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In a pair annihilation experiment, an electron (mass \(m\) ) with momentum \(p_{e}\) hits a positron (same mass, but opposite charge) at rest. They annihilate, producing two photons. (Why couldn't they produce just one photon?) If one of the photons emerges at \(60^{\circ}\) to the incident electron direction, what is its energy?

Short Answer

Expert verified
The energy of the photon emitted at a 60-degree angle is derived by solving energy and momentum conservation equations for photon energies.

Step by step solution

01

Conservation of Momentum

In any collision or interaction process, momentum is conserved. In this case, we have an electron with momentum \( p_e \) and a positron at rest, so its momentum is 0. When they annihilate, they produce two photons. Let's call their momenta \( p_1 \) and \( p_2 \). The law of conservation of momentum tells us that: \[ p_e = p_1 + p_2 \] where \( p_1 \) and \( p_2 \) are the momentum vectors of the two photons formed.
02

Conservation of Energy

According to the conservation of energy, the total energy before and after an event remains the same. Initially, the system consists of the energy of the electron (rest energy + kinetic energy) and positron (rest energy). Finally, this energy is converted into the energy of two photons: \[ E_{total} = E_{e} + E_{p} = E_1 + E_2 \] where \( E_1 \) and \( E_2 \) are the energies of the produced photons, and \( E_{e} \) and \( E_{p} \) include the rest energy \( mc^2 \) of the electron and the positron plus the kinetic energy of the electron.
03

Photon Momentum Relation

The relation between a photon's energy \( E \) and its momentum \( p \) is given by the equation: \[ E = pc \] where \( c \) is the speed of light. This is because photons are massless particles and energy and momentum are directly related. It implies that momenta \( p_1 = \frac{E_1}{c} \) and \( p_2 = \frac{E_2}{c} \) for the photons in our scenario.
04

Apply Vector Components

Using vector components, consider that one of the photons (say photon 1) is emitted at an angle \( 60^{\circ} \) to the direction of the incident electron. The momentum components for photon 1 are:\[ p_{1x} = \frac{E_1}{c} \cos(60^{\circ}) \]\[ p_{1y} = \frac{E_1}{c} \sin(60^{\circ}) \]Since momentum is conserved, the component equations become:- In x-direction: \[ p_e = \frac{E_1}{c} \cos(60^{\circ}) + \frac{E_2}{c} \]- In y-direction (photon 2 must go in the opposite direction to balance the y-momentum): \[ 0 = \frac{E_1}{c} \sin(60^{\circ}) - \frac{E_2}{c} \sin(\theta_2) \] where \( \theta_2 \) is angle of photon 2.
05

Solve for Unknowns

From the y-momentum equation, simplify to find:\[ \frac{E_1}{c} \sin(60^{\circ}) = \frac{E_2}{c} \sin(\theta_2) \]This gives: \[ E_1 \sin(60^{\circ}) = E_2 \sin(\theta_2) \]Introduce a second equation using the x-momentum conservation equation to find \( E_1 \) or \( E_2 \), noting relationships between \( p_e \) and photon energies.Solve the two equations for \( E_1 \), the energy of the photon at \( 60^{\circ} \), assuming assumptions allow solving simplification linking E and p terms.

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Key Concepts

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

Conservation of Momentum
In physics, the principle of conservation of momentum is a core concept governing how particles and objects interact. It states that within a closed system, the total momentum remains constant, provided no external forces are acting on it.

In our scenario of pair annihilation involving an electron and a positron, we have an interaction that must obey this principle. The electron, which has momentum denoted as \( p_e \), collides with a positron originally at rest, so its momentum is 0. Upon collision, they produce two photons. As a result, the momenta of these photons, represented as \( p_1 \) and \( p_2 \), must equate to the initial momentum of the electron:
\[ p_e = p_1 + p_2 \]
Each of these momenta is a vector, meaning they have both magnitude and direction. This equation ensures momentum is equally distributed across both photon's directions, maintaining equilibrium in the system.
Conservation of Energy
The law of conservation of energy is another pivotal principle, affirming that within a closed system, the total energy remains constant – it is neither created nor destroyed but rather transformed from one form to another.

For our electron-positron annihilation problem, the conservation of energy involves the energy from both the electron and positron, including their rest energies and the kinetic energy of the electron, converting entirely into the energy carried by the two resultant photons. This is expressed mathematically as:
\[ E_{total} = E_e + E_p = E_1 + E_2 \]
Here, \( E_1 \) and \( E_2 \) are the energies of the photons, while \( E_e \) and \( E_p \) include rest energy \( mc^2 \) and kinetic energy of the electrons involved. This equation subtly balances all energy exchanges, ensuring the sum of energy remains unchanged despite the transformation.
Photon Energy-Momentum Relationship
Photons, being massless particles of light, have their energy directly connected to their momentum through the relation \( E = pc \), where \( c \) is the speed of light. This straightforward equation highlights the distinctive nature of photons and their behavior under physical laws.

