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.

Riemann tensor in the Schwarzschild metric: observer’s view

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.8.

The component of a four-vector ${A}$ along a basis vector ${\mathbf{o}_{i}}$ is given by the dot product of the four-vector with the basis vector. Since the dot product is a scalar, it can be computed in any reference frame. In particular, if we’re dealing with a locally flat coordinate system with basis vectors ${\mathbf{o}_{i}}$ embedded in the Schwarzschild (S) metric, we can do the calculation in the global S metric since we know the components of ${\mathbf{o}_{i}}$ in the S frame. That is, the components of a four-vector ${A}$ along each of the basis vectors is

 $\displaystyle A_{i,obs}$ $\displaystyle =$ $\displaystyle g_{jm}\left(\mathbf{o}_{i}\right)^{j}A^{m}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\mathbf{o}_{i}\right)^{j}A_{j} \ \ \ \ \ (2)$

The subscript ${obs}$ indicates that the vector component is that seen by an observer in the locally flat frame with basis vectors ${\mathbf{o}_{i}}$.

We can extend this idea to find the components of any tensor in the locally flat frame, since we just apply the same procedure to each index of the tensor. For the Riemann tensor ${R_{\; j\ell m}^{i}}$ we would get

$\displaystyle R_{ij\ell m,obs}=g_{ab}\left(\mathbf{o}_{i}\right)^{a}\left(\mathbf{o}_{j}\right)^{c}\left(\mathbf{o}_{\ell}\right)^{d}\left(\mathbf{o}_{m}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (3)$

To get the observer’s tensor with the first index raised, we need to use the metric to raise the index. The correct metric to use is the metric of the locally flat frame, which is ${\eta^{ij}}$. Therefore we get

 $\displaystyle R_{\; j\ell m,obs}^{i}$ $\displaystyle =$ $\displaystyle \eta^{if}R_{fj\ell m,obs}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{if}g_{ab}\left(\mathbf{o}_{f}\right)^{a}\left(\mathbf{o}_{j}\right)^{c}\left(\mathbf{o}_{\ell}\right)^{d}\left(\mathbf{o}_{m}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (5)$

For example, we can calculate the component ${R_{\; ztz,obs}^{t}}$ for a freely-falling observer by using the components for ${\mathbf{o}_{i}}$ that we worked out earlier.

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}g_{ab}\left(\mathbf{o}_{f}\right)^{a}\left(\mathbf{o}_{z}\right)^{c}\left(\mathbf{o}_{t}\right)^{d}\left(\mathbf{o}_{z}\right)^{e}R_{\; cde}^{b} \ \ \ \ \ (6)$

To save typing, I’ll write the unit vectors in normal type and without the parentheses, so this equation becomes

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}g_{ab}o_{f}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{\; cde}^{b} \ \ \ \ \ (7)$

where sums are implied over all repeated indices except ${t}$ and ${z}$.

Remember that a superscript on a basis vector refers to the component of that vector along the direction given by the superscript, and that this direction is one of those in the S metric. A subscript on a basis vector refers to the axis in the locally flat system along with that basis vector points. Thus ${o_{z}^{r}}$ is the component along the ${r}$ direction in the S metric of the vector ${\mathbf{o}_{z}}$ that points along the ${z}$ direction in the locally flat system.

We’ll reproduce here the basis vectors ${\mathbf{o}_{t}}$ and ${\mathbf{o}_{z}}$ for a freely falling observer (we won’t need the other two vectors):

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \left[\left(1-\frac{2GM}{r}\right)^{-1},-\sqrt{\frac{2GM}{r}},0,0\right]\ \ \ \ \ (8)$ $\displaystyle \mathbf{o}_{z}$ $\displaystyle =$ $\displaystyle \left[-\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}},1,0,0\right] \ \ \ \ \ (9)$

Unfortunately, since both of these vectors have two non-zero components, the sums in 7 give us more than one term. In order to make use of the symmetries of the Riemann tensor, we’ll rewrite 7 using ${R_{acde}=g_{ab}R_{\; cde}^{b}}$:

$\displaystyle R_{\; ztz,obs}^{t}=\eta^{tf}o_{f}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{acde} \ \ \ \ \ (10)$

