Lecture Notes of An Advanced Course in General Relativity by E. Poisson

Lecture Notes of An Advanced Course in GR by E. Poisson

本文档除了包括推导, 疑惑 外,做 读书重点 的记录

Ch.3 Hypersurfaces
P.48 (3.1.4)

容易证明这的确是曲面法向.因为$$\Phi$$对任意曲面坐标参数$$y^\alpha$$导数为零,而这个导数可以表达为(3.1.4)分子上面的偏导乘以一个决定于参数选择的曲面切向方向.而分母仅仅是归一.

P.48 (3.1.5)

对零(法向类光)曲面,上述定义不成立,所以用不归一的(3.1.5)来定义曲面的法向.接着按书中讨论我们得到,$$k^\alpha$$实际上是切向;推导得到(3.1.5)下面一式是一般的测地线方向.比较测地线方程(比如Schutz (6.47))唯一的差别是右边正比于$$k^\alpha$$而非为零.但是其实平行移动要求可以退化为矢量在切向平移下的变化在矢量方向上,这就是这里的测地线方向.换言之,我们发现零曲面其实是由测地线堆砌而成的,而测地线参数$$\lambda$$就可以取为第一个曲面坐标参数.

P.49 (3.1.9)

这就是诱导度规的定义.

P.51 (3.2.1)

这就是用诱导度规来计算曲面面积的公式.但是,实际上,一个常见的特殊情况,就把一个坐标取为常数定义的曲面(且改曲面不类光)对应的曲面面积是很直观的(根本无需讨论诱导度规的定义).而反过来,在曲面类光时,情况反而很不不直观.

这里首先给出一个表面元的洛伦兹标量形式(3.2.2),证明对非类光的表面元(3.2.2)与(3.2.1)一致.接着利用(3.2.2)给出类光表面元的关系(3.2.7).

P.51 (3.2.2)


 * $$\begin{align}

d\Sigma_{\mu}=\varepsilon_{\mu\alpha\beta\gamma}e^{\alpha}_1e^{\beta}_2e^{\gamma}_3d^3y \end{align}$$

In the definition, $$\varepsilon_{\mu\alpha\beta\gamma}$$  is define on the next line,  $$\varepsilon_{\mu\alpha\beta\gamma}=\sqrt{-g}[\mu\alpha\beta\gamma]$$,  $$y$$  (with index $$123$$) is the parameterization of the surface (generally in a curved space), defined at the beginning of section 3.1, (3.1.2),  $$e^{\alpha}_{a}=\frac{\partial x^\alpha}{\partial y^a}$$  is defined in (3.1.7) and a simple example is given for flat Minkowski space on P.50. Now we want to show that $$ d\Sigma_{\mu}$$  defined in this way is indeed a covariant 4-vector.

Before handing out a general proof, let us check out a simple case, a surface element with $$\tau=const.$$  in  $$\{\tau,x,y,\eta\}$$  space. The example is very similar to that on P.51 right below (3.2.3) except in our case now $$\sqrt{-g}=\tau$$. Indices $$\mu,\alpha,\beta,\gamma$$   are chosen from  $$\{\tau,x,y,\eta\}$$ and


 * $$\begin{align}

y^1=x,y^2=y,y^3=\eta \end{align}$$

Straightforward calculation gives


 * $$\begin{align}

d\Sigma_{\mu,B}|_{\tau=const.}= \sqrt{-g}  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right)  dxdyd\eta=  \left( \begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right)  \tau dxdyd\eta \end{align}$$

We want to show explicitly that the module of this surface element remains the same when one transforms the coordinates into flat Minkowski space $$\{t,x,y,z\}$$. Now one has $$\sqrt{-g}=1$$, and the surface can be defined as  $$\Phi(\tau-\sqrt{t^2-z^2})=0$$  (comparing (3.1.1)). Again, three parameters, namely, $$y^1=x,y^2=y,y^3=\eta $$, determine the surface. Therefore one obtains $$d^3y=dy^1dy^2dy^3=dxdyd\eta $$. $$e^{\alpha}_{a}$$ can be obtained from the following $$4\times 4$$ transformation matrix by ripping away the first column.


