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.
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?
证明 .
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 $\Delta t^{\prime}=3600 \mathrm{~s}$.
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 $\pm 0.41 c$.
$$ \begin{aligned} &F_{x} / m=-\frac{v^{2}}{r} \cos \theta \text {, so } \ &F_{x}=-m \frac{v^{2}}{r} \cos \theta . \end{aligned} $$ Since our goal is an equation involving the period, it is natural to eliminate the variable $v=$ circumference $/ T=2 \pi r / T$, giving $$ F_{x}=-\frac{4 \pi^{2} m r}{T^{2}} \cos \theta . $$ The quantity $r \cos \theta$ is the same as $x$, so we have $$ F_{x}=-\frac{4 \pi^{2} m}{T^{2}} x $$
$$ \frac{F_{r}}{F_{t}}=\frac{2 \pi m}{b f}\left(f^{2}-f_{r e s}^{2}\right) $$ This is the ratio of the wasted force to the useful force, and we see that it becomes zero when the system is driven at resonance. The amplitude of the vibrations can be found by attacking the equation $|F|=b v=2 \pi b A f$, which gives $$ A=\frac{\left|F_{\mathrm{t}}\right|}{2 \pi b f} . $$ However, we wish to know the amplitude in terms of $|\boldsymbol{F}|$, not $\left|F_{\mathrm{t}}\right|$. From now on, let’s drop the cumbersome magnitude symbols. With the Pythagorean theorem, it is easily proven that $$ F_{t}=\frac{F}{\sqrt{1+\left(\frac{F_{t}}{F_{t}}\right)^{2}}}, $$
$$ \frac{1}{2} m\left(v_{x}-2 u\right)^{2}-\frac{1}{2} m v_{x}^{2}=-2 m u v_{x} $$ where we have discarded the second order term in $u$. Therefore, the change in internal energy of the ideal gas due to the collision between this class of molecules with the infinitesimal area $d A$ is $$ -2 \operatorname{m\eta ug}\left(v_{x}\right) v_{x}^{2} d v_{x} d A d t $$
证明 .
$$ d E=-2 m \eta d V \int_{0}^{\infty} g\left(v_{x}\right) v_{x}^{2} d v_{x} $$ Since $\int_{0}^{\infty} g\left(v_{x}\right) v_{x}^{2} d v_{x}=\frac{1}{2}\left\langle v_{x}^{2}\right\rangle$ $$ d E=-m \eta\left\langle v_{x}^{2}\right\rangle d V=-\frac{1}{3} m \eta\left\langle v^{2}\right\rangle d V $$
$$ d S-\frac{\delta Q}{T_{e x t}} \geq 0 $$ for the total entropy change of the universe to be non-negative. Applying the first law of thermodynamics $\delta Q=d U+\delta W_{b y}$, where $d U$ is the change in internal energy of the system and $\delta W_{b y}$ is the work done by the system, $$ d U+\delta W_{b y}-T_{e x t} d S \leq 0 $$ If the work done by the system is only that against its external environment, $\delta W_{b y}=p_{e x t} d V$ where $d V$ is the change in volume of the system. Thus, $$ d U+p_{e x t} d V-T_{e x t} d S \leq 0 $$
$$ \begin{aligned} \rho v+\Delta \rho c &=0, \ \mu c v+c_{p} \Delta T &=0 . \end{aligned} $$ $\Delta T$ can be related to $\Delta \rho$ through the adiabatic condition. Since $p^{1-\gamma} T^{\gamma}=$ constant and $\rho \propto \frac{p}{T}$ by the ideal gas law, $$ \rho^{1-\gamma} T=c $$
证明 .
for some constant $c$. Taking the total derivative of the above, $$ \begin{gathered} (1-\gamma) \rho^{-\gamma} T d \rho+\rho^{1-\gamma} d T=0 \ d T=\frac{(\gamma-1) T}{\rho} d \rho . \end{gathered} $$ Since $\Delta \rho$ and $\Delta T$ are small, $$ \Delta T \approx \frac{(\gamma-1) T}{\rho} \Delta \rho $$ Substituting this expression for $\Delta T$ and summarizing our equations, $$ \begin{aligned} \mu c v+\frac{c_{p}(\gamma-1) T}{\rho} \Delta \rho &=0 \ \rho v+c \Delta \rho &=0 \end{aligned} $$
$$ \begin{gathered} \mu c^{2}-c_{p}(\gamma-1) T=0 \ c=\sqrt{\frac{c_{p}(\gamma-1) T}{\mu}} . \end{gathered} $$ Notice that $c_{p}(\gamma-1)=c_{p} \cdot \frac{c_{p}-c_{v}}{c_{v}}=\frac{c_{p}}{c_{v}} \cdot R=\gamma R$ and $\frac{\mu}{R T}=\frac{\rho}{p}$ by the ideal gas law such that the above becomes $$ c=\sqrt{\frac{\gamma p}{\rho}} $$
Advances in physics often enable advances in new technologies. For example, advances in the understanding of electromagnetism, solid-state physics, and nuclear physics led directly to the development of new products that have dramatically transformed modern-day society, such as television, computers, domestic appliances, and nuclear weapons;advances in thermodynamics led to the development of industrialization; and advances in mechanics inspired the development of calculus.
物理课后作业代写
As usual the resonance poles discussed above can be manifested in two ways, either in scattering properties, here of a particle moving along the “wire” $\Sigma$, or through the time evolution of states associated with the “dots” П. By assumption (4.1) there is a nontrivial discrete spectrum of $\bar{H}{\beta}$ embedded in $\left(-\frac{1}{4} \alpha^{2}, 0\right)$. Let us denote the corresponding normalized eigenfunctions $\psi{j}, j=1, \ldots, m$, given by $$ \psi_{j}(x)=\sum_{i=1}^{m} d_{i}^{(j)} \phi_{i}^{(j)}(x), \quad \phi_{i}^{(j)}(x):=\sqrt{-\frac{\epsilon_{j}}{\pi}} K_{0}\left(\sqrt{-\epsilon_{j}}\left|x-y^{(i)}\right|\right) $$ in accordance with [1, Sec. II.3], where the vectors $d^{(j)} \in \mathbb{C}^{m}$ satisfy the equation In particular, if the distances between the points of II are large (the natural length scale is given by $\left.\left(-\epsilon_{j}\right)^{-1 / 2}\right)$, the cross terms are small and $\left|d^{(j)}\right|$ is close to one. Let us now specify the unstable system of our model by identifying its state $P \mathcal{H}$ with the span of the vectors $\psi_{1}, \ldots . \psi_{m}$. Suppose that it is preHilbert space $P \mathcal{H}$ with the span of the vectors $\psi_{1}, \ldots, \psi_{m} .$ Suppose that it is pre- pared at the initial instant $t=0$ at a state $\psi \in P \mathcal{H}$, then the decay law describing the probability of finding the system undecayed at a subsequent measurement performed at $t$, without disturbing it in between $[6]$, is $$ P_{\psi}(t)=\left|P \mathrm{e}^{-i H_{\alpha, \beta} t} \psi\right|^{2} $$
