物理代写|广义相对论代写General relativity代考|PHYS3100

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

物理代写|广义相对论代写General relativity代考|Acceleration

In the simple approach to special relativity in Chap. 1 we studied the Lorentz transformation between uniformly moving systems; this in no way restricts special relativity to uniform motion, and accelerated motion fits nicely into the conceptual and mathematical framework. We first define the 4-vector acceleration of a particle in the obvious way, as the derivative of the 4-vector velocity with respect to the proper time of the particle,
$$a^{\mu}=\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d}^{2} x^{\mu}}{\mathrm{d} \tau^{2}}$$
We may express this in terms of the classical velocity and acceleration, which involve $t$ derivatives, not $\tau$ derivatives. To do this we use the expression for the 4 -velocity in (3.6) and the relation between $\mathrm{d} \tau$ and $\mathrm{d} t$ in (3.4), which implies $\mathrm{d} / \mathrm{d} \tau=\gamma(\mathrm{d} / \mathrm{d} t)$, to obtain

$$a^{\mu}=\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d}}{\mathrm{d} \tau}(\gamma c, \gamma \vec{v})=\gamma \frac{\mathrm{d}}{\mathrm{d} t}(\gamma c, \gamma \vec{v})=\left(\gamma c \frac{\mathrm{d} \gamma}{\mathrm{d} t}, \gamma^{2} \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t}+\gamma \vec{v} \frac{\mathrm{d} \gamma}{\mathrm{d} t}\right)$$
The derivative of the velocity is of course the classical acceleration $\vec{a}=\mathrm{d} \vec{v} / \mathrm{d} t$, while the derivative of $\gamma$ is easy to calculate as
$$\gamma \frac{\mathrm{d} \gamma}{\mathrm{d} t}=\frac{1}{2} \frac{\mathrm{d} \gamma^{2}}{\mathrm{~d} t}=\frac{1}{2}\left(1-\frac{v^{2}}{c^{2}}\right)^{-2} \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{v^{2}}{c^{2}}\right)=\left(\frac{\gamma^{4}}{c^{2}}\right) \vec{v} \cdot \frac{\mathrm{d} \vec{v}}{\mathrm{~d} t}=\left(\frac{\gamma^{4}}{c^{2}}\right) \vec{v} \cdot \vec{a}$$
Thus
$$a^{\mu}=\left(\frac{\gamma^{4}}{c}(\vec{v} \cdot \vec{a}), \vec{v} \frac{\gamma^{4}}{c^{2}}(\vec{v} \cdot \vec{a})+\gamma^{2} \vec{a}\right)$$
In particular, in the proper frame where the velocity vanishes instantaneously, we have
$$a^{\mu}=(0, \vec{a}), \text { proper frame. }$$
This should not be surprising.

物理代写|广义相对论代写General relativity代考|Accelerated Motion

Now we are ready to study the trajectory of an accelerated particle in one space dimension. We will think of the particle as a small rocket, since a rocket is built with internal means of acceleration. In doing this we will see how convenient the concept of rapidity is for such calculations (Misner 1973).

Consider a rocket moving in the $x$ direction as in Fig. 3.3. The proper time $\tau$ provides a convenient parameter for defining the trajectory of the rocket, $c t(\tau), x(\tau)$. At proper time $\tau$ the rocket has velocity $v$ in the lab system $S$, while in its instantaneous rest frame $S^{\prime}$ its velocity is of course zero. A short time $\mathrm{d} \tau$ later its velocity in $S^{\prime}$ is given in terms of the acceleration by
$$\mathrm{d} v^{\prime}=a \mathrm{~d} \tau, \quad \mathrm{d} \beta^{\prime}=(a / c) \mathrm{d} \tau, \quad a=\text { proper acceleration. }$$
The proper acceleration is that measured in the proper frame, where the rocket is instantaneously at rest. In the lab frame $S$ the rocket velocity after the little time interval and velocity change is gotten from the velocity addition relation in Exercise $1.3$
$$\beta(\operatorname{after} \mathrm{d} \tau)=\left(\beta+\mathrm{d} \beta^{\prime}\right) /\left(1+\beta \mathrm{d} \beta^{\prime}\right) .$$
Thus the change in the lab velocity of the rocket to first order in $\mathrm{d} \beta^{\prime}$ is
$$\mathrm{d} \beta=\left(1-\beta^{2}\right) \mathrm{d} \beta^{\prime}=\mathrm{d} \beta^{\prime} / \gamma^{2}=(a / c) \mathrm{d} \tau / \gamma^{2} .$$
This is a differential relation giving $\beta$ as a function of $\tau$ and the proper acceleration since $\gamma$ is a function of $\beta$; if the acceleration were given as a function of $\tau$ we could integrate (3.23) to get $\beta(\tau)$.