 * $$\begin{align}

\left( \begin{array}{cccc} \cosh\eta & 0 & 0 & \tau\sinh\eta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sinh\eta & 0 & 0 & \tau\cosh\eta \end{array}\right) \Rightarrow\left( \begin{array}{cccc} X & 0 & 0 & \tau\sinh\eta\\ X & 1 & 0 & 0\\ X & 0 & 1 & 0\\ X & 0 & 0 & \tau\cosh\eta \end{array}\right) \end{align}$$

Now indices $$\mu,\alpha,\beta,\gamma$$ are chosen from  $$\{t,x,y,z\}$$, the first element is given by


 * $$\begin{align}

d\Sigma_{0}\equiv d\Sigma_{t}= \sqrt{-g}  [0,\alpha,\beta,\gamma]e^{\alpha}_1e^{\beta}_2e^{\gamma}_3 d^3y  =[0,\alpha,\beta,\gamma]e^{\alpha}_1e^{\beta}_2e^{\gamma}_3 dxdyd\eta  =\tau\cosh\eta dxdyd\eta \end{align}$$

since $$[0,\alpha,\beta,\gamma]e^{\alpha}_1e^{\beta}_2e^{\gamma}_3 \ne 0$$  only for  $$\alpha=x,\beta=y,\gamma=z $$. Similar calculation gives $$d\Sigma_{1}= d\Sigma_{2}=0, d\Sigma_{3}=\tau\sinh\eta dxdyd\eta$$. Piecing together one gets


 * $$\begin{align}

d\Sigma_{\mu,C}|_{\tau=const.}= \left( \begin{array}{c} \tau\cosh\eta\\ 0\\ 0\\ \tau\sinh\eta \end{array}\right)  dxdyd\eta \end{align}$$ This is the same as the result obtained naively from the notes "Energy Conservation Problem in SPheRIO". If we do the calculation for the surface $$\Phi(\eta-\frac{1}{2}\ln\frac{t+z}{t-z})=0 $$, similarly one gets


 * $$\begin{align}

&d\Sigma_{\mu,B}|_{\tau=const.}= \sqrt{-g}  \left( \begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right)  dxdyd\tau=  \left( \begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right)  \tau dxdyd\tau \\ &d\Sigma_{\mu,C}|_{\eta=const.}= \left( \begin{array}{c} \sinh\eta\\ 0\\ 0\\ \cosh\eta \end{array}\right)  dxdyd\tau  \\ &|d\Sigma_{\mu,B}|_{\tau=const.}|=\sqrt{g^{\mu\nu}d\Sigma_{\mu B}d\Sigma_{\nu B}}=\sqrt{-1}dxdyd\tau =|d\Sigma_{\mu,C}|_{\tau=const.}| \end{align}$$

Note that the normal direction of the surface is up to a negative sign since we have never explicitly considered it. The two examples show that the module of the surface does not change when the coordinates transform.

We are in the place to deliver a general proof. We are about to show that the module of a surface element is a scale, therefore surface element is a 4-vector. First, consider any surface element in flat Minkowski space $$\{t,x,y,z\}  ( \sqrt{-g}=1 )$$ but parameterized in terms of  $$y^1,y^2,y^3$$. It is noting that the surface element
 * $$d\Sigma_{\mu}=\varepsilon_{\mu\alpha\beta\gamma}e^{\alpha}_1e^{\beta}_2e^{\gamma}_3=\sqrt{-g}[\mu\alpha\beta\gamma]e^{\alpha}_1e^{\beta}_2e^{\gamma}_3$$

can be written in term determinant of an auxiliary matrix $${E_{\alpha}}_{\beta}$$  (whether the indices are superscript or subscript does not enter the question here, since we are dealing with a matrix, though the transform properties of the indices are discussed below)


