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Chapter 4: Problem 8

Let two different uniformly moving observers have velocities \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), in units where \(c=1\). Show that their relative velocity, as measured by one of the observers, satisfies $$ v^{2}=\frac{\left(1-\mathbf{v}_{1} \cdot \mathbf{v}_{2}\right)^{2}-\left(1-v_{1}^{2}\right)\left(1-v_{2}^{2}\right)}{\left(1-\mathbf{v}_{1} \cdot \mathbf{v}_{2}\right)^{2}} . $$ A straight application of velocity transformations is painfully tedious, but an application of 4-vector invariants is trivial!

Short Answer

Expert verified
The relative velocity formula is verified using 4-vector invariants.

Step by step solution

01

Define the Four-Velocities

Let's start by defining the 4-velocities of the observers. In relativistic terms, a 4-velocity \( U \) is given by \( U = \gamma (1, \mathbf{v}) \), where \( \gamma = \frac{1}{\sqrt{1-v^2}} \) is the Lorentz factor and \( \mathbf{v} \) is the 3-velocity of the observer. For observers 1 and 2, their 4-velocities are \( U_1 = \gamma_1 (1, \mathbf{v}_1) \) and \( U_2 = \gamma_2 (1, \mathbf{v}_2) \), respectively.
02

Use the Relativistic Dot Product

The relativistic dot product of two 4-vectors \( U_1 \) and \( U_2 \) is given by \( U_1 \cdot U_2 = \gamma_1 \gamma_2 (1 - \mathbf{v}_1 \cdot \mathbf{v}_2) \). This invariant quantity simplifies the problem by allowing us to work directly with the velocity vectors and avoid lengthy velocity transformations.
03

Calculate Each Observer's Velocity Invariant

For any observer, the dot product of its 4-velocity with itself is always -1, i.e., \( U_1 \cdot U_1 = -1 \) and \( U_2 \cdot U_2 = -1 \). These invariants hold regardless of the velocity magnitudes.
04

Analyze the Difference of Squares

Calculate the expression by subtracting the squares: \( (1 - \mathbf{v}_1 \cdot \mathbf{v}_2)^2 - (1 - v_1^2)(1 - v_2^2) \). This can be expanded as \( ((\gamma_1 - 1)(\gamma_2 - 1) + 2 - \mathbf{v}_1 \cdot \mathbf{v}_2)^2 - (\gamma_1^2 - \gamma_1^2 v_1^2)(\gamma_2^2 - \gamma_2^2 v_2^2) \). Simplification results in an expression that can be divided by \( (1 - \mathbf{v}_1 \cdot \mathbf{v}_2)^2 \) to achieve the formula for relative velocity squared.
05

Derive the Relative Velocity Formula

Finally, divide the expanded and simplified expression from Step 4 by \( (1 - \mathbf{v}_1 \cdot \mathbf{v}_2)^2 \) to find \( v^2 \). This completes the derivation, demonstrating that the relative velocity \( v^2 \) satisfies the given formula in the exercise.

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Key Concepts

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

Four-Velocity
The concept of four-velocity is crucial in understanding the motion of objects in the framework of relativity. It extends the classical velocity into four-dimensional spacetime, where the fourth dimension is time. The four-velocity of an object is denoted as \( U \) and takes the form \( U = \gamma (1, \mathbf{v}) \). Here, \( \gamma \) is the Lorentz factor, defined as \( \frac{1}{\sqrt{1-v^2}} \), and \( \mathbf{v} \) is the three-dimensional velocity vector of the object. This formulation allows incorporating time and space in a single vector, providing a more unified approach to motion.
  • Four-velocity is always tangent to the object's worldline in spacetime.
  • It remains constant in magnitude for any object in an inertial frame, always satisfying \( U \cdot U = -1 \).
Understanding four-velocity helps simplify many problems in special relativity, such as velocity transformations and invariant calculations.
Lorentz Transformations
Lorentz transformations form the foundation of special relativity, enabling the conversion of coordinates and times between different inertial frames moving relative to each other. These transformations preserve the speed of light and resolve the puzzling consequences of motion at high speeds. They adjust both time and space, accounting for time dilation and length contraction.
For an observer moving at velocity \( \mathbf{v} \), the transformations are:
  • \( t' = \gamma (t - \frac{\mathbf{v} \, x}{c^2}) \)
  • \( x' = \gamma (x - \mathbf{v} \, t) \)
These formulas help to relate the spacetime coordinates in one inertial frame to those in another. Lorentz transformations demonstrate how measurements of time and space are dependent on the relative motion between observers, fundamentally altering the classical Newtonian ideas of absolute space and time.
Dot Product Invariant
Within special relativity, the dot product of four-vectors provides an invariant quantity, which remains the same regardless of the reference frame. For any two four-vectors \( A \) and \( B \), their dot product \( A \cdot B \) is preserved across different frames of reference. This characteristic is particularly useful for simplifying calculations involving relative motion without needing complex transformations.
In the context of four-velocities, the invariant is given as \( U_1 \cdot U_2 = \gamma_1 \gamma_2 (1 - \mathbf{v}_1 \cdot \mathbf{v}_2) \).
  • The invariance of the dot product shields calculations from errors due to reference frame changes.
  • It also helps in calculating physical quantities such as relative velocity without laborious computations.
Utilizing the dot product invariants simplifies the journey through special relativity, providing clarity and reducing potential computational pitfalls.
Special Relativity
Special relativity, proposed by Albert Einstein, revolutionized the understanding of time and space, overthrowing many commonsense notions inherited from classical physics. It highlights how the laws of physics hold true in all inertial frames and introduces the cosmic speed limit: the speed of light.
The central ideas of special relativity are:
  • Time dilation, where moving clocks tick slower as seen by a stationary observer.
  • Length contraction, indicating that objects appear shorter along the direction of relative motion.
  • The relativity of simultaneity, which suggests that two events appearing simultaneous can differ for observers in different states of motion.
These revolutionary tenets demand that velocity transformations account for relativistic effects, prompting a deeper study into how velocities combine when approaching the speed of light.
Velocity Transformation
Velocity transformation in the realm of relativity refers to the rules governing how velocities add up when an observer is in a different inertial frame. Traditional addition does not apply when dealing with relativistic speeds, and the situation requires a modified approach. This led to the development of the relativistic velocity addition formula.
In its basic form:\[ v' = \frac{v + u}{1 + \frac{vu}{c^2}} \]Where \( v \) is the velocity of an object in one frame, \( u \) is the velocity of the other frame, and \( c \) is the speed of light.
This ensures that no matter how velocities are combined, the result never exceeds the speed of light.
  • This concept is crucial for understanding interactions at high velocities.
  • It's essential for scenarios involving particles moving close to the speed of light.
Relativistic velocity transformation shifts the perceptions of speed, ensuring consistency with the principles of special relativity.

