A problem in relativity

Solution: Consider an inertial reference frame $S^{\prime }$ with axes $x^{\prime }$ , $y^{\prime }$ , and $z^{\prime }$ coinsiding with $x$ , $y$ , and $z$ at time $t=t^{\prime }=0$ , and moving with constant speed, $v$ , in the positive $x$ direction relative to $S$ . The velocity vector of the point labeled $A$ on the plane, relative to $S$ , has components $(u_{x},u_{y},u_{z})=(0,v,0).$ The velocity vector for point $B$ is also $(u_{x},u_{y},u_{z})=(0,v,0)$ . Using the velocity transformation

$\begin{aligned} u_x^\prime &= \frac{u_{x}-v}{(1-\frac{u_{x}v}{c^{2}})} \\ u_y^\prime &= \frac{u_{y}}{\gamma (1-\frac{u_{x}v}{c^{2}})} \\ u_z^\prime &= \frac{u_{z}}{\gamma (1-\frac{u_{x}v}{c^{2}})}, \end{aligned}$

where $\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}},$ to find the velocity vectors for $A$ and $B$ relative to $S^{\prime }$ gives $(u_{x}^{\prime },u_{y}^{\prime },u_{z}^{\prime })=(-v,\frac{v}{\gamma },0)$ for both $A$ and $B$ . The equations of motion of points $A$ and $B$ relative to $S^{\prime }$ are $\begin{aligned} x_{A}^{\prime } &=&u_{x}^{\prime }t^{\prime }+x_{A0}^{\prime }=-vt^{\prime }+x_{A0}^{\prime } \\ y_{A}^{\prime } &=&u_{y}^{\prime }t^{\prime }+y_{A0}^{\prime }=\frac{v}{ \gamma }t^{\prime }+y_{A0}^{\prime } \\ z_{A}^{\prime } &=&u_{z}^{\prime }t^{\prime }+z_{A0}^{\prime }=z_{A0}^{\prime }\text{,} \end{aligned}$ where $(x_{A0}^{\prime },y_{A0}^{\prime },z_{A0}^{\prime })$ are the coordinates of $A$ , relative to $S^{\prime }$ at time $t^{\prime }=0.$ Similarly, for point $B$ $\begin{aligned} x_{B}^{\prime } &=&u_{x}^{\prime }t^{\prime }+x_{B0}^{\prime }=-vt^{\prime }+x_{B0}^{\prime } \\ y_{B}^{\prime } &=&u_{y}^{\prime }t^{\prime }+y_{B0}^{\prime }=\frac{v}{ \gamma }t^{\prime }+y_{B0}^{\prime } \\ z_{B}^{\prime } &=&u_{z}^{\prime }t^{\prime }+z_{B0}^{\prime }=z_{B0}^{\prime }\text{.} \end{aligned}$ The Lorentz transformation, given in the figure, allows determination of the initial values, $x_{A0}^{\prime },$ etc. The coordinates of $A$ relative to $S$ at time $t=0$ are $(x_{A},y_{A},z_{A})=(-R,0,0)$ and those of $B$ are $(x_{B},y_{B},z_{B})=(R,0,0).$ The Lorentz transformation gives $(x_{A}^{\prime },y_{A}^{\prime },z_{A}^{\prime })=(-R\gamma ,0,0)$ at time $t^{\prime }=\frac{v}{c^{2}}\gamma R.$ and $(x_{B}^{\prime },y_{B}^{\prime },z_{B}^{\prime })=(R\gamma ,0,0)$ at time $t^{\prime }=-\frac{v}{c^{2}}\gamma R\text{.}$ Substitution gives $\begin{aligned} -R\gamma &=&-\frac{v^{2}}{c^{2}}\gamma R+x_{A0}^{\prime } \\ 0 &=&\frac{v^{2}}{c^{2}}R+y_{A0}^{\prime } \\ 0 &=&z_{A0}^{\prime }\text{.} \end{aligned}$ and $\begin{aligned} R\gamma &=&\frac{v^{2}}{c^{2}}\gamma R+x_{B0}^{\prime } \\ 0 &=&-\frac{v^{2}}{c^{2}}R+y_{B0}^{\prime } \\ 0 &=&z_{B0}^{\prime }\text{.} \end{aligned}$ Solving for the initial positions of points $A$ and $B$ relative to $S^{\prime}$ gives $\begin{aligned} x_{A0}^{\prime } &=&-R\gamma (1-\frac{v^{2}}{c^{2}})=-\frac{R}{\gamma } \\ y_{A0}^{\prime } &=&-\frac{v^{2}}{c^{2}}R \\ z_{A0}^{\prime } &=&0\text{.} \end{aligned}$ and $\begin{aligned} x_{B0}^{\prime } &=&\frac{R}{\gamma } \\ y_{B0}^{\prime } &=&\frac{v^{2}}{c^{2}}R \\ z_{B0}^{\prime } &=&0\text{.} \end{aligned}$ At any time $t^{\prime}$ in $S^{\prime },$ $\begin{aligned} x_{A}^{\prime } &=&-vt^{\prime }-\frac{R}{\gamma } \\ y_{A}^{\prime } &=&\frac{v}{\gamma }t^{\prime }-\frac{v^{2}}{c^{2}}R \\ z_{A}^{\prime } &=&0 \end{aligned}$ and $\begin{aligned} x_{B}^{\prime } &=&-vt^{\prime }+\frac{R}{\gamma } \\ y_{B}^{\prime } &=&\frac{v}{\gamma }t^{\prime }+\frac{v^{2}}{c^{2}}R \\ z_{B}^{\prime } &=&0\text{.} \end{aligned}$ Therefore, at any time $t^{\prime}$ in $S^{\prime}$ , the line segment connecting points $A$ and $B$ has slope $m^{\prime }=\frac{\Delta y^{\prime }}{\Delta x^{\prime }}=\frac{ y_{B}^{\prime }-y_{A}^{\prime }}{x_{B}^{\prime }-x_{A}^{\prime }}=\frac{v^{2} }{c^{2}}\gamma \text{.}$ If $v=(\sqrt{3}/2)c$ , as stated in the problem, then $\gamma = 2$ and $m^{\prime} = 3/2$ . If $\theta$ is the angle between the $x^{\prime}$ direction and the line in $S^{\prime}$ connecting $A$ and $B$ in $S^{\prime}$ , then $\tan \theta =m^{\prime }\text{=}\frac{3}{2}.$ Therefore, $\theta = \tan^{-1}(3/2)=56.3^{\circ }.$

