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Chapter 41: Problem 31

Starting with the definition of linear momentum and the relation between mass and energy, prove that \(\mathrm{E}^{2}=p^{2} c^{2}+m^{2} \mathrm{c}^{4}\). Use this relation to show that the translational KE of a particle of mass \(m\) is \(\sqrt{m^{2} \mathrm{c}^{4}+p^{2} \mathrm{c}^{2}}-m \mathrm{c}^{2}\)

Short Answer

Expert verified
E is related to p and m by \(E^2 = p^2 c^2 + m^2 c^4\), leading to KE as \(\sqrt{m^2c^4 + p^2c^2} - mc^2\).

Step by step solution

01

Define Linear Momentum and Energy

Linear momentum \( p \) is defined as \( p = mv \) where \( m \) is the mass and \( v \) is the velocity. The total energy \( E \) of a particle can be expressed using the formula \( E = \sqrt{m^2 c^4 + p^2 c^2} \). This formula derives from Einstein's theory of relativity, combining the rest energy \( m c^2 \) and relativistic momentum.
02

Express Energy-Momentum Relation

In the theory of relativity, the energy-momentum relation is given by \( E^2 = p^2 c^2 + m^2 c^4 \). This expresses the total energy \( E \) as a function of the momentum \( p \) and rest mass energy, \( m c^2 \). The relation shows the balance between kinetic and rest energy.
03

Substitute and Simplify to Derive Translational KE

To find the translational kinetic energy (KE), subtract the rest energy \( m c^2 \) from the total energy \( E \). Thus, \[ KE = \sqrt{m^2 c^4 + p^2 c^2} - m c^2 \]. This formula results from rearranging the energy-momentum relation to emphasize the kinetic energy component.

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Key Concepts

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

linear momentum
Linear momentum is a fundamental concept in physics that describes the motion of an object in a straight line. It is calculated as the product of the mass and velocity of the object. The formula is given by \( p = mv \), where \( p \) is the linear momentum, \( m \) is the mass, and \( v \) is the velocity. Linear momentum is a vector quantity, meaning it has both magnitude and direction.
  • Conservation of Momentum: In an isolated system, linear momentum is conserved. This means that without external forces, the momentum of the system remains constant.
  • Role in Collisions: Linear momentum plays a key role in analyzing collisions. The total momentum before and after the collision remains the same.

Understanding linear momentum helps in grasping how force and mass interact to influence motion, setting the stage for more advanced concepts like energy-momentum relations.
translational kinetic energy
Translational kinetic energy (KE) refers to the energy due to the motion of an object along a path. It is calculated using the formula \( KE = \frac{1}{2} mv^2 \), capturing how mass and speed impact energy.
  • Relation to Motion: Translational kinetic energy depends directly on both the mass and the square of the speed of the object.
  • Conversion from Rest Energy: When a particle is at rest, it has rest energy, \( mc^2 \). As it begins to move, part of this energy is converted into kinetic energy.
  • Link to Relativity: In relativity, kinetic energy becomes a component of the total energy of a particle, linking translational kinetic energy with other forms of energy.

Considering translational kinetic energy in the relativistic framework allows us to see kinetic energy as part of the larger energy expression \( \sqrt{m^2 c^4 + p^2 c^2} - mc^2 \). This highlights the connection between motion and energy beyond classical physics.
relativistic energy
Relativistic energy encompasses both the rest energy of an object and any additional energy due to its motion. In the realm of relativity, total energy \( E \) is described by the formula \( E = \sqrt{m^2 c^4 + p^2 c^2} \). This formula incorporates the effects of motion at speeds close to the speed of light.
  • Rest Energy: This component, \( mc^2 \), is the energy an object possesses due to its rest mass alone. It's the famous equation introduced by Einstein.
  • Kinetic Energy: The relativistic kinetic energy includes additional energy as an object gains speed, contributing to the total energy.
  • Momentum and Energy Relationship: The energy-momentum relation, \( E^2 = p^2 c^2 + m^2 c^4 \), ties these quantities together, revealing how they coexist in a moving system.