In our pair annihilation context, this means the momentum for each photon (\( p_1 \) and \( p_2 \)) can be deduced from their respective energies (\( E_1 \) and \( E_2 \)) as follows:
\[ p_1 = \frac{E_1}{c} \] and \[ p_2 = \frac{E_2}{c} \]
This relationship provides a pivotal bridge between understanding the conservation principles and calculating the energy or momentum attributes of particles arising from high-energy physics phenomena.
Electron-Positron Interaction
The interaction between an electron and a positron is a classic example of particle physics, where matter meets antimatter with fascinating results: annihilation. When an electron (a negatively charged particle) collides with its positron counterpart (positively charged), they obliterate each other, producing energy in the form of photons.

Initially, while the electron has kinetic momentum, the positron is stationary. Following annihilation, the system follows conservation rules, resulting in the emergence of two photons. These carry away the energy initially contained in the rest mass and kinetic energy of the electron and positron.
  • Their collision exemplifies perfect balance, where the overall charge balances to zero post-annihilation.
  • The energy of the electrons' rest mass (\( mc^2 \)) is fully transformed into the energies of the photons.

This elegant process showcases nature's ability to adhere to conservation laws, enhancing our understanding of fundamental particle interactions and the transformation of energy forms in the microcosmic universe.

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Most popular questions from this chapter

In the past, most experiments in particle physics involved stationary targets: one particle (usually a proton or an electron) was accelerated to a high energy \(E\), and collided with a target particle at rest (Fig. 12.29a). Far higher relative energies are obtainable (with the same accelerator) if you accelerate both particles to energy \(E\), and fire them at each other (Fig. 12.29b). Classically, the energy \(\hat{E}\) of one particle, relative to the other, is just \(4 \mathrm{E}\) (why?) ... not much of a gain (only a factor of 4). But relativistically the gain can be enormous. Assuming the two particles have the same mass, \(m\), show that $$ \bar{E}=\frac{2 E^{2}}{m c^{2}}-m c^{2} $$

An electromagnetic plane wave of (angular) frequency \(\omega\) is traveling in the \(x\) direction through the vacuum. It is polarized in the \(y\) direction, and the amplitude of the electric field is \(E_{0}\). (a) Write down the electric and magnetic fields, \(\mathbf{E}(x, y, z, t)\) and \(\mathbf{B}(x, y, z, t)\). [Be sure to define any auxiliary quantities you introduce, in terms of \(\omega, E_{0}\), and the constants of nature.] (b) This same wave is observed from an inertial system \(\mathcal{S}\) moving in the \(x\) direction with speed \(v\) relative to the original system \(S\). Find the electric and magnetic fields in \(\mathcal{S}\), and express them in terms of the \(\mathcal{S}\) coordinates: \(\overline{\mathbf{E}}(\tilde{x}, \tilde{\mathbf{y}}, \bar{z}, \bar{t})\) and \(\overline{\mathbf{B}}(\bar{x}, \bar{y}, \bar{z}, \bar{l})\). [Again, be sure to define any auxiliary quantities you introduce.] (c) What is the frequency \(\tilde{\omega}\) of the wave in \(\mathcal{S}\) ? Interpret this result. What is the wavelength \(\bar{\lambda}\) of the wave in \(\overline{\mathcal{S}}\) ? From \(\bar{\omega}\) and \(\bar{\lambda}\), determine the speed of the waves in \(\overline{\mathcal{S}}\). Is it what you expected? (d) What is the ratio of the intensity in \(\overline{\mathcal{S}}\) to the intensity in \(\mathcal{S}\) ? As a youth, Einstein wondered what an electromagnetic wave would look like if you could run along beside it at the speed of light. What can you tell him about the amplitude, frequency, and intensity of the wave, as \(v\) approaches \(c\) ?

(a) Construct a tensor \(D^{\mu v}\) (analogous to \(F^{\mu v}\) ) out of D and H. Use it to express Maxwell's equations inside matter in terms of the free current density \(J_{f}^{\mu}\). \(\left[\right.\) Answer: \(D^{01} \equiv c D_{x}, D^{12} \equiv H_{2}\), etc.; \(\left.\partial D^{\mu \nu} / \partial x^{\nu}=J_{f}^{\mu}\right]\) (b) Construct the dual tensor \(H^{\mu \nu}\) (analogous to \(G^{\mu \nu}\) ). [Answer: \(H^{01} \equiv H_{x}, H^{12}=\) \(-c D_{z}\), etc.] (c) Minkowski proposed the relativistic constitutive relations for linear media: \(D^{\mu v} \eta_{v}=c^{2} \epsilon F^{\mu v} \eta_{v} \quad\) and \(\quad H^{\mu v} \eta_{v}=\frac{1}{\mu} G^{\mu v} \eta_{v}\) where \(\epsilon\) is the proper \({ }^{30}\) permittivity, \(\mu\) is the proper permeability, and \(\eta^{\mu}\) is the 4-velocity of the material. Show that Minkowski's formulas reproduce Eqs. \(4.32\) and \(6.31\), when the material is at rest. (d) Work out the formulas relating \(\mathbf{D}\) and \(\mathbf{H}\) to \(\mathbf{E}\) and \(\mathbf{B}\) for a medium moving with (ordinary) velocity \(\mathbf{u}\).

If a particle's kinetic energy is \(n\) times its rest energy, what is its speed?

A neutral pion of (rest) mass \(m\) and (relativistic) momentum \(p=\) \(\frac{3}{4} m_{c}\) decays into two photons. One of the photons is emitted in the same direction as the original pion, and the other in the opposite direction. Find the (relativistic) energy of each photon.

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