物理代写| Orbits in Schwarzschild Spacetime 相对论代考

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6.4 Orbits in Schwarzschild Spacetime
6.4.1 The First Integrals
A test particle-a particle of small mass such that its gravitational field can be safely ignored in the given physical scenario-under no external forces (except gravitation-recall that in GR, gravity is not looked upon as a force) moves along a geodesic of the spacetime. If the particle has non-zero rest-mass then it moves along a timelike geodesic-the tangent to the geodesic at any point $u^{i}=d x^{i} / d s$ is a timelike vector, and we have $u^{i} u_{i}>0-$ in fact if $s$ is the arclength or $c$ times the proper time then $u^{i} u_{i}=1$. If the test particle has zero rest-mass like a photon, then the tangent to the geodesic is null, that is, if we define the tangent as $u^{i}=d x^{i} / d \lambda$ We take the parameter $\lambda$ to be an affine parameter which has been explained in Chapter $4 .$

Normally, to compute the geodesics, we need to solve the geodesic equations:
$$\frac{\mathrm{d}^{2} x^{i}}{\mathrm{~d} \lambda^{2}}+\Gamma^{i}{ }_{j k} \frac{\mathrm{d} x^{j}}{\mathrm{~d} \lambda} \frac{\mathrm{d} x^{k}}{\mathrm{~d} \lambda}=0,$$
where we have generically used $\lambda$. We may set $\lambda=s$ for the timelike geodesics. We need to solve these equations with boundary conditions or initial conditions.

However, for the Schwarzschild spacetime a first principle approach is more useful, because it has several symmetries-it is static, the metric is time independentand also it is spherically symmetric. Therefore, it makes sense to start from the variational principle and integrate the geodesic equations step by step, by first obtaining the first integrals and identifying the integration constants with physical quantities. We proceed as follows.

First of all, from spherical symmetry, the orbit (geodesic) must be confined to a plane. Because if the particle starts of from a given point and in a given direction, the particle must remain in the plane determined by the, initial space-time point, the tangent vector and $r=0$-otherwise one hemisphere would be chosen over the other, violating symmetry. We will orient the Schwarzschild coordinates in such a and obtain its extremum; hence we write:
$$\delta \int\left(e^{v} c^{2} \dot{t}^{2}-e^{-v} \dot{r}^{2}-r^{2} \dot{\phi}^{2}\right) d s=0$$
where the ‘dot’ denotes differentiation with respect to $s$. Thus $\dot{t}=d t / d s$ and so on. We have also considered timelike geodesics first because they are easier to understand. Note that the $\dot{\theta}=0$, because $\theta$ is set to be a constant, namely, $\pi / 2$, a consequence of spherical symmetry. Note that this can be rigorously proved from the geodesic equation for $\theta$. If we set $\theta=\pi / 2$ and $\dot{\theta}=0$ in the geodesic equation for $\theta$,
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6 Schwarzschild Solution and Black Holes
we find that $\ddot{\theta}=0$, establishing that the particle does not leave the $\theta=\pi / 2$ plane. See Exercise $3 .$

We now use the Euler-Lagrange equations to obtain the orbits. Consider a function $F\left(\dot{x}^{i}, x^{k}\right)$ and we require:
$$\delta \int F\left(\dot{x}^{i}, x^{k}\right) d s=0$$
where the dot represents $d / d s$. This condition on the integral is converted into a differential condition by the Euler-Lagrange equations:
$$\frac{d}{d s}\left(\frac{\partial F}{\partial \dot{x}^{i}}\right)-\frac{\partial F}{\partial x^{i}}=0$$
Here we have $F=e^{v} c^{2} \dot{t}^{2}-e^{-v} \dot{r}^{2}-r^{2} \dot{\phi}^{2}$ and $x^{0}=c t$. If $F$ does not depend on explicitly on $x^{i}$ for some specific $i$, then,
$$\frac{d}{d s}\left(\frac{\partial F}{\partial \dot{x}^{i}}\right)=0 \Rightarrow \frac{\partial F}{\partial \dot{x}^{i}}=\text { const. } \equiv p_{i}$$

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6.4 史瓦西时空中的轨道
6.4.1 第一积分

$$\frac{\mathrm{d}^{2} x^{i}}{\mathrm{~d} \lambda^{2}}+\Gamma^{i}{ }_{jk} \frac{\mathrm {d} x^{j}}{\mathrm{~d} \lambda} \frac{\mathrm{d} x^{k}}{\mathrm{~d} \lambda}=0,$$

$$\delta \int\left(e^{v} c^{2} \dot{t}^{2}-e^{-v} \dot{r}^{2}-r^{2} \dot {\phi}^{2}\right) ds=0$$

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6 史瓦西解和黑洞

$$\delta \int F\left(\dot{x}^{i}, x^{k}\right) d s=0$$

$$\frac{d}{ds}\left(\frac{\partial F}{\partial \dot{x}^{i}}\right)-\frac{\partial F}{\partial x^{i} }=0$$

$$\frac{d}{ds}\left(\frac{\partial F}{\partial \dot{x}^{i}}\right)=0 \Rightarrow \frac{\partial F}{\partial \dot {x}^{i}}=\text { 常量。 } \equiv p_{i}$$

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