# Plane symmetric spacetime

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

A static, plane-symmetric spacetime is one in which spacetime is independent of time (static) and is composed of a set of planes, where each plane is labelled by a coordinate ${x}$. Within each plane, points are labelled by coordinates ${y}$ and ${z}$ and because the spacetime is static, the distance between two points depends only on these two coordinates:

$\displaystyle \left[ds^{2}\right]_{x}=dy^{2}+dz^{2} \ \ \ \ \ (1)$

where the subscript ${x}$ denotes the plane with coordinate ${x}$.

If the ${x}$ basis vector ${\mathbf{e}_{x}}$ is everywhere perpendicular to ${\mathbf{e}_{y}}$ and ${\mathbf{e}_{z}}$ (and ${\mathbf{e}_{y}\perp\mathbf{e}_{z}}$), then the spatial off-diagonal components of the metric are zero

 $\displaystyle g_{ij}$ $\displaystyle \equiv$ $\displaystyle \mathbf{e}_{i}\cdot\mathbf{e}_{j}\ \ \ \ \ (2)$ $\displaystyle g_{xy}$ $\displaystyle =$ $\displaystyle g_{xz}=g_{yz}=0 \ \ \ \ \ (3)$

The general metric between any two spacetime points is then

$\displaystyle ds^{2}=g_{tt}dt^{2}+2g_{tx}dt\;dx+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (4)$

Because the spacetime is static, a displacement forward in time by ${dt}$ should give the same separation as a displacement backwards by the same amount ${-dt}$. Because of this symmetry, the ${2g_{tx}dt\;dx}$ term should remain unchanged when ${dt}$ is replaced by ${-dt}$. However, since the metric is independent of time, ${g_{tx}\left(t\right)=g_{tx}\left(-t\right)}$, so the only way we can satisfy the symmetry requirement is if ${g_{tx}=0}$. Thus the plane-symmetric metric is symmetric:

$\displaystyle ds^{2}=g_{tt}dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (5)$

Further, ${g_{tt}}$ can depend at most on ${x}$ alone.

To work out the consequences of this metric, we need to evaluate the Christoffel symbols and Ricci tensor. The Christoffel symbol worksheet is:

 ${\Gamma_{00}^{0}=\frac{1}{2A}A_{0}}$ ${\Gamma_{10}^{0}=\Gamma_{01}^{0}=\frac{1}{2A}A_{1}}$ ${\Gamma_{20}^{0}=\Gamma_{02}^{0}=\frac{1}{2A}A_{2}}$ ${\Gamma_{30}^{0}=\Gamma_{03}^{0}=\frac{1}{2A}A_{3}}$ ${\Gamma_{11}^{0}=\frac{1}{2A}B_{0}}$ ${\Gamma_{22}^{0}=\frac{1}{2A}C_{0}}$ ${\Gamma_{33}^{0}=\frac{1}{2A}D_{0}}$ other ${\Gamma_{\mu\nu}^{0}=0}$ ${\Gamma_{01}^{1}=\Gamma_{10}^{1}=\frac{1}{2B}B_{0}}$ ${\Gamma_{11}^{1}=\frac{1}{2B}B_{1}}$ ${\Gamma_{12}^{1}=\Gamma_{21}^{1}=\frac{1}{2B}B_{2}}$ ${\Gamma_{13}^{1}=\Gamma_{31}^{1}=\frac{1}{2B}B_{3}}$ ${\Gamma_{00}^{1}=\frac{1}{2B}A_{1}}$ ${\Gamma_{22}^{1}=-\frac{1}{2B}C_{1}}$ ${\Gamma_{33}^{1}=-\frac{1}{2B}D_{1}}$ other ${\Gamma_{\mu\nu}^{1}=0}$ ${\Gamma_{02}^{2}=\Gamma_{20}^{2}=\frac{1}{2C}C_{0}}$ ${\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{2C}C_{1}}$ ${\Gamma_{22}^{2}=\frac{1}{2C}C_{2}}$ ${\Gamma_{32}^{2}=\Gamma_{23}^{2}=\frac{1}{2C}C_{3}}$ ${\Gamma_{00}^{2}=\frac{1}{2C}A_{2}}$ ${\Gamma_{11}^{2}=-\frac{1}{2C}B_{2}}$ ${\Gamma_{33}^{2}=-\frac{1}{2C}D_{2}}$ other ${\Gamma_{\mu\nu}^{2}=0}$ ${\Gamma_{03}^{3}=\Gamma_{30}^{3}=\frac{1}{2D}D_{0}}$ ${\Gamma_{13}^{3}=\Gamma_{31}^{3}=\frac{1}{2D}D_{1}}$ ${\Gamma_{23}^{3}=\Gamma_{32}^{3}=\frac{1}{2D}D_{2}}$ ${\Gamma_{33}^{3}=\frac{1}{2D}D_{3}}$ ${\Gamma_{00}^{3}=\frac{1}{2D}A_{3}}$ ${\Gamma_{11}^{3}=-\frac{1}{2D}B_{3}}$ ${\Gamma_{22}^{3}=-\frac{1}{2D}C_{3}}$ other ${\Gamma_{\mu\nu}^{3}=0}$

In this case ${\left(x^{0},x^{1},x^{2},x^{3}\right)=\left(t,x,y,z\right)}$ and

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -g_{tt}\left(x\right)\ \ \ \ \ (6)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (7)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (8)$ $\displaystyle D$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (9)$

Thus the only nonzero symbols will be those involving ${A_{1}}$, since all other derivatives are zero. These are

 $\displaystyle \Gamma_{10}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{01}^{0}=\frac{1}{2A}A_{1}=\frac{1}{2A}\frac{dA}{dx}\ \ \ \ \ (10)$ $\displaystyle \Gamma_{00}^{1}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{1}=\frac{1}{2}\frac{dA}{dx} \ \ \ \ \ (11)$

[We can use the total derivative rather than partial because ${A}$ depends only on ${x}$.]

From the Ricci tensor worksheet, the only nonzero components of ${R_{\mu\nu}}$ are those involving ${A_{11}}$ or ${A_{1}}$ only, so we see that

 $\displaystyle R_{00}$ $\displaystyle =$ $\displaystyle \frac{1}{2B}A_{11}-\frac{1}{4BA}A_{1}^{2}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{d^{2}A}{dx^{2}}-\frac{1}{4A}\left(\frac{dA}{dx}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle R_{11}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\frac{d^{2}A}{dx^{2}}+\frac{1}{4A}\left(\frac{dA}{dx}\right)^{2} \ \ \ \ \ (14)$

with all other ${R_{\mu\nu}=0}$. In flat space, all components satisfy ${R_{\mu\nu}=0}$ so these two components both give the same condition on ${A}$:

$\displaystyle \frac{d^{2}A}{dx^{2}}=\frac{1}{2A}\left(\frac{dA}{dx}\right)^{2} \ \ \ \ \ (15)$

To examine the structure of the spacetime, we need the full Riemann tensor, which is defined in terms of the Christoffel symbols:

$\displaystyle R_{\epsilon\nu\lambda\sigma}=g_{\epsilon\mu}R_{\;\nu\lambda\sigma}^{\mu}=g_{\epsilon\mu}\left[-\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}-\Gamma_{\;\lambda\nu}^{\kappa}\Gamma_{\;\kappa\sigma}^{\mu}+\Gamma_{\;\sigma\nu}^{\kappa}\Gamma_{\;\lambda\kappa}^{\mu}\right] \ \ \ \ \ (16)$

