Chapter 1, Problem 14P

A rod of length L₀ moves with a speed v along the horizontal direction. The rod makes an angle of θ₀ with respect to the x′-axis. (a) Show that the length of the rod as measured by a stationary observer is given by $L=L_0{\left[1-\left(v^2/c^2\right){\cos }^2{\theta }_0\right]}^{1/2}$. (b) Show that the angle that the rod makes with the x-axis is given by the expression tan θ = γ tan θ₀. These results show that the rod is both contracted and rotated. (Take the lower end of the rod to be at the origin of the primed coordinate system.)

To determine

To show that the length of the rod as measured by a stationary observer is given by $L=L_0{\left[1-{\left(v/c\right)}^2{\cos }^2{\theta }_0\right]}^{1/2}$

Answer to Problem 14P

The length of the rod as measured by a stationary observer is given by $L=L_0{\left[1-{\left(v/c\right)}^2{\cos }^2{\theta }_0\right]}^{1/2}$ is shown.

Explanation of Solution

The rod of length $L_0$ moves along the horizontal axis and hence the horizontal component of the rod gets contracted.

The horizontal component of the rod is given by,

$L_{x^{\prime }}=L_0\cos {\theta }_0$

Here, $L_{x^{\prime }}$ is the horizontal component of the rod and ${\theta }_0$ is the angle made by the rod.

The horizontal component of the rod as measured by the stationary observer is,

$\begin{array}{c}{L}_x=\frac{L_{x^{\prime }}}{{\left[1-{\left(v/c\right)}^2\right]}^{1/2}}\\ =\frac{L_0\cos {\theta }_0}{{\left[1-{\left(v/c\right)}^2\right]}^{1/2}}\end{array}$

Here, $L_x$ is horizontal component of the rod as measured by the stationary observer

The vertical component of the rod is given by,

$L_{y^{\prime }}=L_0\sin {\theta }_0$

Here, $L_{y^{\prime }}$ is the vertical component of the rod.

The vertical component of the rod as measured by the stationary observer is,

$\begin{array}{c}{L}_y=\frac{L_{y^{\prime }}}{{\left[1-{\left(v/c\right)}^2\right]}^{1/2}}\\ =\frac{L_0\sin {\theta }_0}{{\left[1-{\left(v/c\right)}^2\right]}^{1/2}}\\ =\frac{L_0\sin {\theta }_0}{{\left[1-{\left(0/c\right)}^2\right]}^{1/2}}\left(v=0\text{ along vertical direction}\right)\\ =L_0\sin {\theta }_0\end{array}$

Here, $L_y$ is the vertical component of the rod as measured by the stationary observer.

Conclusion:

The length of the rod as measured by a stationary observer is $\begin{array}{c}L={\left[{\left(L_x\right)}^2+{\left(L_y\right)}^2\right]}^{1/2}\\ ={\left[{\left(\frac{L_0\cos {\theta }_0}{{\left[1-{\left(v/c\right)}^2\right]}^{1/2}}\right)}^2+{\left(L_0\sin {\theta }_0\right)}^2\right]}^{1/2}\\ =L_0{\left[{\cos }^2{\theta }_0\left(1-\frac{v^2}{c^2}\right)+{\sin }^2{\theta }_0\right]}^{1/2}\\ =L_0{\left[1-\frac{v^2}{c^2}{\cos }^2{\theta }_0\right]}^{1/2}\end{array}$

Therefore, the length of the rod as measured by a stationary observer is given by $L=L_0{\left[1-{\left(v/c\right)}^2{\cos }^2{\theta }_0\right]}^{1/2}$ is shown.

To determine

To show that the angle the rod makes with the x axis is $\tan \theta =\gamma \tan {\theta }_0$

Answer to Problem 14P

The angle the rod makes with the x axis is $\tan \theta =\gamma \tan {\theta }_0$ is shown.

Explanation of Solution

The angle the rod makes with the x axis is,

$\begin{array}{c}\tan \theta =\frac{L_y}{L_x}\\ =\frac{L_0\sin {\theta }_0}{L_0\cos {\theta }_0/\gamma }\because \left(\gamma ={\left[1-{\left(v/c\right)}^2\right]}^{-1/2}\right)\\ =\gamma \tan {\theta }_0\end{array}$

Conclusion:

Therefore, the angle the rod makes with the x axis is $\tan \theta =\gamma \tan {\theta }_0$ is shown.