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Chapter 2: Problem 62

In a particle collider experiment, particle I is moving to the right at \(0.99 c\) and particle 2 to the left at \(0.99 c\), both relative to the laboratory. What is the relative velocity of the two particles according to (an observer moving with) particle \(2 ?\)

Short Answer

Expert verified
The relative velocity of the two particles, as seen from Particle 2's reference frame, can be found by applying Einstein's Velocity Addition Theorem. After applying the theorem and simplifying the equation, we find that the relative velocity is approximately \( -1.982c/1.9601 = -1.0112c\). The actual value would be limited by the speed of light to be \( -c\).

Step by step solution

01

Understand the scenario and the given

We have two particles, Particle 1 and Particle 2, moving in opposite directions at a velocity of \(0.99c\) relative to the laboratory. The question is to find the relative velocity of Particle 1 as seen from Particle 2's reference frame. Use Einstein's Velocity Addition Theorem, which states that if a particle moves with velocity \(v_1\) in frame \(S_1\) and \(S_1\) moves with velocity \(v_2\) relative to \(S_2\), then the velocity \(v\) of the particle in \(S_2\) frame is given by \(v = (v_1 + v_2) / (1 + v_1 * v_2/ c^2)\). Here \(v_1\) and \(v_2\) are the velocities of particles 1 and 2 respectively.
02

Apply the formula

Apply Einstein's Velocity Addition Theorem. Set \(v_1 = 0.99c\), \(v_2 = -0.99c\), where \(c\) is the speed of light. Input these values in the formula: \(v = (0.99c - 0.99c) / (1 + (0.99c * -0.99c) / c^2)\).
03

Simplify the equation

Simplifying the equation we get \(v = -0.99c / (1 - 0.99^2)\). The negative sign indicates that the direction is opposite to the direction of Particle 2, which is in concordance with the given scenario.
04

Solve for the relative velocity

Solving the above equation gives the relative velocity of the two particles according to an observer moving with Particle 2.

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Key Concepts

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

Understanding Relativistic Velocity
In classical physics, you might be familiar with the idea of simply adding velocities. However, when dealing with objects moving at speeds close to the speed of light, classical addition no longer applies. This is where the concept of relativistic velocity comes into play.
Einstein's Velocity Addition Theorem is crucial in understanding how velocities combine in a relativistic framework.
  • It ensures that the calculated velocity never exceeds the speed of light, adhering to the limits set by relativity.
  • According to this theorem, if two objects are moving at high speeds, their combined speed is found using the formula: \[v = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}}\] Here, \(v_1\) and \(v_2\) are the velocities of the two particles and \(c\) is the speed of light.
This formula helps us compute the relative velocity when observing from one of the moving objects' frames, giving us a more accurate picture of their motion in the universe.
Role of Particle Colliders
Particle colliders are fascinating devices used in physics to study fundamental particles. These machines accelerate particles to high speeds, often close to the speed of light, and bring them to collide with each other.
This creates opportunities to observe the results of such high-energy interactions which are crucial in the world of particle physics.
  • These collisions provide evidence of various particles' behaviors and interactions.
  • They enable validation and discovery of new theories and particles, such as the Higgs boson.
  • Understanding the energies and velocities involved in collisions is key to making new scientific discoveries.
In this context, using relativistic velocity calculations are vital as the colliding particles are relativistic, meaning their velocities are approaching that of light. Such calculations help physicists accurately predict and understand the outcomes of collisions.
Importance of Reference Frames
A reference frame is essentially your viewpoint or perspective from which you are seeing and measuring motion. In physics, especially when involving high speeds and relativistic conditions, selecting an appropriate reference frame is vital.
For instance, in the given exercise, you are asked to find the velocity of one particle as seen from another particle moving in the opposite direction.
  • Reference frames allow different observers to agree on the laws of physics while perhaps observing different values for speeds.
  • An observer moving with one of the particles will see the velocities differently compared to a stationary observer.
  • This concept is key in understanding relativity and ensures all computations remain consistent and conform to Einstein's theories.
In essence, using reference frames helps us interpret differing observations of the same physical phenomena depending on an observer's state of motion.

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Most popular questions from this chapter

Equation \((2-30)\) is an approximation correct only if the gravitational time- dilation effect is small. In this exercise, it is also assumed to be small. but we still allow for a nonuniform gravitational field. We stan with (2-29), based on the Doppler effect in the acceleraing frame. Consider two elevations, the lower at \(r_{1}\) and the upper at \(r_{1}+d r\). Equation \((2-29)\) becomes $$ \frac{f\left(r_{1}+d r\right)}{f\left(r_{i}\right)}=\left(1-\frac{g\left(r_{1}\right) d r}{c^{2}}\right) $$ Similarly, if we consider elevations \(r_{1}+d r\) and \(r_{1}+\) \(2 d r\). we have $$ \frac{f\left(r_{1}+2 d r\right)}{f\left(r_{1}+d r\right)}=\left(1-\frac{g\left(r_{1}+d r\right) d r}{c^{2}}\right) $$ We continue the process, incrementing \(r\) by \(d r\), until we reach \(r_{2}\) $$ \frac{f\left(r_{2}\right)}{f\left(r_{2}-d r\right)}=\left(1-\frac{g\left(r_{2}-d r\right) d r}{c^{2}}\right) $$ Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply \(f\left(r_{2}\right) / f\left(r_{1}\right)\). (b) Assuming that the term \(g d r / c^{2}\) in each individual equation is very small. so that products of such terms can be ignored, argue that the right side of the product is $$ 1-\frac{1}{c^{2}} \int g(r) d r $$ (c) Deduce \(g(r)\) from Newton's universal law of gravitation, then argue that equation \((2-31)\) follows from the result, just as \((2 \cdot 30)\) does from \((2-29)\).

An experimenter determines that a particle created at one end of the laboratory apparatus moved at \(0.94 c\) and survived for 0.032 \mus, decaying just as it reached the other end. (a) According to the experimenter. how far did the particle move? (b) In its own frame of reference, how long did the particle survive? (c) According to the particle, what was the length of the laboratory apparatus?

With reckless disregard for safety and the law. you set your high-perfomance rocket cycle on course to streak through an intersection at top speed. Approaching the intersection, you observe grcen \((540 \mathrm{nm})\) light from the traffic signal. After passing through, you look back to observe red \((650 \mathrm{nm})\) light. Actually, the traffic signal never changed color wit didn't have time! What is the top speed of your rocket cycle, and what was the color of the traffic signal (according to an appalled bystander)?

The light from galaxy NGC 22 I consists of a recognizable spectrum of wavelengths. However, all are shifted tow ard the shorter-wavelength end of the spectrum. In particular, the calcium "line" ordinarily observed at \(396.85 \mathrm{nm}\) is observed at \(396.58 \mathrm{nm}\). Is this galaxy moving toward or away from Earth? At what speed?

Prove that if \(v\) and \(u^{\prime}\) are less than \(c\), it is impossible for a speed \(u\) greater than \(c\) to result from equation \((2-19 b)\). [Hint: The product (c \(\left.-u^{\prime}\right)(c-v)\) is positive.]
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