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# Ricci tensor for a spherically symmetric metric: the worksheet

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.2.

In the last post, we developed the general form for the metric in a spherically symmetric situation:

$\displaystyle ds^{2}=g_{tt}dt^{2}+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

The next step is to use the Einstein equation in the form

$\displaystyle R_{ij}=8\pi G\left(T_{ij}-\frac{1}{2}g_{ij}T\right) \ \ \ \ \ (2)$

to generate a system of differential equations that can be solved to find ${g_{tt}}$ and ${g_{rr}}$. This involves calculating the components of the Ricci tensor ${R^{ij}}$. Recall that the Ricci tensor is a contraction of the Riemann tensor:

$\displaystyle R_{ij}=R_{\;iaj}^{a} \ \ \ \ \ (3)$

and the Riemann tensor is defined in terms of Christoffel symbols:

$\displaystyle R_{\;j\ell m}^{i}\equiv-\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\;mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\;km}^{i}-\Gamma_{\;mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (4)$

The Christoffel symbols are, in turn, calculated from the metric tensor and its derivatives:

$\displaystyle \Gamma_{\;ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (5)$

Each component of ${R_{ij}}$ is therefore ultimately a differential equation involving components of the metric tensor ${g_{ij}}$, so if we know the stress-energy tensor ${T_{ij}}$, 2 gives us a set of PDEs that can, in principle at least, be solved to find the metric tensor. Since ${R_{ij}}$ is symmetric, it has 10 independent components, each of which is a sum of terms involving the components of ${g_{ij}}$. For a general (non-diagonal) metric, this can get very messy and, even for a diagonal metric such as we have for the spherically symmetric case, things are bad enough. In the appendix to Moore’s book, he gives a worksheet for calculating the Christoffel symbols and the independent components of ${R_{ij}}$ for a general diagonal metric. In the worksheet, all the work of expanding ${R_{ij}}$ in terms of ${g_{ij}}$ has been done, so we just need to fill in the results for our specific metric, such as that given in 1.

The worksheets are given for the generic diagonal metric, written as

$\displaystyle ds^{2}=-A\left(dx^{0}\right)^{2}+B\left(dx^{1}\right)^{2}+C\left(dx^{2}\right)^{2}+D\left(dx^{3}\right)^{2} \ \ \ \ \ (6)$

where ${x^{0}}$ is the time coordinate and the other three are space coordinates. Note the minus sign in the first term: this makes explicit the fact that the metric component for time should be negative. Thus we have ${g_{00}=-A}$, ${g_{11}=B}$, ${g_{22}=C}$ and ${g_{33}=D}$.

Derivatives with respect to coordinates are written as subscripts, so that ${A_{01}=\frac{\partial^{2}A}{\partial x^{0}\partial x^{1}}}$ and so on. It’s important not to confuse this notation with tensor notation; ${A_{01}}$ is not the 01 component of a tensor.

The 10 independent components of ${R_{ij}}$ are:

