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Show that any two zero-mass particles have a CM frame, provided their three- momenta are not parallel. [Hint: As you should explain, this is equivalent to showing that the sum of two forward light-like vectors is forward time-like, unless the spatial parts are parallel.]

Short Answer

Expert verified
A CM frame exists because the momenta's sum becomes time-like if they're not parallel.

Step by step solution

01

Understanding the scenario

We have two zero-mass (massless) particles with momenta. We want to show that they have a Center of Mass (CM) frame if the spatial parts of their momenta are not parallel. This question leads us to consider properties of light-like vectors.
02

Define momentum vectors

Since the particles are zero-mass, their momentum vectors can be expressed as light-like vectors, i.e., if p and q are momenta of particles, then \( p^2 = 0 \) and \( q^2 = 0 \), implying that these particles move at the speed of light.
03

Analyze the sum of momentum vectors

Determine the sum of the two momentum vectors, \( r = p + q \), and calculate \( r^2 = (p + q)^2 \).
04

Expand the square of the sum

Using the property of the dot product, we have \( r^2 = p^2 + 2(p \cdot q) + q^2 \). Since \( p^2 = q^2 = 0 \), this simplifies to \( r^2 = 2(p \cdot q) \).
05

Consider the dot product of momentum vectors

If the spatial parts of the momenta are not parallel, then the dot product \( p \cdot q > 0 \). Thus, \( r^2 = 2(p \cdot q) > 0 \), making \( r \) a time-like vector.
06

Conclusion on the CM frame

Since the sum of momenta \( r = p + q \) is time-like when not parallel, there is a reference frame (CM frame) where these momenta can be viewed as having opposite directions and equal and opposite magnitudes. This verifies the existence of the CM frame.

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Key Concepts

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

Zero-Mass Particles
Zero-mass particles, also known as massless particles, are unique in the field of physics because they always travel at the speed of light. Examples of such particles include photons and gluons. These particles do not have rest mass, which can lead to interesting consequences when discussing their momentum and interactions with other particles.
In our exercise, zero-mass particles have momentum vectors that align with their direction of travel as they do not experience rest. The essence of this is expressed in the equation that the magnitude of their momentum vector squared is zero (\(p^2 = 0\)). This property becomes crucial when determining the behavior of such particles in various physical frames, like the Center of Mass (CM) frame, and influences how their interactions are understood.
When studying such particles, we often leverage their properties to simplify complex physical scenarios and use them as a baseline comparison for understanding phenomena that involve particles with non-zero mass.
Light-Like Vectors
Light-like vectors are an essential concept in understanding the interactions of zero-mass particles. A vector is considered light-like if its squared magnitude is zero, similar to the property of the momentum of massless particles. In mathematical terms, a vector \(p\) is light-like if \(p^2 = 0\).
This property is significant because it directly influences how energy and momentum are conserved in particle interactions. For zero-mass particles, these vectors describe paths that lie on the light cone in spacetime diagrams. They connect points across spacetime that photons or other massless particles can traverse.
Understanding light-like vectors is crucial in demonstrating that two zero-mass particles can have a Center of Mass frame. When considering two such particles with three-momenta that are not parallel, their momentum vectors can be combined to form a sum that diverges from the light cone, suggesting the existence of a frame where their sum behaves time-like.
Momentum Vectors
Momentum vectors are a fundamental tool in describing motion in physics. For massless particles, these vectors reflect how these particles carry momentum and energy through space. A momentum vector encapsulates both the direction of travel and magnitude of a particle's energy, crucial for predicting how particles interact and move.
In the context of zero-mass particles, momentum vectors are special because they follow the relation \(p^2 = 0\). Unlike particles with mass, whose momentum is influenced by their rest mass, zero-mass particles derive all their momentum from their energy due to traveling at light speed.
These vectors become particularly interesting when we look at their sums; combining them leads to different outcomes that provide insights into possible frames, like the CM frame. When adding light-like vectors (zero-mass particle momentum vectors), if their spatial parts are not parallel, the resultant vector becomes time-like, revealing new properties and enabling the study of different frames where these particles might interact.
Time-Like Vector
Time-like vectors differ from light-like vectors in a fundamental way: a time-like vector has a positive squared magnitude, \(r^2 > 0\). This indicates that the vector points in a direction in spacetime allowing for the definition of a reference frame where time-like separation exists.
This property is crucial in the context of the exercise where we explore combining momentum vectors of two zero-mass particles. When their spatial parts are not parallel, their sum results in a time-like vector, allowing us to establish a CM frame—a frame where particles previously traveling at speed of light appear to move with defined velocities less than light speed, suggesting interactions and relative momentum balances.
Time-like vectors serve as a key understanding aid in developing physical theories and models that incorporate the conservation of momentum and energy in relativistic physics, especially when considering referencing frames in which interactions are studied, such as the CM frame that balances these elements perfectly.

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Most popular questions from this chapter

As an observer moves through space with position \(\mathbf{x}(t),\) the four- vector \((\mathbf{x}(t), c t)\) traces a path through space-time called the observer's world line. Consider two events that occur at points \(P\) and \(Q\) in space-time. Show that if, as measured by the observer, the two events occur at the same time \(t,\) then the line joining \(P\) and \(Q\) is orthogonal to the observer's world line at the time \(t\); that is, \(\left(x_{P}-x_{Q}\right) \cdot d x=0,\) where \(d x\) joins two neighboring points on the world line at times \(t\) and \(t+d t\).

Show that the four-velocity of any object has invariant length squared \(u \cdot u=-c^{2}\)

The neutral pion \(\pi^{0}\) is an unstable particle (mass \(m=135\ \mathrm{MeV} / c^{2}\) ) that can decay into two photons, \(\pi^{0} \rightarrow \gamma+\gamma .\) (a) If the pion is at rest, what is the energy of each photon? (b) Suppose instead that the pion is traveling along the \(x\) axis and that the photons are observed also traveling along the \(x\) axis, one forward and one backward. If the first photon has three times the energy of the second, what was the pion's original speed \(v ?\)

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.]

Let \(\Lambda_{\mathrm{B}}(\theta)\) denote the 4 \(\times 4\) matrix that gives a pure boost in the direction that makes an angle \(\theta\) with the \(x_{1}\) axis in the \(x_{1} x_{2}\) plane. Explain why this can be found as \(\Lambda_{\mathrm{B}}(\theta)=\Lambda_{\mathrm{R}}(-\theta) \Lambda_{\mathrm{B}}(0) \Lambda_{\mathrm{R}}(\theta)\) where \(\Lambda_{\mathrm{R}}(\theta)\) denotes the matrix that rotates the \(x_{1} x_{2}\) plane through angle \(\theta\) and \(\Lambda_{\mathrm{B}}(0)\) is the standard boost along the \(x_{1}\) axis. Use this result to find \(\Lambda_{\mathrm{B}}(\theta)\) and check your result by finding the motion of the spatial origin of the frame \(\mathcal{S}\) as observed in \(\mathcal{S}^{\prime}\).

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