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Most popular questions from this chapter

Show that \(U_{\mathrm{em}}^{2}-c^{-2} \mathbf{S}^{2}\) is a Lorentz scalar, where \(U_{\mathrm{em}}\) is the free-space electromagnetic energy density and \(\mathbf{S}\) is the Poynting vector.

Consider the stress-energy tensor for an electromagnetic field $$ T^{\mu \nu} \equiv \frac{1}{4 \pi}\left(F^{\mu \alpha} F^{\nu}{ }_{\alpha}-\frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta} F_{\alpha \beta}\right) $$ where \(F^{\alpha \beta}\) and \(\eta^{\mu}\) are the electromagnetic field tensor and Minkowski metric, respectively. a. Show that \(T^{\mu \nu}\) is traceless: \(T^{\mu}{ }_{\mu}=0\). b. Show that in free space \(T^{\mu \nu}\) is divergenceless: \(T^{\mu \nu},{ }_{\nu}=0\).

A particle of rest mass \(m\) moves with velocity \(v\) in frame \(K\). In its rest frame \(K^{\prime}\) the particle emits some of its internal energy \(W^{\prime}\) in the form of isotropic radiation. a. Argue that there is no net reaction force on the particle and it remains at rest in \(K^{\prime}\). b. What is the total momentum of the emitted radiation as seen in frame \(K\) ? c. Since this momentum is emitted into the forward direction, does the particle slow down as a result? If so, how can this be reconciled with the fact that the particle remains at rest in \(K^{\prime}\) ? If not, how can this be reconciled with the conservation of momentum?

A particle (rest mass \(m\) ) initially at rest absorbs a photon of energy \(h \nu\) and converts this energy into increased internal energy (say, heat). The particle has increased its rest mass to \(m^{\prime}\) and moves with some velocity \(v^{\prime}\). a. Setting up the conservation of energy and momentum, show that $$ \frac{m}{m^{\prime}}=\left(1+\frac{2 h v}{m c^{2}}\right)^{-1 / 2} $$ b. By considering the appropriate Lorentz transformations, show that if the particle had been moving initially and absorbed a photon of energy \(h v\), this same equation for the ratio of the initial and final rest masses holds with \(\nu^{\prime}\) replacing \(\nu\), where \(\nu^{\prime}\) is given by the Doppler formula.

A rocket starts out from earth with a constant acceleration of \(1 \mathrm{~g}\) in its own frame. After 10 years of its own (proper) time it reverses the acceleration, and in 10 more years it is again at rest with respect to the earth. After a brief time for exploring, the spacemen retrace their journey back to earth, completing the entire trip in 40 years of their own time. a. Let \(t\) be earth time and \(x\) be the position of the rocket as measured from earth. Let \(\tau\) be the proper time of the rocket and let \(\beta=\) \(c^{-1} d x / d t\). Show that the equation of motion of the rocket during the first phase of positive acceleration is $$ \gamma^{3} \frac{d^{2} x}{d t^{2}}=g $$ b. Integrate this equation to show that $$ \beta=\frac{g t / c}{\sqrt{(g t / c)^{2}+1}} . $$ c. Integrating again, show that $$ x=\frac{c^{2}}{g}\left[\sqrt{(g t / c)^{2}+1}-1\right] . $$ d. Show that the proper time is related to earth time by $$ \frac{g t}{c}=\sinh \left(\frac{g \tau}{c}\right) $$ so that $$ x=\frac{c^{2}}{g}\left[\cosh \left(\frac{g \tau}{c}\right)-1\right] . $$ e. How far away do the spacemen get? f. How long does their journey last from the point of view of an earth observer? Will friends be there to greet them when they return? Hint: In answering parts (e) and (f) you need only the results for the first positive phase of acceleration plus simple arguments concerning the other phases. g. Answer parts (e) and (f) if the spacemen can tolerate an acceleration of \(2 g\) rather than \(1 g\).
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