 ${R_{00}=0}$ ${+\frac{1}{2B}A_{11}}$ ${+\frac{1}{2C}A_{22}}$ ${+\frac{1}{2D}A_{33}}$ ${+0}$ ${-\frac{1}{2B}B_{00}}$ ${-\frac{1}{2C}C_{00}}$ ${-\frac{1}{2D}D_{00}}$ ${+0}$ ${+\frac{1}{4B^{2}}B_{0}^{2}}$ ${+\frac{1}{4C^{2}}C_{0}^{2}}$ ${+\frac{1}{4D^{2}}D_{0}^{2}}$ ${+0}$ ${+\frac{1}{4AB}A_{0}B_{0}}$ ${+\frac{1}{4AC}A_{0}C_{0}}$ ${+\frac{1}{4AD}A_{0}D_{0}}$ ${-\frac{1}{4BA}A_{1}^{2}}$ ${-\frac{1}{4B^{2}}A_{1}B_{1}}$ ${+\frac{1}{4BC}A_{1}C_{1}}$ ${+\frac{1}{4BD}A_{1}D_{1}}$ ${-\frac{1}{4CA}A_{2}^{2}}$ ${+\frac{1}{4CB}A_{2}B_{2}}$ ${-\frac{1}{4C^{2}}A_{2}C_{2}}$ ${+\frac{1}{4CD}A_{2}D_{2}}$ ${-\frac{1}{4DA}A_{3}^{2}}$ ${+\frac{1}{4DB}A_{3}B_{3}}$ ${+\frac{1}{4DC}A_{3}C_{3}}$ ${-\frac{1}{4D^{2}}A_{3}D_{3}}$
 ${R_{11}=\frac{1}{2A}B_{00}}$ ${+0}$ ${-\frac{1}{2C}B_{22}}$ ${-\frac{1}{2D}B_{33}}$ ${-\frac{1}{2A}A_{11}}$ ${+0}$ ${-\frac{1}{2C}C_{11}}$ ${-\frac{1}{2D}D_{11}}$ ${+\frac{1}{4A^{2}}A_{1}^{2}}$ ${+0}$ ${+\frac{1}{4C^{2}}C_{1}^{2}}$ ${+\frac{1}{4D^{2}}D_{1}^{2}}$ ${-\frac{1}{4A^{2}}B_{0}A_{0}}$ ${-\frac{1}{4AB}B_{0}^{2}}$ ${+\frac{1}{4AC}B_{0}C_{0}}$ ${+\frac{1}{4AD}B_{0}D_{0}}$ ${+\frac{1}{4BA}B_{1}A_{1}}$ ${+0}$ ${+\frac{1}{4BC}B_{1}C_{1}}$ ${+\frac{1}{4BD}B_{1}D_{1}}$ ${-\frac{1}{4CA}B_{2}A_{2}}$ ${+\frac{1}{4CB}B_{2}^{2}}$ ${+\frac{1}{4C^{2}}B_{2}C_{2}}$ ${-\frac{1}{4CD}B_{2}D_{2}}$ ${-\frac{1}{4DA}B_{3}A_{3}}$ ${+\frac{1}{4DB}B_{3}^{2}}$ ${-\frac{1}{4DC}B_{3}C_{3}}$ ${+\frac{1}{4D^{2}}B_{3}D_{3}}$
 ${R_{22}=\frac{1}{2A}C_{00}}$ ${-\frac{1}{2B}C_{11}}$ ${+0}$ ${-\frac{1}{2D}C_{33}}$ ${-\frac{1}{2A}A_{22}}$ ${-\frac{1}{2B}B_{22}}$ ${+0}$ ${-\frac{1}{2D}D_{22}}$ ${+\frac{1}{4A^{2}}A_{2}^{2}}$ ${+\frac{1}{4B^{2}}B_{2}^{2}}$ ${+0}$ ${+\frac{1}{4D^{2}}D_{2}^{2}}$ ${-\frac{1}{4A^{2}}C_{0}A_{0}}$ ${+\frac{1}{4AB}C_{0}B_{0}}$ ${-\frac{1}{4AC}C_{0}^{2}}$ ${+\frac{1}{4AD}C_{0}D_{0}}$ ${-\frac{1}{4BA}C_{1}A_{1}}$ ${+\frac{1}{4B^{2}}C_{1}B_{1}}$ ${+\frac{1}{4BC}C_{1}^{2}}$ ${-\frac{1}{4BD}C_{1}D_{1}}$ ${+\frac{1}{4CA}C_{2}A_{2}}$ ${+\frac{1}{4CB}C_{2}B_{2}}$ ${+0}$ ${+\frac{1}{4CD}C_{2}D_{2}}$ ${-\frac{1}{4DA}C_{3}A_{3}}$ ${-\frac{1}{4DB}C_{3}B_{3}}$ ${+\frac{1}{4DC}C_{3}^{2}}$ ${+\frac{1}{4D^{2}}C_{3}D_{3}}$
 ${R_{33}=\frac{1}{2A}D_{00}}$ ${-\frac{1}{2B}D_{11}}$ ${-\frac{1}{2C}D_{22}}$ ${+0}$ ${-\frac{1}{2A}A_{33}}$ ${-\frac{1}{2B}B_{33}}$ ${-\frac{1}{2C}C_{33}}$ ${+0}$ ${+\frac{1}{4A^{2}}A_{3}^{2}}$ ${+\frac{1}{4B^{2}}B_{3}^{2}}$ ${+\frac{1}{4C^{2}}C_{3}^{2}}$ ${+0}$ ${-\frac{1}{4A^{2}}D_{0}A_{0}}$ ${+\frac{1}{4AB}D_{0}B_{0}}$ ${+\frac{1}{4AC}D_{0}C_{0}}$ ${-\frac{1}{4AD}D_{0}^{2}}$ ${-\frac{1}{4BA}D_{1}A_{1}}$ ${+\frac{1}{4B^{2}}D_{1}B_{1}}$ ${-\frac{1}{4BC}D_{1}C_{1}}$ ${+\frac{1}{4BD}D_{1}^{2}}$ ${-\frac{1}{4CA}D_{2}A_{2}}$ ${-\frac{1}{4CB}D_{2}B_{2}}$ ${+\frac{1}{4C^{2}}D_{2}C_{2}}$ ${+\frac{1}{4CD}D_{2}^{2}}$ ${+\frac{1}{4DA}D_{3}A_{3}}$ ${+\frac{1}{4DB}D_{3}B_{3}}$ ${+\frac{1}{4DC}D_{3}C_{3}}$ ${+0}$
 ${R_{01}=-\frac{1}{2C}C_{01}}$ –${\frac{1}{2D}D_{01}}$ ${+\frac{1}{4C^{2}}C_{0}C_{1}}$ ${+\frac{1}{4D^{2}}D_{0}D_{1}}$ ${+\frac{1}{4AC}A_{1}C_{0}}$ ${+\frac{1}{4AD}A_{1}D_{0}}$ ${+\frac{1}{4BC}B_{0}C_{1}}$ ${+\frac{1}{4BD}B_{0}D_{1}}$
 ${R_{02}=-\frac{1}{2B}B_{02}}$ –${\frac{1}{2D}D_{02}}$ ${+\frac{1}{4B^{2}}B_{0}B_{2}}$ ${+\frac{1}{4D^{2}}D_{0}D_{2}}$ ${+\frac{1}{4AB}A_{2}B_{0}}$ ${+\frac{1}{4AD}A_{2}D_{0}}$ ${+\frac{1}{4BC}B_{2}C_{0}}$ ${+\frac{1}{4CD}C_{0}D_{2}}$
 ${R_{03}=-\frac{1}{2B}B_{03}}$ –${\frac{1}{2C}C_{03}}$ ${+\frac{1}{4B^{2}}B_{0}B_{3}}$ ${+\frac{1}{4C^{2}}C_{0}C_{3}}$ ${+\frac{1}{4AB}A_{3}B_{0}}$ ${+\frac{1}{4AC}A_{3}C_{0}}$ ${+\frac{1}{4BD}B_{3}D_{0}}$ ${+\frac{1}{4CD}C_{3}D_{0}}$
 ${R_{12}=-\frac{1}{2A}A_{12}}$ –${\frac{1}{2D}D_{12}}$ ${+\frac{1}{4A^{2}}A_{1}A_{2}}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}}$ ${+\frac{1}{4AB}A_{1}B_{2}}$ ${+\frac{1}{4BD}B_{2}D_{1}}$ ${+\frac{1}{4AC}A_{2}C_{1}}$ ${+\frac{1}{4CD}C_{1}D_{2}}$
 ${R_{13}=-\frac{1}{2A}A_{13}}$ –${\frac{1}{2C}C_{13}}$ ${+\frac{1}{4A^{2}}A_{1}A_{3}}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}}$ ${+\frac{1}{4AB}A_{1}B_{3}}$ ${+\frac{1}{4BC}B_{3}C_{1}}$ ${+\frac{1}{4DA}D_{1}A_{3}}$ ${+\frac{1}{4CD}C_{3}D_{1}}$
 ${R_{23}=-\frac{1}{2A}A_{23}}$ –${\frac{1}{2B}B_{23}}$ ${+\frac{1}{4A^{2}}A_{2}A_{3}}$ ${+\frac{1}{4B^{2}}B_{2}B_{3}}$ ${+\frac{1}{4AC}A_{2}C_{3}}$ ${+\frac{1}{4BC}B_{2}C_{3}}$ ${+\frac{1}{4DA}D_{2}A_{3}}$ ${+\frac{1}{4BD}B_{3}D_{2}}$

To apply these tables to the specific metric 1, we observe that ${x^{3}=\phi}$ does not appear in any component of ${g_{ij}}$ so all terms with a subscript 3 are zero. Also, ${x^{2}=\theta}$ appears only in ${g_{\phi\phi}=D}$, so any subscript 2 on ${A}$, ${B}$ or ${C}$ also gives zero. Finally, ${x^{0}=t}$ doesn’t appear in ${C}$ or ${D}$, so a subscript 0 there also gives zero. After using these simplifications, we have:

 ${R_{tt}=0}$ ${+\frac{1}{2B}A_{11}}$ ${+\frac{1}{2C}A_{22}=0}$ ${+\frac{1}{2D}A_{33}=0}$ ${+0}$ ${-\frac{1}{2B}B_{00}}$ ${-\frac{1}{2C}C_{00}=0}$ ${-\frac{1}{2D}D_{00}=0}$ ${+0}$ ${+\frac{1}{4B^{2}}B_{0}^{2}}$ ${+\frac{1}{4C^{2}}C_{0}^{2}=0}$ ${+\frac{1}{4D^{2}}D_{0}^{2}=0}$ ${+0}$ ${+\frac{1}{4AB}A_{0}B_{0}}$ ${+\frac{1}{4AC}A_{0}C_{0}=0}$ ${+\frac{1}{4AD}A_{0}D_{0}=0}$ ${-\frac{1}{4BA}A_{1}^{2}}$ ${-\frac{1}{4B^{2}}A_{1}B_{1}}$ ${+\frac{1}{4BC}A_{1}C_{1}=\frac{1}{4r^{2}B}A_{1}\left(2r\right)}$ ${+\frac{1}{4BD}A_{1}D_{1}=\frac{1}{4Br^{2}\sin^{2}\theta}A_{1}\left(2r\sin^{2}\theta\right)}$ ${-\frac{1}{4CA}A_{2}^{2}=0}$ ${+\frac{1}{4CB}A_{2}B_{2}=0}$ ${-\frac{1}{4C^{2}}A_{2}C_{2}=0}$ ${+\frac{1}{4CD}A_{2}D_{2}=0}$ ${-\frac{1}{4DA}A_{3}^{2}=0}$ ${+\frac{1}{4DB}A_{3}B_{3}=0}$ ${+\frac{1}{4DC}A_{3}C_{3}=0}$ ${-\frac{1}{4D^{2}}A_{3}D_{3}=0}$

