Chapter 26, Problem 63P

(a)

To determine

To show that $\gamma -1<<1$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Answer to Problem 63P

It is showed that $\gamma -1<<1$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Explanation of Solution

Write the statement given in the problem.

  $\gamma -1<<1$

Here, $\gamma$ is the Lorentz factor.

Rearrange above inequality to get range of $\gamma$.

  $\gamma <<2$                                                                                                                    (I)

Write the expression for $\gamma$.

  $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $v$ is the relativistic velocity and $c$ is the speed of light in vacuum.

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in expression (I) to expand relation to show that $v<<c$.

  $\begin{array}{c}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}<<2\\ \frac{1}{\left(1-\frac{v^2}{c^2}\right)}<<4\\ 1<<4-4\frac{v^2}{c^2}\end{array}$

Rearrange above inequality to get $v<<c$.

  $\begin{array}{l}{v}^2<<\frac{3c^2}{4}<c^2\\ v<<c\end{array}$

Conclusion:

Thus, it is showed that, $\gamma -1<<1$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

(b)

To determine

To show that $K<<m{c}^2$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Answer to Problem 63P

It is showed that $K<<m{c}^2$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Explanation of Solution

Write the statement given in the problem.

  $K<<m{c}^2$                                                                                                                (II)

Here, $K$ is the kinetic energy, $m$ is the rest mass of the particle.

Write expression for the relativistic kinetic energy of a particle

  $K=\left(\gamma -1\right)m{c}^2$                                                                                                      (III)

Write $\left(\gamma -1\right)m{c}^2$ for $K$ in inequality (II) to expand the equation

  $\left(\gamma -1\right)m{c}^2<<m{c}^2$

Divide both side of above inequality by $m{c}^2$ to show $\gamma -1<<1$.

  $\left(\gamma -1\right)<<1$

In part (a) it is showed that if $\left(\gamma -1\right)<<1$, then $v<<c$. Thus, $K<<m{c}^2$ .

Conclusion:

Therefore, it is showed that $K<<m{c}^2$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

(c)

To determine

To show that $p<<mc$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Answer to Problem 63P

It is showed that $p<<mc$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Explanation of Solution

Write the statement given in the problem.

  $p<<mc$                                                                                                                 (IV)

Here, $p$ is the relativistic momentum.

Write expression for the relativistic momentum of a particle.

  $p=\gamma mv$                                                                                                                  (V)

Write $\gamma mv$ for $p$ in inequality (IV) to expand the equation

  $\gamma mv<<mc$

Divide both side of above inequality by $m\gamma$ to show $v<<c$.

  $v<<\frac{c}{\gamma }$                                                                                                                  (VI)

Since $v<c$ , $\frac{1}{\gamma }<1$.

Rearrange inequality (VI) to get $v<<c$.

  $\begin{array}{l}v<<\frac{c}{\gamma }<c\\ v<<c\end{array}$

Conclusion:

Therefore, it is showed that $p<<mc$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

(d)

To determine

To show that $K\approx \frac{p^2}{2m}$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Answer to Problem 63P

It is showed that $K\approx \frac{p^2}{2m}$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.

Explanation of Solution

Write the statement given in the problem.

  $K\approx \frac{p^2}{2m}$                                                                                                               (VII)

Rearrange above expression to get $p^2$.

  $p^2\approx 2mK$                                                                                                           (VIII)

Multiply both sides of above expression with $c^2$.

  $p^2{c}^2\approx 2Km{c}^2$

Replace $p^2{c}^2$ with $K^2+2K{E}_0$ and $m{c}^2$ with $E_0$ to get $K$.

  $\begin{array}{c}{K}^2+2K{E}_0\approx 2K{E}_0\\ K\approx 0\end{array}$                                                                                               (IX)

From equation (III) $K=\left(\gamma -1\right)m{c}^2$.

Substitute $\left(\gamma -1\right)m{c}^2$ for $K$ in expression (IX) to get $\gamma$.

  $\begin{array}{c}\left(\gamma -1\right)m{c}^2\approx 0\\ \gamma \approx 1\end{array}$

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in above expression to show that $v<<c$.

  $\begin{array}{c}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx 1\\ \frac{v^2}{c^2}\approx 0\\ v<<c\end{array}$

Conclusion:

Therefore ,it is showed that $K\approx \frac{p^2}{2m}$ implies that $v<<c$, which means that $v$ can be considered a nonrelativistic speed.