# How to find a curvature of the space-time by having $g^{\alpha \beta}$ in the following case without cumbersome calculations?

The metric tensor for Fock-Lorentz space-time, $$\mathbf r_{||}{'} = \frac{\gamma (u)(\mathbf r_{||} - \mathbf u t)}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r) + \lambda c^{2} (1 - \gamma (u))t + 1},$$ $$\mathbf r_{\perp}{'} = \frac{\mathbf r_{\perp}}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r) + \lambda c^{2} (1 - \gamma (u))t + 1},$$ $$t' = \frac{\gamma (u)(t - \frac{(\mathbf u \cdot \mathbf r)}{c^{2}})}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r ) + \lambda c^{2} (1 - \gamma (u))t + 1},$$ is given by $$g^{\alpha \beta} = \begin{bmatrix} \frac{1}{(1 + c^{2}\lambda t)^{4}} & \frac{c\lambda x}{(1 + c^{2}\lambda t)^{3}} & \frac{c\lambda y}{(1 + c^{2}\lambda t)^{3}} & \frac{c\lambda z}{(1 + c^{2}\lambda t)^{3}} \\ \frac{c\lambda x}{(1 + c^{2}\lambda t)^{3}} & -\frac{1}{(1 + c^{2}\lambda t)^{2}} & 0 & 0 \\ \frac{c\lambda y}{(1 + c^{2}\lambda t)^{3}} & 0 & -\frac{1}{(1 + c^{2}\lambda t)^{2}} & 0 \\ \frac{c\lambda z}{(1 + c^{2}\lambda t)^{3}} & 0 & 0 & -\frac{1}{(1 + c^{2}\lambda t)^{2}} \end{bmatrix},$$

where $c, \lambda$ are constant.

Is there a slick way to find the Ricci scalar without cumbersome calculations with Christoffel symbols (if at all possible)?

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Does "the scalar" refer to a particular curvature invariant such as the Ricci scalar? Personally I'd just crank out the calculation using a computer algebra system. There are good open-source systems that can do this problem. I'd use maxima with ctensor. – Ben Crowell Mar 31 '13 at 20:01
I read that a curvature (Ricci scalar) for this case is equal to $\frac{1}{\lambda c}$. It's very simple answer (for a homogeneous space-time with constant curvature). Maybe, there is some simple method... – user8817 Mar 31 '13 at 21:24
A reference for the metric would also be nice... – Alex Nelson Apr 1 '13 at 0:14
@PhysiXxx: honestly, if the answer is that simple, you should be able to do a coordinate change and have it obviously be something like hyper-spherical coordinates or de Sitter space or something. Right away, looking at that, I can see that if you go to spherical coordinates, you won't have $g^{t\phi}$ or $g^{t\theta}$ terms. I'd also do some sort of change of coordinates on your $t$ coordinate to get rid of or simplify those $\left(1 + \lambda t\right)^{n}$ terms – Jerry Schirmer Apr 1 '13 at 1:10
And also, please stop carrying those factors of c around. They just clutter your notation – Jerry Schirmer Apr 1 '13 at 1:39

Cleaning up the notation a bit by rescaling coordinates to get rid of $c$ and $\lambda$ and pulling out a common factor gives: $$g^{\alpha \beta} = \frac{1}{(1 + t)^{2}} \begin{bmatrix} \frac{1}{(1 + t)^{2}} & \frac{x}{1 + t} & \frac{y}{1 + t} & \frac{z}{1 + t} \\ \frac{x}{1 + t} & -1 & 0 & 0 \\ \frac{y}{1 + t} & 0 & -1 & 0 \\ \frac{z}{1 + t} & 0 & 0 & -1 \end{bmatrix}.$$

You can use the properties of the Ricci scalar under conformal transformations (google them) to forget about the overall factor by performing a conformal (Weyl) rescaling. You can change time coordinates $t\to\tau$ by integrating

$$\mathrm{d}\tau = (1+t) \mathrm{d}t,\ \partial_\tau = \frac{1}{1+t} \partial_t.$$

This removes $t$ completely from the conformally rescaled metric:

$$\tilde{g}^{\alpha \beta} = \begin{bmatrix} 1 & x & y & z \\ x & -1 & 0 & 0 \\ y & 0 & -1 & 0 \\ z & 0 & 0 & -1 \end{bmatrix}.$$

Then going to spherical coordinates (scroll to the bottom of the page for the relevant formulae) simplifies the off diagonal part (check this, I haven't been careful!):

$$\tilde{g}^{\alpha \beta} = \begin{bmatrix} 1 & r & 0 & 0 \\ r & -1 & 0 & 0 \\ 0 & 0 & -\frac{1}{r^2} & 0 \\ 0 & 0 & 0 & -\frac{1}{r^2 \sin^2 \theta} \end{bmatrix}.$$

You can do some more coordinate transformation mixing $r$ and $\tau$ to diagonalise the metric if you want but I'm getting tired of this. It is straightforward now to compute the Ricci scalar for this metric (and significantly simpler than the original form of the metric). You can probably look up formulae for the curvature tensors of metrics in this form.

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Why there is no multipliyer $/frac{1}{1 + t}^{2}$ near the tensor after time coordinate change? – user8817 Apr 1 '13 at 10:33
You mean the overall factor out the front? I got rid of that by a conformal or "Weyl" transformation of the metric. This is not a coordinate transformation, it is an actual physical change of the metric to get rid of an overall factor. The various curvature tensors have known properties under conformal transformations, so you can often use these to simplify problems. (The notation on the wiki page is a little odd to me, hopefully I can find an easier reference for you.) – Michael Brown Apr 1 '13 at 11:46
Are you sure, that "Weyl" transformation can simplify finding scalar curvature? The expression for $g^{\mu \nu}$ will be simplified, but the expression for Ricci tensor will be complicated, isn't it? – user8817 Apr 1 '13 at 20:24

That metric is diagonal in the spatial coordinates, and its constant-time sections are conformally flat. It is easy enough to derive what the curvature of a conformally flat metric is in terms of the conformal factor.

Once you have this, plug this solution into the ADM equations. You can find references that let you reconstruct the Einstein equations from the ADM equations, which obviously then gives you the Ricci scalar. In particular, $\rho + \gamma^{ab}S_{ab}$, where $\rho$ is the matter density, $\gamma_{ab}$ is your 3-metric, and $S_{ab}$ is the 3-pressure defined by $\gamma_{ac}\gamma_{bd}T^{cd}$ will give you a quantity proportional to the Ricci scalar.

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