## Vacuum stress-energy and the cosmological constant

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.9.

The Einstein equation is

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

Up to now, we’ve usually taken ${\Lambda=0}$, since we know from the Newtonian limit that ${\Lambda}$ must be very small. If ${\Lambda\ne0}$, the Newtonian limit becomes

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

so ${\Lambda}$ acts as a negative mass density, that is, it adds a repulsive term into the gravitational force. Einstein originally introduced it to counter the attractive force of gravity on a cosmological scale, since at the time it was believed that the universe was static (neither expanding nor contracting) and if gravity were purely attractive, the universe would be contracting.

At the moment, the universe is believed to be expanding so ${\Lambda}$ is believed to be non-zero and positive, although still small enough that its effects are not noticeable on the scale of the solar system (or indeed on a galactic scale). Because of this, ${\Lambda}$ is called the cosmological constant.

We can include ${\Lambda}$ within the stress-energy tensor by defining a vacuum stress-energy as

$\displaystyle T_{vac}^{ij}=-\frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (3)$

We can define a vacuum stress-energy scalar:

 $\displaystyle T_{vac}$ $\displaystyle \equiv$ $\displaystyle g_{ij}T_{vac}^{ij}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g_{ij}g^{ij}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4\Lambda}{8\pi G} \ \ \ \ \ (6)$

Therefore

 $\displaystyle T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g^{ij}+\frac{2\Lambda}{8\pi G}g^{ij}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (8)$

and we can write 1 as

 $\displaystyle R^{ij}$ $\displaystyle =$ $\displaystyle 8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T+T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle 8\pi G\left(T_{all}^{ij}-\frac{1}{2}g^{ij}T_{all}\right) \ \ \ \ \ (10)$

where ${T_{all}^{ij}}$ includes the stress-energy from the mass-energy density and the vacuum.

The dominant energy condition is a constraint placed on the stress-energy tensor so that observers in any local orthogonal frame will measure the fluid’s speed to be less than the speed of light. The condition is that if ${a^{i}}$ is any four-vector that is causal, that is, it satisfies the conditions

 $\displaystyle \mathbf{a}\cdot\mathbf{a}$ $\displaystyle \le$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle a^{t}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (12)$

then we require the stress-energy tensor ${T^{ij}}$ to satisfy the condition that if

$\displaystyle b^{i}=-T^{ij}g_{jk}a^{k} \ \ \ \ \ (13)$

then ${\mathbf{b}}$ is also a causal four-vector. For the vacuum stress-energy this condition says

 $\displaystyle b^{i}$ $\displaystyle =$ $\displaystyle -T_{vac}^{ij}g_{jk}a^{k}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij}g_{jk}a^{k}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}\delta_{\; k}^{i}a^{k}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}a^{i} \ \ \ \ \ (17)$

That is, ${b^{i}}$ is just a positive (if ${\Lambda>0}$) constant multiplied by ${a^{i}}$, so if ${a^{i}}$ is causal, then ${b^{i}}$ must also be causal. Thus ${T_{vac}^{ij}}$ satisfies the dominant energy condition.

## 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

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

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

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

In a perfect fluid at rest, these components represent the pressure of the fluid in the ${y}$ and ${z}$ directions, so this result implies that the pressure would have to be negative, which doesn’t make sense.

## Einstein equation on the surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.7.

According to the Einstein equation, the Riemann tensor in 2D must be zero in empty space, implying that gravitational fields cannot exist in 2D. Another consequence of the Einstein equation is that the stress-energy must be zero on the surface of a sphere. That is, even though a 2D surface is manifestly curved, the curvature is not the result of any mass or energy. This is another example of how general relativity breaks down in two dimensions.

The Einstein equation is

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

$\displaystyle R^{ij}=\left[\begin{array}{cc} \frac{1}{r^{4}} & 0\\ 0 & \frac{1}{r^{4}\sin^{2}\theta} \end{array}\right] \ \ \ \ \ (2)$

The metric for a sphere is (in both forms):

 $\displaystyle g^{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \frac{1}{r^{2}} & 0\\ 0 & \frac{1}{r^{2}\sin^{2}\theta} \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} r^{2} & 0\\ 0 & r^{2}\sin^{2}\theta \end{array}\right] \ \ \ \ \ (4)$

Since the off-diagonal elements of ${g^{ij}}$ and ${R^{ij}}$ are all zero, 1 tells us that

$\displaystyle T^{\theta\phi}=T^{\phi\theta}=0 \ \ \ \ \ (5)$

To deal with the diagonal elements, we first need the stress-energy scalar.