We can work out the terms in ${R_{\;\nu\lambda\sigma}^{\mu}}$ using 10 and 11. First, we’ll expand the implied sums and label the terms:

 $\displaystyle -\Gamma_{\;\lambda\nu}^{\kappa}\Gamma_{\;\kappa\sigma}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{-\Gamma_{\;\lambda\nu}^{0}\Gamma_{\;0\sigma}^{\mu}}^{\left[1\right]}\overbrace{-\Gamma_{\;\lambda\nu}^{1}\Gamma_{\;1\sigma}^{\mu}}^{\left[2\right]}\ \ \ \ \ (17)$ $\displaystyle \Gamma_{\;\sigma\nu}^{\kappa}\Gamma_{\;\lambda\kappa}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{\Gamma_{\;\sigma\nu}^{0}\Gamma_{\;\lambda0}^{\mu}}^{\left[3\right]}\overbrace{+\Gamma_{\;\sigma\nu}^{1}\Gamma_{\;\lambda1}^{\mu}}^{\left[4\right]}\ \ \ \ \ (18)$ $\displaystyle -\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}$ $\displaystyle =$ $\displaystyle \overbrace{-\partial_{\sigma}\Gamma_{\;\lambda\nu}^{\mu}}^{\left[5\right]}\overbrace{+\partial_{\lambda}\Gamma_{\;\sigma\nu}^{\mu}}^{\left[6\right]} \ \ \ \ \ (19)$

Next, we’ll identify the index combinations that give (potentially) nonzero values for components of ${R_{\;\nu\lambda\sigma}^{\mu}}$ in each term, using the fact that only ${\Gamma_{10}^{0}}$ and ${\Gamma_{00}^{1}}$ are nonzero, and that only the derivative with respect to ${x}$ (index 1) is nonzero.

• Term 1:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1
• Term 2:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 0 0
• Term 3:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 1 0 0 1 0 1 1 0 1 1 0 0
• Term 4:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 0 0
• Term 5:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 0 1 0 1 1 0 0 1
• Term 6:
 ${\mu}$ ${\nu}$ ${\lambda}$ ${\sigma}$ 0 0 1 1 0 1 1 0 1 0 1 0

From these tables, we see that there are 7 unique index combinations that can potentially give nonzero Riemann tensor components ${R_{\;\nu\lambda\sigma}^{\mu}}$. We have (remember that the Christoffel symbols are symmetric in their lower 2 indices: ${\Gamma_{\nu\lambda}^{\mu}=\Gamma_{\lambda\nu}^{\mu}}$):

 $\displaystyle R_{\;100}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\Gamma_{01}^{0}\Gamma_{00}^{1}=0\ \ \ \ \ (20)$ $\displaystyle R_{\;011}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}+\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}+\partial_{1}\Gamma_{01}^{0}=0\ \ \ \ \ (21)$ $\displaystyle R_{\;000}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{00}^{1}\Gamma_{10}^{0}+\Gamma_{00}^{1}\Gamma_{10}^{0}=0\ \ \ \ \ (22)$ $\displaystyle R_{\;010}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\partial_{1}\Gamma_{00}^{1}\ \ \ \ \ (23)$ $\displaystyle R_{\;001}^{1}$ $\displaystyle =$ $\displaystyle +\Gamma_{01}^{0}\Gamma_{00}^{1}-\partial_{1}\Gamma_{00}^{1}=-R_{\;010}^{1}\ \ \ \ \ (24)$ $\displaystyle R_{\;101}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}\ \ \ \ \ (25)$ $\displaystyle R_{\;110}^{0}$ $\displaystyle =$ $\displaystyle \Gamma_{01}^{0}\Gamma_{01}^{0}+\partial_{1}\Gamma_{01}^{0}=-R_{\;101}^{0} \ \ \ \ \ (26)$

Thus only the last 4 can potentially be nonzero. To go further, we need the derivative terms:

 $\displaystyle \partial_{x}\Gamma_{\;10}^{0}$ $\displaystyle =$ $\displaystyle -\frac{1}{2A^{2}}\left(\frac{dA}{dx}\right)^{2}+\frac{1}{2A}\frac{d^{2}A}{dx^{2}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2A^{2}}A_{1}^{2}+\frac{1}{2A}A_{11}\ \ \ \ \ (28)$ $\displaystyle \partial_{x}\Gamma_{\;00}^{1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{d^{2}A}{dx^{2}}=\frac{1}{2}A_{11} \ \ \ \ \ (29)$

Now we can use 10 and 11 to write these components in terms of ${A}$:

 $\displaystyle R_{\;010}^{1}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{00}^{1}+\partial_{1}\Gamma_{00}^{1}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (30)$ $\displaystyle R_{\;001}^{1}$ $\displaystyle =$ $\displaystyle -R_{\;010}^{1}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11}\ \ \ \ \ (31)$ $\displaystyle R_{\;101}^{0}$ $\displaystyle =$ $\displaystyle -\Gamma_{01}^{0}\Gamma_{01}^{0}-\partial_{1}\Gamma_{01}^{0}=\frac{1}{4A^{2}}A_{1}^{2}-\frac{1}{2A}A_{11}\ \ \ \ \ (32)$ $\displaystyle R_{\;110}^{0}$ $\displaystyle =$ $\displaystyle -R_{\;101}^{0}=-\frac{1}{4A^{2}}A_{1}^{2}+\frac{1}{2A}A_{11} \ \ \ \ \ (33)$

To get the Riemann tensor with all 4 indices lowered, we multiply by the metric:

$\displaystyle R_{\epsilon\nu\lambda\sigma}=g_{\epsilon\mu}R_{\;\nu\lambda\sigma}^{\mu} \ \ \ \ \ (34)$

Here, the only two metric components we need are ${g_{00}=-A}$ and ${g_{11}=1}$ so

 $\displaystyle R_{1010}$ $\displaystyle =$ $\displaystyle g_{11}R_{\;010}^{1}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (35)$ $\displaystyle R_{1001}$ $\displaystyle =$ $\displaystyle g_{11}R_{\;001}^{1}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11}\ \ \ \ \ (36)$ $\displaystyle R_{0101}$ $\displaystyle =$ $\displaystyle g_{00}R_{\;101}^{0}=-\frac{1}{4A}A_{1}^{2}+\frac{1}{2}A_{11}\ \ \ \ \ (37)$ $\displaystyle R_{0110}$ $\displaystyle =$ $\displaystyle g_{00}R_{\;110}^{0}=\frac{1}{4A}A_{1}^{2}-\frac{1}{2}A_{11} \ \ \ \ \ (38)$

Note that in this lowered form, the symmetries of the Riemann tensor are obeyed: ${R_{\mu\nu\lambda\sigma}=-R_{\nu\mu\lambda\sigma}=-R_{\mu\nu\sigma\lambda}}$.