Since ${\eta^{ij}}$ is diagonal, we must have ${f=t}$ and since ${\eta^{tt}=-1}$ we have

$\displaystyle R_{\; ztz,obs}^{t}=-o_{t}^{a}o_{z}^{c}o_{t}^{d}o_{z}^{e}R_{acde} \ \ \ \ \ (11)$

Because of the symmetries, we must have ${a\ne c}$ and ${d\ne e}$, so there are four possibilities for these four indices:

$\displaystyle \left[a,c,d,e\right]=\left[t,r,t,r\right],\left[r,t,r,t\right],\left[r,t,t,r\right],\left[t,r,r,t\right] \ \ \ \ \ (12)$

The sum in 11 thus expands to

$\displaystyle R_{\; ztz,obs}^{t}=-R_{trtr}o_{t}^{t}o_{z}^{r}o_{t}^{t}o_{z}^{r}-R_{rtrt}o_{t}^{r}o_{z}^{t}o_{t}^{r}o_{z}^{t}-R_{rttr}o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}-R_{trrt}o_{t}^{t}o_{z}^{r}o_{t}^{r}o_{z}^{t} \ \ \ \ \ (13)$

Using the symmetries when swapping the first two or the last two indices in the Riemann tensor in this lowered form, we can rewrite this as

 $\displaystyle R_{\; ztz,obs}^{t}$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[o_{t}^{t}o_{z}^{r}o_{t}^{t}o_{z}^{r}+o_{t}^{r}o_{z}^{t}o_{t}^{r}o_{z}^{t}-o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}-o_{t}^{t}o_{z}^{r}o_{t}^{r}o_{z}^{t}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[\left(o_{t}^{t}o_{z}^{r}\right)^{2}+\left(o_{t}^{r}o_{z}^{t}\right)^{2}-2o_{t}^{r}o_{z}^{t}o_{t}^{t}o_{z}^{r}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left(o_{t}^{t}o_{z}^{r}-o_{t}^{r}o_{z}^{t}\right)^{2}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left(\left(1-\frac{2GM}{r}\right)^{-1}-\left[-\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}}\right]\left[-\sqrt{\frac{2GM}{r}}\right]\right)^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr}\left[\left(1-\frac{2GM}{r}\right)^{-1}\left(1-\frac{2GM}{r}\right)\right]^{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{trtr} \ \ \ \ \ (19)$

We now need to find ${R_{trtr}}$, which is

$\displaystyle R_{trtr}=g_{ta}R_{\; rtr}^{a}=g_{tt}R_{\; rtr}^{t} \ \ \ \ \ (20)$

We’ve worked out the last tensor component earlier, so plugging this in, we get

 $\displaystyle R_{\; ztz,obs}^{t}$ $\displaystyle =$ $\displaystyle -g_{tt}R_{\; rtr}^{t}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{2GM}{r}\right)\left[\frac{2GM}{r^{3}}\left(1-\frac{2GM}{r}\right)^{-1}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r^{3}} \ \ \ \ \ (23)$

Thus this component of the Riemann tensor has no singularity at ${r=2GM}$ in the observer’s local frame.

Riemann tensor in a 2-d curved space

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.6.

Here’s another example of the Riemann tensor in a 2-d coordinate system. The tensor is

$\displaystyle R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (1)$

As usual, we need the Christoffel symbols, which we can get by comparing the two forms of the geodesic equation. These equations are

 $\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \ddot{x}^{a}+\Gamma_{\; ij}^{a}\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (3)$

The metric is

$\displaystyle ds^{2}=dp^{2}+e^{2p/p_{0}}dq^{2} \ \ \ \ \ (4)$

so ${g_{pp}=1}$ and ${g_{qq}=e^{2p/p_{0}}}$. For the two coordinates, 2 gives us

 $\displaystyle \ddot{p}-\frac{1}{p_{0}}e^{2p/p_{0}}\dot{q}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle e^{2p/p_{0}}\ddot{q}+\frac{2}{p_{0}}e^{2p/p_{0}}\dot{p}\dot{q}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Dividing through by the coefficient of the second derivative in the second equation case gives:

 $\displaystyle \ddot{p}-\frac{1}{p_{0}}e^{2p/p_{0}}\dot{q}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle \ddot{q}+\frac{2}{p_{0}}\dot{p}\dot{q}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

Comparing with 3 we get

 $\displaystyle \Gamma_{\; qq}^{p}$ $\displaystyle =$ $\displaystyle -\frac{1}{p_{0}}e^{2p/p_{0}}\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\; pq}^{q}$ $\displaystyle =$ $\displaystyle \Gamma_{\; qp}^{q}=\frac{1}{p_{0}} \ \ \ \ \ (10)$

with all other Christoffel symbols equal to zero.