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}T^{\theta\theta}+r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (7)$

We have

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\theta\theta}-\frac{1}{2r^{2}}T\right)+\frac{\Lambda}{r^{2}}\ \ \ \ \ (9)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T\right)+\frac{\Lambda}{r^{2}\sin^{2}\theta} \ \ \ \ \ (11)$

Combining these we get

 $\displaystyle \frac{R^{\theta\theta}}{\sin^{2}\theta}-R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \kappa\left(\frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}\right)=0\ \ \ \ \ (12)$ $\displaystyle \frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle T^{\theta\theta}$ $\displaystyle =$ $\displaystyle T^{\phi\phi}\sin^{2}\theta\ \ \ \ \ (14)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 2r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (15)$

Therefore

 $\displaystyle T^{\theta\theta}-\frac{1}{2r^{2}}T$ $\displaystyle =$ $\displaystyle T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T=0 \ \ \ \ \ (16)$

holds identically. Thus the stress-energy contribution to the Einstein equation is always zero on a sphere (although the stress-energy tensor may have two non-zero components, these two components always combine to give zero contribution to the Einstein equation).

## 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.

## Einstein equation in the Newtonian limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Problem 21.1.

The Einstein equation is

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

where we have yet to determine the constant ${\kappa}$. To do this, we need to show that the Einstein equation reduces to Newton’s law of gravity for weak gravitational fields. Actually, there are three conditions that should hold in the Newtonian limit. First, as we’ve said, the gravitational field is weak, meaning that spacetime is nearly flat. Second, objects should travel with a speed much less than the speed of light (the spatial four-velocity components ${u^{i}\ll1}$ for ${i=x,y,z}$). The second condition implies that the only non-negligible component of the stress-energy tensor ${T^{ij}}$ is ${T^{tt}}$. For example, for a perfect fluid, we can assume that it’s effectively at rest, so the off-diagonal elements are all zero. For the diagonal spatial compoments, we have (for ${T^{zz}}$; the other 2 components have the analogous formulas):

$\displaystyle T^{zz}=\frac{1}{L^{3}}\int dp^{x}\int dp^{y}\int\left(p^{z}\right)^{2}\frac{N\left(p\right)}{p^{t}}dp^{z} \ \ \ \ \ (2)$

This equation is for a cubic volume of side length ${L}$ containing ${N\left(p\right)}$ particles of momentum ${p}$. Since ${p^{z}=mu^{z}}$ and ${u^{z}\ll1}$, ${T^{zz}\approx0}$ (the requirement of a weak gravitational field means that ${N\left(p\right)}$ can’t be very large, since we can’t have that much mass). As the spatial diagonal elements ${T^{ii}=P}$, the pressure and ${T^{tt}=\rho}$, the energy density, this condition translates to ${\rho\gg P}$.

With these assumptions, we can try to show that the relativistic equation of geodesic deviation reduces to the Newtonian version. That is, we want to show that

$\displaystyle \ddot{\mathbf{n}}^{i}=-R_{\; j\ell m}^{i}u^{m}u^{j}n^{\ell} \ \ \ \ \ (3)$

reduces to

$\displaystyle \ddot{n}^{i}=-\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right)n^{k}\mbox{ (for }i,j,k=x,y,z\mbox{)} \ \ \ \ \ (4)$

where ${n^{i}}$ is the separation of two infinitesimally close geodesics (this is the tidal force.) and ${\Phi}$ is the Newtonian gravitational potential.

Starting with 3, we can eliminate terms where ${m\ne t}$ or ${j\ne t}$ (because ${u^{i}\approx0}$ for ${i\ne t}$) to get

$\displaystyle \ddot{n}^{i}=-R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (5)$

where we’re now considering only the spatial components: ${i,\ell=x,y,z}$. Note that summing ${\ell}$ over ${x,y,z}$ is the same as summing it over ${t,x,y,z}$ since due to the anti-symmetry of the Riemann tensor under interchange of its last two indices, ${R_{\; ttt}^{i}=-R_{\; ttt}^{i}=0}$. Also, because we’re in the non-relativistic limit, the proper time and coordinate time are essentially the same thing: ${\tau\approx t}$, so the time derivative is with respect to ${t}$.