Finally, if we impose the condition 15 in the form ${A_{11}=\frac{1}{2A}A_{1}^{2}}$, we find that all four of these components are zero, thus making the entire Riemann tensor zero, indicating that spacetime is completely flat. [There are a lot of indices flying about here, so I’m hoping I got them all right…]

# Riemann and Ricci tensors in the weak field limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Boxes 22.2-22.4.

The Einstein equation is

$\displaystyle G^{ij}+\Lambda g^{ij}=8\pi GT^{ij} \ \ \ \ \ (1)$

where ${\Lambda}$ is the cosmological constant and ${T^{ij}}$ is the stress-energy tensor. The Einstein tensor ${G^{ij}}$ is defined in terms of the Ricci tensor and curvature scalar as

$\displaystyle G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R \ \ \ \ \ (2)$

The Einstein equation can be written in the alternative form

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (3)$

where

$\displaystyle T\equiv g_{ij}T^{ij} \ \ \ \ \ (4)$

In preparation for examining the behaviour of this equation in the weak field limit, we need to see how the Ricci tensor behaves for metrics that are nearly flat. Since the Ricci tensor is a contraction of the Riemann tensor, we need first to see how the Riemann tensor behaves in this limit.

The metric for a weak field can be written as

$\displaystyle g_{ij}=\eta_{ij}+h_{ij} \ \ \ \ \ (5)$

where ${\eta_{ij}}$ is the flat space metric and ${h_{ij}}$ is a small perturbation, where small means ${\left|h_{ij}\right|\ll1}$. Note that since ${g_{ij}}$ and ${\eta_{ij}}$ are both symmetric, we must have

$\displaystyle h_{ij}=h_{ji} \ \ \ \ \ (6)$

As an approximation, we’ll discard all terms of order ${\left|h_{ij}\right|^{2}}$ or higher. First, we’ll get an approximation for the inverse metric ${g^{ij}}$. Since ${\eta_{ij}}$ is diagonal with elements ${\left[-1,1,1,1\right]}$ it is its own inverse so

$\displaystyle \eta^{ij}=\eta_{ij} \ \ \ \ \ (7)$

Therefore it’s reasonable to assume that

$\displaystyle g^{ij}=\eta^{ij}+b^{ij} \ \ \ \ \ (8)$

where ${\left|b^{ij}\right|\ll1}$. We can find ${b^{ij}}$ from the condition that

$\displaystyle g^{ij}g_{jk}=\delta_{k}^{i} \ \ \ \ \ (9)$

Therefore

 $\displaystyle g^{ij}g_{jk}$ $\displaystyle =$ $\displaystyle \eta^{ij}\eta_{jk}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk}+b^{ij}h_{jk}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \delta_{k}^{i}+\eta^{ij}h_{jk}+b^{ij}\eta_{jk} \ \ \ \ \ (11)$

to first order (since we’re assuming both ${b^{ij}}$ and ${h_{jk}}$ are very small). Therefore

 $\displaystyle b^{ij}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{ij}h_{jk}\ \ \ \ \ (12)$ $\displaystyle b^{ij}\eta^{kl}\eta_{jk}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (13)$ $\displaystyle b^{ij}\delta_{j}^{l}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (14)$ $\displaystyle b^{il}$ $\displaystyle =$ $\displaystyle -\eta^{kl}\eta^{ij}h_{jk}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -h^{il} \ \ \ \ \ (16)$

where

$\displaystyle h^{il}\equiv\eta^{kl}\eta^{ij}h_{jk} \ \ \ \ \ (17)$

Thus the perturbation ${h^{ij}}$ in the inverse metric is of the same order as the perturbation ${h_{jk}}$ in the original metric which means that, to first order, we can raise an index of a tensor whose components are all of order ${h_{jk}}$ by simply multiplying by the flat space metric ${\eta^{ij}}$ rather than the full metric ${g^{ij}}$. That is, for some tensor ${A_{kl}}$ where ${\left|A_{kl}\right|\ll1}$ we have, to first order

$\displaystyle A_{\; l}^{j}=g^{jk}A_{kl}=\left(\eta^{jk}-h^{jk}\right)A_{kl}=\eta^{jk}A_{kl}-\mathcal{O}\left(\left|h\right|^{2}\right)\approx\eta^{jk}A_{kl} \ \ \ \ \ (18)$

From 3 with ${\Lambda=0}$ and indices lowered, the Einstein equation is

$\displaystyle R_{ij}=8\pi G\left(T_{ij}-\frac{1}{2}g_{ij}T\right) \ \ \ \ \ (19)$

so we need to see what ${R_{ij}}$ looks like in the weak field limit. We’ll start with the Christoffel symbols

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}g_{il}+\partial_{i}g_{lj}-\partial_{l}g_{ji}\right) \ \ \ \ \ (20)$

In the weak field limit

$\displaystyle \partial_{j}g_{il}=\partial_{j}\left(\eta_{il}+h_{il}\right)=\partial_{j}h_{il} \ \ \ \ \ (21)$

since ${\partial_{j}\eta_{il}=0}$. Therefore,

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}g^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (22)$

This result is exact so far. Now if we want only first order terms, we can replace ${g^{ml}=\eta^{ml}-h^{ml}}$ by just ${\eta^{ml}}$ since the ${h^{ml}}$ multiplied into the derivatives gives only second order terms. Therefore, to first order

$\displaystyle \Gamma_{\; ij}^{m}=\frac{1}{2}\eta^{ml}\left(\partial_{j}h_{il}+\partial_{i}h_{lj}-\partial_{l}h_{ji}\right) \ \ \ \ \ (23)$

Now the Riemann tensor is defined as

$\displaystyle R_{\; jlm}^{i}\equiv-\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}+\Gamma_{\; lj}^{k}\Gamma_{\; km}^{i}-\Gamma_{\; mj}^{k}\Gamma_{\; lk}^{i}\right] \ \ \ \ \ (24)$

Since the Christoffel symbols are all of order ${h_{ij}}$ the last two terms in the Riemann tensor are of order ${\left|h_{ij}\right|^{2}}$ so we can drop them. With the first index lowered, we get

 $\displaystyle R_{njlm}$ $\displaystyle =$ $\displaystyle -g_{ni}\left[\partial_{m}\Gamma_{\; lj}^{i}-\partial_{l}\Gamma_{\; mj}^{i}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\eta_{ni}\left[\eta^{ik}\partial_{m}\left(\partial_{j}h_{lk}+\partial_{l}h_{kj}-\partial_{k}h_{jl}\right)-\eta^{ik}\partial_{l}\left(\partial_{j}h_{mk}+\partial_{m}h_{kj}-\partial_{k}h_{jm}\right)\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\partial_{m}\left(\partial_{j}h_{ln}+\partial_{l}h_{nj}-\partial_{n}h_{jl}\right)-\partial_{l}\left(\partial_{j}h_{mn}+\partial_{m}h_{nj}-\partial_{n}h_{jm}\right)\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right] \ \ \ \ \ (28)$

where we used 6 to get the last line.

