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Two particles \(a\) and \(b\) with masses \(m_{a}=0\) and \(m_{b}>0\) approach one another. Prove that they have a CM frame (that is, a frame in which their total three-momentum is zero). [Hint: As you should explain, this is equivalent to showing that the sum of two four-vectors, one of which is forward light-like and one forward time-like, is itself forward time-like.]

Short Answer

Expert verified
Two particles have a CM frame because the sum of their four-vectors is forward time-like.

Step by step solution

01

Understanding the Problem

We are asked to prove that two particles, one with zero mass \( m_a = 0 \) and another with mass \( m_b > 0 \), have a center-of-mass (CM) frame, where the total momentum is zero. This involves using concepts from special relativity, specifically four-vectors.
02

Four-Vector Definitions

For particle \(a\), the four-momentum vector is light-like: \( p_{a}^\mu = (E_{a}, \vec{p}_{a}) \), with \( E_{a} = |\vec{p}_{a}| \). For particle \(b\), it is time-like: \( p_{b}^\mu = (E_{b}, \vec{p}_{b}) \), with \( E_{b} > |\vec{p}_{b}| \). The goal is to add these vectors and show that their sum is also a four-vector.
03

Summing the Four-Vectors

Calculate the sum of the four-vectors: \( P^\mu = p_{a}^\mu + p_{b}^\mu = (E_{a} + E_{b}, \vec{p}_{a} + \vec{p}_{b}) \). We need to demonstrate that \( P^\mu \) is forward time-like.
04

Condition for Forward Time-like Vector

A vector is forward time-like if its energy component is greater than its spatial component: \( E_{a} + E_{b} > |\vec{p}_{a} + \vec{p}_{b}| \). Since \( E_{b} > |\vec{p}_{b}| \) and \( E_{a} = |\vec{p}_{a}| \), combining these inequalities shows that the sum \( E_{a} + E_{b} \) is greater than the magnitude of the spatial component \(|\vec{p}_{a} + \vec{p}_{b}|\).
05

Conclusion: Existence of CM Frame

Since the sum of these four-vectors is forward time-like, there exists a reference frame, the center-of-mass frame, where the sum of the spatial components \( \vec{p}_{a} + \vec{p}_{b} = 0 \). Thus, the total momentum is zero in this frame.

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Key Concepts

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

Four-momentum Vector
In special relativity, the four-momentum vector is a crucial concept. It extends the idea of momentum from the familiar three-dimensional space into four-dimensional spacetime. This vector includes both energy and momentum components. Specifically, the four-momentum vector for a particle is expressed as: \( p^\mu = (E, \vec{p}) \), where \( E \) is the energy and \( \vec{p} \) is the three-dimensional momentum of the particle.

The energy in this context is often linked to the mass and velocity of the particle. For massive particles, energy is given by the relativistic energy-momentum relationship \( E^2 = (pc)^2 + (m_0c^2)^2 \), where \( m_0 \) is the rest mass of the particle and \( c \) is the speed of light. For a massless particle, such as a photon, the relation simplifies to \( E = pc \).

In exercises involving four-momentum vectors, like the current one, understanding how to add and interpret these vectors is crucial. Summing four-momentum vectors can indicate properties like collisions or transformations between different frames of reference.
Special Relativity
Special relativity, formulated by Albert Einstein, revolutionized our understanding of physics. It asserts that the laws of physics are the same in all inertial frames of reference. One of the key postulates of special relativity is the constancy of the speed of light, which remains the same for all observers, regardless of their motion relative to the light source.

This theory introduces important consequences such as time dilation and length contraction. Time dilation implies that a moving clock ticks slower compared to a stationary one, while length contraction indicates that an object's length shortens along the direction of motion as its speed approaches the speed of light.

Special relativity also provides the framework for analyzing four-momentum vectors, which are essential for understanding particle physics phenomena. These vectors appropriately account for the relativistic effects acting on moving particles, enabling precise calculations and predictions.
Light-like and Time-like Vectors
Light-like and time-like vectors are terms used to describe the nature of four-vectors within spacetime. A vector is light-like if it represents the path of light in spacetime; it satisfies the condition \( E = |\vec{p}| \). Light-like vectors are characteristic of massless particles such as photons.

Time-like vectors, on the other hand, pertain to particles that have a rest mass. These vectors satisfy the condition \( E > |\vec{p}| \). For such particles, there exists a reference frame where the particle is at rest, and the energy component of the four-vector is associated with the rest mass.

In the context of this exercise, different combinations of these vectors help determine if a center-of-mass frame exists. Adding a light-like vector and a time-like vector can result in a forward time-like vector, allowing for a zero total three-momentum in some frame, confirming the presence of a CM frame.
Reference Frame
In physics, a reference frame is essentially a perspective relative to which measurements are made. It is a coordinate system that specifies the position and motion of objects. Einstein’s theory of relativity emphasizes the significance of choosing a suitable reference frame for analysis.

Reference frames can be inertial, meaning they are either at rest or in uniform motion, and non-inertial, which involves acceleration. In an inertial reference frame, Newton's laws of motion hold true. The center-of-mass (CM) frame, often used in particle physics, is one such inertial frame where the total momentum is zero.

Finding the CM frame for two particles involves demonstrating that their combined momentum can be zero in this perspective, as seen in the current task. Such frames simplify calculations, particularly in collision problems, by transforming the complexity of motion into a manageable form.

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Most popular questions from this chapter

Consider two events that occur simultaneously at \(t=0\) in frame \(\mathcal{S},\) both on the \(x\) axis at \(x=0\) and \(x=a\). (a) Find the times of the two events as measured in a frame \(\mathcal{S}^{\prime}\) traveling in the positive direction along the \(x\) axis with speed \(V\). (b) Do the same for a second frame \(\mathcal{S}^{\prime \prime}\) traveling at speed \(V\) but in the negative direction along the \(x\) axis. Comment on the time ordering of the two events as seen in the three different frames. This startling result is discussed further in Section 15.10 .

(a) What is a mass of \(1 \mathrm{MeV} / c^{2}\) in kilograms? ( \(\mathbf{b}\) ) What is a momentum of \(1 \mathrm{MeV} / c\) in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} ?\)

A particle of unknown mass \(M\) decays into two particles of known masses \(m_{a}=0.5 \mathrm{GeV} / c^{2}\) and \(m_{b}=1.0 \mathrm{GeV} / c^{2},\) whose momenta are measured to be \(\mathbf{p}_{a}=2.0 \mathrm{GeV} / \mathrm{c}\) along the \(x_{2}\) axis and \(\mathbf{p}_{b}=1.5 \mathrm{GeV} / c\) along the \(x_{1}\) axis. \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV} .\right)\) Find the unknown mass \(M\) and its speed.

If one defines a variable mass \(m_{\text {var }}=\gamma m\), then the relativistic momentum \(\mathbf{p}=\gamma m \mathbf{v}\) becomes \(m_{\text {var }} \mathbf{v}\) which looks more like the classical definition. Show, however, that the relativistic kinetic energy is not equal to \(\frac{1}{2} m_{\mathrm{var}} v^{2}\)

A neutral pion (Problem 15.86) is traveling with speed \(v\) when it decays into two photons, which are seen to emerge at equal angles \(\theta\) on either side of the original velocity. Show that \(v=c \cos \theta\)

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