Collecting terms, we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{11}-\frac{1}{2B}B_{00}+\frac{1}{4B^{2}}B_{0}^{2}+\frac{1}{4AB}A_{0}B_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle$ $\displaystyle -\frac{1}{4BA}A_{1}^{2}-\frac{1}{4B^{2}}A_{1}B_{1}+\frac{1}{2rB}A_{1}+\frac{1}{2rB}A_{1}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2B}\left[\partial_{rr}^{2}A-\partial_{tt}^{2}B+\frac{\left(\partial_{t}B\right)^{2}}{2B}+\frac{\left(\partial_{t}A\right)\left(\partial_{t}B\right)-\left(\partial_{r}A\right)^{2}}{2A}-\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)}{2B}+\frac{2\partial_{r}A}{r}\right] \ \ \ \ \ (8)$
 ${R_{rr}=\frac{1}{2A}B_{00}}$ ${+0}$ ${-\frac{1}{2C}B_{22}=0}$ ${-\frac{1}{2D}B_{33}=0}$ ${-\frac{1}{2A}A_{11}}$ ${+0}$ ${-\frac{1}{2C}C_{11}=-\frac{1}{r^{2}}}$ ${-\frac{1}{2D}D_{11}=-\frac{1}{r^{2}}}$ ${+\frac{1}{4A^{2}}A_{1}^{2}}$ ${+0}$ ${+\frac{1}{4C^{2}}C_{1}^{2}=\frac{1}{r^{2}}}$ ${+\frac{1}{4D^{2}}D_{1}^{2}=\frac{1}{r^{2}}}$ ${-\frac{1}{4A^{2}}B_{0}A_{0}}$ ${-\frac{1}{4AB}B_{0}^{2}}$ ${+\frac{1}{4AC}B_{0}C_{0}=0}$ ${+\frac{1}{4AD}B_{0}D_{0}=0}$ ${+\frac{1}{4BA}B_{1}A_{1}}$ ${+0}$ ${+\frac{1}{4BC}B_{1}C_{1}=\frac{B_{1}}{2rB}}$ ${+\frac{1}{4BD}B_{1}D_{1}=\frac{B_{1}}{2rB}}$ ${-\frac{1}{4CA}B_{2}A_{2}=0}$ ${+\frac{1}{4CB}B_{2}^{2}=0}$ ${+\frac{1}{4C^{2}}B_{2}C_{2}=0}$ ${-\frac{1}{4CD}B_{2}D_{2}=0}$ ${-\frac{1}{4DA}B_{3}A_{3}=0}$ ${+\frac{1}{4DB}B_{3}^{2}=0}$ ${-\frac{1}{4DC}B_{3}C_{3}=0}$ ${+\frac{1}{4D^{2}}B_{3}D_{3}=0}$
 $\displaystyle R_{rr}$ $\displaystyle =$ $\displaystyle \frac{1}{2A}B_{00}-\frac{1}{2A}A_{11}+\frac{1}{4A^{2}}A_{1}^{2}-\frac{1}{4A^{2}}B_{0}A_{0}-\frac{1}{4AB}B_{0}^{2}+\frac{1}{4BA}B_{1}A_{1}+\frac{B_{1}}{rB}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2A}\left[\partial_{tt}^{2}B-\partial_{rr}^{2}A+\frac{\left(\partial_{r}A\right)^{2}-\left(\partial_{t}A\right)\left(\partial_{t}B\right)}{2A}+\frac{\left(\partial_{r}A\right)\left(\partial_{r}B\right)-\left(\partial_{t}B\right)^{2}}{2B}+\frac{2A\partial_{r}B}{rB}\right] \ \ \ \ \ (10)$
 ${R_{\theta\theta}=\frac{1}{2A}C_{00}=0}$ ${-\frac{1}{2B}C_{11}=-\frac{1}{B}}$ ${+0}$ ${-\frac{1}{2D}C_{33}=0}$ ${-\frac{1}{2A}A_{22}=0}$ ${-\frac{1}{2B}B_{22}=0}$ ${+0}$ ${-\frac{1}{2D}D_{22}=-\frac{2\cos2\theta}{2\sin^{2}\theta}=\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin^{2}\theta}}$ ${+\frac{1}{4A^{2}}A_{2}^{2}=0}$ ${+\frac{1}{4B^{2}}B_{2}^{2}=0}$ ${+0}$ ${+\frac{1}{4D^{2}}D_{2}^{2}=\frac{\sin^{2}2\theta}{4\sin^{4}\theta}=\frac{\cos^{2}\theta}{\sin^{2}\theta}}$ ${-\frac{1}{4A^{2}}C_{0}A_{0}=0}$ ${+\frac{1}{4AB}C_{0}B_{0}=0}$ ${-\frac{1}{4AC}C_{0}^{2}=0}$ ${+\frac{1}{4AD}C_{0}D_{0}=0}$ ${-\frac{1}{4BA}C_{1}A_{1}=-\frac{A_{1}r}{2AB}}$ ${+\frac{1}{4B^{2}}C_{1}B_{1}=\frac{rB_{1}}{2B^{2}}}$ ${+\frac{1}{4BC}C_{1}^{2}=\frac{1}{B}}$ ${-\frac{1}{4BD}C_{1}D_{1}=-\frac{1}{B}}$ ${+\frac{1}{4CA}C_{2}A_{2}=0}$ ${+\frac{1}{4CB}C_{2}B_{2}=0}$ ${+0}$ ${+\frac{1}{4CD}C_{2}D_{2}=0}$ ${-\frac{1}{4DA}C_{3}A_{3}=0}$ ${-\frac{1}{4DB}C_{3}B_{3}=0}$ ${+\frac{1}{4DC}C_{3}^{2}=0}$ ${+\frac{1}{4D^{2}}C_{3}D_{3}=0}$
 $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle -\frac{1}{B}+\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin^{2}\theta}+\frac{\cos^{2}\theta}{\sin^{2}\theta}-\frac{A_{1}r}{2AB}+\frac{rB_{1}}{2B^{2}}+\frac{1}{B}-\frac{1}{B}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{A_{1}r}{2AB}+\frac{rB_{1}}{2B^{2}}+1-\frac{1}{B}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B} \ \ \ \ \ (13)$
 ${R_{\phi\phi}=\frac{1}{2A}D_{00}=0}$ ${-\frac{1}{2B}D_{11}=-\frac{\sin^{2}\theta}{B}}$ ${-\frac{1}{2C}D_{22}=1-2\cos^{2}\theta}$ ${+0}$ ${-\frac{1}{2A}A_{33}=0}$ ${-\frac{1}{2B}B_{33}=0}$ ${-\frac{1}{2C}C_{33}=0}$ ${+0}$ ${+\frac{1}{4A^{2}}A_{3}^{2}=0}$ ${+\frac{1}{4B^{2}}B_{3}^{2}=0}$ ${+\frac{1}{4C^{2}}C_{3}^{2}=0}$ ${+0}$ ${-\frac{1}{4A^{2}}D_{0}A_{0}=0}$ ${+\frac{1}{4AB}D_{0}B_{0}=0}$ ${+\frac{1}{4AC}D_{0}C_{0}=0}$ ${-\frac{1}{4AD}D_{0}^{2}=0}$ ${-\frac{1}{4BA}D_{1}A_{1}=-\frac{r\sin^{2}\theta}{2BA}A_{1}}$ ${+\frac{1}{4B^{2}}D_{1}B_{1}=\frac{r\sin^{2}\theta}{2B^{2}}B_{1}}$ ${-\frac{1}{4BC}D_{1}C_{1}=-\frac{\sin^{2}\theta}{B}}$ ${+\frac{1}{4BD}D_{1}^{2}=\frac{\sin^{2}\theta}{B}}$ ${-\frac{1}{4CA}D_{2}A_{2}=0}$ ${-\frac{1}{4CB}D_{2}B_{2}=0}$ ${+\frac{1}{4C^{2}}D_{2}C_{2}=0}$ ${+\frac{1}{4CD}D_{2}^{2}=\cos^{2}\theta}$ ${+\frac{1}{4DA}D_{3}A_{3}=0}$ ${+\frac{1}{4DB}D_{3}B_{3}=0}$ ${+\frac{1}{4DC}D_{3}C_{3}=0}$ ${+0}$
 $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle -\frac{\sin^{2}\theta}{B}+1-2\cos^{2}\theta-\frac{r\sin^{2}\theta}{2BA}A_{1}+\frac{r\sin^{2}\theta}{2B^{2}}B_{1}-\frac{\sin^{2}\theta}{B}+\frac{\sin^{2}\theta}{B}+\cos^{2}\theta\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin^{2}\theta\left[-\frac{r\partial_{r}A}{2AB}+\frac{r\partial_{r}B}{2B^{2}}+1-\frac{1}{B}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{\theta\theta}\sin^{2}\theta \ \ \ \ \ (16)$
 ${R_{tr}=-\frac{1}{2C}C_{01}=0}$ –${\frac{1}{2D}D_{01}=0}$ ${+\frac{1}{4C^{2}}C_{0}C_{1}=0}$ ${+\frac{1}{4D^{2}}D_{0}D_{1}=0}$ ${+\frac{1}{4AC}A_{1}C_{0}=0}$ ${+\frac{1}{4AD}A_{1}D_{0}=0}$ ${+\frac{1}{4BC}B_{0}C_{1}=\frac{B_{0}}{2rB}}$ ${+\frac{1}{4BD}B_{0}D_{1}=\frac{B_{0}}{2rB}}$