Now to get the Ricci tensor in the weak field limit. We have, to first order

 $\displaystyle R_{jm}$ $\displaystyle =$ $\displaystyle g^{nl}R_{njlm}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\eta^{nl}\left[\partial_{j}\partial_{l}h_{nm}+\partial_{n}\partial_{m}h_{jl}-\partial_{n}\partial_{l}h_{jm}-\partial_{j}\partial_{m}h_{nl}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\left(\partial_{j}\partial_{l}h_{nm}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)+\eta^{nl}\left(\partial_{n}\partial_{m}h_{jl}-\frac{1}{2}\partial_{j}\partial_{m}h_{nl}\right)\right]-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{jl}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{ln}\partial_{j}\left(\partial_{l}h_{nm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\eta^{nl}\partial_{j}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right)+\eta^{nl}\partial_{m}\left(\partial_{n}h_{lj}-\frac{1}{2}\partial_{j}h_{nl}\right)-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (35)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (36)$

The index magic we performed above includes using ${\eta^{nl}=\eta^{ln}}$ and ${h_{nl}=h_{ln}}$ in 33 and then swapping the bound indices ${l}$ and ${n}$ and then using ${h_{ln}=h_{nl}}$ again in the first term in 34. The Ricci tensor in the weak field limit is thus

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (37)$

# Einstein equation for an exponential metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.8.

Consider the metric:

$\displaystyle ds^{2}=-e^{2gx}dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (1)$

We’ll have a look at what the Einstein equation has to say about gravity in a spacetime using this metric. First, we’ll find the Christoffel symbols using the method of comparing the two forms of the geodesic equation. The geodesic equation is

$\displaystyle \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (2)$

The following equation is formally equivalent to this (where a dot above a symbol means the derivative with respect to ${\tau}$):

$\displaystyle \ddot{x}^{m}+\Gamma_{\;ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (3)$

Since the metric is diagonal and none of its components depends on ${y}$ or ${z}$, the LHS of 2 is identically zero for ${a=y}$ or ${a=z}$, so all Christoffel symbols with any index being ${y}$ or ${z}$ is zero.

Now look at ${a=t}$. We get from 2, since ${g_{ij}}$ doesn’t depend on ${t}$:

 $\displaystyle \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle -2ge^{2gx}\dot{x}\dot{t}-e^{2gx}\ddot{t}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \ddot{t}+2g\dot{x}\dot{t}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Comparing with 3 we see that

$\displaystyle \Gamma_{\;tx}^{t}=\Gamma_{\;xt}^{t}=g \ \ \ \ \ (7)$

Now for ${a=x}$:

 $\displaystyle \ddot{x}-\frac{1}{2}\left(\partial_{x}g_{tt}\right)\dot{t}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \ddot{x}+ge^{2gx}\dot{t}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\;tt}^{x}$ $\displaystyle =$ $\displaystyle ge^{2gx} \ \ \ \ \ (10)$

These are the only non-zero Christoffel symbols. We can get an expression for the acceleration of a particle in its rest frame by noting that

$\displaystyle \ddot{x}=-ge^{2gx}\dot{t}^{2} \ \ \ \ \ (11)$

The four velocity of a particle at rest (so ${dx=dy=dz=0}$) must satisfy

 $\displaystyle u^{i}u_{i}$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}u^{i}u^{j}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{tt}\dot{t}^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{2gx}\dot{t}^{2}\ \ \ \ \ (15)$ $\displaystyle \dot{t}^{2}$ $\displaystyle =$ $\displaystyle e^{-2gx} \ \ \ \ \ (16)$

Plugging into 11 we get

$\displaystyle \ddot{x}=-g \ \ \ \ \ (17)$

That is, the acceleration in the ${x}$ direction is a constant, so this would seem to indicate a uniform gravitational field. However, let’s apply the Einstein equation and see what we get. We’ll need the Riemann tensor in order to get the Ricci tensor, so we’ll use the definition of the former:

$\displaystyle R_{\;j\ell m}^{i}\equiv-\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\;mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\;km}^{i}-\Gamma_{\;mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (18)$

Since all Christoffel symbols with a ${y}$ or ${z}$ index are zero, the only possibly non-zero components of ${R_{ij\ell m}}$ are those containing only ${x}$ or ${t}$, and due to the symmetry relations, the only independent component is

 $\displaystyle R_{txtx}$ $\displaystyle =$ $\displaystyle g_{ti}R_{\;xtx}^{i}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{2gx}\left[\partial_{x}\Gamma_{\;tx}^{t}-\partial_{t}\Gamma_{\;xx}^{t}+\Gamma_{\;tx}^{k}\Gamma_{\;kx}^{t}-\Gamma_{\;xx}^{k}\Gamma_{\;tk}^{t}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{2gx}\left[0-0+\Gamma_{\;tx}^{t}\Gamma_{\;tx}^{t}-0\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{2}e^{2gx} \ \ \ \ \ (22)$

Now for the Ricci tensor. We have

 $\displaystyle R_{ab}$ $\displaystyle =$ $\displaystyle R_{\;aib}^{i}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ij}R_{jaib} \ \ \ \ \ (24)$

Since the metric is diagonal, the upstairs components are just the reciprocals of the downstairs ones, so

 $\displaystyle g^{tt}$ $\displaystyle =$ $\displaystyle -e^{-2gx}\ \ \ \ \ (25)$ $\displaystyle g^{xx}$ $\displaystyle =$ $\displaystyle g^{yy}=g^{zz}=1 \ \ \ \ \ (26)$

and we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle g^{xx}R_{xtxt}=g^{2}e^{2gx}\ \ \ \ \ (27)$ $\displaystyle R_{xx}$ $\displaystyle =$ $\displaystyle g^{tt}R_{txtx}=g^{tt}R_{xtxt}=-g^{2}\ \ \ \ \ (28)$ $\displaystyle R_{xt}=R_{tx}$ $\displaystyle =$ $\displaystyle g^{ab}R_{bxat}=0 \ \ \ \ \ (29)$

where in the last line we see that ${R_{bxat}}$ can be non-zero only if ${b=t}$ and ${a=x}$ but since ${g^{ab}}$ is diagonal, ${g^{xt}=0}$. All components of ${R_{ab}}$ involving ${y}$ or ${z}$ indices are zero because all components of ${R_{abcd}}$ involving ${y}$ or ${z}$ indices are zero. To use the Ricci tensor in the Einstein equation, we need the upstairs version, which is

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{ta}g^{tb}R_{ab}=e^{-4gx}\left(g^{2}e^{2gx}\right)=g^{2}e^{-2gx}\ \ \ \ \ (30)$ $\displaystyle R^{xx}$ $\displaystyle =$ $\displaystyle g^{xa}g^{xb}R_{ab}=R_{xx}=-g^{2} \ \ \ \ \ (31)$

In a vacuum (assuming ${\Lambda=0}$) the stress-energy tensor ${T^{ij}=0}$ and the Einstein equation says that

$\displaystyle R^{ab}=0 \ \ \ \ \ (32)$

The only way this can be true is if ${g=0}$, meaning that there is no gravitational field.

