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)

The Ricci tensor for a spherical surface is

\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)

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)

To do this, we start with the Bianchi identity

\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; Box 21.1.

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.

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