$\displaystyle R_{tr}=\frac{\partial_{t}B}{rB} \ \ \ \ \ (17)$

 ${R_{r\theta}=-\frac{1}{2A}A_{12}=0}$ –${\frac{1}{2D}D_{12}=-\frac{2\cos\theta}{r\sin\theta}}$ ${+\frac{1}{4A^{2}}A_{1}A_{2}=0}$ ${+\frac{1}{4D^{2}}D_{1}D_{2}=\frac{\cos\theta}{r\sin\theta}}$ ${+\frac{1}{4AB}A_{1}B_{2}=0}$ ${+\frac{1}{4BD}B_{2}D_{1}=0}$ ${+\frac{1}{4AC}A_{2}C_{1}=0}$ ${+\frac{1}{4CD}C_{1}D_{2}=\frac{\cos\theta}{r\sin\theta}}$

$\displaystyle R_{r\theta}=-\frac{2\cos\theta}{r\sin\theta}+\frac{\cos\theta}{r\sin\theta}+\frac{\cos\theta}{r\sin\theta}=0 \ \ \ \ \ (18)$

Looking at the worksheet tables above, we can see that applying the rules stated earlier makes all entries in ${R_{02}}$, ${R_{03}}$, ${R_{13}}$ and ${R_{23}}$ zero, so these components of the Ricci tensor are all zero.

# Spherically symmetric solution to the Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.1.

We’ll return now to general relativity, and build up to a derivation of the Schwarzschild metric. As a quick review, the problem is to find a solution to the Einstein equation in the form

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (1)$

where ${R^{ij}}$ is the Ricci tensor, itself a contraction of the Riemann tensor, ${T^{ij}}$ is the stress-energy tensor and ${T=g_{ij}T^{ij}}$ is the stress-energy scalar.

In a practical problem, ${T^{ij}}$ will be given, and the problem is to determine the metric ${g^{ij}}$ from the Ricci tensor. The Ricci tensor is specified in terms of Christoffel symbols, which are in turn defined in terms of the metric and its derivatives, so the Einstein equation becomes a system of coupled, non-linear partial differential equations in the components of the metric tensor.

A good starting point is to look at spacetime around a source with spherical symmetry. We can picture this spacetime as a set of nested spherical shells, on the surface of which the usual 2-d spherical metric applies:

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)$

This gives us several of the metric components already, in that

 $\displaystyle g_{\theta\theta}$ $\displaystyle =$ $\displaystyle r^{2}\ \ \ \ \ (3)$ $\displaystyle g_{\phi\phi}$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (4)$ $\displaystyle g_{\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\phi\theta}=0 \ \ \ \ \ (5)$

We can set up the coordinates on each shell such that any line with fixed values of ${\theta}$ and ${\phi}$ is perpendicular to all the surfaces. This means that the basis vectors ${\mathbf{e}_{\theta}}$ and ${\mathbf{e}_{\phi}}$ are perpendicular to the third spatial basis vector ${\mathbf{e}_{r}}$, so that ${g_{r\theta}=g_{r\phi}=0}$. With spatial symmetry, there should be no difference in the way the metric treats motions in different directions of ${\theta}$ or ${\phi}$, so we’d expect the terms ${g_{r\theta}drd\theta}$, ${g_{r\phi}drd\phi}$, ${g_{t\theta}dtd\theta}$ and ${g_{t\phi}dtd\phi}$ to all be zero, which gives us four more (well, eight, actually, since ${g_{ij}=g_{ji}}$) metric components.

We’re left with ${g_{tt}}$, ${g_{rr}}$ and ${g_{rt}=g_{tr}}$. If ${t}$ is a time coordinate, we must have ${g_{tt}<0}$ and likewise, if ${r}$ is a spatial coordinate, then ${g_{rr}>0}$. We can, in fact, eliminate ${g_{rt}}$ by making a coordinate transformation as follows:

$\displaystyle t'=t+f\left(r,t\right) \ \ \ \ \ (6)$

where ${f}$ is some function of the original ${r}$ and ${t}$ coordinates (unknown at present). We can, in principle, always determine ${f}$ so that ${g_{rt'}=0}$ and then use ${t'}$ as our new time coordinate. [Note that we can’t use the symmetry argument to claim that ${g_{rt}=0}$, since in a spherically symmetric situation, it does make a difference whether you are travelling in the plus or minus ${r}$ direction, so it isn’t necessarily so that ${g_{rt}=0}$ in all cases.]

Take the differential of this equation to get

 $\displaystyle dt'$ $\displaystyle =$ $\displaystyle dt+\partial_{r}f\; dr+\partial_{t}f\; dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\partial_{t}f\right)dt+\partial_{r}f\; dr\ \ \ \ \ (8)$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle \frac{dt'-\partial_{r}f\; dr}{1+\partial_{t}f}\equiv\alpha\left(dt'-\partial_{r}f\; dr\right) \ \ \ \ \ (9)$

With the deductions above, our original metric equation is

$\displaystyle ds^{2}=g_{tt}dt^{2}+2g_{rt}dr\; dt+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (10)$

Substituting 9 into the first two terms on the RHS, we get

 $\displaystyle g_{tt}dt^{2}+2g_{rt}dr\; dt$ $\displaystyle =$ $\displaystyle g_{tt}\alpha^{2}\left(dt'-\partial_{r}f\; dr\right)^{2}+2\alpha g_{rt}dr\left(dt'-\partial_{r}f\; dr\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dr^{2}\left(g_{tt}\alpha^{2}\left(\partial_{r}f\right)^{2}-2\alpha g_{rt}\partial_{r}f\right)+\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle dr\; dt'\left(2\alpha g_{rt}-2\alpha^{2}g_{tt}\partial_{r}f\right)+\left(dt'\right)^{2}g_{tt}\alpha^{2}\nonumber$

We can now set the coefficient of ${dr\; dt'}$ to zero to get

 $\displaystyle g_{rt}$ $\displaystyle =$ $\displaystyle \alpha g_{tt}\partial_{r}f\ \ \ \ \ (13)$ $\displaystyle \frac{g_{rt}}{g_{tt}}\left(1+\partial_{t}f\right)$ $\displaystyle =$ $\displaystyle \partial_{r}f \ \ \ \ \ (14)$

Assuming this partial differential equation for ${f\left(r,t\right)}$ can be solved (which we won’t be able to do a priori, since we don’t know ${g_{rt}}$ or ${g_{tt}}$, but in principle, the equation can be solved), it is always possible to find a time coordinate ${t'}$ such that ${g_{rt'}=0}$, so we might as well use that time coordinate from the start. Relabelling this time coordinate from ${t'}$ back to ${t}$, the spherically symmetric metric is then

$\displaystyle ds^{2}=g_{tt}dt^{2}+g_{rr}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (15)$

We therefore have only two metric components that need to be found by solving the Einstein equation 1, which we’ll get to in the next post.