The only independent Riemann tensor component in 2-d is ${R_{\; qpq}^{p}}$ :

 $\displaystyle R_{\; qpq}^{p}$ $\displaystyle =$ $\displaystyle \partial_{p}\Gamma_{\; qq}^{p}-\partial_{q}\Gamma_{\; pq}^{p}+\Gamma_{\; kp}^{p}\Gamma_{\; qq}^{k}-\Gamma_{\; qk}^{p}\Gamma_{\; pq}^{k}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{p}\Gamma_{\; qq}^{p}-0+0-\Gamma_{\; qq}^{p}\Gamma_{\; pq}^{q}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2}{p_{0}^{2}}e^{2p/p_{0}}+\frac{1}{p_{0}^{2}}e^{2p/p_{0}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{p_{0}^{2}}e^{2p/p_{0}} \ \ \ \ \ (14)$

Any non-zero component indicates that the space is curved, so this metric represents a curved space.

Riemann tensor in 2-d flat space

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem P19.5.

Another example of the Riemann tensor in a 2-d space. The metric is

$\displaystyle ds^{2}=dp^{2}+\frac{dq^{2}}{b^{2}q^{2}} \ \ \ \ \ (1)$

where ${b}$ is a constant. The metric tensor is therefore ${g_{pp}=1}$, ${g_{qq}=1/b^{2}q^{2}}$. By comparing the two forms of the geodesic equation, we can calculate the Christoffel symbols.

 $\displaystyle g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \ddot{x}^{a}+\Gamma_{\; ij}^{a}\dot{x}^{j}\dot{x}^{i}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (3)$

With ${a=p}$, we get from 2

$\displaystyle \ddot{p}=0 \ \ \ \ \ (4)$

From 3, we see that ${\Gamma_{\; ij}^{p}=0}$ for all ${i}$ and ${j}$.

With ${a=q}$, we have

 $\displaystyle \frac{1}{b^{2}q^{2}}\ddot{q}-\frac{2}{b^{2}q^{3}}\dot{q}^{2}+\frac{1}{b^{2}q^{3}}\dot{q}^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \ddot{q}-\frac{1}{q}\dot{q}^{2}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Comparing with 3 we find

 $\displaystyle \Gamma_{qq}^{q}$ $\displaystyle =$ $\displaystyle -\frac{1}{q}\ \ \ \ \ (7)$ $\displaystyle \Gamma_{pq}^{q}$ $\displaystyle =$ $\displaystyle \Gamma_{qp}^{q}=\Gamma_{pp}^{q}=0 \ \ \ \ \ (8)$

The only independent component of the Riemann tensor in 2-d is ${R_{\; qpq}^{p}}$ :

 $\displaystyle R_{\; qpq}^{p}$ $\displaystyle =$ $\displaystyle \partial_{p}\Gamma_{\; qq}^{p}-\partial_{q}\Gamma_{\; pq}^{p}+\Gamma_{\; kp}^{p}\Gamma_{\; qq}^{k}-\Gamma_{\; qk}^{p}\Gamma_{\; pq}^{k}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (10)$

since all terms involve components of the form ${\Gamma_{ij}^{p}}$. Therefore, the Ricci tensor is also zero: ${R_{ij}=0}$ as is the curvature scalar ${R=0}$. The space is flat.

Ricci tensor and curvature scalar

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.5, Problem P19.1.

We can form contractions over the indices of the Riemann tensor to get some other useful quantities.

