- #1

- 804

- 436

The observer who is stationary relative to the desktop thinks that the length of the rigid rod ##~ l=l_o \sqrt{1-\beta^2}=\frac {l_0}{2}~##, which is only one-half of the hole width. Therefore, when the rigid rod passes through the hole, it will fall into the hole due to gravity.

However, for the observer who is stationary relative to the rigid rod, since the tabletop moves relative to him at the speed of ## ~\frac {\sqrt{3}}{2}c~ ## , the width of the hole he measured is ##~ l'=l_o \sqrt{1-\beta^2}=\frac {l_0}{2}~##, so he thinks that the the rigid rod with length of ##~l_0~## will not fall into the hole with width only ##~\frac {l_0}{2}~##.

Since the length contraction of the theory of relativity is not an illusion but a real effect, there is only one answer to whether the rigid rod falls into the hole, so how do we use the simplest way to explain this contradiction and find the answer?