Relativistic energy is crucial for understanding how energy and mass transform and are conserved in high-speed phenomena, illustrating the depth of Einstein's insights.
Einstein's theory of relativity
Einstein's theory of relativity revolutionized how we understand space, time, and energy. Specifically, his theories are split into the Special and General relativity.
  • Special Relativity: This focuses on the physics of objects moving at constant speeds, particularly those near the speed of light. Key insights include the constancy of the speed of light and the relative nature of time and space.
  • Energy-Mass Equivalence: The famous equation \( E=mc^2 \) arises, quantifying the idea that mass can be viewed as a concentrated form of energy.
  • General Relativity: Extending the insights of special relativity, this theory addresses gravitation's effect on the fabric of space-time, thereby bending both space and time around massive objects.

Einstein's theory of relativity connects deeply with concepts like linear momentum, translational kinetic energy, and relativistic energy, offering a comprehensive framework to explore the dynamics of high-energy physics and cosmic phenomena.

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Most popular questions from this chapter

A \(2.0\) -kg object is lifted from the floor to a tabletop \(30 \mathrm{~cm}\) above the floor. By how much did the mass of the system consisting of the Earth and the object increase because of this increased \(\mathrm{PE}_{\mathrm{G}}\) ? We use \(\Delta \mathrm{E}_{\mathrm{O}}=(\Delta m) \mathrm{c}^{2}\), with \(\Delta \mathrm{E}_{\mathrm{O}}=m g h .\) Therefore, $$ \Delta m=\frac{\Delta E_{0}}{c^{2}}=\frac{m g h}{c^{2}}=\frac{(2.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.30 \mathrm{~m})}{\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=6.5 \times 10^{-17} \mathrm{Kg} $$

Find the speed and momentum of a proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has been accelerated through a potential difference of 2000 MV. (We call this a \(2 \mathrm{GeV}\) proton.) Give your answers to three significant figures.

A spacecraft moving at \(0.95 \mathrm{c}\) travels from the Earth to the star Alpha Centauri, which is \(4.5\) light years away. How long will the trip take according to \((a)\) Earth clocks and \((b)\) spacecraft clocks? (c) How far is it from Earth to the star according to spacecraft occupants? ( \(d\) ) What do they compute their speed to be? A light year is the distance light travels in 1 year, namely 1 light year \(=\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\left(3.16 \times 10^{7} \mathrm{~s}\right)=9.47 \times 10^{15} \mathrm{~m}\) Hence the distance to the star (according to earthlings) is $$ \begin{array}{l} d_{e}=(4.5)\left(9.47 \times 10^{15} \mathrm{~m}\right)=4.3 \times 10^{16} \mathrm{~m} \\ \text { (a) } \Delta t_{e}=\frac{d_{e}}{v}=\frac{4.3 \times 10^{16} \mathrm{~m}}{(0.95)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=1.5 \times 10^{8} \mathrm{~s} \end{array} $$ (b) Because clocks on the moving spacecraft run slower, $$ \Delta t_{\text {cnth }}=\Delta t_{\varepsilon} \sqrt{1-(v / \mathrm{c})^{2}}=\left(1.51 \times 10^{8} \mathrm{~s}\right)(0.312)=4.7 \times 10^{7} \mathrm{~s} $$ (c) For the spacecraft occupants, the Earth-star distance is moving past them with speed 0.95c. Therefore, that distance is shortened for them; they find it to be $$ d_{\text {cath }}=\left(4.3 \times 10^{16} \mathrm{~m}\right) \sqrt{1-(0.95)^{2}}=1.3 \times 10^{16} \mathrm{~m} $$ (d) For the spacecraft occupants, their relative speed is $$ v=\frac{d_{\text {cnft }}}{\Delta t_{\text {caff }}}=\frac{1.34 \times 10^{16} \mathrm{~m}}{4.71 \times 10^{7} \mathrm{~s}}=2.8 \times 10^{8} \mathrm{~m} / \mathrm{s} $$ which is 0.95c. Both Earth and spacecraft observers measure the same relative speed.

The fastest vehicle leaving Earth so far was NASA's New Horizon Pluto mission. The craft attained a speed of \(16.26 \mathrm{~km} / \mathrm{s}\). Determine the corresponding value of \(y\) using \(\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{3}}\) first; then use Eq. (41.2). [ Hint: Work in \(\mathrm{m} / \mathrm{s}\) and use \(\mathrm{c}=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

A certain light source sends out \(2 \times 10^{15}\) pulses each second. As a spaceship travels parallel to the Earth's surface with a speed of \(0.90 \mathrm{c}\), it uses this source to send pulses to the Earth. The pulses are sent perpendicular to the path of the ship. How many pulses are recorded on Earth each second?
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