# Extrasolar planet detection with light curves: OGLE-TR-56b

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.18.

The TwoStars program can also be used to generate light curves in a binary system where one component is a star and the other is an exoplanet.

One example of this is the planet OGLE-TR-56b, which is a “hot Jupiter”, in that it’s a gas giant but with a surface temperature of around 1973 K. It was discovered in 2002 by observing a dip in the light received from its parent star, OGLE-TR-56, by a magnitude drop of only around 0.01. [OGLE stands for Optical Gravitational Lensing Experiment, and is a project based in Poland.]

Although the actual parameters of the planet are now believed to be quite different from those quoted in Carroll & Ostlie, we’ll use their values in the TwoStars program to generate a light curve. They are ${P=29\mbox{ hours}}$, ${a=0.023\mbox{ AU}}$, ${m_{p}=0.9M_{J}}$, ${T_{p}=1000\mbox{ K}}$ and ${R_{p}=R_{J}}$, where a subscript ${p}$ indicates the planet and ${J}$ indicates Jupiter. For the parent star, we are given that ${R_{s}=\frac{0.023\mbox{ AU}}{4.5}}$, and we can work out its mass using the same technique as before, by solving the formula:

$\displaystyle \frac{m_{s}^{3}}{\left(m_{p}+m_{s}\right)^{2}}\sin^{3}i=\frac{P}{2\pi G}v_{p,r}^{3} \ \ \ \ \ (1)$

The radial velocity ${v_{p,r}}$ of the planet can be found by assuming the orbit is circular:

$\displaystyle v_{p,r}=\frac{2\pi a}{P}=207\mbox{ km s}^{-1} \ \ \ \ \ (2)$

Taking ${i=90^{\circ}}$ and plugging in the numbers, and then solving for ${m_{s}}$ (using Maple to get an exact value), we find

$\displaystyle m_{s}=1.11M_{S} \ \ \ \ \ (3)$

where ${M_{S}}$ is the mass of the Sun. The temperature of the parent star is 3000 K. I used ${\phi=90^{\circ}}$ to position the light dips in the middle of the graph.

We now have all the data we need to run TwoStars. Carroll & Ostlie suggest using ${i=90^{\circ}}$ but this gives a spurious spike in magnitude at the centre of minimum. I’m not sure exactly what causes this, but it’s probably something to do with the transformation to the plane of the sky coordinate system, which involves ${\cos i}$, thus giving a value near zero. In any case, if we use ${i=89^{\circ}}$ we get a more sensible light curve:

The drop in magnitude during a transit of the planet is around 0.012 magnitudes. We can also see a tiny drop between ${t/P}$ values of 0.7 and 0.8 when the planet goes behind the star.

The radial velocity curves look like this:

The flat red line is the curve for the star, so we see that it’s practically unaffected by the planet. Just to verify that it does actually change, here’s a magnified view of the star on its own:

The maximum radial velocity of the star is only around 0.16 km/sec.

# Extrasolar planets: some detection methods

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problems 7.10 – 7.14.

The search for extrasolar planets (planets orbiting stars other than the Sun) is a major area of research in astronomy at the moment, with close to 2000 planets (probably more than 2000 by the time you read this) being discovered so far. Many of the techniques used to discover them are the same as those used in the analysis of binary star systems. Variations in the radial velocity of the parent star can lead to the discovery of an invisible companion. As the star’s mass is likely to be much larger than that of the planet, this technique is most effective if the mass of the planet is as large as possible, so that it exerts a larger perturbation in the star’s velocity. Shorter periods also aid discovery, since several periods can be observed relatively quickly. For example, if we tried to detect Jupiter from a distant solar system, we’d have to wait 11.86 years for one complete period.

As with binary stars, if we don’t know the inclination ${i}$ of the orbit of a planet, all we can calculate is a lower limit on the planet’s (and the star’s) mass, since all we can measure is ${\left(m_{p}\sin i\right)^{3}/\left(m_{p}+m_{s}\right)^{2}}$, where ${m_{p,s}}$ are the masses of the planet and star.

A couple of the earliest discoveries of exoplanets are those orbiting the stars 51 Peg and HD 168443. For 51 Peg, the lower limit of the planet’s mass is ${0.45M_{J}}$ (${M_{J}}$ is Jupiter’s mass), with a period of 4.23077 days and a semimajor axis of 0.051 AU. Taking the lower limit as the actual mass, and the orbit as circular, we can get an estimate for the mass of the star 51 Peg from the formula

$\displaystyle \frac{m_{2}^{3}}{\left(m_{1}+m_{2}\right)^{2}}\sin^{3}i=\frac{P}{2\pi G}v_{1,r}^{3} \ \ \ \ \ (1)$

We take ${m_{1}=0.45M_{J}=8.541\times10^{26}\mbox{ kg}}$, ${\sin i=1}$, ${P=4.23077\mbox{ days}}$ and

$\displaystyle v_{1,r}=\frac{2\pi a}{P}=\frac{2\pi\left(0.051\mbox{ AU}\left(1.496\times10^{11}\mbox{ m AU}^{-1}\right)\right)}{4.23077\mbox{ days}\left(24\times3600\mbox{ s day}^{-1}\right)}=1.31\times10^{5}\mbox{ m s}^{-1} \ \ \ \ \ (2)$

The mass of the star is then found from

 $\displaystyle \frac{m_{2}^{3}}{\left(8.541\times10^{26}+m_{2}\right)^{2}}$ $\displaystyle =$ $\displaystyle \left(1.31\times10^{5}\right)^{3}\frac{\left(4.23077\right)\left(24\right)\left(3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.96\times10^{30}\mbox{ kg} \ \ \ \ \ (4)$

Taking ${m_{2}\gg m_{1}}$, this result is thus an approximate value for the mass of 51 Peg (solving the cubic equation exactly in Maple gives the same result, to 2 decimal places, anyway). This is ${0.986M_{S}}$ so 51 Peg is effectively the mass of the Sun. (The currently accepted value is ${1.11M_{S}}$ so this estimate isn’t too bad.)

For planet HD 168443c, the values are ${m_{1}=16.96M_{J}=3.22\times10^{28}\mbox{ kg}}$, ${P=1770\mbox{ days}=1.53\times10^{8}\mbox{ s}}$ and ${a=2.87\mbox{ AU}}$.

 $\displaystyle v_{1,r}$ $\displaystyle =$ $\displaystyle 1.763\times10^{4}\mbox{ m s}^{-1}\ \ \ \ \ (5)$ $\displaystyle \frac{m_{2}^{3}}{\left(3.22\times10^{28}+m_{2}\right)^{2}}$ $\displaystyle =$ $\displaystyle \left(1.763\times10^{4}\right)^{3}\frac{1.53\times10^{8}}{2\pi\left(6.67\times10^{-11}\right)}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.00\times10^{30}\mbox{ kg} \ \ \ \ \ (7)$

Solving exactly gives

$\displaystyle m_{2}=2.06\times10^{30}\mbox{ kg}\approx1.04M_{S} \ \ \ \ \ (8)$

In the case of the planet eclipsing the star, we can get an idea of how much the radiant flux received from the star is diminished by considering someone on a distant solar system observing a transit of Jupiter across the Sun. The fraction decrease in flux is just the ratio of the cross sectional areas of Jupiter and the Sun, so (we can neglect the radius of Jupiter’s orbit since for an observer sufficiently far away, the Sun and Jupiter appear at essentially the distance):

$\displaystyle \Delta F=\frac{R_{J}^{2}}{R_{S}^{2}}=0.01 \ \ \ \ \ (9)$

That is, the Sun’s brightness would decrease by only 1%. [Carroll & Ostlie give the temperature of the Sun in the problem, but I can’t see that that is relevant to the calculation, since the fractional decrease in flux depends only on how much of the Sun’s surface is covered. Doing the same calculation for HD 209458 in their Figure 7.12 gives a reduction in flux of 1.34% which matches the light curve shown in the figure.]