 * $$\begin{align}

d\Sigma_{\mu}= \sqrt{-g}  \begin{vmatrix} \begin{array}{cccc} \hat{e}_0 & e^0_1 & e^0_2 & e^0_3\\ \hat{e}_1 & e^1_1 & e^1_2 & e^1_3\\ \hat{e}_2 & e^2_1 & e^2_2 & e^2_3\\ \hat{e}_3 & e^3_1 & e^3_2 & e^3_3 \end{array}\end{vmatrix}_C =  \begin{vmatrix} \begin{array}{cccc} \hat{e}_0 & e^0_1 & e^0_2 & e^0_3\\ \hat{e}_1 & e^1_1 & e^1_2 & e^1_3\\ \hat{e}_2 & e^2_1 & e^2_2 & e^2_3\\ \hat{e}_3 & e^3_1 & e^3_2 & e^3_3 \end{array}\end{vmatrix}_C =\det ({E_{\alpha}}_{\beta C}) \end{align}$$


 * $$\begin{align}

{E_{\alpha} \beta C }\equiv \left( \begin{array}{cccc} \hat{e}_0 & e^0_1 & e^0_2 & e^0_3\\ \hat{e}_1 & e^1_1 & e^1_2 & e^1_3\\ \hat{e}_2 & e^2_1 & e^2_2 & e^2_3\\ \hat{e}_3 & e^3_1 & e^3_2 & e^3_3 \end{array}\right)_C \end{align}$$

where the first column elements are the unitary vectors and $$e^{\alpha}_{a}=\frac{\partial x^\alpha}{\partial y^a}$$. As a second step, we want to calculate the same surface element in a different coordinate system $$\{\tau,x,y,\eta\}$$. The word "same" is mathematically implemented when that one adopts the same set of parameters, namely, $$y^1,y^2,y^3$$  to describe the surface, except now new coordinates are composite functions of the parameters. The surface element, according to its definition can be written down as


 * $$\begin{align}

&d\Sigma_{\mu}=\varepsilon_{\mu\alpha\beta\gamma}e^{\alpha}_1e^{\beta}_2e^{\gamma}_3=\sqrt{-g}[\mu\alpha\beta\gamma]e^{\alpha}_1e^{\beta}_2e^{\gamma}_3 =\sqrt{-g}  \begin{vmatrix} \begin{array}{cccc} \hat{e}_0 & e^0_1 & e^0_2 & e^0_3\\ \hat{e}_1 & e^1_1 & e^1_2 & e^1_3\\ \hat{e}_2 & e^2_1 & e^2_2 & e^2_3\\ \hat{e}_3 & e^3_1 & e^3_2 & e^3_3 \end{array}\end{vmatrix}_E \\ &=\sqrt{-g}\det ({E_{\alpha {\beta} E}}) =\sqrt{-g}\det ({\Lambda^{\alpha}}_{\lambda}{E_{\lambda{\beta}C}})\\ &{E_{\alpha{\beta}E}} \equiv \left( \begin{array}{cccc} \hat{e}_0 & e^0_1 & e^0_2 & e^0_3\\ \hat{e}_1 & e^1_1 & e^1_2 & e^1_3\\ \hat{e}_2 & e^2_1 & e^2_2 & e^2_3\\ \hat{e}_3 & e^3_1 & e^3_2 & e^3_3 \end{array}\right)_E \end{align}$$

The last step was based on two facts. Firstly, for the three spatial columns, chain rule of derivative gives


 * $$\begin{align}

e^{\alpha}_{aE}=\frac{\partial x_E^{\alpha}}{\partial y^a}=\frac{\partial x_E^{\alpha}}{\partial x_C^{\beta}}\frac{\partial x_C^{\beta}}{\partial y^a}={\Lambda^{\alpha}}_{\beta}\frac{\partial x_C^{\beta}}{\partial y^a} = {\Lambda^{\alpha}}_{\beta}e^{\beta}_{a C}   \end{align}$$

and secondly for the zeroth column, the transformation properties of unitary vectors read