First, we can contract the first and third indices to get the Ricci tensor ${R_{bc}}$:

 $\displaystyle R_{\; bac}^{a}$ $\displaystyle =$ $\displaystyle g^{ad}R_{dbac}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle R_{bc} \ \ \ \ \ (2)$

Using the symmetry relation ${R_{dbac}=R_{acdb}}$ and the symmetry of the metric, we have

$\displaystyle R_{bc}=g^{ad}R_{dbac}=g^{da}R_{acdb}=R_{cb} \ \ \ \ \ (3)$

so the Ricci tensor is symmetric.

We can contract the Ricci tensor in turn to get the curvature scalar ${R}$:

 $\displaystyle R_{\; c}^{b}$ $\displaystyle =$ $\displaystyle g^{ab}R_{ac}\ \ \ \ \ (4)$ $\displaystyle R_{\; b}^{b}$ $\displaystyle =$ $\displaystyle g^{ab}R_{ab}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle R \ \ \ \ \ (6)$

Since the Riemann tensor is identically zero in flat spacetime, the Ricci tensor and curvature scalar are also both zero there. However, although the Riemann tensor always has at least one non-zero component in curved spacetime, the Ricci tensor and curvature scalar can both be zero in curved spacetime. Thus we can say that if the Ricci tensor or the curvature tensor are non-zero, the spacetime is curved, but we can’t draw any conclusions if they are zero; we then need to work out the full Riemann tensor.

One other contraction of the Riemann tensor is over its first and second indices:

$\displaystyle R_{\; abc}^{a}=g^{ad}R_{dabc}=-g^{da}R_{adbc}=0 \ \ \ \ \ (7)$

In the second equation we’ve used the antisymmetry relation ${R_{dabc}=-R_{adbc}}$ and the symmetry of the metric. Thus this contraction doesn’t tell us anything useful.

The Bianchi identity for the Riemann tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.4.

Another relation of the Riemann tensor involves the covariant derivative of the tensor, and is known as the Bianchi identity (actually the second Bianchi identity; the first identity is the symmetry relation ${R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0}$ that we saw earlier). The identity is easiest to derive at the origin of a locally inertial frame (LIF), where the first derivatives of the metric tensor, and thus the Christoffel symbols, are all zero. At this point, we have

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

If the Christoffel symbols are all zero, then the covariant derivative becomes the ordinary derivative

$\displaystyle \nabla_{j}A^{k}\equiv\partial_{j}A^{k}+A^{i}\Gamma_{\; ij}^{k}=\partial_{j}A^{k} \ \ \ \ \ (2)$

Therefore, we get, at the origin of a LIF:

 $\displaystyle \nabla_{k}R_{nj\ell m}$ $\displaystyle =$ $\displaystyle \partial_{k}R_{nj\ell m}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{k}\partial_{\ell}\partial_{j}g_{mn}+\partial_{k}\partial_{m}\partial_{n}g_{j\ell}-\partial_{k}\partial_{\ell}\partial_{n}g_{jm}-\partial_{k}\partial_{m}\partial_{j}g_{\ell n}\right) \ \ \ \ \ (4)$

By cyclically permuting the index of the derivative with the last two indices of the tensor, we get

 $\displaystyle \nabla_{\ell}R_{njmk}$ $\displaystyle =$ $\displaystyle \partial_{\ell}R_{njmk}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{\ell}\partial_{m}\partial_{j}g_{kn}+\partial_{\ell}\partial_{k}\partial_{n}g_{jm}-\partial_{\ell}\partial_{m}\partial_{n}g_{jk}-\partial_{\ell}\partial_{k}\partial_{j}g_{mn}\right)\ \ \ \ \ (6)$ $\displaystyle \nabla_{m}R_{njk\ell}$ $\displaystyle =$ $\displaystyle \partial_{m}R_{njk\ell}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\partial_{m}\partial_{k}\partial_{j}g_{\ell n}+\partial_{m}\partial_{\ell}\partial_{n}g_{jk}-\partial_{m}\partial_{k}\partial_{n}g_{j\ell}-\partial_{m}\partial_{\ell}\partial_{j}g_{kn}\right) \ \ \ \ \ (8)$

By adding up 4, 6 and 8 and using the commutativity of partial derivatives, we see that the terms cancel in pairs, so we get

$\displaystyle \boxed{\nabla_{k}R_{nj\ell m}+\nabla_{\ell}R_{njmk}+\nabla_{m}R_{njk\ell}=0} \ \ \ \ \ (9)$

As usual we can use the argument that since we can set up a LIF with its origin at any non-singular point in spacetime, this equation is true everywhere and since the covariant derivative is a tensor, this is a tensor equation and is thus valid in all coordinate systems. This is the Bianchi identity.