Comparing this to 4, we have (renaming ${\ell}$ to ${k}$ in the last equation):

$\displaystyle R_{\; tkt}^{i}\approx\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right) \ \ \ \ \ (6)$

Newton’s law of gravity in differential form is

 $\displaystyle \nabla^{2}\Phi$ $\displaystyle =$ $\displaystyle 4\pi G\rho\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{ij}\partial_{i}\partial_{j}\Phi\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle R_{\; tit}^{i}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{tt} \ \ \ \ \ (10)$

(Again, the contraction over index ${i}$ in the second to last line can be taken over 3 or 4 coordinates, since ${R_{\; ttt}^{i}=0}$.) We can raise both indices on the Ricci tensor in the usual way:

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle g^{ti}g^{tj}R_{ij}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \eta^{ti}\eta^{tj}R_{ij}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{2}R_{tt}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{tt} \ \ \ \ \ (14)$

so we can write 10 in upper index form as

$\displaystyle \nabla^{2}\Phi\approx R^{tt} \ \ \ \ \ (15)$

Now looking back at 1 and using the condition that ${T^{tt}=\rho}$ is the only significant entry in the stress-energy tensor, we have

$\displaystyle T=g_{ij}T^{ij}=\eta_{ij}T^{ij}=-\rho \ \ \ \ \ (16)$

so

 $\displaystyle R^{tt}$ $\displaystyle =$ $\displaystyle \kappa\left(T^{tt}-\frac{1}{2}\eta^{tt}T\right)+\Lambda\eta^{tt}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\kappa}{2}\rho-\Lambda\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \nabla^{2}\Phi\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4\pi G\rho \ \ \ \ \ (20)$

Comparing the second and fourth lines, we see that ${\Lambda\approx0}$ and

$\displaystyle \kappa=8\pi G \ \ \ \ \ (21)$

and the Einstein equation becomes

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (22)$

or in its original form

$\displaystyle G^{ij}=8\pi GT^{ij} \ \ \ \ \ (23)$

Actually, we can’t take ${\Lambda=0}$; all we can say is that for gravitational systems on the scale of the solar system (where Newtonian theory works well, except in the case of the orbit of Mercury) ${\Lambda\ll4\pi G\rho}$. To get a feel for how small ${\Lambda}$ needs to be, suppose we have a spherical gravitational potential in empty space (${\rho=0}$). Then in spherical coordinates

 $\displaystyle \nabla^{2}\Phi$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle -\Lambda\ \ \ \ \ (25)$ $\displaystyle \frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right)$ $\displaystyle \approx$ $\displaystyle -\Lambda r^{2}\ \ \ \ \ (26)$ $\displaystyle \frac{d\Phi}{dr}$ $\displaystyle \approx$ $\displaystyle -\frac{\Lambda}{3}r \ \ \ \ \ (27)$

This is the radial component (the only non-zero component) of the gradient of the potential, and the negative gradient of the gravitational potential is the gravitational field, which is the acceleration of gravity. In the solar system, Newton’s theory says that the acceleration due to the sun is

$\displaystyle g=\frac{GM_{s}}{r^{2}} \ \ \ \ \ (28)$

so if ${\Lambda\ne0}$ but its effect is not felt in Newton’s theory, we must have

$\displaystyle \frac{\Lambda}{3}r\ll\frac{GM_{s}}{r^{2}} \ \ \ \ \ (29)$

in order for Newton’s theory to be valid in the solar system. Distances in the solar system are around ${r\approx10^{12}\mbox{ m}}$, ${GM_{s}\approx1500\mbox{ m}}$ and ${G}$ in general relativistic units is ${7.426\times10^{-28}\mbox{ m kg}^{-1}}$ so this means

$\displaystyle \frac{\Lambda}{8\pi G}\ll\frac{1500}{8\pi\left(7.426\times10^{-28}\right)\left(10^{12}\right)^{3}}=2.4\times10^{-7}\mbox{ kg m}^{-3} \ \ \ \ \ (30)$

## Gravity can’t exist in 3 spacetime dimensions either

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Problem 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; Problem 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 of zero implies a zero Ricci tensor

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

We wish to prove that ${G^{ij}=0}$ if and only if ${R^{ij}=0}$. The Einstein tensor is defined as

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

Clearly if the Ricci tensor ${R^{ij}=0}$ then ${G^{ij}=0}$ (since the curvature scalar is the contraction of the Ricci tensor: ${R=g_{ij}R^{ij}}$). To prove the converse, suppose ${G^{ij}=0}$. Then we can multiply both sides by ${g_{ij}}$ to get

 $\displaystyle 0$ $\displaystyle =$ $\displaystyle g_{ij}R^{ij}-\frac{1}{2}g_{ij}g^{ij}R\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R-2R\ \ \ \ \ (3)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

since ${g_{ij}g^{ij}=4}$. With ${G^{ij}=0}$ and ${R=0}$, 1 tells us that ${R^{ij}=0}$. QED.

## Einstein equation for a perfect fluid

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Problem 21.3.