If the planet eclipses its parent star, we can estimate its radius by measuring the times at which the eclipse starts (the edge of the planet just touches the edge of the star) and becomes total (the entire planet is in front of the star). The planet orbiting the star HD 209458 eclipses the star, and the light curves for a couple of transits are shown in Carroll & Ostlie’s Figure 7.12. The data are somewhat scattered, but we can estimate that the eclipse starts about 0.05 days before the midpoint and becomes total around 0.03 days before the midpoint. From the given period of ${P=3.525\mbox{ days}}$ and semimajor axis of ${a=0.0467\mbox{ AU}}$, and assuming a circular orbit, its velocity is

$\displaystyle v=\frac{2\pi a}{P}=1.441\times10^{5}\mbox{ m s}^{-1} \ \ \ \ \ (10)$

As it takes about 0.01 days for the radius to move across the edge of the star, the radius of the planet is

$\displaystyle R=1.441\times10^{5}\left(0.01\right)\left(24\right)\left(3600\right)=1.25\times10^{8}\mbox{ m}\approx1.8R_{J} \ \ \ \ \ (11)$

This compares with the value of ${1.27R_{J}}$ given in the text, but it’s difficult to estimate the eclipse times from such a scattered plot, so I suppose this isn’t bad.

# YY Sgr: computer model of an eclipsing binary star

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.17.

So far, we’ve used the TwoStars program (included in Carroll & Ostlie’s Appendix K, with my own Maple translation for this problem available here) to model non-eclipsing binary star systems. It can also generate light curves for eclipsing binary systems, so as an example, we’ll revisit YY Sgr. [If you want to run the Maple code to generate your own light curve, note that the calculation of the eclipsing portions of the curve can take quite a while (a few minutes on my machine, running on an Intel i7 processor) to run, so be patient; if the program appears to hang, just give it a while longer.]

The data for YY Sgr are ${M_{1}=5.9M_{S}}$, ${R_{1}=3.2R_{S}}$, ${T_{1}=15200\mbox{ K}}$, ${M_{2}=5.6M_{S}}$, ${R_{2}=2.9R_{S}}$, ${T_{2}=13700\mbox{ K}}$, ${P=2.6284734\mbox{ days}}$, ${e=0.1573}$, ${i=88.89^{\circ}}$, ${\phi=214.6^{\circ}}$ and velocity of the centre of mass is zero. The light curve is

The minima don’t have flat bottoms, indicating that the eclipses are only partial, as we’d expect given that the radii of the two stars are similar and the inclination angle is far enough from ${90^{\circ}}$ that only a partial eclipse occurs.

The radial velocities of the two components give this plot:

As the eccentricity is quite small, the radial velocities are nearly sinusoidal.

# Apparent motion of a binary star: computer model

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.16.

The TwoStars program included in Carroll & Ostlie’s Appendix K also produces the positions of the two components of a binary star system as seen by an observer on Earth, that is, as projected onto the plane of the sky, taken to be the ${y'z'}$ plane. The output of the program gives these positions in metres, measured from the stars’ positions at ${t=0}$. To convert this to seconds of arc, we need to know the distance ${d}$ (in parsecs) to the system. We then get these positions in seconds of arc from

$\displaystyle \left[y,z\right]=\frac{180\times3600}{\pi}\frac{\left[y',z'\right]}{\left(3.08567758\times10^{16}\mbox{ m pc}^{-1}\right)d} \ \ \ \ \ (1)$

I’ve produced a Maple version of this program that generates these positions in seconds of arc, and produces plots of the motion of the system relative to the origin, which corresponds to the position of the centre of mass at ${t=0}$. The Maple file is available here (with the usual caution that the code is a direct translation of the C++ original, and is therefore rather ugly!).

We’ll use the same binary system as in the previous post, with the additional information that this system is 3.2 parsecs from Earth. The parameters are ${M_{1}=0.5M_{S}}$, ${R_{1}=1.8R_{S}}$, ${T_{1}=8190\mbox{ K}}$, ${M_{2}=2.0M_{S}}$, ${R_{2}=0.63R_{S}}$, ${T_{2}=3840\mbox{ K}}$, ${P=1.8\mbox{ yr}}$, ${\phi=0}$, and ${i=30^{\circ}}$. First, we’ll check the program in the case where the eccentricity is ${e=0.4}$, the inclination is ${i=0}$ (so we’re viewing the system face on) and the centre of mass velocity is zero. In that case, the apparent paths of the stars looks like this:

This looks OK, since the two paths are ellipses with their focus at the origin.

Now suppose we make the inclination ${i=30^{\circ}}$ and give the centre of mass a velocity of

$\displaystyle \mathbf{v}_{cm}=\left[30,42,-15.3\right]\mbox{ km s}^{-1} \ \ \ \ \ (2)$

Over one period, the motion looks like this (the axes of the plot are chosen so that the scale is the same in both directions, giving an actual view of the paths):

This is actually quite a large apparent motion (it corresponds to a proper motion of the centre of mass of around ${3.5^{\prime\prime}\mbox{ yr}^{-1}}$, which is comparable to that of ${\alpha\mbox{ Cen}}$, the closest star to our solar system. This is due to the proximity of this fictional system to Earth and to its large velocity through space.

# Radial velocities of a binary system: computer model

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.15.

In their Appendix K, Carroll & Ostlie describe a computer program called TwoStars, which computes radial velocities and magnitudes for an eclipsing binary system. We won’t repeat the description here; rather we’ll just summarize how it works.

The required input to the program is the mass, radius and temperature of each of the two stars, the orbital period, the eccentricity and inclination of the orbit, the angle of periastron (the angle between the minor axis and line of sight) and the velocity of the centre of mass. The program then uses these data along with Kepler’s laws to calculate the orbits of the two stars as a function of time, using numerical simulation. At each time interval, a check is made to see if one star eclipses the other and, if so, the change in received flux from the system is calculated by subtracting the flux from the obscured portion of the eclipsed star, thus generating a light curve.

They provide their source code in Fortran and C++, but in order to facilitate the generation of plots, I translated the code into Maple. The code is too long to reproduce here (and it’s also kind of ugly), but if you’re interested, you can download the Maple code for this problem from here.

As an example, we’ll look at the radial velocities of the components of a binary star with the parameters ${M_{1}=0.5M_{S}}$, ${R_{1}=1.8R_{S}}$, ${T_{1}=8190\mbox{ K}}$, ${M_{2}=2.0M_{S}}$, ${R_{2}=0.63R_{S}}$, ${T_{2}=3840\mbox{ K}}$, ${P=1.8\mbox{ yr}}$, and ${i=30^{\circ}}$. We’ll assume that the angle of periastron is ${\phi=0}$ and that the velocity of the centre of mass is zero.

With this inclination, there are no eclipses, so the light curve is constant. We can calculate the radial velocities for several eccentricities of the orbit. For ${e=0}$, the orbits are circular, and the velocities follow sine curves:

For higher eccentricities, the curves become more and more lopsided. Here are the graphs for ${e=0.2,0.4,0.5}$:

The eccentricity of a binary system can thus be estimated by comparing the velocity curves with those from a model such as this.