 * $$\begin{align}

\hat e_{\alpha E}=\frac{\partial {x_\beta}_C}{\partial x_{\alpha E}}\hat {e_{\beta C}}={({\Lambda^{-1})}_{\beta}}^{\alpha}\hat e_{\beta C}={{\Lambda}^{\alpha}}_{\beta}\hat {{e_\beta}_C} \end{align}$$

above one has made use of the property of transformation matrix
 * $$\begin{align}

{\Lambda^{\nu}}_{\sigma}{\Lambda_{\nu}}^{\mu}={g_{\sigma}}^{\mu}=\delta_{\sigma}^{\mu}  \Rightarrow   {\Lambda^{\nu}}_{\sigma}={(\Lambda^{-1})_{\sigma}}^{\nu} \end{align}$$ One sees explicitly that the first index $$\alpha$$  of the auxiliary determinant  $${E_{\alpha}}_{\beta}$$  transforms like a superscripted contravariant 4-vector, meanwhile its second index  $$\beta$$  only depends on the parameterization of the surface element, therefore it does not depend on coordinate transformation. To show that the module of the surface element remain unchange, one need to evalute the determinant. The trick is, if one replace the first column with its algebraic complement, and taking into account the metric, one gets the module. For Minkowski case, the module of the surface element is


 * $$\begin{align}

\end{align}$$
 * d\Sigma_{\mu}|^2=\sum_{\mu\nu}\eta^{\mu\nu}A_{{\mu}1C}A_{{\nu}1C} =\det({E_{\alpha\beta C}})^2\sum_{\mu\nu}\eta^{\mu\nu}({E^{-1}_C})_{1\mu}({E^{-1}_C})_{1\nu}

where $$A_{\mu 1}$$  is the algebraic complement of  $$(\mu,1)$$th element of matrix  $${E_{\alpha}}_{\beta}$$, therefore  $$A_{\mu 1}=\det({E_{\alpha\beta}})({E^{-1}})_{1\mu}$$. For general case


 * $$\begin{align}

&|d\Sigma_{\mu}|^2=(\sqrt{-g})^2\sum_{\mu\nu}g^{\mu\nu}A_{{\mu}1E}A_{{\nu}1E} \\ &=(\sqrt{-g})^2\det({E_{\rho\sigma E}})^2\sum_{\mu\nu}g^{\mu\nu}({E^{-1}_E})_{1\mu}({E^{-1}_E})_{1\nu} \\ &=(\sqrt{-g})^2 \det({\Lambda^{\rho}}_{\lambda}{E_{\lambda}}_{\sigma C})^2\sum_{\mu\nu\alpha\beta}g^{\mu\nu}(({E_1}_{\beta C})^{-1}{(\Lambda^{-1})^{\beta}}_{\mu})((E_{1\alpha C})^{-1}{(\Lambda^{-1})^{\alpha}}_{\nu}) \\ &=(\sqrt{-g})^2 \det({\Lambda^{\rho}}_{\lambda})^2\det(E_{\lambda\sigma C})^2\sum_{\mu\nu\alpha\beta}g^{\mu\nu}{(\Lambda^{-1})^{\beta}}_{\mu}{(\Lambda^{-1})^{\alpha}}_{\nu}({E}_{1\beta C})^{-1}({E}_{1\alpha C})^{-1}\\ &=\det(E_{\lambda\sigma C})^2\sum_{\alpha\beta}\eta^{\beta\alpha}({E}_{1\beta C})^{-1}({E}_{1\alpha C})^{-1} \end{align}$$