A million visitors

At around 7 AM UK time today physicspages got its one millionth visitor. I don’t know who this was so I can’t give them any prizes, but I’d like to express my appreciation for all the visitors I’ve had over the past 2 or 3 years. I’ve learned a lot writing all the posts and it’s gratifying that others find the information useful as well.

On to 2 million now…

Riemann tensor for surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problem 19.4.

As an example of the Riemann tensor in 2-d curved space we can use our old standby of the surface of a sphere. 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 a sphere, the interval is

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

Note that ${r}$ (the radius of the sphere) is a constant here.

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

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

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

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

With ${a=\phi}$ we have

 $\displaystyle 2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}+r^{2}\sin^{2}\theta\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle 2\cot\theta\dot{\theta}\dot{\phi}+\ddot{\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}$ $\displaystyle =$ $\displaystyle \cot\theta\ \ \ \ \ (9)$ $\displaystyle \Gamma_{\theta\theta}^{\phi}=\Gamma_{\phi\phi}^{\phi}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (10)$

We can use these results to get the Riemann tensor. Unfortunately, in the form ${R_{\; bcd}^{a}}$, the Riemann tensor doesn’t have all the symmetries of the form ${R_{abcd}}$, so if we want the latter form, we need to work out the former form first and then use

 $\displaystyle R_{abcd}$ $\displaystyle =$ $\displaystyle g_{af}R_{\; bcd}^{f}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{af}\left(\partial_{c}\Gamma_{\; db}^{f}-\partial_{d}\Gamma_{\; cb}^{f}+\Gamma_{\; db}^{k}\Gamma_{\; ck}^{f}-\Gamma_{\; cb}^{k}\Gamma_{\; kd}^{f}\right) \ \ \ \ \ (12)$

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.

 $\displaystyle R_{\theta\phi\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\theta f}R_{\;\phi\theta\phi}^{f}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\sin^{2}\theta-\cos^{2}\theta-0+0+\cos^{2}\theta\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\ \ \ \ \ (17)$ $\displaystyle R_{\theta\phi\phi\theta}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\phi\phi\theta}^{\theta}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}-\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}+\Gamma_{\;\theta\phi}^{k}\Gamma_{\;\phi k}^{\theta}-\Gamma_{\;\phi\phi}^{k}\Gamma_{\; k\theta}^{\theta}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{\theta\phi\theta\phi}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -r^{2}\sin^{2}\theta\ \ \ \ \ (21)$ $\displaystyle R_{\phi\theta\theta\phi}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}R_{\;\theta\theta\phi}^{\phi}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(\partial_{\theta}\Gamma_{\;\phi\theta}^{\phi}-\partial_{\phi}\Gamma_{\;\theta\theta}^{\phi}+\Gamma_{\;\phi\theta}^{k}\Gamma_{\;\theta k}^{\phi}-\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\phi}^{\phi}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(-\frac{1}{\sin^{2}\theta}-0+0+\frac{\cos^{2}\theta}{\sin^{2}\theta}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -r^{2}\sin^{2}\theta\ \ \ \ \ (25)$ $\displaystyle R_{\phi\theta\phi\theta}$ $\displaystyle =$ $\displaystyle g_{\phi\phi}R_{\;\theta\phi\theta}^{\phi}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta\left(\partial_{\phi}\Gamma_{\;\theta\theta}^{\phi}-\partial_{\theta}\Gamma_{\;\phi\theta}^{\phi}+\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\phi}^{\phi}-\Gamma_{\;\phi\theta}^{k}\Gamma_{\;\theta k}^{\phi}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{\phi\theta\theta\phi}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\sin^{2}\theta \ \ \ \ \ (29)$

Finally, we can calculate one of the other components to verify that it’s zero.