We can check the results for ${e=0}$ by inverting the process we used to estimate stellar masses to calculate the maximum radial velocities from the parameters given above. We have

 $\displaystyle \frac{v_{2,r}}{v_{1,r}}$ $\displaystyle =$ $\displaystyle \frac{M_{1}}{M_{2}}=0.25\ \ \ \ \ (1)$ $\displaystyle v_{1,r}+v_{2,r}$ $\displaystyle =$ $\displaystyle \left(\frac{2\pi G\left(M_{1}+M_{2}\right)}{P}\right)^{1/3}\sin i\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 16.6\mbox{ km s}^{-1}\ \ \ \ \ (3)$ $\displaystyle v_{1,r}$ $\displaystyle =$ $\displaystyle 13.3\mbox{ km s}^{-1}\ \ \ \ \ (4)$ $\displaystyle v_{2.r}$ $\displaystyle =$ $\displaystyle 3.32\mbox{ km s}^{-1} \ \ \ \ \ (5)$

These values agree well with the heights of the sine curves in the first plot above.

# YY Sagittarii: an eclipsing binary star

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.7.

The variable star YY Sagittarii is an eclipsing binary system. The V (visual) light curve is shown in Carroll & Ostlie’s Fig. 7.2, and shows a maximum magnitude of 10.03, a primary minimum of 10.78 and a secondary minimum of 10.67 (values estimated from the graph). However, there is no flat bottom on the minima, which would indicate that the eclipse isn’t total, that is, neither star totally obscures the other during the eclipsing periods. The inclination is given in the figure as ${i=88.89^{\circ}}$which is around the value we might expect for a partial eclipse (that is, it’s near the minimum inclination at which a partial eclipse just occurs).

We can still estimate the temperature ratio of the two stars from the formula

$\displaystyle \frac{T_{B}}{T_{A}}=\left[\frac{1-B_{p}/B_{0}}{1-B_{s}/B_{0}}\right]^{1/4} \ \ \ \ \ (1)$

where ${B_{0}}$, ${B_{p}}$ and ${B_{s}}$ are the observed maximum, primary and secondary brightnesses. Since the eclipses aren’t total, the observed values of ${B_{p}}$ and ${B_{s}}$ are probably larger than what they would be if the eclipses were total, so the temperature ratio won’t be that accurate.

We first convert the magnitudes into brightness ratios:

 $\displaystyle \frac{B_{p}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(10.03-10.78\right)/5}=0.50\ \ \ \ \ (2)$ $\displaystyle \frac{B_{s}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(10.03-10.67\right)/5}=0.55 \ \ \ \ \ (3)$

This gives a temperature ratio of

$\displaystyle \frac{T_{B}}{T_{A}}=\left(\frac{0.5}{0.45}\right)^{1/4}=1.03 \ \ \ \ \ (4)$

As there isn’t much difference between the two minima, the temperatures of the two stars are almost the same.

# Spectroscopic, eclipsing binary stars: mass, radius and temperature

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.6.

Here’s another example of finding the parameters of the two stars in a spectroscopic, eclipsing binary system. The measured values are

 $\displaystyle P$ $\displaystyle =$ $\displaystyle 6.31\mbox{ yr}\ \ \ \ \ (1)$ $\displaystyle v_{A,r}$ $\displaystyle =$ $\displaystyle 5.4\mbox{ km s}^{-1}\ \ \ \ \ (2)$ $\displaystyle v_{B,r}$ $\displaystyle =$ $\displaystyle 22.4\mbox{ km s}^{-1} \ \ \ \ \ (3)$

The ratio of masses is

$\displaystyle \frac{m_{A}}{m_{B}}=\frac{v_{B,r}}{v_{A,r}}=4.15 \ \ \ \ \ (4)$

The sum of masses is

$\displaystyle m_{1}+m_{2}=\frac{P}{2\pi G}\left(\frac{v_{A,r}+v_{B,r}}{\sin i}\right)^{3} \ \ \ \ \ (5)$

As the binary is eclipsing, we’ll assume that ${i\approx90^{\circ}}$. We get

 $\displaystyle m_{A}+m_{B}$ $\displaystyle =$ $\displaystyle \frac{\left(6.31\right)\left(365.25\times24\times3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\left(5.4\times10^{3}+22.4\times10^{3}\right)^{3}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.02\times10^{31}\mbox{ kg} \ \ \ \ \ (7)$

We can now get the individual masses:

 $\displaystyle m_{A}$ $\displaystyle =$ $\displaystyle 4.15m_{B}\ \ \ \ \ (8)$ $\displaystyle 5.15m_{B}$ $\displaystyle =$ $\displaystyle 1.02\times10^{31}\mbox{ kg}\ \ \ \ \ (9)$ $\displaystyle m_{B}$ $\displaystyle =$ $\displaystyle 1.98\times10^{30}\mbox{ kg}\approx M_{S}\ \ \ \ \ (10)$ $\displaystyle m_{A}$ $\displaystyle =$ $\displaystyle 8.23\times10^{30}\mbox{ kg}=4.14M_{S} \ \ \ \ \ (11)$

Thus the smaller star has a mass roughly that of the Sun.

The radii of the two stars in an eclipsing system can be found by measuring the times of the various stages of the eclipse. Suppose ${t_{a}}$ is the time when the eclipse starts, that is, it’s the time when one star just starts to go behind the other one. At this point, the light curve just begins to dip. When the first star is completely behind the other one (we’re assuming ${i=90^{\circ}}$ so all eclipses are total) at time ${t_{b}}$, the light curve reaches its minimum. If we know the velocity ${v}$ of one star relative to the other, this time interval tells us the radius of the eclipsed star:

$\displaystyle r_{B}=\frac{v}{2}\left(t_{b}-t_{a}\right) \ \ \ \ \ (12)$

The radius of the other star can be found from how long it takes the eclipsed star to re-emerge. The eclipsed star first touches the edge of the other star (as seen from Earth) at ${t_{a}}$ and begins to emerge at ${t_{c}}$ so the radius of the eclipsing star (the one in front) is

 $\displaystyle r_{A}$ $\displaystyle =$ $\displaystyle \frac{v}{2}\left(t_{c}-t_{a}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v}{2}\left(t_{c}-t_{b}\right)+r_{B} \ \ \ \ \ (14)$

For our example, the measured times are ${t_{b}-t_{a}=0.58\mbox{ day}}$ and ${t_{c}-t_{b}=0.64\mbox{ day}}$. The relative velocity is ${v=v_{A,r}+v_{B,r}}$ (since the velocities are non-relativistic), so

 $\displaystyle r_{B}$ $\displaystyle =$ $\displaystyle \frac{\left(22.4+5.4\right)\times10^{3}}{2}\left(0.58\times24\times3600\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.97\times10^{8}\mbox{ m}=R_{S}\ \ \ \ \ (16)$ $\displaystyle r_{A}$ $\displaystyle =$ $\displaystyle \frac{0.64+0.58}{0.58}r_{B}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.47\times10^{9}\mbox{ m}=2.1R_{S} \ \ \ \ \ (18)$

The ratio of the temperatures of the two stars can also be estimated by measuring how much the flux received from the binary star dips in the two minima, if we assume the stars behave as blackbodies. If the smaller star B is hotter, then when it is eclipsed by star A, the light curve drops farther than when star B goes in front of star A, since B is brighter than A, and the same area of stellar surface is eclipsed in both cases. The amount of light received from the binary system when both stars are fully visible is

$\displaystyle B_{0}=k\left(L_{A}+L_{B}\right) \ \ \ \ \ (19)$

where ${L_{j}}$ is the luminosity of star ${j}$ and ${k}$ is a constant that depends on the distance and other factors such as the amount of dust between the binary and Earth, and so on.

