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

Doug I. Jones

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

The general geodesics for photons can be viewed schematically using Eq. (8.70). Very interesting orbits occur when $J=J_c$, the value allowing circular orbits of $r=R_c$, and the light starts towards the black hole from $r>R_c$. Equation (8.70) can be rewritten with the aid of Eq. (8.76), to transparently show the importance of $R_c$,
\begin{aligned} \pm \frac{d \phi}{d r}= & -\frac{-J_c+\left(2 R^{\prime} / r\right)\left(J_c-a\right)}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \frac{1}{\left[1+\left(2 R^{\prime} / r^3\right)\left(J_c-a\right)^2-\left(J_c^2-a^2\right) / r^2\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{1 / 2}\left(J_c-a\right)\left[2 / r^3-3 / R_c / r^2+\left(\left(J_c-a\right)^2 R^{\prime}\right)^{-1}\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{\prime 1 / 2}\left(J_c-a\right)\left[2 / r^3-3 / R_c / r^2+1 / R_c^3\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{\prime 1 / 2}\left(J_c-a\right)\left(2 / r+1 / R_c\right)^{1 / 2}\left(1 / r-1 / R_c\right)} \ = & \frac{R^{\prime 1 / 2}}{\left(J_c+a\right)} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \frac{1}{\left(2 / r+1 / R_c\right)^{1 / 2}\left(1 / r-1 / R_c\right)} . \end{aligned}

In Figs. 8.5 and $8.6, a= \pm 0.4 R$ is used, in order to check these results with those of Chandrasekhar. In Fig. 8.6, the orbits are sketched. Note, the circles are not drawn to scale, so the photon path is more clearly shown. The spin of the black hole points out of the page. In order to start the direct, retrograde orbit in the same (counter clockwise), opposite (clockwise) sense as the spin, the derivatives, $\pm \frac{d \phi}{d r}$, shown to scale in Fig. 8.5, require opposite signs. The magnitude of the derivative becomes infinite as $r \rightarrow R_c$ from outside. It spends an indeterminate amount of time moving in this circle and the horizon circles, but some instability will force the orbit to smaller $r$ and eventually to the singularity.

In the case of direct motion, the derivative doesn’t change sign as $R_c$ is traversed. The light travel continues in the same sense. The derivative decreases in magnitude, and then increases to infinite magnitude, when $r \rightarrow R_{+}$. So the direct photon goes around the horizon, in the same sense as the spin, as you would expect from frame-drag.

物理代写|广义相对论代写General relativity代考|An Interstellar Example

The movie “Interstellar” illustrates some wonderful effects concerning GR and black holes. One such effect, showing a distinct difference between a spinning, and a static black hole, is discussed. Astronauts, from a dying earth, punch through a worm hole. They find themselves in a different part of the universe, that otherwise is unreachable, due to the large proper distance from earth. They explore the planets of a solar system, whose star is a huge rotating black hole, aptly named Gargantua. They hope to find a planet that can serve as a new home. One planet is in the minimum, stable, circular orbit, almost as close as possible to the event horizon. This is possible because Gargantua is rotating extremely close to its maximum value $a=\left(1-1.3 \times 10^{-14}\right) R / 2$ (Thorne, 2014). An enormous time dilation factor of $\approx 6 \times 10^4$, as required by the movie director, is experienced. The pilot astronaut spends a short time – hour(s) – on the planet, and after the entire trip, has hardly aged. However, when he returns to earth, he finds the young daughter he left behind, is now a very elderly woman.

This example shows how such phenomena can be possible with a spinning, but not a static black hole. The radial coordinate of the planet’s stable circular orbit is $r=R_c(\min )$. In Section 8.5 , the following was found:
$$R_c(\min )= \begin{cases}3 R, & a=0, \ R / 2, & a=R / 2 .\end{cases}$$
Thus, for a static black hole, Eqs. (8.65)-(8.67) and (8.88) yield
\begin{aligned} \frac{1}{E^{\prime 1 / 2}} & =\left(\frac{8}{9}\right)^{1 / 2} \ \frac{d t}{d \tau} & =\frac{3}{2}\left(\frac{8}{9}\right)^{1 / 2}=1.4 \end{aligned}

广义相对论代考

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

\begin{aligned} \pm \frac{d \phi}{d r}= & -\frac{-J_c+\left(2 R^{\prime} / r\right)\left(J_c-a\right)}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \frac{1}{\left[1+\left(2 R^{\prime} / r^3\right)\left(J_c-a\right)^2-\left(J_c^2-a^2\right) / r^2\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{1 / 2}\left(J_c-a\right)\left[2 / r^3-3 / R_c / r^2+\left(\left(J_c-a\right)^2 R^{\prime}\right)^{-1}\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{\prime 1 / 2}\left(J_c-a\right)\left[2 / r^3-3 / R_c / r^2+1 / R_c^3\right]^{1 / 2}} \ = & \frac{R_c}{3} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \ & \times \frac{1}{R^{\prime 1 / 2}\left(J_c-a\right)\left(2 / r+1 / R_c\right)^{1 / 2}\left(1 / r-1 / R_c\right)} \ = & \frac{R^{\prime 1 / 2}}{\left(J_c+a\right)} \frac{-3 J_c / R_c+2\left(J_c+a\right) / r}{\left(r-R_{+}\right)\left(r-R_{-}\right)} \frac{1}{\left(2 / r+1 / R_c\right)^{1 / 2}\left(1 / r-1 / R_c\right)} . \end{aligned}

物理代写|广义相对论代写General relativity代考|An Interstellar Example

$$R_c(\min )= \begin{cases}3 R, & a=0, \ R / 2, & a=R / 2 .\end{cases}$$

\begin{aligned} \frac{1}{E^{\prime 1 / 2}} & =\left(\frac{8}{9}\right)^{1 / 2} \ \frac{d t}{d \tau} & =\frac{3}{2}\left(\frac{8}{9}\right)^{1 / 2}=1.4 \end{aligned}

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