More generally, suppose that there is some fluid in the region so that ${T^{ij}\ne0}$. In that case

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{2}e^{-2gx}=8\pi G\left(T^{tt}+\frac{1}{2}e^{-2gx}T\right)\ \ \ \ \ (33)$ $\displaystyle R^{xx}$ $\displaystyle =$ $\displaystyle -g^{2}=8\pi G\left(T^{xx}-\frac{1}{2}T\right)\ \ \ \ \ (34)$ $\displaystyle R^{yy}$ $\displaystyle =$ $\displaystyle 0=8\pi G\left(T^{yy}-\frac{1}{2}T\right)\ \ \ \ \ (35)$ $\displaystyle R^{zz}$ $\displaystyle =$ $\displaystyle 0=8\pi G\left(T^{zz}-\frac{1}{2}T\right) \ \ \ \ \ (36)$

where the stress-energy scalar is

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{2gx}T^{tt}+T^{xx}+T^{yy}+T^{zz} \ \ \ \ \ (38)$

From 35 and 36 we have

$\displaystyle T^{yy}=T^{zz}=\frac{T}{2} \ \ \ \ \ (39)$

so from 38 we have

$\displaystyle -e^{2gx}T^{tt}+T^{xx}=0 \ \ \ \ \ (40)$

However, if we multiply 33 by ${e^{2gx}}$ and add it to 34 we get

$\displaystyle 0=8\pi G\left(T^{tt}e^{2gx}+T^{xx}\right) \ \ \ \ \ (41)$

Combining the last two equations we get

$\displaystyle T^{tt}=T^{xx}=0 \ \ \ \ \ (42)$

Plugging 39 into 34 we get

$\displaystyle T^{yy}=T^{zz}=\frac{g^{2}}{8\pi G} \ \ \ \ \ (43)$

This result agrees with the correction to Moore’s problem 21.8 given in the errata.

In a perfect fluid at rest, ${T^{tt}}$ is the energy density and the diagonal components ${T^{ii}}$ for ${i=x,y,z}$ are the pressures in each of the 3 directions. If these results are to be believed, the fluid has zero energy (${T^{tt}=0}$) and no pressure in the ${x}$ direction, but a non-zero pressure in the ${y}$ and ${z}$ directions. It’s hard to imagine how a fluid could have zero energy and yet still exert pressure, so this would appear to be why the results are absurd, as stated by Moore.

# Ricci tensor and curvature scalar for a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.6.

As an example of calculating the Ricci tensor and curvature scalar we’ll find them for the 2-d surface of a sphere. The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we’ll need them first. It’s easiest to find them from the geodesic equation

$\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (1)$

which is formally equivalent to

$\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)$

The metric for a sphere is

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$

where ${r}$ is the constant radius of the sphere, so

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

We have two equations arising from 1. For ${a=\theta}$

$\displaystyle r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)$

Comparing with 2 we get, after dividing out the ${r^{2}}$:

$\displaystyle \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)$

For ${a=\phi}$:

$\displaystyle r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)$

Dividing through by ${r^{2}\sin^{2}\theta}$ and comparing with 2 we get (remember that the second term is ${\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}}$ and that ${\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}}$):

$\displaystyle \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)$

All other Christoffel symbols are zero.

In 2D, the Riemann tensor has only one independent component, which we can take to be ${R_{\theta\phi\theta\phi}}$, which can be calculated from

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)$

Lowering the first index, we have

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (16)$

We can now find the Ricci tensor.

$\displaystyle R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)$

Since the metric is diagonal and ${g^{ab}}$ is the inverse of ${g_{ab}}$, we have

 $\displaystyle g^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)$ $\displaystyle g^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)$

so, using the symmetries of the Riemann tensor,

 $\displaystyle R_{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle R_{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin^{2}\theta\ \ \ \ \ (26)$ $\displaystyle R_{\theta\phi}$ $\displaystyle =$ $\displaystyle R_{\phi\theta}=0 \ \ \ \ \ (27)$

We can get the upstairs version of the Ricci tensor as well:

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (30)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)$

The curvature scalar is

 $\displaystyle R$ $\displaystyle =$ $\displaystyle g_{ij}R^{ij}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{r^{2}} \ \ \ \ \ (35)$

As we would expect, the curvature of a sphere decreases as its radius gets larger.

# Gravity can't exist in 3 spacetime dimensions either

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; 6.

We’ve seen that the Einstein equation doesn’t allow gravity to exist in 2 spacetime dimensions. Here we’ll look at a demonstration that gravity also cannot exist in 3 spacetime dimensions of ${t}$, ${x}$ and ${y}$.

Because of the symmetries of the Riemann tensor, there are six independent, (possibly) non-zero components in 3 dimensions. We can take these components to be ${R_{xtxt}}$, ${R_{ytyt}}$, ${R_{xtyt}}$, ${R_{txxy}}$, ${R_{txyt}}$ and ${R_{tyxy}}$. As before, we use the Einstein equation

$\displaystyle R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

with ${\Lambda=0}$ and consider a vacuum so that ${T^{ij}=T=0}$, meaning that ${R^{ij}=0}$. We’ll look at a local inertial frame (LIF), where the metric is ${g^{ij}=\eta^{ij}}$. Then we get

 $\displaystyle R_{ij}$ $\displaystyle =$ $\displaystyle R_{\; iaj}^{a}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{ak}R_{kiaj}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{titj}+R_{xixj}+R_{yiyj} \ \ \ \ \ (4)$

Now we look at the 6 independent components of ${R_{ij}}$. Because ${R_{ijkm}=-R_{jikm}=-R_{ijmk}}$,
any component with either the
first two indices or last two indices equal is zero, so we get

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle R_{xtxt}+R_{ytyt}=0\ \ \ \ \ (5)$ $\displaystyle R_{xx}$ $\displaystyle =$ $\displaystyle -R_{txtx}+R_{yxyx}=0\ \ \ \ \ (6)$ $\displaystyle R_{yy}$ $\displaystyle =$ $\displaystyle -R_{tyty}+R_{xyxy}=0\ \ \ \ \ (7)$ $\displaystyle R_{tx}$ $\displaystyle =$ $\displaystyle R_{ytyx}=0\ \ \ \ \ (8)$ $\displaystyle R_{ty}$ $\displaystyle =$ $\displaystyle R_{xtxy}=0\ \ \ \ \ (9)$ $\displaystyle R_{xy}$ $\displaystyle =$ $\displaystyle -R_{txty}=0 \ \ \ \ \ (10)$

The last 3 equations show that 3 of the Riemann components are zero. The first 3 equations can be rewritten using the symmetries of the Riemann tensor:

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle R_{xtxt}+R_{ytyt}=0\ \ \ \ \ (11)$ $\displaystyle R_{xx}$ $\displaystyle =$ $\displaystyle -R_{xtxt}+R_{xyxy}=0\ \ \ \ \ (12)$ $\displaystyle R_{yy}$ $\displaystyle =$ $\displaystyle -R_{ytyt}+R_{xyxy}=0 \ \ \ \ \ (13)$

Solving these equations gives

$\displaystyle R_{xtxt}=R_{ytyt}=R_{xyxy}=0 \ \ \ \ \ (14)$

Thus all 6 components of the Riemann tensor are zero, showing that 3d spacetime must be flat and gravity cannot exist in 3 spacetime dimensions. (As usual, a tensor equation valid in a LIF is valid in all coordinate systems, so the conclusion is general.)

# Gravity can't exist in 2 spacetime dimensions

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; 5.

One consequence of the Einstein equation is that gravity cannot exist in a vacuum in a universe with fewer than 4 dimensions (3 space and 1 time). Here we’ll look at a demonstration of this for 2 dimensions of ${t}$ and ${x}$.