 $\displaystyle R_{\theta\theta\theta\theta}$ $\displaystyle =$ $\displaystyle g_{\theta\theta}R_{\;\theta\theta\theta}^{\theta}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}\left(\partial_{\theta}\Gamma_{\;\theta\theta}^{\theta}-\partial_{\theta}\Gamma_{\;\theta\theta}^{\theta}+\Gamma_{\;\theta\theta}^{k}\Gamma_{\; k\theta}^{\theta}-\Gamma_{\;\theta\theta}^{k}\Gamma_{\;\theta k}^{\theta}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (32)$

Riemann tensor: counting components in general

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Problems 19.2, 19.3.

We can generalize the method for counting the number of independent components in the Riemann tensor to ${n}$-dimensional spacetime. As before, we know that the first pair and last pair of indices must both consist of different values in order for the component to be (possibly) non-zero. With ${n}$ components to choose from, this gives us ${\binom{n}{2}^{2}=\left[\frac{1}{2}n\left(n-1\right)\right]^{2}}$ components. If we arrange these components in a ${\frac{1}{2}n\left(n-1\right)\times\frac{1}{2}n\left(n-1\right)}$ matrix with the rows and columns labelled by the first and second pairs of indices, respectively, then due to the condition

$\displaystyle R_{nj\ell m}=R_{\ell mnj} \ \ \ \ \ (1)$

the lower triangle of this matrix is a mirror of the upper triangle, so the possible number of independent components is reduced to at most ${\sum_{i=1}^{n\left(n-1\right)/2}=\frac{1}{2}\left(\frac{1}{2}n\left(n-1\right)\right)\left(\frac{1}{2}n\left(n-1\right)+1\right)}$. This gives

$\displaystyle \frac{1}{2}\left(\frac{1}{2}n\left(n-1\right)\right)\left(\frac{1}{2}n\left(n-1\right)+1\right)=\frac{1}{8}n\left(n-1\right)\left[n\left(n-1\right)+2\right] \ \ \ \ \ (2)$

We now need to apply the final symmetry condition, which is

$\displaystyle R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0 \ \ \ \ \ (3)$

As we saw in the last post, this equation gives new constraints only if all four indices are different, and the order in which these indices appear in the first term doesn’t matter. Therefore, this equation provides a total of ${\binom{n}{4}=\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(n-3\right)}$ constraints, so the total number of independent components is

 $\displaystyle N\left(n\right)$ $\displaystyle =$ $\displaystyle \frac{1}{8}n\left(n-1\right)\left[n\left(n-1\right)+2\right]-\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(n-3\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{24}\left(n^{2}-n\right)\left(n^{2}-n+2\right)-\frac{1}{24}\left(n^{2}-n\right)\left(n^{2}-5n+6\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{24}\left(n^{2}-n\right)\left(2n^{2}+2n\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}\left(n^{2}-n\right)\left(n^{2}+n\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}\left(n^{4}-n^{2}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{12}n^{2}\left(n^{2}-1\right) \ \ \ \ \ (9)$

This formula works even if ${n<4}$, since the second term in 4 is zero in this case.

The numbers of independent components for the first few dimensions are

 ${n}$ ${N\left(n\right)}$ 2 1 3 6 4 20 5 50 6 105

\

Riemann tensor: counting independent components

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.3.

The symmetries of the Riemann tensor mean that only some of its components are independent. The two conditions

 $\displaystyle R_{jn\ell m}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}\ \ \ \ \ (1)$ $\displaystyle R_{njm\ell}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m} \ \ \ \ \ (2)$

show that all components where either the first and second indices, or the third and fourth indices are equal must be zero. In four dimensional spacetime, this means that at most ${\binom{4}{2}^{2}=36}$ components can be non-zero, since we can choose two distinct values for both the first and last pairs of indices.

The condition

$\displaystyle R_{nj\ell m}=R_{\ell mnj} \ \ \ \ \ (3)$

means that ${R_{nl\ell m}}$ is symmetric with respect to its two pairs of indices. If we arrange the 36 non-zero components in a ${6\times6}$ matrix where the rows and columns are labelled by the distinct pairs of values 01, 02, 03, 12, 13, 23, then the lower triangle of this matrix is the mirror image of the upper triangle, meaning we can eliminate ${\sum_{i=1}^{5}i=15}$ more components, leaving ${36-15=21}$ possibly independent components.

There is one final symmetry condition for the Riemann tensor, and it is the trickiest to handle.

