General Relativity

From Smithnet

1 Introduction

Einstein's General Relativity is a theory of gravity which is based on the Equivance Principle that asserts that a uniform gravitational field (next to an infinite flat massive body) is indistinguishable from a uniform accelleration.

$\nabla \cdot \mathbf{A} = - 4 \pi G\rho$

2 Coordinate Systems

We will use the Einstein Notation. Consider Cartesian space in $\mathbb{R}^2$ , covered by Cartisian coordinates $(x^1, x^2)\,$ . Pythagorous can be used to give the squared differential displacement:

$ds^2 = (dx^1)^2 + (dx^2)^2\,$

Now consider a general curvilinear coordinate system (CS) $(y^1, y^2\,)$ . Since the lines of the CS are neither parallel nor perpendicular, we form the same quantity:

$ds^2=g_{11}(dy^1)^2 + 2g_{12} dx^1 dx^2 + g_{22}(dy^2)^2\,$

In general the metric tensor $\mathbf{g}\,$ is a function of position. The terms $g_{11}\,$ and $g_{22}\,$ describe the relative spread of the axes, while $g_{12}\,$ describes how the axes are not perpendicular.

A particular geometry is called Flat if it can be described by a Euclidian geometry. In general, curved manifolds (such as the surface of a sphere) require a more complex metric.

Example: Metric Tensor in Polar Coordinates

Consider polar coordinates in Euclidean space:

$x = r \cos (\theta),\;y = r \sin (\theta)$

The differential distance is given by:

$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 = \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2\,$

which shows the components of the metric tensor are:

$g_{rr} = 1,\;g_{\theta\theta} = r^2,\; g_{r\theta} = 0$

Transformation

Now consider a general transformation between two general CS $x,$ and $y,$ , and a scalar field $φ,$ . The scalar field does not depend on the CS used, thus:

φ(y) = φ(x);

.

Now,

Contravariant vector transformation:

$V^n(y) = \frac{\partial y^n}{\partial x^m} V^m(x)$

Covariant vector transformation:

$V_n(y) = \frac{\partial x^m}{\partial y^n} V_m(x)$

Derivative of a Tensor

Scalar

Consider a scalar field $\phi\,$ . Clearly the value of the field at any point P is independent of the CS used to label P. If the field is constant on one CS, will be constant in any CS. Note that:

$\frac{\partial \phi}{\partial x^m}$

is a vector.

Vector

Now consider a constant vector field $\mathbf{V}$ in flat space, in the Cartesian CS. The covariant components of $\mathbf{V}$ are also constant - that is, the projections on the axes:

$\frac{\partial V_m}{\partial x^n} = 0$ .

If the vector field is now labelled by a general curvilinear CS, we see that the components of the vectors now change over position, even if the vector is constant, becasue the CS is not "constant":

$\frac{\partial V_m}{\partial x^n} \neq 0$ .

Moreover:

$\frac{\partial V_m(x)}{\partial x^n} = 0 \nRightarrow \frac{\partial V_m(y)}{\partial y^n} = 0$

which states that vectors do not transform as tensors under a CS change.

Covariant Derivative

We would like to find a new definition of a derivative that maps a vector to a vector, and a tensor to a tensor. "Covariant" in this context means the quantity is independent of the CS chosen. Suppose there exists:

$T_{mn}(x) = \frac{\partial V_m}{\partial x^n(x)}$

So:

$T_{mn}(y) = \frac{\partial x^r}{\partial y^m} \frac{\partial x^s}{\partial y^n} T_{sr}(x) = \frac{\partial x^r}{\partial y^m} \frac{\partial x^s}{\partial y^n} \frac{\partial V_r(x)}{\partial x^s}$

Now compare with (expanding):

$\frac{\partial V_m(y)}{\partial y^n} = \frac{\partial}{\partial y^n} \frac{\partial x^r}{\partial y^m} V_r(x) = \frac{\partial x^r}{\partial y^m} \frac{\partial V_r(x)}{\partial y^n} + \frac{\partial}{\partial y^n} \frac{\partial x^r}{\partial y^m} V_r(x)$

The first term is equal to $T_{mn}(y)\,$ , but we have a second term, which we denote by $\Gamma^r{}_{mn}\,$

Hence in gernal we need to differentiate in a way to cancel the second term to get back a tensor:

$\nabla_p T_{mn} = \frac{\partial T_{mn}}{\partial y^p} + \Gamma^r{}_{pm} T_{rn} + \Gamma^r{}_{pm} T_{mr}$

The Christoffel Symbols can be found from the metric tensor:

$\Gamma^a{}_{bc}(y) = \frac{1}{2} g^{ad} \left ( \frac{\partial g_{dc}}{\partial y^b} + \frac{\partial g_{db}}{\partial y^c} - \frac{\partial g_{bc}}{\partial y^d} \right )$

Parallel Transport and Curvature

Space is called "flat" (like Cartesian space) when it is possible to choose a CS such that the covariant derivavtive of the metric tensor is everywhere zero:

$\nabla_r g^{mn} = 0$ that is the metric tensor reduces to the delta function: $g_{mn} = \delta_{mn}\,$

A space may be flat in some areas, but there may be curvature at other points. It is always possible to choose a CS such that at a given point, the derivavtive of the metric is zero - that is in, a small enough neighbourhood of any point of a continuous, differentiable space looks flat.

Consider a vector V moving from point A to B over a curve, whilst remaining parallel to itself. This is notion of parallel transport, and can be extended to general nonflat spaces. The covariant derivavtive:

$\nabla_s V^m = \frac{dV^m}{ds} + \Gamma^m{}_{np} V^n \frac{dx^p}{ds}$

If a vector is parallel transported such that the covariant derivative is always zero, the vector follows a geodesic; geodesics are in some sense the "straightest" lines that exist in the manifold. For example: all straight lines in flat 2D space are geodesics; on the surface of a sphere, all great circles are geodesics.

The geodesic condition is:

$\frac{d^2x^m}{ds^2} + \Gamma^m{}_{np} \frac{dx^n}{ds} \frac{dx^p}{ds} = 0$

General Relativity

From Smithnet

Contents

1 Introduction

2 Coordinate Systems

Example: Metric Tensor in Polar Coordinates

Transformation

Derivative of a Tensor

Scalar

Vector

Covariant Derivative

Parallel Transport and Curvature

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