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Chapter 37: Problem 59

In an experiment, two protons are shot directly toward each other, each moving at half the speed of light relative to the laboratory. (a) What speed does one proton measure for the other proton? (b) What would be the answer to part (a) if we used only nonrelativistic Newtonian mechanics? (c) What is the kinetic energy of each proton as measured by (i) an observer at rest in the laboratory and (ii) an observer riding along with one of the protons? (d) What would be the answers to part (c) if we used only nonrelativistic Newtonian mechanics?

Short Answer

Expert verified
(a) 0.8c (b) c (c) (i) 0.155mc虏, (ii) 0 (d) (i) 0.125mc虏, (ii) 0

Step by step solution

01

Define the Problem

We need to calculate the relative speed and kinetic energy of protons moving towards each other at relativistic speeds. We will consider both relativistic and non-relativistic (Newtonian) physics scenarios.
02

Relativistic Speed Calculation

The relativistic velocity addition formula is used when dealing with speeds close to the speed of light. If two objects are moving towards each other with speeds \(u\) and \(v\) with respect to an observer, the relative speed \(s_r\) is given by \[ s_r = \frac{u + v}{1 + \frac{uv}{c^2}} \] where \(c\) is the speed of light. Here, \(u = v = 0.5c\). Substituting,\[ s_r = \frac{0.5c + 0.5c}{1 + \frac{(0.5c)(0.5c)}{c^2}} = \frac{c}{1.25} = 0.8c \]
03

Newtonian Speed Calculation

In a non-relativistic frame (Newtonian mechanics), velocities simply add up. Thus, the speed of one proton as measured by the other is\[ v = 0.5c + 0.5c = c \]
04

Relativistic Kinetic Energy - Laboratory Frame

Relativistic kinetic energy \(K_r\) is calculated using \[ K_r = (\gamma - 1)mc^2 \] where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) and \(m\) is the rest mass. For a proton, \(v = 0.5c\), so \(\gamma = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} = 1.155\). Thus, \[ K_r = (1.155 - 1)mc^2 = 0.155mc^2 \]
05

Relativistic Kinetic Energy - Riding Observer

For an observer riding with one of the protons, its own speed is 0 (\(\gamma = 1\)), so its kinetic energy is 0.
06

Newtonian Kinetic Energy - Laboratory Frame

In Newtonian mechanics, kinetic energy \(K_n\) is \( K_n = \frac{1}{2}mv^2 \). Hence, for a proton, \(v = 0.5c\), \[ K_n = \frac{1}{2}m(0.5c)^2 = \frac{1}{2}m \times 0.25c^2 = 0.125mc^2 \]
07

Newtonian Kinetic Energy - Riding Observer

For an observer moving with one of the protons (like in Step 5), the speed of that proton is zero. Thus, the kinetic energy is 0 in this frame as well.

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Key Concepts

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

Velocity Addition
When dealing with objects moving at significant fractions of the speed of light, as is the case with the protons in our experiment, we must use the relativistic velocity addition formula. Standard addition of velocities fails to account for the relativistic effects that occur when objects move at such high speeds.
  • In our exercise, both protons are moving at half the speed of light toward each other. Using the formula \[ s_r = \frac{u + v}{1 + \frac{uv}{c^2}} \] we calculate the relative speed. Here, each proton moves at 0.5c relative to the laboratory.
  • Substituting into the equation gives us \[ s_r = \frac{0.5c + 0.5c}{1 + \frac{(0.5c)(0.5c)}{c^2}} = \frac{c}{1.25} = 0.8c \]. So the protons see each other moving at 0.8 times the speed of light, not 1c as a straightforward total would suggest.
This calculation shows the necessity of considering relativistic mechanics to understand motion correctly when velocities are high.
Relativistic Kinetic Energy
Kinetic energy in relativistic mechanics differs from the classical Newtonian expression due to the introduction of the Lorentz factor (\( \gamma \)). This factor accounts for the variation in observed mass as an object's velocity approaches the speed of light.
  • The relativistic kinetic energy formula is \[ K_r = (\gamma - 1)mc^2 \], where\( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).
  • In our example, each proton moving at a velocity of 0.5c has \( \gamma = 1.155 \), leading to a kinetic energy of \( 0.155mc^2 \).
For someone riding along with one of the protons, the kinetic energy reduces to zero because, in their frame, the proton is at rest. This result highlights how kinetic energy is frame-dependent when relativistic speeds are involved.The correct treatment of kinetic energy is crucial to accurately describe the energy dynamics of particles moving at relativistic speeds.
Newtonian Physics
In contrast to relativistic mechanics, Newtonian physics applies when velocities are much smaller than \( c \), the speed of light. In these conditions, the complexities of relativistic formulas reduce significantly.
  • Newtonian velocity addition tells us that the velocities simply sum up:\[ v = 0.5c + 0.5c = c \]. While straightforward, this does not hold true at relativistic speeds.
  • The kinetic energy in Newtonian terms is calculated using \[ K_n = \frac{1}{2}mv^2 \]. For each proton moving at 0.5c, this gives \( 0.125mc^2 \), slightly underestimating the energy compared to the relativistic calculation.
For rides along an object (like an observer alongside a proton), the Newtonian approach also reflects zero kinetic energy as the object is perceived to be at rest. Newtonian physics provides an excellent approximation for everyday experiences but falls short when velocities near the speed of light, at which point relativistic mechanics become indispensable.

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Most popular questions from this chapter

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9910 \(\mathrm{c}\) . An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?

(a) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(m v ?\) Express your answer in terms of the speed of light. (b) A force is apphed to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an observer on Tatooine, the cruiser is traveling away from the planet with a speed of 0.600 \(\mathrm{c}\) . The pursuit ship is traveling at a speed of 0.800 \(\mathrm{c}\) relative to Tatooine, in the same direction as the cruiser. (a) For the pursuit ship to catch the cruiser, should the speed of the cruiser relative to the pursuit ship be positive or negative? (b) What is the speed of the cruiser relative to the pursuit ship?

An enemy spaceship is moving toward your starfighter with a speed, as measured in your frame, of 0.400 \(\mathrm{c}\) . The enemy ship fires a missile toward you at a speed of 0.700\(c\) relative to the enemy ship (Fig. 37.28\()\) . (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure that the enemy ship is \(8.00 \times 10^{6} \mathrm{km}\) away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?
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