$\displaystyle R_{nj\ell m}+R_{n\ell mj}+R_{nmj\ell}=0 \ \ \ \ \ (4)$

The first thing we need to show about this condition is that if any two indices are equal, then 4 follows from the other three conditions and tells us nothing new. To see this, consider the various ways in which two indices can be equal.

First, suppose ${n=j}$. Then ${R_{nn\ell m}=0}$ from 1, so we’re left with

 $\displaystyle R_{n\ell mn}+R_{nmn\ell}$ $\displaystyle =$ $\displaystyle R_{n\ell mn}+R_{n\ell nm}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R_{n\ell mn}-R_{n\ell mn}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

The first line uses 3 and the second line uses 2. Because 4 cyclically permutes the last three indices, the same argument also applies to the cases ${n=\ell}$ and ${n=m}$.

Now suppose ${j=\ell}$. Then the third term in 4 is ${R_{nmjj}=0}$ using 2, so we’re left with

$\displaystyle R_{njjm}+R_{njmj}=R_{njjm}-R_{njjm}=0 \ \ \ \ \ (8)$

using 2.

For ${j=m}$ the second term in 4 is ${R_{n\ell jj}=0}$ so

$\displaystyle R_{nj\ell j}+R_{njj\ell}=R_{nj\ell j}-R_{nj\ell j}=0 \ \ \ \ \ (9)$

Finally, if ${\ell=m}$, the first term in 4 is ${R_{njmm}=0}$ and

$\displaystyle R_{nmmj}+R_{nmjm}=R_{nmmj}-R_{nmmj}=0 \ \ \ \ \ (10)$

Now that we know that 4 gives us new information only if all the indices are different, how many ways can we choose these indices? Given that we have four indices, we might think that there are ${4!=24}$ possibilities, depending on the ordering of the indices. In fact, for any set of four indices, the condition gives us only one independent constraint, as the four different indices can be placed in any order in the first term. Starting with 4 with all indices different, suppose we put ${j}$ first instead of ${n}$. We can do this by swapping ${n}$ and ${j}$ to get

 $\displaystyle R_{jn\ell m}+R_{j\ell mn}+R_{jmn\ell}$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}+R_{mnj\ell}+R_{n\ell jm}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{nj\ell m}-R_{nmj\ell}-R_{n\ell mj}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

We used the first three symmetries in the first and second lines and then 4 to get the zero in the third line.

If we swap ${n}$ with ${\ell}$ we get

 $\displaystyle R_{\ell jnm}+R_{\ell nmj}+R_{\ell mjn}$ $\displaystyle =$ $\displaystyle R_{nm\ell j}-R_{n\ell mj}+R_{jn\ell m}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{nmj\ell}-R_{n\ell mj}-R_{nj\ell m}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (16)$

If we swap ${n}$ and ${m}$ we get

 $\displaystyle R_{mj\ell n}+R_{m\ell nj}+R_{mnj\ell}$ $\displaystyle =$ $\displaystyle R_{\ell nmj}+R_{njm\ell}-R_{nmj\ell}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -R_{n\ell mj}-R_{nj\ell m}-R_{nmj\ell}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (19)$

Swapping any two of the last three indices gives the same result, since these three indices are present in a cyclic permutation, so we need to consider only one such case, say swapping ${j}$ with ${m}$:

 $\displaystyle R_{nm\ell j}+R_{n\ell jm}+R_{njm\ell}$ $\displaystyle =$ $\displaystyle -R_{nmj\ell}-R_{n\ell mj}-R_{nj\ell m}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (21)$

Therefore, the condition 4 can give us only ${\binom{4}{4}=1}$ extra constraint. The total number of independent components in four-dimensional spacetime is therefore ${21-1=20}$.

Example As another example, we can apply this reasoning to find the number of independent components in two dimensions. First, the number of possible pairs with distinct values is ${\binom{2}{2}=1}$, so the matrix referred to above is only ${1\times1}$. To verify that the condition 4 doesn’t reduce the number of independent components any further, note that with only 2 possible values for indices, we must repeat at least one of them when choosing the indices in ${R_{nj\ell m}}$, so this condition doesn’t in fact tell us anything new. Thus there is only one independent component in two dimensions.