I'm self-studying the properties of a Schwarzchild geometry, with line element $$ds^2 = \left(1-\frac{2m}{r}\right)dt^2-\left(1-\frac{2m}{r}\right)^{-1}dr^2-r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right),$$ In particular, I'm looking at the problem of the tidal acceleration on an extended object that is at rest at a very large distance from the black hole. The geodesic deviation equation is $$\frac{D^2\eta^\mu}{Ds^2}+R^\mu_{\alpha \beta \gamma} u^\alpha u^\gamma \eta^\beta = 0$$ where $\eta^\mu$ is the seperation vector between geodesics for two points on the extended object, and $u^\mu = dx^\mu/ds$. Using the standard Schwarzchild coordinates $(t,r,\theta,\phi)$, the object starts at rest, so $u^\mu = (1,0,0,0).$ The geodesic deviation equation thus reduces to $$\frac{d^2\eta^\mu}{ds^2} +R^\mu_{0\beta0}\eta^\beta=0.$$ Solving for the case when $\mu=3$, I see that the only non-zero component of the curvature tensor is $$R^3_{030}=\frac{m}{r^3}-\frac{2m^2}{r^4},$$ so it follows that $$\frac{d^2\eta^3}{ds^2}=\left(\frac{2m^2}{r^4}-\frac{m}{r^3}\right)\eta^3$$ is the tidal acceleration. For $r>>2m$, this further simplifies to $$\frac{d^2\eta^3}{ds^2}=-\frac{m}{r^3}\eta^3$$ which is equivalent to the Newtonian result. What I'm trying to do is obtain this result directly from the first integrals of the geodesic equations (the notes I'm studying say to try this as an exercise), to show that $$\frac{1}{\eta^3}\frac{d^2\eta^3}{ds^2}=-\frac{m}{r^3}$$ holds. How does one tackle this problem from this approach? The first integrals of the geodesic equations lead to the relations $$r^2 \frac{d\phi}{ds}=h$$ $$\frac{dt}{ds}=\left(1-\frac{2m}{r}\right)^{-1/2}$$ $$\left(1-\frac{2m}{r}\right)^{-1}\frac{d^2r}{ds^2} +\frac{m}{r^2}\left(\frac{dt}{ds}\right)^2-\left(1-\frac{2m}{r}\right)^{-2} \frac{m}{r^2}\left(\frac{dr}{ds}\right)^2-r\left(\frac{d\phi}{ds}\right)^2=0$$ where $h$ is the angular momentum per unit mass. Since the object is at rest, we have $h=0$, so the first equation implied $\frac{d\phi}{ds}=0$. Similarlay, since the object starts from rest, we have $\frac{dr}{ds}=0$. Plugging all of this into the last equation, we get that $$\frac{d^2 r}{ds^2}= -\frac{m}{r^2}.$$

How can I proceed from here to find the tidal acceleration in the $\phi$ direction?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.