/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Expand the equation for relativi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Expand the equation for relativistic kinetic energy \(K=m(\gamma-1)\) in a Taylor series, and find the first two nonvanishing terms. Show that the first term is the nonrelativistic expression.

Short Answer

Expert verified
The first term is \( \frac{1}{2}mv^2 \), the nonrelativistic kinetic energy. The second term is \( \frac{3}{8}m\frac{v^4}{c^2} \).

Step by step solution

01

Understanding the equation

The relativistic kinetic energy is given by \( K = m(\gamma - 1) \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, \( m \) is the mass of the object, \( v \) is its velocity, and \( c \) is the speed of light. We will expand \( \gamma \) as a Taylor series for \( \frac{v^2}{c^2} \ll 1 \), which is a common approximation for low velocities.
02

Expanding the Lorentz factor

Expand \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) using the binomial series, \( (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2}x^2 + \cdots \). Here, \( x = \frac{v^2}{c^2} \) and \( n = \frac{1}{2} \). So, \( \gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\left(\frac{v^2}{c^2}\right)^2 + \cdots \).
03

Substituting into the kinetic energy equation

Substitute the expanded \( \gamma \) into \( K = m(\gamma - 1) \). This gives \( K \approx m\left(\frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\left(\frac{v^2}{c^2}\right)^2\right)\).
04

Simplifying and identifying the first two terms

Simplify the expression for kinetic energy: \( K \approx \frac{1}{2}mv^2 + \frac{3}{8}m\frac{v^4}{c^2} \). The first term, \( \frac{1}{2}mv^2 \), is the nonrelativistic kinetic energy expression. The second term, \( \frac{3}{8}m\frac{v^4}{c^2} \), is the first relativistic correction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Lorentz Factor
The Lorentz factor, denoted by \( \gamma \), plays a critical role in the realm of special relativity. It quantifies the effects of time dilation, length contraction, and the increase of relativistic mass as an object's speed approaches the speed of light. The formula for the Lorentz factor is \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where:
  • \( v \) is the velocity of the object.
  • \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \).
When speeds are much slower than the speed of light, specifically when \( \frac{v}{c} \ll 1 \), the effects of relativity become minimal. Here, \( \gamma \) approaches 1, which implies that relativistic effects are negligible, and classical mechanics holds. As an object's velocity gets closer to \( c \), \( \gamma \) increases significantly, reflecting the profound changes in perception and physical properties as observed in relativistic scenarios.
Taylor Series Expansion
The Taylor series provides a way to approximate complex functions using simpler polynomial expressions. Particularly, we use it to expand functions like the Lorentz factor when dealing with small quantities. Let's break down how this works for \( \gamma \).When expanding \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), we treat \( \frac{v^2}{c^2} \) as a small variable \( x \). The binomial series formula \((1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2}x^2 + \cdots \) assists us in this approximation. For the Lorentz factor, \( n = \frac{1}{2} \), so we get:
\[ \gamma \approx 1 + \frac{1}{2} \frac{v^2}{c^2} + \frac{3}{8} \left( \frac{v^2}{c^2} \right)^2 + \cdots \]
This series expansion is particularly valuable because it provides a straightforward way to see the influence of increasing powers of velocity divided by the speed of light, with the terms becoming increasingly insignificant as velocity decreases.
Nonrelativistic Kinetic Energy
Kinetic energy is a measure of an object's motion and, in classical physics, is given by the formula \( K = \frac{1}{2}mv^2 \). This nonrelativistic kinetic energy formula applies when an object's speed is significantly lower than the speed of light.
In the context of the exercise, by expanding the relativistic kinetic energy \( K = m(\gamma - 1) \) and substituting the Taylor series expansion of \( \gamma \), we find:
\[ K \approx \frac{1}{2}mv^2 + \frac{3}{8}m\frac{v^4}{c^2} \]
The term \( \frac{1}{2}mv^2 \) represents the nonrelativistic kinetic energy expression. It reflects the familiar classical interpretation of kinetic energy when relativistic effects are negligible. As velocities increase and approach relativistic speeds, additional terms, like \( \frac{3}{8}m\frac{v^4}{c^2} \), introduce corrections that account for the nuances of relativity, altering how motion is perceived and measured at high velocities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In section \(4.5\) we proved the work-energy relation \(d E / d x=F\) in the context of relativity. Recapitulate the derivation in the context of pure Newtonian mechanics.

In nuclear beta decay, an electron or antielectron is typically emitted with an energy on the order of \(1 \mathrm{MeV} .\) In alpha decay, the alpha particle typically has an energy of about \(5 \mathrm{MeV} .\) In each case, do a rough estimate of whether the particle is nonrelativistic, relativistic, or ultrarelativistic.

This problem assumes you know some basic quantum physics. The point of this problem is to estimate whether or not a neutron or proton in an an atomic nucleus is highly relativistic. Nuclei typically have diameters of a few \(f m\) \(1 \mathrm{fm}=10^{-15} \mathrm{~m}\) ). Take a neutron or proton to be a particle in a box of this size. In the ground state, half a wavelength would fit in the box. Use the de Broglie relation to estimate its typical momentum and thus its typical speed. How relativistic is it?

Expand the equation \(p=m \gamma v\) in a Taylor series, and find the first two nonvanishing terms. Show that the first term is the classical expression.

Criticize the following reasoning. Temperature is a measure of the energy per atom. In nonrelativistic physics, there is a minimum temperature, which corresponds to zero energy per atom, but no maximum. In relativity, there should be a maximum temperature, which would be the temperature at which all the atoms are moving at \(\mathrm{c.}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.