Chapter 39, Problem 41PQ

To determine

The magnitude of velocity of an electron in the laboratory frame.

Answer to Problem 41PQ

The magnitude of velocity of the electron measured in the laboratory frame is $8.50\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the magnitude of velocity measured from the laboratory frame.

    $v=\sqrt{v_{\text{x}}{}^2+v_{\text{y}}{}^2+v_{\text{z}}{}^2}$                                                                                                    (I)

Here, $v$ is the resultant velocity, $v_{\text{x}}$ is the velocity transformation in the $x$ direction, $v_{\text{y}}$ is the velocity transformation in the $y$ direction, and $v_{\text{z}}$ the velocity transformation in the $z$ direction.

Write the expression for velocity components in each direction,

    $v_{\text{x}}=\frac{v_{\text{x}}{}^{\prime }+v_{\text{rel}}}{\left[1+\frac{v_{\text{rel}}\times v_{\text{x}}{}^{\prime }}{c^2}\right]}$                                                                                                    (II)

    $v_{\text{y}}=\frac{v_{\text{y}}{}^{\prime }}{\gamma \left[1+\frac{v_{\text{rel}}\times v_{\text{x}}{}^{\prime }}{c^2}\right]}$                                                                                                (III)

    $v_{\text{z}}=\frac{v_{\text{z}}{}^{\prime }}{\gamma \left[1+\frac{v_{\text{rel}}\times v_{\text{x}}{}^{\prime }}{c^2}\right]}$                                                                                                (IV)

Here, $v_{\text{rel}}$ is the relative velocity, $c$ is the speed of light, $\gamma$ is the Lorentz constant, $v_{\text{x}}{}^{\prime }$ is the inverse velocity transformation in $x$ direction, $v_{\text{y}}{}^{\prime }$ is the inverse velocity transformation in $y$ direction and $v_{\text{z}}{}^{\prime }$ is the inverse velocity transformation in $z$ direction.

Write the expression for Lorentz constant,

    $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}$                                                                                                       (V)

Conclusion:

Substitute $2.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (V) to find $\gamma$.

    $\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(\frac{2.00\times {10}^7\text{m}/\text{s}}{3.00\times {10}^8\text{m}/\text{s}}\right)}^2}}\\ =\frac{1}{\sqrt{1-{\left(0.06\right)}^2}}\\ =\frac{1}{0.998}\\ =1.002\end{array}$

Substitute $5.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}{}^{\prime }$, $2.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (III) to find $v_{\text{x}}$.

    $\begin{array}{c}{v}_{\text{x}}=\frac{5.00\times {10}^7\text{m}/\text{s}+2.00\times {10}^7\text{m}/\text{s}}{\left[1+\frac{5.00\times {10}^7\text{m}/\text{s}\times 2.00\times {10}^7\text{m}/\text{s}}{{\left(3.00\times {10}^8\right)}^2{\text{m}}^2/{\text{s}}^2}\right]}\\ =\frac{7.00\times {10}^7\text{m}/\text{s}}{\left[1+\frac{{10}^{15}}{9\times {10}^{16}}\right]}\\ =\frac{7.00\times {10}^7\text{m}/\text{s}}{1.011}\\ =6.92\times {10}^7\text{m}/\text{s}\end{array}$

Substitute $4.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{y}}{}^{\prime }$, $1.002$ for $\gamma$, $5.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}{}^{\prime }$, $2.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (IV) to find $v_{\text{y}}$.

    $\begin{array}{c}{v}_{\text{y}}=\frac{4.00\times {10}^7\text{m}/\text{s}}{1.002\left[1+\frac{2.00\times {10}^7\text{m}/\text{s}\times 5.00\times {10}^7\text{m}/\text{s}}{{\left(3.00\times {10}^8\right)}^2{\text{m}}^2/{\text{s}}^2}\right]}\\ =\frac{4.00\times {10}^7\text{m}/\text{s}}{1.002\left[1.011\right]}\\ =\frac{4.00\times {10}^7\text{m}/\text{s}}{1.013}\\ =3.95\times {10}^7\text{m}/\text{s}\end{array}$

Substitute $3.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{z}}{}^{\prime }$, $1.002$ for $\gamma$, $5.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}{}^{\prime }$, $2.00\times {10}^7\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (V) to find $v_{\text{z}}$.

    $\begin{array}{c}{v}_{\text{z}}=\frac{3.00\times {10}^7\text{m}/\text{s}}{1.002\left[1+\frac{2.00\times {10}^7\text{m}/\text{s}\times 5.00\times {10}^7\text{m}/\text{s}}{{\left(3.00\times {10}^8\right)}^2{\text{m}}^2/{\text{s}}^2}\right]}\\ =\frac{3.00\times {10}^7\text{m}/\text{s}}{1.002\left[1.011\right]}\\ =\frac{3.00\times {10}^7}{1.013}\\ =2.96\times {10}^7\text{m}/\text{s}\end{array}$

Substitute $6.92\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}$, $3.95\times {10}^7\text{m}/\text{s}$ for $v_{\text{y}}$ and $2.96\times {10}^7\text{m}/\text{s}$ for $v_{\text{z}}$ in the equation (I) to obtain the velocity of the electron.

    $\begin{array}{c}v=\sqrt{v_{\text{x}}{}^2+v_{\text{y}}{}^2+v_{\text{z}}{}^2}\\ =\sqrt{{\left(6.92\times {10}^7\text{m}/\text{s}\right)}^2+{\left(3.95\times {10}^7\text{m}/\text{s}\right)}^2+{\left(2.96\times {10}^7\text{m}/\text{s}\right)}^2}\\ =\sqrt{72.2505\times {10}^{14}{\text{m}}^2/{\text{s}}^2}\\ =8.50\times {10}^7\text{m}/\text{s}\end{array}$

Therefore, the magnitude of velocity of electron is $8.50\times {10}^7\text{m}/\text{s}$.