This subject is designed for students with a sound background in physics, and aims to provide a strong understanding of a broad range of physics principles.
这是一份unimelb墨尔本大学PHYS1149的成功案例
An atomic clock is placed in a jet airplane. The clock measures a time interval of $3600 \mathrm{~s}$ when the jet moves with speed $400 \mathrm{~m} / \mathrm{s}$.
- How much larger a time interval does an identical clock held by an observer at rest on the ground measure?
We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be the rest frame of the atomic clock. It follows from that
and from that
$$
\gamma \simeq 1+\beta^{2} / 2
$$
It follows that $\delta t=3.2 \mathrm{~ns}$ when $v=400 \mathrm{~m} / \mathrm{s}$ and $\Delta t^{\prime}=3600 \mathrm{~s}$.
PHYC10003 COURSE NOTES ：
Two spaceships approach each other, each moving with the same speed as measured by a stationary observer on the Earth. Their relative speed is $0.70 c$,
- Determine the velocities of each spaceship as measured by the stationary observer on Earth.
Solution
Text Eq. (1.32) gives the Lorentz velocity transformation:
$$
u_{x}^{\prime}=\frac{u_{x}-v}{1-u_{x} v / c^{2}}
$$
where $u_{x}$ is the velocity of an object measured in the $S$ frame, $u_{x}^{\prime}$ is the velocity of the object measured in the $S^{\prime}$ frame and $v$ is the velocity of the $S^{\prime}$ frame along the $x$ axis of $S$.
We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be attached to the spaceship moving to the right with velocity $v$. The other spaceship has velocity $u_{x}=-v$ in $S$ and velocity $u_{x}^{\prime}=-0.70 c$ in $S^{\prime}$.
It follows fromthat
$$
0.70=\frac{2 \beta}{1+\beta^{2}}
$$
solving which yields $\beta=0.41$. As measured by the stationary observer on Earth, the spaceships are moving with velocities $\pm 0.41 c$.