The Einstein equation can be written as

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

For a perfect fluid the stress-energy tensor is

$\displaystyle T^{ij}=\left(\rho_{0}+P_{0}\right)u^{i}u^{j}+P_{0}g^{ij} \ \ \ \ \ (2)$

where ${u^{i}}$ is the four-velocity, ${\rho_{0}}$ is the energy density and ${P_{0}}$ is the pressure.
We have for the stress-energy scalar:

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \left(\rho_{0}+P_{0}\right)g_{ij}u^{i}u^{j}+P_{0}g_{ij}g^{ij}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\rho_{0}+P_{0}\right)+4P_{0}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3P_{0}-\rho_{0} \ \ \ \ \ (5)$

since ${g_{ij}g^{ij}=4}$ and ${g_{ij}u^{i}u^{j}=-1}$ in any coordinate system.

The stress energy term in 1 is therefore

 $\displaystyle T^{ij}-\frac{1}{2}g^{ij}T$ $\displaystyle =$ $\displaystyle \left(\rho_{0}+P_{0}\right)u^{i}u^{j}+P_{0}g^{ij}-\frac{1}{2}g^{ij}\left(3P_{0}-\rho_{0}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho_{0}+P_{0}\right)u^{i}u^{j}+\frac{1}{2}g^{ij}\left(\rho_{0}-P_{0}\right) \ \ \ \ \ (7)$

In a local orthogonal frame (LOF) in which the fluid is at rest ${u^{t}=1}$, ${u^{i}=0}$ for ${i=x,y,z}$ and ${g^{ij}=\eta^{ij}}$, so

 $\displaystyle T^{tt}-\frac{1}{2}g^{tt}T$ $\displaystyle =$ $\displaystyle \rho_{0}+P_{0}-\frac{1}{2}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\rho_{0}+3P_{0}\right)\ \ \ \ \ (9)$ $\displaystyle T^{xx}-\frac{1}{2}g^{xx}T$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (10)$ $\displaystyle T^{yy}-\frac{1}{2}g^{yy}T$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (11)$ $\displaystyle T^{zz}-\frac{1}{2}g^{zz}T$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\rho_{0}-P_{0}\right) \ \ \ \ \ (12)$

## Einstein equation: alternative form

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Box 21.3.

The general relativistic generalization of Newton’s law of gravity is

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

where the Einstein tensor is defined in terms of the Ricci tensor and the curvature scalar as

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

We can write this in a different form that is sometimes easier to use in calculations. Eliminating ${G^{ij}}$ we have

$\displaystyle R^{ij}-\frac{1}{2}g^{ij}R+\Lambda g^{ij}=\kappa T^{ij} \ \ \ \ \ (3)$

Multiplying both sides by ${g_{ij}}$ we get

$\displaystyle g_{ij}R^{ij}-\frac{1}{2}g_{ij}g^{ij}R+\Lambda g_{ij}g^{ij}=\kappa g_{ij}T^{ij} \ \ \ \ \ (4)$

Because the tensor ${g_{ij}}$ is the inverse of ${g^{ij}}$, their product gives the identity matrix of rank 4 (this can be seen by doing the calculation in a local inertial frame where ${g_{ij}=\eta_{ij}}$ and noting that since it’s a tensor equation, it’s valid in all coordinate systems). That is

$\displaystyle g_{ij}g^{jk}=\delta_{\; i}^{k} \ \ \ \ \ (5)$

so if we contract the ${\delta_{\; j}^{k}}$ tensor we just sum up its diagonal elements and since these are all 1 (and there are four rows), we get

$\displaystyle \delta_{\; k}^{k}=4 \ \ \ \ \ (6)$

Returning to 4 we get

 $\displaystyle g_{ij}R^{ij}-2R+4\Lambda$ $\displaystyle =$ $\displaystyle \kappa g_{ij}T^{ij} \ \ \ \ \ (7)$

Since the curvature scalar is given by

$\displaystyle R\equiv g_{ij}R^{ij} \ \ \ \ \ (8)$

and the stress-energy scalar is

$\displaystyle T\equiv g_{ij}T^{ij} \ \ \ \ \ (9)$

we get

$\displaystyle -R+4\Lambda=\kappa T \ \ \ \ \ (10)$

Multiplying this by ${-\frac{1}{2}g^{ij}}$ and subtracting from 3 we have

 $\displaystyle R^{ij}-\Lambda g^{ij}$ $\displaystyle =$ $\displaystyle \kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (11)$

Isolating the Ricci tensor gives us the alternative form of the Einstein equation:

$\displaystyle \boxed{R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij}} \ \ \ \ \ (12)$