However there is a more elegant way to analyze (3.23) in terms of rapidity. From the definition of rapidity $\theta$ in (1.28) we have
$$\begin{gathered} \beta=\tanh \theta, \quad \mathrm{d} \beta=\operatorname{sech}^{2} \theta \mathrm{d} \theta=\mathrm{d} \theta / \cosh ^{2} \theta=\mathrm{d} \theta / \gamma^{2} \ \mathrm{~d} \theta=\gamma^{2} \mathrm{~d} \beta \end{gathered}$$

物理代写|广义相对论代写General relativity代考|Curves and Arc Lengths

The lengths of lines and curves in the spacetime of special relativity have some peculiar and interesting properties. Let us study the time elapsed for travelers aboard rocket ships having diverse trajectories, curves in spacetime, using the geometric view that we have developed. The proper time interval for such a traveler is equal to the square root of the line element divided by $c$, as in (3.2),
$$c \mathrm{~d} \tau=\mathrm{d} s=\sqrt{c^{2} \mathrm{~d} t^{2}-\mathrm{d} \vec{x}^{2}}=c \mathrm{~d} t \sqrt{1-(\mathrm{d} \vec{x} / c \mathrm{~d} t)^{2}}=\sqrt{1-\beta^{2}} c \mathrm{~d} t$$
where the space and time intervals are measured in some inertial system such as our lab. Notice that this has meaning only so long as the velocity of the rocket ship is less than $c$, for otherwise the proper time becomes imaginary. That is the trajectory must always have a slope of over $45^{\circ}$. The time elapsed for a traveler is thus simply the integral of the line element along the trajectory, or the arc length of the curve between initial and final points in spacetime,
$$c \tau=s=\int_{i}^{f} \sqrt{1-\beta^{2}} c \mathrm{~d} t .$$
It is obvious from the integrand in (3.31) that this arc length is largest when the velocity of the rocket remains small along the trajectory. In particular the longest curve in spacetime for the roundtrips shown in Fig. $3.4$ is the straight line along the time axis; any other curve is shorter, and as the curve approaches the $45^{\circ}$ lines (light cone) its length approaches zero!

A straight line of this type is the longest distance between 2 points in spacetime, whereas it is the shortest distance between two points in Euclidean space. The minus sign in the line element (3.2) produces this profoundly different behavior. A physical consequence of this is that someone who leaves earth and travels at high velocity, say to a nearby star, and returns to earth will be younger than indicated by an earthbound clock. The infamous “twin paradox” is based on this peculiar behavior. See Exercise 3.6.

物理代写|广义相对论代写General relativity代考|Acceleration

CdCd吨=12dC2 d吨=12(1−在2C2)−2dd吨(在2C2)=(C4C2)在→⋅d在→ d吨=(C4C2)在→⋅一个→

物理代写|广义相对论代写General relativity代考|Accelerated Motion

d在′=一个 dτ,db′=(一个/C)dτ,一个= 适当的加速。

b(后⁡dτ)=(b+db′)/(1+bdb′).

db=(1−b2)db′=db′/C2=(一个/C)dτ/C2.

b=腥⁡θ,db=秒2⁡θdθ=dθ/科什2⁡θ=dθ/C2  dθ=C2 db

物理代写|广义相对论代写General relativity代考|Curves and Arc Lengths

C dτ=ds=C2 d吨2−dX→2=C d吨1−(dX→/C d吨)2=1−b2C d吨

Cτ=s=∫一世F1−b2C d吨.

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