# 经典力学和狭义相对论|PHYS2113 Classical Mechanics and Special Relativity代写

At what speed does a meter stick move if its length is observed to shrink to $0.5 \mathrm{~m}$ ?

Solution:

We assume that the meter stick is at rest in $S^{\prime}$. As observed by stationary observers in $S$, the meter stick moves in the positive $x$-direction with speed $v$. Text ：
$$x^{\prime}=\gamma(x-v t)$$
relates the position $x^{\prime}$ measured in $S^{\prime}$ with the position $x$ measured in $S$.
Let $\Delta x^{\prime}$ be the length of the meter stick measured by an observer at rest in $S^{\prime}$. ( $\Delta x^{\prime}$ is the proper length of the meter stick.)

The meter stick is moving with speed $v$ along the $x$ axis in $S$. To determine its length in $S$, the positions of the front and back of the meter stick are observed by two stationary observers in $S$ at the same time. The length of the meter stick as measured in $S$ is the distance $\Delta x$ between the two stationary observers at $\Delta t=0$.
Then
$$\Delta x^{\prime}=\gamma \Delta x$$
where $\gamma$ is given. It follows that
$$\beta=\sqrt{1-\left(\frac{\Delta x}{\Delta x^{\prime}}\right)^{2}}$$
It follows from that $\beta=0.866$ when $\Delta x=\Delta x^{\prime} / 2$. indicates that the length $\Delta x$ of an object measured by observers at rest in $S$ is smaller than the length $\Delta x^{\prime}$ measured by an observer at rest with respect to the meter stick. That is, “moving rods contract”.
It is important to note that compares an actual length measurement in $S^{\prime}$ (a proper length) with a length measurement determined at equal times on two separate clocks in $S$.\text { The average lifetime of a } \pi \text { meson in its own frame of reference is } 26.0 \mathrm{~ns} \text {. (This is its proper lifetime.) }

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