# A few tricks for tensor quantities

A rather tedious task in general relativity is the computation of the various tensor quantities. While there are no general methods to make this task easier (except of course for the use of numeric methods), given metrics of specific properties, it is possible to simplify it greatly.

For a start, a fairly well-known property is the fact that given a diagonal metric, its inverse metric is simply its inverse, component-wise.

$$g^{ab} g_{ac} = \delta^b_c \to g^{aa} g_{aa} = 1 \to g^{aa} = (g_{aa})^{-1}$$Now, given a diagonal metric, let's consider the Christoffel symbols :

$${ \Gamma^a }_{bc} = \frac{1}{2} g^{ad} (g_{db,c} + g_{dc,b} - g_{bc,d})$$Then, for a given coordinate $a = x$, we have

$${ \Gamma^x }_{bc} = \frac{1}{2} g^{xx} (g_{xb,c} + g_{xc,b} - g_{bc,x})$$The first two terms are only non-zero if $b$ or $c$ are equal to $x$, and the last term only if $b = c$. This means that the only type of symbols we have are

\begin{eqnarray} { \Gamma^x }_{bb} &=& -\frac{1}{2} g^{xx} g_{bb,x} & (b \neq x)\\ { \Gamma^x }_{bx} &=& \frac{1}{2} g^{xx} g_{xx,b} &(b \neq x)\\ { \Gamma^x }_{xx} &=& \frac{1}{2} g^{xx} g_{xx,x} \end{eqnarray}If both $b$ and $c$ are different from $x$ and each other, the symbol will be zero. For a given $x$, there are $(n-1)$ symbols of the first kind, $(n-1)$ of the second kind, and $1$ of the third kind. This reduce the number of symbols from $n^2 (\frac{n + 1}{2})$ to $n (2n-1)$, or for instance, in $4$ dimensions, from $40$ components to $28$.

We can also note some similarities between the different symbols. That is,

$${ \Gamma^x }_{bb} = - \frac{g^{bb}}{g^{xx}} { \Gamma^b }_{bx} = - \frac{g_{xx}}{g_{bb}} { \Gamma^b }_{bx} $$Occasionally, we find that the metric has the property that, in addition to being diagonal, every component is independent from the coordinate involved. That is, given a coordinate $x$, we have $g_{xx,x} = 0$ (This does not mean that the metric doesn't depend on that coordinate in other components). This is for instance true of the Morris-Thorne metric :

$$ds^2 = -f(l) dt^2 + dl^2 + r^2(l) (d\theta^2 + \sin^2(\theta) d\varphi^2)$$In this case, things simplify further to just the first and second kind of symbols, leaving only $2n(n-1)$ symbols (or $24$ in $4$ dimensions).

While not the actual definition of symmetry, it is very common for metrics to be completely independent of a coordinate, ie, $g_{ab,x} = 0$. There are up to $n$ such symmetries at most, and given a diagonal metric, this leads to, for $x$ a symmetry, the Christoffel symbols of the first and third type to be zero, as well as those of the second type for when $b$ is such a coordinate. We then get, for $p$ such symmetries, that there are $n(n-p)$ symbols of the first type, $n(n-p)$ of the second type, and $(n-p)$ of the third type, for a total of $(2n+1)(n-p)$. This indeed leads to $0$ symbols for $n = p$ (which is simply Minkowski space).

We can indeed check that, for the Morris-Thorne metric, where we have $p = 2$ (the metric is independant from $t$ and $\varphi$), we have up to $18$ components (actually less so than that due to the lack of dependence on $\varphi$ for most components).

Another property one may encounter from the metric is if it is both diagonal, has a symmetry $g_{ab,x} = 0$, and is such that $g_{xx,a} = 0$ for every coordinate $a$. This is for instance true of the Ellis-Bronnikov metric

$$ds^2 = -dt^2 + dl^2 + (l^2 + a^2) (d\theta^2 + \sin^2 \theta d\varphi^2)$$where this is true for the coordinate $t$. In such a case, for $x$ the symmetry, we have that all three types of symbols are zero, which gets rid of quite a lot of components (at this point the spacetime is basically a trivial warped product $ds^2 = dx + h$, and we simply need to get the connection components of the metric $h$, similarly to what happens for

Once we have those symbols, we can tackle the Riemann tensor

$${R^a}_{bcd} = {\Gamma^a}_{bd,c} - {\Gamma^a}_{bc,d} + {\Gamma^a}_{ce} {\Gamma^e}_{bd} - {\Gamma^a}_{de} {\Gamma^e}_{bc}$$Due to its symmetries, there are

$$C_n = \frac{1}{12} n^2(n^2 - 1)$$independent components. There again, we can use properties from the metric to reduce the number of components to compute.

$${R^x}_{bcd} = {\Gamma^x}_{bd,c} - {\Gamma^x}_{bc,d} + {\Gamma^x}_{ce} {\Gamma^e}_{bd} - {\Gamma^x}_{de} {\Gamma^e}_{bc}$$Posted on

*2019-05-14 13:55:45*