The world-lines of the points $A$ and $B$ have coordinates $\big(ct,-R,vt,0\big) \hspace{2cm} \big(ct, R, vt, 0\big)$ respectively. In the frame $S'$ these can be written $\big(\gamma(ct + \tfrac{vR}{c}),\gamma(-R - vt),vt,0\big) \hspace{2cm} \big( \gamma(ct - \tfrac{vR}{c}),\gamma(R - vt),vt,0\big)$ respectively. Changing the time coordinates of both of these so that $A$ and $B$ can be considered simultaneously in $S'$ , we have worldlines $\big(c\tau,-v\tau -\gamma^{-1}R,\tfrac{v\tau}{\gamma} - \tfrac{v^2}{c^2}R,0\big) \hspace{2cm} \big(c\tau,-v\tau +\gamma^{-1}R,\tfrac{v\tau}{\gamma} + \tfrac{v^2}{c^2}R,0\big)$ Thus the displacement vector $\overrightarrow{AB}$ has coordinates (in $S'$ ) $\left(\begin{array}{c} 2\gamma^{-1}R \\ 2\tfrac{v^2}{c^2}R \\ 0 \end{array} \right)$ Thus the angle $\theta$ satisfies $\tan\theta \; =\; \tfrac{v^2}{c^2}\gamma$ For this problem, we have $\tan\theta = \tfrac32$ , and so $\theta = \boxed{56.3^\circ}$ .

Although this is not necessary to solve the problem, it is worth noting how we can show that the rod does indeed slip through the hole successfully. In $S'$ , we have $\overrightarrow{AB} \; = \; \left(\begin{array}{c} R \\ \tfrac32R \\ 0 \end{array} \right)$ while the common velocity of $A$ and $B$ is $\left(\begin{array}{c} -v \\ \tfrac12v \\ 0 \end{array}\right)$ Thus the gap between $A$ and $B$ is available (for the rod to pass through) for a time of $\frac{\frac32R}{\frac12v} = \frac{3R}{v}$ , which is precisely the time that a rod of length $3R$ will take to travel through the gap.