Chapter 39, Problem 10PQ

(a)

To determine

The coordinates $\left(x,y,z,t\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $400\text{m}/\text{s}$.

Answer to Problem 10PQ

The coordinates $\left(12.2\text{m},4.00\text{m},6.00\text{m},4.00\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $400\text{m}/\text{s}$.

Explanation of Solution

Write the expression to obtain the Lorentz factor.

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

Here, $\gamma$ is the Lorentz factor, $v_{\text{rel}}$ is the relative velocity of the object and $c$ is the speed of light.

Write the expression to obtain the expression of time interval in the laboratory frame.

    $t=\gamma \left(t\text{'}+\frac{v_{\text{rel}}}{c^2}x\text{'}\right)$                                                                                                 (II)

Here, $t$ is the time interval in the laboratory frame, $x\text{'}$ is the position of the object along $x$ axis in primed frame, $v_{\text{rel}}$ is the relative velocity of the object and $t\text{'}$ is the time interval in primed frame.

Write the expression to obtain the position of the object along $x$ axis in laboratory frame.

    $x=\gamma \left(x\text{'}+v_{\text{rel}}t\text{'}\right)$                                                                                                   (III)

Here, $x$ is the position of the object along $x$ axis in laboratory frame, $\gamma$ is the Lorentz factor, $x\text{'}$ is the position of the object along $x$ axis in primed frame, $v_{\text{rel}}$ is the relative velocity of the object and $t\text{'}$ is the time interval in primed frame.

As the object is moving with a relative velocity along the x-axis only, thus the coordinate in y-axis and $y\text{'}$ axis remain same and similarly the coordinate in z-axis and $z\text{'}$ axis remain same

Write the expression to obtain the $y$ coordinate in laboratory frame.

    $y=y\text{'}$                                                                                                                    (IV)

Here, $y$ and $y\text{'}$ are coordinates in laboratory frame and primed frame respectively.

Write the expression to obtain the $z$ coordinate in laboratory frame.

    $z=z\text{'}$                                                                                                                    (V)

Here, $z$ and $z\text{'}$ are coordinates in laboratory frame and primed frame respectively.

Conclusion:

Substitute $400.0\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I) to calculate $\gamma$.

    $\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(\frac{400.0\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\\ =\frac{1}{\sqrt{1-0}}\\ =1\end{array}$

Substitute $1$ for $\gamma$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $400.0\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (II) to calculate $t$.

    $\begin{array}{c}t=1\left(\left(4.00\times {10}^{-4}\text{s}\right)+\frac{400.0\text{m}/\text{s}}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\left(12.0\text{m}\right)\right)\\ =\left(\left(4.00\times {10}^{-4}\text{s}\right)+\left(5.33\times {10}^{-14}\text{s}\right)\right)\\ =4.00\times {10}^{-4}\text{s}\end{array}$

Substitute $1$ for $\gamma$, $400.0\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (III) to calculate $x$.

    $\begin{array}{c}x=1\left(12.0\text{m}+\left(400.0\text{m}/\text{s}\right)\left(4.00\times {10}^{-4}\text{s}\right)\right)\\ =12.0\text{m }+\text{0}\text{.16}\text{m}\\ =12.16\text{m}\\ \approx 12.2\text{m}\end{array}$

Substitute $4.00\text{m}$ for $y\text{'}$ in equation (III) to calculate $y$.

    $y=4.00\text{m}$

Substitute $6.00\text{m}$ for $z\text{'}$ in equation (IV) to calculate $z$.

    $z=6.00\text{m}$

Therefore, the coordinates $\left(12.2\text{m},4.00\text{m},6.00\text{m},4.00\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $400\text{m}/\text{s}$.

(b)

To determine

The coordinates $\left(x,y,z,t\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $-400\text{m}/\text{s}$.

Answer to Problem 10PQ

The coordinates $\left(11.8\text{m},4.00\text{m},6.00\text{m},4.00\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $-400\text{m}/\text{s}$.

Explanation of Solution

Conclusion:

Substitute $-400.0\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I) to calculate $\gamma$.

