# General relativity

## Curved space-time

Curvature of the four-dimensional space-time is the basis of general relativity. A curved space is difficult to conceive particularly the fourth dimension is peculiar. Einstein calls it t=x4. It seems simpler to consider it as an imaginary number ict where i is the quadratic root of -1 and c the speed of light. Then the space-time has the following four dimensions: (x,y,z,w=ict).

## Riemann coordinates

Understanding of general relativity, like restricted relativity, will be easier by using two dimensions (x, y=ict) instead of four. With this representation, we will have a riemannian instead of pseudo-riemannian space. Cartesian coordinates are the most common reference system. The Earth, being spherical, is not a flat space and the Pythagorean theorem is valid only locally. The cartesian frame changes its orientation from place to place but the law of gravity is the same in Paris or in Valparaiso. The Riemann coordinates are local cartesian coordinates. They are such that the Pythagorean theorem is valid even on a curved surface. It is not necessary to know the transformation from curved coordinates to use them. They are not always suitable, for example, it is necessary to compute the Riemann tensor in Gauss (e.g. spherical) coordinates in order to obtain the Schwarzschild metric.

## The metric

The metric of a euclidean space represents, in the plane, the Pythagorean theorem.

$\mathrm ds^2= \mathrm dx^2 + \mathrm dy^2\,$
The metric of a curved surface is, according to Gauss:

 $\mathrm ds^2= g_{xx} \, \mathrm dx^2 + 2 g_{xy} \,\mathrm dx \mathrm dy + g_{yy} \mathrm dy^2$

where the gij are the coefficients of the metric. Every curved surface may be approximated, locally, by the osculating paraboloid, becoming the tangent plane z=0 when the principal curvatures kx et ky cancel:

$z= \frac12 \left(k_{x} x^2 + k_{y} y^2 \right)$

Indeed, in the frame used, the axes Ox and Oy are in the tangent plane z=0, the origin of the coordinates, x=0, y=0 being at the contact point. The Gauss curvature is, by definition, the product of the principal curvatures:

$K=k_{x}k_{y}= \frac{\partial ^2 z}{\partial x^2} \frac{\partial ^2 z}{\partial y^2}$

In order to be in Riemann coordinates, it remains to orientate the axes Ox and Oy in such a manner that the metric be diagonal (the computation is given in Bernard Schaeffer, Relativités et quanta clarifiés, Publibook, 2007):

$\mathrm ds^2= \mathrm dx^2 + \left[1 - K\left( x^2 + y^2\right)\right] \, \mathrm dy^2$
where K= kxky is the Gaussian curvature. In this expression, we have gxx=1, gxy=0 and

$g_{yy} =1 - K\left( x^2 + y^2\right)$

Riemann tensorIt is not necessary to determine the principal directions to work with the Riemann coordinates since the laws of physics are invariant under a frame change. It is also not necessary to change the scales of the coordinate axes to get a metric with coefficients equal to one. It only assumed that it is always possible to change the coordinates in such a way that the Pythagorean theorem is verified locally, at the contact point, taken as the origin of the coordinates. In Riemann coordinates, all the paraboloids, including the sphere, locally, have the same metric, provided thet have the same Gaussian curvature.

Gauss found a formula of the curvature K of a surface with a computation, complicated in Gaussian coordinates but much simpler in Riemannian coordinates where the curvature and the Riemann tensor are equal (in two dimensions):

$R_{xyxy}= -\frac12 \left(\frac{\partial^2 g_{xx}}{\partial y^2} + \frac{\partial^2 g_{yy}}{\partial x^2}\right)$
Let us check that the Riemann tensor is equal to the total Gauss curvature:

$R_{xyxy}= -\frac12 (0-2K)=K$

We have also, by partial derivation of the coefficients of the metric:

$\frac{\partial ^2 g_{xx}}{\partial x^2} + \frac{\partial ^2 g_{xx}}{\partial y^2}= 0$

The same for gyy

$\frac{\partial ^2 g_{yy}}{\partial x^2} + \frac{\partial ^2 g_{yy}}{\partial y^2}= -4K$

Einstein equations in vacuumWe have obtained a Laplace equation and a Poisson equation.