The brightness ${B_{p}}$ of the primary minimum, when star B is fully eclipsed, is just star A on its own:

$\displaystyle B_{p}=kL_{A} \ \ \ \ \ (20)$

The brightness ${B_{s}}$ of the secondary minimum is the brightness of star B plus that of star A minus the bit that is obscured by star B:

 $\displaystyle B_{s}$ $\displaystyle =$ $\displaystyle kL_{B}+k\frac{\pi r_{A}^{2}-\pi r_{B}^{2}}{\pi r_{A}^{2}}L_{A}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B_{0}-\frac{r_{B}^{2}}{r_{A}^{2}}B_{p} \ \ \ \ \ (22)$

Comparing the two light minima, we get

$\displaystyle \frac{B_{0}-B_{p}}{B_{0}-B_{s}}=\frac{L_{B}}{L_{A}}\frac{r_{A}^{2}}{r_{B}^{2}} \ \ \ \ \ (23)$

The luminosity is given in terms of the temperature by the Stefan-Boltzmann law:

$\displaystyle L=4\pi r^{2}\sigma T^{4} \ \ \ \ \ (24)$

where ${\sigma=5.670373\times10^{-8}\mbox{ W m}^{-2}\mbox{K}^{-4}}$ is the Stefan-Boltzmann constant and ${T}$ is the temperature. We thus have

$\displaystyle \frac{B_{0}-B_{p}}{B_{0}-B_{s}}=\left(\frac{T_{B}}{T_{A}}\right)^{4} \ \ \ \ \ (25)$

For our example, the brightnesses are measured in magnitudes, with the maximum brightness at magnitude 5.40, the primary minimum at 9.20 and the secondary minimum at 5.44. We can convert these into the flux of light:

 $\displaystyle \frac{B_{p}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(5.40-9.20\right)/5}=0.030\ \ \ \ \ (26)$ $\displaystyle \frac{B_{s}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(5.40-5.44\right)/5}=0.964 \ \ \ \ \ (27)$

The temperature ratio is

 $\displaystyle \frac{T_{B}}{T_{A}}$ $\displaystyle =$ $\displaystyle \left[\frac{1-B_{p}/B_{0}}{1-B_{s}/B_{0}}\right]^{1/4}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.28 \ \ \ \ \ (29)$

As we’d expect, the smaller, brighter star B is significantly hotter than star A.

# Spectroscopic binary stars: Zeta Phe

References: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.5.

Andersen, J. (1983) Spectroscopic observations of eclipsing binaries, Astron. Astrophys. 118, 255-261.

As an example of estimating the masses of the components of a spectroscopic binary star, we’ll look at ${\zeta\mbox{ Phe}}$ in the constellation Phoenix. Its observed period is ${P=1.67\mbox{ days}}$ and the maximum radial velocities observed from Doppler shifts are (we’re assuming the orbits are circular, so the velocities of the stars are constant throughout their orbits):

 $\displaystyle v_{1,r}$ $\displaystyle =$ $\displaystyle 121.4\mbox{ km s}^{-1}\ \ \ \ \ (1)$ $\displaystyle v_{2,r}$ $\displaystyle =$ $\displaystyle 247\mbox{ km s}^{-1} \ \ \ \ \ (2)$

The formula for the sum of masses is

$\displaystyle m_{1}+m_{2}=\frac{P}{2\pi G}\left(\frac{v_{1,r}+v_{2,r}}{\sin i}\right)^{3} \ \ \ \ \ (3)$

As we don’t know the inclination angle ${i}$, the best we can do is find:

 $\displaystyle \left(m_{1}+m_{2}\right)\sin^{3}i$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}\left(v_{1,r}+v_{2,r}\right)^{3}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(1.67\right)\left(24\times3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\left(1.214\times10^{5}+2.47\times10^{5}\right)^{3}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.72\times10^{31}\mbox{ kg} \ \ \ \ \ (6)$

To find the mass ratio, we need the ratio of semimajor axes (or just radii, since we’re assuming the orbits are circular), which we can get from velocities and period.

 $\displaystyle r_{i}$ $\displaystyle =$ $\displaystyle \frac{v_{i,r}}{2\pi}P\ \ \ \ \ (7)$ $\displaystyle \frac{r_{1}}{r_{2}}$ $\displaystyle =$ $\displaystyle \frac{v_{1,r}}{v_{2,r}}=0.491\ \ \ \ \ (8)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle \frac{r_{1}}{r_{2}}m_{1}=0.491m_{1}\ \ \ \ \ (9)$ $\displaystyle 1.491m_{1}\sin^{3}i$ $\displaystyle =$ $\displaystyle 1.72\times10^{31}\mbox{ kg}\ \ \ \ \ (10)$ $\displaystyle m_{1}\sin^{3}i$ $\displaystyle =$ $\displaystyle 1.153\times10^{31}\mbox{ kg}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.8M_{S}\ \ \ \ \ (12)$ $\displaystyle m_{2}\sin^{3}i$ $\displaystyle =$ $\displaystyle 5.66\times10^{30}\mbox{ kg}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.85M_{S} \ \ \ \ \ (14)$

Using the average value of ${\left\langle \sin^{3}i\right\rangle =\frac{3\pi}{16}}$ that takes into account the Doppler shift selection effect (the fact that the larger the inclination angle, the more likely it is that a spectroscopic binary will be observed), this gives us mass estimates for the components of ${\zeta\mbox{ Phe}}$:

 $\displaystyle m_{1}$ $\displaystyle =$ $\displaystyle 9.85M_{S}\ \ \ \ \ (15)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle 4.84M_{S} \ \ \ \ \ (16)$

These values are much higher than the actual values of ${m_{1}\sin^{3}i=3.92M_{S}}$ and ${m_{2}\sin^{3}i=2.55M_{S}}$ given in the paper by Andersen. The radial velocity values given by Carroll & Ostlie don’t seem to take into account the overall radial velocity of the binary star system relative to Earth, which is ${R_{v}=+15.4\mbox{ km s}^{-1}}$. Even after subtracting this out, however, the ratio of the radial velocities of the two components is significantly different from those given in Andersen’s paper. If we use the radial velocities in Andersen’s Table 5 (he calls them ${K_{A}}$ and ${K_{B}}$), which are ${v_{1,r}=131.5\mbox{ km s}^{-1}}$ and ${v_{2,r}=202.6\mbox{ km s}^{-1}}$, we get

 $\displaystyle \left(m_{1}+m_{2}\right)\sin^{3}i$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}\left(v_{1,r}+v_{2,r}\right)^{3}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(1.67\right)\left(24\times3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\left(1.315\times10^{5}+2.026\times10^{5}\right)^{3}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.28\times10^{31}\mbox{ kg}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.46M_{S}\ \ \ \ \ (20)$ $\displaystyle \frac{r_{1}}{r_{2}}$ $\displaystyle =$ $\displaystyle \frac{v_{1,r}}{v_{2,r}}=0.649\ \ \ \ \ (21)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle 0.649m_{1}\ \ \ \ \ (22)$ $\displaystyle m_{1}\sin^{3}i$ $\displaystyle =$ $\displaystyle 3.92M_{S}\ \ \ \ \ (23)$ $\displaystyle m_{2}\sin^{3}i$ $\displaystyle =$ $\displaystyle 2.54M_{S} \ \ \ \ \ (24)$

Since ${\zeta\mbox{ Phe}}$ is an eclipsing binary, ${i}$ must be very close to ${90^{\circ}}$ so these values are likely very close to the actual masses.