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Show that Equation \(2-31\) can be written \(E=m c^{2}\left(1+p^{2} / m^{2} c^{2}\right)^{1 / 2}\) and use the binomial expansion to show that, when \(p c\) is much less than \(m c^{2}, E \approx m c^{2}+p^{2} / 2 m\).

Short Answer

Expert verified
By rearranging the energy-momentum relation and applying the binomial expansion for small \(\frac{p^2}{m^2c^2}\), we deduce that \(E \approx mc^2 + \frac{p^2}{2m}\) when \(pc \ll mc^2\).

Step by step solution

01

Recall Equation 2-31

Equation 2-31 is given by the relation for energy, which describes the energy of an object as a function of its mass and momentum. This is known as the energy-momentum relation: \(E = \sqrt{m^2c^4 + p^2c^2}\).
02

Rearrange Equation 2-31

Rearrange Equation 2-31 to show the expression in the given form. To do this, factor out \(m^2c^4\) from the square root to get the given form: \(E = mc^2\sqrt{1 + \frac{p^2}{m^2c^2}}\). This demonstrates that Equation 2-31 can indeed be written in the given form.
03

Use the binomial expansion

Apply the binomial expansion to the square root term \(\left(1+\frac{p^2}{m^2c^2}\right)^{1/2}\) for the case when \(pc \ll mc^2\). The binomial expansion of \((1+x)^n\) for small \(x\) and \(n = 1/2\) is approximated as \(1 + \frac{n}{1!}x\). Use this approximation to expand the square root term and ignore higher-order terms.
04

Simplify the approximation

After applying the binomial expansion, the equation simplifies to \(E \approx mc^2 \left(1 + \frac{1}{2} \frac{p^2}{m^2c^2}\right)\). Multiply the terms within the parentheses and simplify to obtain the approximate relation \(E \approx mc^2 + \frac{p^2}{2m}\), which is valid when \(pc \ll mc^2\).

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Key Concepts

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

Special Relativity
At the heart of modern physics lies the theory of special relativity, formulated by Albert Einstein in 1905. This groundbreaking concept revolutionized our understanding of space, time, and how they relate to each other. Special relativity is based on two postulates: first, the laws of physics are the same for all observers, no matter their relative speed or direction; and second, the speed of light in a vacuum is constant and will always be measured as the same value, which is approximately 299,792 kilometers per second, regardless of the motion of the light source or observer.

One of the most remarkable outcomes of special relativity is the realization that time and space are interwoven into a single continuum known as spacetime. Events that occur are described by four coordinates - three of space and one of time - changing how we think about motion and the propagation of light. As objects move closer to the speed of light, time dilation and length contraction occur, leading to counterintuitive phenomena such as time passing at different rates for observers in relative motion.
Mass-Energy Equivalence
A consequence of special relativity is the famous equation E=mc^2, which conveys the principle of mass-energy equivalence. This equation tells us that mass can be converted into energy and vice versa; where 'E' represents energy, 'm' stands for mass, and 'c' is the speed of light in a vacuum. The square of the speed of light, c^2, is a large number indicating that even a small amount of mass can be converted into a significant amount of energy.

This principle is not just theoretical but has practical implications ranging from nuclear power generation, where the binding energy of atoms is converted to usable power, to the understanding of how the Sun generates energy through nuclear fusion. In the context of our original textbook exercise, mass-energy equivalence allows us to relate an object's rest mass to its total energy when that object is either at rest or in motion.
Binomial Expansion
Binomial expansion is a mathematical tool used to expand expressions that are raised to a power. The binomial theorem explains how to expand expressions of the form \( (a + b)^n \) where 'n' is any positive integer. This series expansion is significant because it can simplify calculations, especially when dealing with small perturbations or quantities.

In the context of our exercise, we apply the binomial expansion to approximate terms within square roots – an approach often used in physics when dealing with small quantities. The first two terms of the binomial expansion \( (1 + x)^{1/2} \) are \( 1 + \frac{1}{2}x \) for small 'x', which in physics often represents a term that is much smaller in comparison to another term it's being added to or subtracted from.
Momentum
Momentum, a fundamental concept in physics, is a measure of the motion of an object and is defined as the product of the object's mass and velocity. Represented by the symbol 'p', momentum takes into account both how much matter is moving and how fast it's moving. In equations, we say that momentum \( p = mv \) where 'm' is the mass and 'v' is the velocity.

In the realm of special relativity, momentum is also affected by an object's velocity relative to the speed of light. The relativistic momentum is more complex than the classical definition; it increases without bound as an object's speed approaches the speed of light. The correct relativistic relation between energy, momentum, and mass is what led to the modified equation provided in our textbook exercise.

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Most popular questions from this chapter

An unstable particle of mass \(M\) decays into two identical particles, each of mass \(m\). Obtain an expression for the velocities of the two decay particles in the lab frame ( \(a\) ) if \(M\) is at rest in the lab and \((b)\) if \(M\) has total energy \(4 m c^{2}\) when it decays and the decay particles move along the direction of \(M\).

What is the speed of a particle that is observed to have momentum \(500 \mathrm{MeV} / \mathrm{c}\) and energy \(1746 \mathrm{MeV} ?\) What is the particle's mass (in \(\mathrm{MeV} / c^{2}\) )?

The rest energy of a proton is about \(938 \mathrm{MeV}\). If its kinetic energy is also \(938 \mathrm{MeV}\), find \((a)\) its momentum and \((b)\) its speed.

At what value of \(u / c\) does the measured mass of a particle exceed its rest mass by (a) 10 percent, (b) a factor of \(5,\) and \((c)\) a factor of \(20 ?\)

A pion spontaneously decays into a muon and a muon antineutrino according to (among other processes) \(\pi^{-} \rightarrow \mu^{-}+\bar{v}_{\mu}\). Recent experimental evidence indicates that the mass \(m\) of the \(\bar{v}_{\mu}\) is no larger than about \(190 \mathrm{keV} / \mathrm{c}^{2}\) and may be as small as zero. Assuming that the pion decays at rest in the laboratory, compute the energies and momenta of the muon and muon antineutrino \((a)\) if the mass of the antineutrino were zero and \((b)\) if its mass were \(190 \mathrm{keV} / \mathrm{c}^{2}\). The mass of the pion is \(139.56755 \mathrm{MeV} / \mathrm{c}^{2}\) and the mass of the muon is \(105.65839 \mathrm{MeV} / \mathrm{c}^{2}\). (See Chapters 11 and 12 for more on the neutrino mass.)

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