Chapter 25, Problem 26E

(a)

To determine

The fractional error when $v=0.1c$ .

Answer to Problem 26E

The fractional error when $v=0.1c$ is $0.0075$ .

Explanation of Solution

Write the expression for the nonrelativistic approximation for kinetic energy of the particle,

    $K_{approx}=\frac{1}{2}m{v}^2$        (I)

Here, $K_{approx}$ is the nonrelativistic approximation for kinetic energy of the particle, $v$ is the speed of the particle and $m$ is the mass of the particle.

Write the expression for the relativistic exact kinetic energy of the particle,

    $K_{exact}=m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)$        (II)

Here, $K_{exact}$ is the relativistic exact kinetic energy of the particle, $m$ is the mass of the particle, $v$ is the speed of the particle and $c$ is the speed of the light.

Write the expression for the fractional error,

    $f_{error}=\frac{\left|K_{exact}-K_{approx}\right|}{K_{exact}}$        (III)

Here, $f_{error}$ is the fractional error.

Substitute (I) and (II) in (III),

    $f_{error}=\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)-\frac{1}{2}m{v}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)}$        (IV)

Conclusion:

Substitute $0.1c$ for $v$ in (IV),

    $\begin{array}{c}{f}_{error}=\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.1c/c\right)}^2}}-1\right)-\frac{1}{2}m{\left(0.1c\right)}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.1c/c\right)}^2}}-1\right)}\\ =\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.1\right)}^2}}-1\right)-\left(0.005\right)m{c}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.1\right)}^2}}-1\right)}\\ =\frac{\left|\left(0.0050378\right)-\left(0.005\right)\right|}{\left(0.0050378\right)}\\ =0.0075\end{array}$

Therefore, the fractional error when $v=0.1c$ is $0.0075$ .

(b)

To determine

The fractional error when $v=0.5c$ .

Answer to Problem 26E

The fractional error when $v=0.5c$ is $0.192$ .

Explanation of Solution

Write the expression for the nonrelativistic approximation for kinetic energy of the particle,

    $K_{approx}=\frac{1}{2}m{v}^2$        (I)

Here, $K_{approx}$ is the nonrelativistic approximation for kinetic energy of the particle, $v$ is the speed of the particle and $m$ is the mass of the particle.

Write the expression for the relativistic exact kinetic energy of the particle,

    $K_{exact}=m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)$        (II)

Here, $K_{exact}$ is the relativistic exact kinetic energy of the particle, $m$ is the mass of the particle, $v$ is the speed of the particle and $c$ is the speed of the light.

Write the expression for the fractional error,

    $f_{error}=\frac{\left|K_{exact}-K_{approx}\right|}{K_{exact}}$        (III)

Here, $f_{error}$ is the fractional error.

Substitute (I) and (II) in (III),

    $f_{error}=\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)-\frac{1}{2}m{v}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}-1\right)}$        (IV)

Conclusion:

Substitute $0.5c$ for $v$ in (IV),

    $\begin{array}{c}{f}_{error}=\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.5c/c\right)}^2}}-1\right)-\frac{1}{2}m{\left(0.5c\right)}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.5c/c\right)}^2}}-1\right)}\\ =\frac{\left|m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.5\right)}^2}}-1\right)-\left(0.125\right)m{c}^2\right|}{m{c}^2\left(\frac{1}{\sqrt{1-{\left(0.5\right)}^2}}-1\right)}\\ =\frac{\left|\left(0.1547\right)-\left(0.125\right)\right|}{\left(0.1547\right)}\\ =0.192\end{array}$

Therefore, the fractional error when $v=0.5c$ is $0.192$.