Einstein's hypothesis is that the curvature of space-time is zero in the vacuum which is thus a flat space. This is true in two dimensions where the Gaussian curvature is zero. In higher dimensions, only the Ricci tensor is zero according to the Einstein equation. In matter, the Ricci tensor is different from zero. We shall not consider this case, here, but it should be considered to describe the universe which contains matter. The Einstein equations are, in the vacuum:

$R_{ik}= 0 \,$

Rik is a complicated function of the various componants of the Riemann tensorRijkl and of the metric gik. The Ricci tensor, like the Riemann tensor dépends only on the coefficients of the metric. The Christoffel symbolshttp://en.wikipedia.org/wiki/Christoffel_symbol are then unnecessary intermediaries. In two dimensions, the Ricci tensor has two components each proportional to the single component of the Riemann tensor. Therefore there is only one Einstein equation in two dimensions:

In two dimensions and in Riemann coordinates, the Riemann tensor is equal to the Gaussian curvature K, which is zero in the vacuum. Then the coefficients of the metric have to satisfy the Laplace equation Δgxx=0 and Δgyy=0. But, in two dimensions, the Laplace equation diverges unless the coefficients of the metric are constants, corresponding to a pseudo-euclidean space. In three and four dimensions, the Ricci tensor has to be zero, the corresponding space is called Ricci flat. The calculation is too complicated to be given here.

## Gravitational waves

Replacing y by ict in the Laplace equation, one obtains the d'Alembert equation of the plane gravitational waves for the coefficients of the metric:

$\frac{\partial ^2 g_{xx}(x,t)}{\partial x^2} - \frac{1}{c^2(x)}\frac{\partial ^2 g_{xx}(x,t)}{\partial t^2} = 0$

$\frac{\partial ^2 g_{tt}(x,t)}{\partial x^2} - \frac{1}{c^2(x)}\frac{\partial ^2 g_{tt}(x,t)}{\partial t^2} = 0$
The gravitational waves have not yet been detected.

## The two-dimensional Laplace equation may be extrapolated in higher spaces with small curvature. In three dimensions, spherical symmetry and time independent metric, the Einstein equations reduce to the radial laplacian:

$\frac{1}{r^2}\frac{\mathrm d}{\mathrm dr}\left(r^2\frac{\mathrm dg_{rr}}{\mathrm dr}\right)= 0$

## Its solution is the Coulomb potential in 1/r:

$\mathrm ds^2= g_{rr} \,\mathrm dr^2 - g_{tt} c^2 \, \mathrm dt^2= \left(A+ \frac{B}{r}\right) \, \mathrm dr^2 -c^2 \left(A'+ \frac{B'}{r}\right) \, \mathrm dt^2$

The correspondence principle with special relativity will give us the integration constants A and A'. For r=∞, we have:

$\mathrm ds^2= A \, \mathrm dr^2 -c^2 A' \, \mathrm dt^2$

It should be the Minkowski metric:

$\mathrm ds^2= -c^2 \, \mathrm d \tau^2= \mathrm dr^2-c^2 \mathrm dt^2$
where dr/dt = v is the velocity of the particle. Identifying these two metrics, we get A=A'=1. To obtain B', we apply the correspondence principle with the newtonian gravitation of a light particle on a circular trajectory around a highly attracting star similar to a black hole. Then dr=0, the metric is simplified:

$\mathrm ds^2= -c^2 g_{tt} \, \mathrm dt^2 = -c^2 \left( 1+ \frac{B'}{R}\right) \, \mathrm dt^2$

For a photon, v=c, ds=0: the length of a light trajectory is zero. It is the shortest way possible. Assuming that this remains true in general relativity, we have the condition:

$1+ \frac{B'}{R}=0$

which gives R=-B'. The trajectory being a circle and the curvature of space small, we may apply newtonian mechanics. The kinetic energy is equal to the newtonian gravitation potential:

$\frac12 mc^2=\frac{GmM}{R}$

where G is the gravitation constant, M the mass of the attracting star and c the speed of light. Replacing R with -B' and v with c, we get:

$B'= -\frac{2GM}{c^2}$

According to Einstein, the determinant (or its trace for low gravitation) of the metric should be equal to one. This can be shown by solving the four-dimensional Einstein equations for a static and spherically symmetric gravitational field. Therefore we may write B=-B' and obtain an approximation of the Schwarzschid metric:

$\mathrm ds^2= \left(1+ \frac{2GM}{c^2r}\right) \, \mathrm dr^2 -c^2 \left(1- \frac{2GM}{c^2r}\right) \, \mathrm dt^2$

This metric gives a light deviation by the sun twice as predicted by the newtonian theory or by the first Einstein theory of 1911 where time is dilated by gravitation. In his 1916 theory, gravitation dilates time and contracts space.