    $\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(\frac{-400.0\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\\ =\frac{1}{\sqrt{1-0}}\\ =1\end{array}$

Substitute $1$ for $\gamma$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $-400.0\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (II) to calculate $t$.

    $\begin{array}{c}t=1\left(\left(4.00\times {10}^{-4}\text{s}\right)+\frac{-400.0\text{m}/\text{s}}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\left(12.0\text{m}\right)\right)\\ =\left(\left(4.00\times {10}^{-4}\text{s}\right)-\left(5.33\times {10}^{-14}\text{s}\right)\right)\\ =4.00\times {10}^{-4}\text{s}\end{array}$

Substitute $1$ for $\gamma$, $-400.0\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (III) to calculate $x$.

    $\begin{array}{c}x=1\left(12.0\text{m}+\left(-400.0\text{m}/\text{s}\right)\left(4.00\times {10}^{-4}\text{s}\right)\right)\\ =12.0\text{m}-\text{0}\text{.16 m}\\ =11.\text{84 m}\\ \approx 11.\text{8 m}\end{array}$

Substitute $4.00\text{m}$ for $y\text{'}$ in equation (III) to calculate $y$.

    $y=4.00\text{m}$

Substitute $6.00\text{m}$ for $z\text{'}$ in equation (IV) to calculate $z$.

    $z=6.00\text{m}$

Therefore, the coordinates $\left(11.8\text{m},4.00\text{m},6.00\text{m},4.00\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $-400\text{m}/\text{s}$.

(c)

To determine

The coordinates $\left(x,y,z,t\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $2.0\times {10}^8\text{m}/\text{s}$.

Answer to Problem 10PQ

The coordinates $\left(1.1\times {10}^5\text{m},4.00\text{m},6.00\text{m},5.368\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $2.0\times {10}^8\text{m}/\text{s}$.

Explanation of Solution

Conclusion:

Substitute $2.0\times {10}^8\text{m}/\text{s}$ for $v_{\text{rel}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I) to calculate $\gamma$.

    $\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\left(\frac{2.0\times {10}^8\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\\ =1.342\end{array}$

Substitute $1.342$ for $\gamma$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $2.0\times {10}^8\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (II) to calculate $t$.

    $\begin{array}{c}t=1.342\left(\left(4.00\times {10}^{-4}\text{s}\right)+\frac{2.0\times {10}^8\text{m}/\text{s}}{{\left(3\times {10}^8\text{m}/\text{s}\right)}^2}\left(12.0\text{m}\right)\right)\\ =1.342\left[\left(4.00\times {10}^{-4}\text{s}\right)-\left(2.667\times {10}^{-8}\text{s}\right)\right]\\ =5.368\times {10}^{-4}\text{s}\end{array}$

Substitute $1.342$ for $\gamma$, $2.0\times {10}^8\text{m}/\text{s}$ for $v_{\text{rel}}$, $\left(4.00\times {10}^{-4}\text{s}\right)$ for $t\text{'}$ and $12.0\text{m}$ for $x\text{'}$ in equation (III) to calculate $x$.

    $\begin{array}{c}x=1.342\left(12.0\text{m}+\left(2.0\times {10}^8\text{m}/\text{s}\right)\left(4.00\times {10}^{-4}\text{s}\right)\right)\\ =1.342\left(12.0\text{m}+8\times {10}^4\text{m}\right)\\ =1.072\times {10}^5\text{m}\\ \approx 1.1\times {10}^5\text{m}\end{array}$

Substitute $4.00\text{m}$ for $y\text{'}$ in equation (III) to calculate $y$.

    $y=4.00\text{m}$

Substitute $6.00\text{m}$ for $z\text{'}$ in equation (IV) to calculate $z$.

    $z=6.00\text{m}$

Therefore, the coordinates $\left(1.1\times {10}^5\text{m},4.00\text{m},6.00\text{m},5.368\times {10}^{-4}\text{s}\right)$ as measured in the laboratory frame if the particle is moving along the $x$ and $x\text{'}$ axes with relative speed of $2.0\times {10}^8\text{m}/\text{s}$.