spectrum of emitted photons, such as the maximum energy of a bremsstrahlung photon.
Special Relativity
Special relativity is a fundamental theory introduced by Albert Einstein that revolutionized the understanding of space, time, and energy. At the heart of this theory lies the invariance of the speed of light and the relativity of simultaneity. It significantly affects how we calculate the energy and momentum of particles moving at velocities close to the speed of light.

Key implications of special relativity include time dilation, length contraction, and the equivalence of mass and energy, encapsulated in the famous equation E=mc^2. When we deal with the energy of a particle, such as the electron emitting a photon in bremsstrahlung, we must account for relativistic effects to accurately describe its kinetic energy and momentum.
Lorentz Factor
The Lorentz factor, commonly denoted as \(\gamma\), plays a central role in the equations of special relativity. It quantifies the amount of time dilation, length contraction, and the increase of mass a moving body experiences as its speed approaches the speed of light. Mathematically, it is expressed as \(\gamma = \frac{1}{{\sqrt{1 - \frac{v^2}{c^2}}}}\), with \(v\) being the velocity of the moving object and \(c\) the speed of light.

The Lorentz factor allows us to adjust Newtonian mechanics to high-velocity scenarios. For instance, to determine the range of photon energy in bremsstrahlung, applying the Lorentz factor to the kinetic energy of the electron lets us predict the phenomena accurately within the relativistic regime.
Kinetic Energy in Relativity
In the realm of high-speed particles, kinetic energy deviates from the classical \(\frac{1}{2}m v^2\) formula. Relativistically, the total energy (E) of a particle with rest mass (m) moving at velocity (v) is given by \(E = \gamma m c^2\), where \(\gamma\) is the Lorentz factor and \(c\) is the speed of light. The kinetic energy (K) can then be isolated as \(K = E - mc^2\), simplifying to \(K = (\gamma - 1)mc^2\).

This formulation is critical when predicting the energy of photons emitted through bremsstrahlung, as the maximum photon energy corresponds to the initial kinetic energy of the electron before the emission, adjusted for relativistic speeds.
Collision of Particles
In particle physics, collisions are events where two or more particles come together and interact, leading to the potential exchange of energy and momentum. These interactions are governed by conservation laws, including the conservation of four-momentum. A collision can be elastic, where kinetic energy is conserved, or inelastic, where kinetic energy is not conserved but total energy still is.

During the bremsstrahlung process, the electron, upon colliding with a heavy nucleus, decelerates and emits a photon. By applying the conservation of four-momentum, we conclude that the emitted photon's energy must account for the loss in the electron's kinetic energy. The maximum photon energy, hence, would be when the electron gives up all its kinetic energy, minus its rest mass energy, to the photon in a perfectly elastic collision.

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Most popular questions from this chapter

If one neutron and one pi-meson are to ernerge from the collision of a photon with a stationary proton, find the threshold frequency of the photon in terms of the rest mass \(n\) of a proton o neutron (here assumed equal) and that, \(m\), of a pi-meson. [Answer \(\left.c^{2}\left(m^{2}+2 m n\right) / 2 h n .\right]\)

A particle of rest mass \(m\) decays from rest into a particle of rest mass \(m\) ' and a photon. Find the separate energies of these end products. [Answer: \(c^{2}\left(m^{2} \pm m^{\prime 2}\right) / 2 m\). Hint: use a four-vector argument.]

. Show that a photon cannot spontaneously disintegrate into an electron- positron pair. [Hint: four-momentum conservation.] But in the presence of a stationary nucleus (acting as a kind of catalyst) it can. If the rest mass of the nucleus is \(N\), and that of the electron (and positron) is \(m\), what is the threshold frequency of the photon? Verify that for large \(N\) the efficiency is \(\sim 100\) per cent (cf. the preceding problem), so that the nucleus then comes close to being a pure catalyst.

If \(\Delta x^{*}=(c \Delta t, \Delta \mathbf{r})\) is the four-vector join of two events on the worldline of a uniformly moving particle (or photon), prove that the frequency vector of its de Broglie wave is given by $$ N^{\mu}=\frac{v}{c \Delta t} \Delta x^{\mu}, $$ whence \(v / \Delta t\) is invariant. Compare with Exercise III \((7)\).

Uniform parallel radiation is observed in two arbitrary inertial frames \(\mathbf{S}\) and \(\mathbf{S}^{\prime}\) in which it has frequencies \(v\) and \(v^{\prime}\) respectively. If \(p, g, \sigma\) denote, respectively, the radiation pressure, momentum density, and energy density of the radiation in \(S\), and primed symbols denote corresponding quantities in \(S^{\prime}\), prove \(p^{\prime} / p=g^{\prime} / g=\sigma^{\prime} / \sigma\) \(=v^{\prime 2} / v^{2} .[\) Hint: Exercise III \((17) .]\)
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