Because of the symmetries of the Riemann tensor, there is only one independent, (possibly) non-zero component in 2 dimensions. We can take this component to be ${R_{xtxt}}$. We start with the Einstein equation

$\displaystyle R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

For this argument, we’ll assume ${\Lambda=0}$ (whether or not it is actually zero is still a subject of debate, but we do know it’s very small). In a vacuum, the stress-energy tensor is zero (since there is no matter or energy), so in a vacuum ${R^{ij}=0}$. We’ll now see what this implies about the Riemann tensor (which is the only thing we can use to show conclusively whether spacetime is flat or curved). If ${R^{ij}=0}$ then its lowered form is also zero:

$\displaystyle R_{ij}=g_{ia}g_{jb}R^{ab}=0 \ \ \ \ \ (2)$

From the definition of the Ricci tensor

 $\displaystyle R_{ij}$ $\displaystyle =$ $\displaystyle R_{\; iaj}^{a}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ak}R_{kiaj}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{tt}R_{titj}+g^{tx}R_{tixj}+g^{xt}R_{xitj}+g^{xx}R_{xixj} \ \ \ \ \ (5)$

Because ${R_{ijkm}=-R_{jikm}=-R_{ijmk}}$, any component with either the first two indices or last two indices equal is zero. Therefore

 $\displaystyle R_{tt}$ $\displaystyle =$ $\displaystyle 0+0+0+g^{xx}R_{xtxt}=0\ \ \ \ \ (6)$ $\displaystyle R_{tx}$ $\displaystyle =$ $\displaystyle 0+0+g^{xt}R_{xttx}+0=0\ \ \ \ \ (7)$ $\displaystyle R_{xt}$ $\displaystyle =$ $\displaystyle 0+g^{tx}R_{txxt}+0+0=0\ \ \ \ \ (8)$ $\displaystyle R_{xx}$ $\displaystyle =$ $\displaystyle g^{tt}R_{txtx}+0+0+0=0 \ \ \ \ \ (9)$

All four of the Riemann components in these equations are either equal to ${R_{xtxt}}$ or to ${-R_{xtxt}}$ so we see that this component must be zero (since all four components of the metric can’t be zero). Therefore, the Riemann tensor is identically zero and space is flat in the vacuum. Flat space means no gravitational field, so gravity can’t exist in two dimensions.

# Einstein tensor and Einstein equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Box 21.2.

Our first attempt at constructing a tensor generalization of Newton’s law of gravitation failed because the total derivative of the Ricci tensor is not, in general, zero. Recall that the general idea is to look for an equation of form

$\displaystyle G^{ij}=\kappa T^{ij} \ \ \ \ \ (1)$

where ${G^{ij}}$ depends on the Riemann tensor and the metric. At this stage, there’s no particular reason to select any special form for this tensor, but if possible, we’d like it to be linear in the Riemann tensor. Also remember that ${G^{ij}}$ must satisfy the conditions:

1. It must be symmetric: (${G^{ij}=G^{ji}}$).
2. It must be of rank 2.
3. Its total derivative must be zero: ${\nabla_{j}G^{ij}=0}$.

To that end, let’s try a formula as follows:

$\displaystyle R^{ij}+bg^{ij}R+\Lambda g^{ij} \ \ \ \ \ (2)$

where ${R}$ is the curvature scalar

$\displaystyle R=g^{ab}R_{ab} \ \ \ \ \ (3)$

and ${b}$ and ${\Lambda}$ are constants. Note that although ${R}$ is a scalar, it is not a constant, so we can’t just merge the last two terms.

This form for ${G^{ij}}$ satisfies the first two conditions above, since everything on the RHS is of rank 2 and is also symmetric. So it just remains to show that the total derivative is zero. Since ${\nabla_{j}g^{ij}=0}$ always, the problem reduces to showing that

$\displaystyle \nabla_{j}\left(R^{ij}+bg^{ij}R\right)=0 \ \ \ \ \ (4)$

$\displaystyle \nabla_{s}R_{abmn}+\nabla_{n}R_{absm}+\nabla_{m}R_{abns}=0 \ \ \ \ \ (5)$

Multiplying through by ${g^{gs}g^{am}g^{bn}}$ we get (the metrics can be taken inside the derivatives since their derivatives are zero, so they act as constants):

$\displaystyle \nabla_{s}g^{gs}g^{am}g^{bn}R_{abmn}+\nabla_{n}g^{gs}g^{am}g^{bn}R_{absm}+\nabla_{m}g^{gs}g^{am}g^{bn}R_{abns}=0 \ \ \ \ \ (6)$

In the first term, we have

 $\displaystyle g^{am}g^{bn}R_{abmn}$ $\displaystyle =$ $\displaystyle g^{bn}R_{\;bmn}^{m}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{bn}R_{bn}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R \ \ \ \ \ (9)$

so we get

$\displaystyle \nabla_{s}g^{gs}R+\nabla_{n}g^{gs}g^{am}g^{bn}R_{absm}+\nabla_{m}g^{gs}g^{am}g^{bn}R_{abns}=0 \ \ \ \ \ (10)$

We can now use the symmetries of the Riemann tensor to simplify the last two terms. In the second term, we use ${R_{absm}=R_{smab}=-R_{msab}}$ and in the third term we use ${R_{abns}=R_{nsab}=-R_{nsba}}$:

$\displaystyle \nabla_{s}g^{gs}R-\nabla_{n}g^{gs}g^{am}g^{bn}R_{msab}-\nabla_{m}g^{gs}g^{am}g^{bn}R_{nsba}=0 \ \ \ \ \ (11)$

The Ricci tensor with lowered indices is

$\displaystyle R_{ab}=g^{cd}R_{dacb} \ \ \ \ \ (12)$

so to raise both its indices we have

$\displaystyle R^{gs}=g^{ga}g^{sb}g^{mn}R_{namb} \ \ \ \ \ (13)$

Comparing this with the second term in 11 we can map the indices in that term as follows: ${m\rightarrow n}$, ${s\rightarrow a}$, ${a\rightarrow m}$, ${n\rightarrow s}$ and ${b\rightarrow b}$. Thus the second term is the same as

 $\displaystyle \nabla_{n}g^{gs}g^{am}g^{bn}R_{msab}$ $\displaystyle =$ $\displaystyle \nabla_{s}g^{ga}g^{sb}g^{mn}R_{namb}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{s}R^{gs} \ \ \ \ \ (15)$

Similarly, in the third term we can map ${n\rightarrow n}$, ${s\rightarrow a}$, ${b\rightarrow m}$, ${a\rightarrow b}$ and ${m\rightarrow s}$ to get

 $\displaystyle \nabla_{m}g^{gs}g^{am}g^{bn}R_{nsba}$ $\displaystyle =$ $\displaystyle \nabla_{s}g^{ga}g^{sb}g^{mn}R_{namb}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla_{s}R^{gs} \ \ \ \ \ (17)$

Thus 11 becomes

$\displaystyle \nabla_{s}g^{gs}R-2\nabla_{s}R^{gs}=0 \ \ \ \ \ (18)$

or

$\displaystyle \nabla_{s}\left(R^{gs}-\frac{1}{2}g^{gs}R\right)=0 \ \ \ \ \ (19)$

Thus from 2 we can satisfy ${\nabla_{j}\left(R^{ij}+bg^{ij}R+\Lambda g^{ij}\right)=0}$ if ${b=-\frac{1}{2}}$. In practice, ${G^{ij}}$ is defined as just the first two terms:

$\displaystyle \boxed{G^{ij}\equiv R^{ij}-\frac{1}{2}g^{ij}R} \ \ \ \ \ (20)$

This is known as the Einstein tensor and the equation

$\displaystyle \boxed{G^{ij}+\Lambda g^{ij}=\kappa T^{ij}} \ \ \ \ \ (21)$

is the Einstein equation. This is the general relativistic replacement for Newton’s law of gravity.