where one has taken into consideration that


 * $$\begin{align}

&\sqrt{-g}=\sqrt{-\det(g_{\mu\nu}) } \\ &{\Lambda^{\mu}}_{\lambda}g_{\mu\nu}{\Lambda^{\nu}}_{\sigma}=\eta_{\lambda\sigma} \\ &(\sqrt{-g})^2\det({\Lambda^{\mu}}_{\nu})^2 =-\det({\Lambda^{\mu}}_{\lambda}g_{\mu\nu}{\Lambda^{\nu}}_{\sigma})=-\det(\eta_{\lambda\sigma})=1 \\ &\eta^{\lambda\sigma}={\Lambda_{\mu}}^{\lambda}g^{\mu\nu}{\Lambda_{\nu}}^{\sigma} ={(\Lambda^{-1})^{\lambda}}_{\mu}g^{\mu\nu}{(\Lambda^{-1})^{\sigma}}_{\nu} \end{align}$$

since


 * $$\begin{align}

{\Lambda^{\nu}}_{\sigma}{\Lambda_{\nu}}^{\mu}={g_{\sigma}}^{\mu}=\delta_{\sigma}^{\mu}  \Rightarrow   {\Lambda_{\nu}}^{\mu}={(\Lambda^{-1})^{\mu}}_{\nu} \end{align}$$

This concludes our proof that the module of d\Sigma_\mu  defined by (3.2.2) is indeed a scalar, therefore  $$d\Sigma_\mu$$  is a four-vector.

P.55 (3.3.1)


 * $$\begin{align}

\int_{V}{A^{\alpha}}_{;\alpha}\sqrt{-g}d^4x=\oint_{\partial V}A^{\alpha}d\Sigma_{\alpha} \end{align}$$

这是高斯定义,其中的表面元d\Sigma_{\alpha} 的定义就是(3.2.2).

这里对高斯定理的证明与普通场论中非广义相对论情况下的证明很不同.首先利用矢量的协变微分的形式把等式写成普通散度的体积分的形式.考虑任意有限大小的封闭曲线$$\Sigma$$,包络了一个有限体积.接着我们选取的是一个特殊的坐标系,其"时间"或者准确的说"径向"坐标对应一个描写表面的函数$$\Phi$$(3.1.1),它像洋葱一样的一层一层,使得在中心处$$\Phi=0$$,在外表面处$$\Phi=1$$.选择了这个径向坐标以后,余下的其他坐标都是角度坐标,即任何物理量(这里的被积函数)是余下坐标中的至少一个坐标的周期函数(考虑极坐标的两个角度部分坐标的情况).接着把散度径向和角度部分的导数分开写,这时角度坐标部分的散度的体积分用高斯定理化为面积分后为零,因为被积函数是至少一个坐标的周期函数.剩余的径向坐标的积分仅有在表面上积分的贡献.注意到上述三个角度坐标正是(3.1.2)中标记曲面的三个$$y$$坐标.在这个情况下,从全空间$$(\Phi,y^1,y^2,y^3)$$到$$y^1,y^2,y^3$$的诱导度规对应的行列式平庸的有$$\sqrt{-g}=\sqrt{h}$$,而按(3.2.2)表面元只有零分量不为零.所以最后表达式又能写成协变的形式.

在下一节,表面积分被写成沿着时间方向的圆柱表面的积分.和普通场论中又守恒流得到守恒荷的证明过程类似,在给定固定时刻的在任意空间坐标为无限大的表面上的积分取为零,这是因为考虑感兴趣的实际体系仅仅局限在有限的空间中.最后的守恒荷的表达式为(3.3.2).
 * $$\begin{align}

Q \equiv \int_{\Sigma}{j^{\alpha}}d\Sigma_{\alpha} = \int_{\Sigma}{j^{\alpha}}n_{\alpha}\sqrt{h}d^3y \end{align}$$

一个具体计算的例子,参见Research Paper Notes on Gravitational Collapse使用$$(u,v)$$坐标和$$(u,r)$$计算电荷的过程.