# Einstein equation: trying the Ricci tensor as a solution

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; a.

We can now start looking at a derivation of the Einstein equation, which is the generalization of Newton’s formula for the gravitational force. In Newtonian theory, gravity is an attractive, conservative, inverse-square force so (apart from the sign) it is mathematically identical to the electrostatic force, which means we can write a differential form of Newton’s gravitational theory using Gauss’s law. That is

$\displaystyle \nabla\cdot\mathbf{g}=-4\pi G\rho \ \ \ \ \ (1)$

where ${\mathbf{g}}$ is the gravitational field, ${\rho}$ is the mass density and ${G}$ is the gravitational constant. The minus sign occurs because gravity is attractive, whereas the electric force for like charges is repulsive. Because the force is conservative, ${\mathbf{g}}$ can be written as the gradient of a potential so an alternative form of the equation is

$\displaystyle \nabla\cdot\left(-\nabla\Phi\right)=-\nabla^{2}\Phi=-4\pi G\rho \ \ \ \ \ (2)$

or

$\displaystyle \boxed{\nabla^{2}\Phi=4\pi G\rho} \ \ \ \ \ (3)$

The derivation of the Einstein equation is, like so many derivations in relativity, based on a plausibility argument. We want to find a tensor equation that generalizes Newton’s equation, and we want this tensor equation to reduce to Newton’s equation in the weak field limit.

First, the generalization of Newtonian mass density ${\rho}$ is the stress-energy tensor ${T^{ij}}$ we can try replacing the RHS of 3 by ${\kappa T^{ij}}$ where ${\kappa}$ is a scalar constant. Since the RHS is now a rank-2 tensor, the LHS must also be a rank-2 tensor, so we must have an equation like

$\displaystyle G^{ij}=\kappa T^{ij} \ \ \ \ \ (4)$

where the form of ${G^{ij}}$ needs to be determined. To do this, think about what we want the theory to do. The idea behind general relativity is that the energy density in a region of space should determine the curvature of the space in that region. The Riemann
tensor
and the metric tensor
describe the curvature of space-time, so it makes sense that ${G^{ij}}$ could depend on these two tensors.

Suppose we try to express ${G^{ij}}$ solely in terms of the Riemann tensor. What other constraints can we impose to narrow things down? First, since the Riemann tensor is rank 4 and ${G^{ij}}$ is rank 2, we’ll need to contract the Riemann tensor to get rid of 2 of its indices. One candidate is the Ricci tensor, defined as the contraction of the Riemann tensor over its first and third indices:

 $\displaystyle R_{\; bac}^{a}$ $\displaystyle =$ $\displaystyle g^{ad}R_{dbac}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle R_{bc} \ \ \ \ \ (6)$

To use ${R_{bc}}$, we need to raise both its indices, so we get

 $\displaystyle R^{ij}$ $\displaystyle =$ $\displaystyle g^{ib}g^{jc}R_{bc}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ib}g^{jc}R_{cb}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ic}g^{jb}R_{bc}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{ji} \ \ \ \ \ (10)$

where in the third line, we’ve swapped the dummy indices ${b}$ and ${c}$. Since ${T^{ij}}$ is symmetric, ${G^{ij}}$ must also be symmetric, but since ${R^{ij}=R^{ji}}$, this condition is satisfied.

From conservation of energy and momentum, we know that ${\nabla_{i}T^{ij}=0}$, so we must also have ${\nabla_{i}G^{ij}=0}$ (since ${\kappa}$ is a constant). This is where we run into a snag. The condition ${\nabla_{i}G^{ij}=0}$ must apply everywhere, in every reference frame, so it must apply the origin of a locally inertial frame (LIF). In a LIF, the Riemann tensor reduces to

$\displaystyle R_{nj\ell m}=\frac{1}{2}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (11)$

The Ricci tensor is then

 $\displaystyle R^{ik}$ $\displaystyle =$ $\displaystyle g^{ij}g^{km}R_{jm}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g^{ij}g^{km}g^{\ell n}R_{nj\ell m}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}g^{ij}g^{km}g^{\ell n}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (14)$

In a LIF, the first derivatives of ${g^{ij}}$ are all zero (by definition of the LIF), so

$\displaystyle \nabla_{i}R^{ik}=\frac{1}{2}g^{ij}g^{km}g^{\ell n}\nabla_{i}\left(\partial_{\ell}\partial_{j}g_{mn}+\partial_{m}\partial_{n}g_{j\ell}-\partial_{\ell}\partial_{n}g_{jm}-\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (15)$

In a LIF, the total derivative ${\nabla_{i}}$ reduces to the ordinary derivative ${\partial_{i}}$ so we get

$\displaystyle \nabla_{i}R^{ik}=\frac{1}{2}g^{ij}g^{km}g^{\ell n}\left(\partial_{i}\partial_{\ell}\partial_{j}g_{mn}+\partial_{i}\partial_{m}\partial_{n}g_{j\ell}-\partial_{i}\partial_{\ell}\partial_{n}g_{jm}-\partial_{i}\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (16)$

The indexes in the first term can be relabelled by swapping ${i}$ with ${\ell}$ and ${j}$ with ${n}$ to give

$\displaystyle \frac{1}{2}g^{ij}g^{km}g^{\ell n}\partial_{i}\partial_{\ell}\partial_{j}g_{mn}=\frac{1}{2}g^{\ell n}g^{km}g^{ij}\partial_{\ell}\partial_{i}\partial_{n}g_{mj} \ \ \ \ \ (17)$

which is the negative of the third term (since ${g_{mj}=g_{jm}}$ and the order of the partial derivatives doesn’t matter), so these two terms cancel and we’re left with

$\displaystyle \nabla_{i}R^{ik}=\frac{1}{2}g^{ij}g^{km}g^{\ell n}\left(\partial_{i}\partial_{m}\partial_{n}g_{j\ell}-\partial_{i}\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (18)$

The only way this can be identically zero is if we could swap ${n}$ with ${j}$ in the first term and have it equal the negative of the second term. However, if we try this, we get

$\displaystyle \frac{1}{2}g^{ij}g^{km}g^{\ell n}\partial_{i}\partial_{m}\partial_{n}g_{j\ell}=\frac{1}{2}g^{in}g^{km}g^{\ell j}\partial_{i}\partial_{m}\partial_{j}g_{n\ell} \ \ \ \ \ (19)$

The partial derivatives match up with those in the second term, but the product of the three metric tensors doesn’t, so in general this isn’t zero, meaning that

$\displaystyle \nabla_{i}R^{ik}\ne0 \ \ \ \ \ (20)$

Thus setting ${G^{ij}=R^{ij}}$ won’t work, and we’ll need to try something else.

# Riemann tensor for 3-d spherical coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem 19.7.

We saw that the Riemann tensor for the surface of a sphere had a non-zero component, indicating that this is a curved space. If we use spherical coordinates in 3-d space, however, the Riemann tensor should be zero, since this is a flat space.

As usual, we need the Christoffel symbols and we get them by comparing the two forms of the geodesic equation.

 $\displaystyle \frac{d}{d\tau}\left(g_{aj}\dot{x}^{j}\right)-\frac{1}{2}\partial_{a}g_{ij}\dot{x}^{i}\dot{x}^{j}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (2)$

where as usual a dot denotes a derivative with respect to proper time ${\tau}$.

For spherical coordinates, the interval is

$\displaystyle ds^{2}=dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)$

Note that ${r}$ is now a variable, rather than the constant radius of the sphere in the 2-d system.

From 1 we get, with ${a=\theta}$:

$\displaystyle 2r\dot{r}\dot{\theta}+r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (4)$

Dividing through by ${r^{2}}$ and comparing with 2 we get

 $\displaystyle \Gamma_{\phi\phi}^{\theta}$ $\displaystyle =$ $\displaystyle -\sin\theta\cos\theta\ \ \ \ \ (5)$ $\displaystyle \Gamma_{r\theta}^{\theta}$ $\displaystyle =$ $\displaystyle \Gamma_{\theta r}^{\theta}=\frac{1}{r}\ \ \ \ \ (6)$ $\displaystyle \Gamma_{\theta\phi}^{\theta}$ $\displaystyle =$ $\displaystyle \Gamma_{\phi\theta}^{\theta}=\Gamma_{\theta\theta}^{\theta}=0 \ \ \ \ \ (7)$

With ${a=\phi}$ we have

 $\displaystyle 2r\sin^{2}\theta\dot{r}\dot{\phi}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}+r^{2}\sin^{2}\theta\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \frac{2}{r}\dot{r}\dot{\phi}+2\cot\theta\dot{\theta}\dot{\phi}+\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}$ $\displaystyle =$ $\displaystyle \cot\theta\ \ \ \ \ (10)$ $\displaystyle \Gamma_{r\phi}^{\phi}=\Gamma_{\phi r}^{\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\ \ \ \ \ (11)$ $\displaystyle \Gamma_{\theta\theta}^{\phi}=\Gamma_{\phi\phi}^{\phi}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

For ${a=r}$, we have

 $\displaystyle \ddot{r}-\frac{1}{2}\left(2r\dot{\theta}^{2}+2r\sin^{2}\theta\dot{\phi}^{2}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \ddot{r}-r\dot{\theta}^{2}-r\sin^{2}\theta\dot{\phi}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle \Gamma_{\theta\theta}^{r}$ $\displaystyle =$ $\displaystyle -r\ \ \ \ \ (15)$ $\displaystyle \Gamma_{\phi\phi}^{r}$ $\displaystyle =$ $\displaystyle -r\sin^{2}\theta \ \ \ \ \ (16)$

We can use these results to get the Riemann tensor.

 $\displaystyle R_{abcd}$ $\displaystyle =$ $\displaystyle g_{af}R_{\; bcd}^{f}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{af}\left(\partial_{c}\Gamma_{\; db}^{f}-\partial_{d}\Gamma_{\; cb}^{f}+\Gamma_{\; db}^{k}\Gamma_{\; ck}^{f}-\Gamma_{\; cb}^{k}\Gamma_{\; kd}^{f}\right) \ \ \ \ \ (18)$

Although we know there is only one independent component in 2-d, we can work out all four non-zero components to see how the calculations go.

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta f}R_{\;\phi\theta\phi}^{f}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta-0-\frac{r\sin^{2}\theta}{r}+\cos^{2}\theta\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (23)$

The other components of the Riemann tensor can be evaluated in the same way, and they all come out to zero, so spherical coordinates represent flat 3-d space.

# Falling into a black hole: tidal forces

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.9.

Using the Riemann tensor, we can get an idea of the force felt by someone falling into a black hole. Recall the original definition of the Riemann tensor was in terms of the equation of geodesic deviation:

$\displaystyle \ddot{\mathbf{n}}^{i}=-R_{\; j\ell m}^{i}u^{m}u^{j}n^{\ell} \ \ \ \ \ (1)$

where ${\mathbf{n}}$ is the four-vector separating two infinitesimally close geodesics, and ${u}$ is the four-velocity of an object in freefall along one of the geodesics.

For an object, such as a person, that has a large enough size that different parts of the object would, if they weren’t connected, follow different geodesics, a tension force is felt as the various geodesics that pass through different parts of the object diverge during the object’s journey. Suppose our unfortunate person is falling feet first into a black hole (we’ll assume that the person started at rest very far away from the black hole). If we set up a locally inertial frame (LIF) at the person’s centre of mass and align the person’s local ${z}$ axis with the radial direction in the Schwarzschild (S) metric, then we’ve seen that we can write the geodesic deviation as

$\displaystyle \frac{d^{2}n^{i}}{dt^{2}}=-R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (2)$

In the case of the falling person, we can look at the ${z}$ direction, since this is where most of the tidal effects will be felt. In that case, we get, using the person’s LIF as the reference frame:

$\displaystyle \ddot{n}^{z}=-R_{\; t\ell t}^{z}n^{\ell} \ \ \ \ \ (3)$

If we neglect separations in the ${x}$ and ${y}$ directions, this becomes

$\displaystyle \ddot{n}^{z}=-R_{\; tzt}^{z}n^{z} \ \ \ \ \ (4)$

To get the Riemann component, we can use the symmetry of the tensor:

$\displaystyle g_{za}R_{\; tzt}^{a}=R_{ztzt}=R_{tztz}=g_{ta}R_{\; ztz}^{a} \ \ \ \ \ (5)$

In the LIF, ${g_{ij}=\eta_{ij}}$ so this equation becomes

$\displaystyle R_{\; tzt}^{z}=-R_{\; ztz}^{t} \ \ \ \ \ (6)$

and we worked out the RHS in the last post, so we have

$\displaystyle R_{\; tzt}^{z}=-\frac{2GM}{r^{3}} \ \ \ \ \ (7)$

The acceleration of the ${z}$ separation of the two geodesics is then

$\displaystyle \ddot{n}^{z}=\frac{2GM}{r^{3}}n^{z} \ \ \ \ \ (8)$

which we can rewrite with ${r}$ as a function of the acceleration felt at that distance from the black hole:

$\displaystyle r=\left[\frac{2GMn^{z}}{\ddot{n}^{z}}\right]^{1/3} \ \ \ \ \ (9)$

To see how long it takes the person to fall from this distance to ${r=0}$, we can use the formula we derived earlier:

$\displaystyle \Delta\tau=\frac{\pi r^{3/2}}{\sqrt{8GM}} \ \ \ \ \ (10)$

So the time measured by the observer is

$\displaystyle \Delta\tau=\frac{\pi}{2}\sqrt{\frac{n^{z}}{\ddot{n}^{z}}} \ \ \ \ \ (11)$

To put this in practical terms, a typical person can handle up to about 5g (five times the acceleration due to gravity at the Earth’s surface) along their vertical direction before losing consciousness. If we take ${n^{z}\approx1\mbox{ m}}$ (about half a person’s height) and ${\ddot{n}^{z}\approx50\mbox{ m s}^{-2}}$ then the time from first experiencing this force to annihilation at the singularity at the centre of the black hole is about

$\displaystyle \Delta\tau\approx0.2\mbox{ sec} \ \ \ \ \ (12)$

Using Moore’s estimate of the speed of pain impulses (around 1 m/sec) any pain resulting from this probably wouldn’t be felt before the person gets annihilated.