Spacetime metrics

## Special relativity

Every event takes place in spacetime and thus is labeled by a 4-vector $\left( t , x, y , z \right)$ marking its spatial location and its timestamp with respect to some inertial frame of reference.

Inertial frames have to obey the three postulates:

1. A particle free to move under no forces has constant velocity in any inertial frame

2. The speed of light relative to any inertial frame is $c$ in all directions

3. The geometry of space is Euclidean in any inertial frame.

and Lorentz transformations are designed to respect these defining traits of inertial frames. The spacetime interval is invariant under Lorentz transformations and is given by

(1)
\begin{align} s^2 = - c^2 \left(t- t_0 \right)^2 + \left( x-x_0 \right)^2 + \left( y-y_0 \right)^2 + \left( z-z_0 \right)^2, \end{align}

or if we conveniently agree to choose units where the speed of light is 1, it is just

(2)
\begin{align} s^2 = - \left(t- t_0 \right)^2 + \left( x-x_0 \right)^2 + \left( y-y_0 \right)^2 + \left( z-z_0 \right)^2. \end{align}

Note that if we want to think of a vector in spacetime with its tail at the event $\left( t_0,x_0,y_0,z_0 \right)$ and its head at $\left( t,x,y,z \right)$, the spacetime interval is actually the norm of that vector, taken with the inner product induced by the matrix

(3)
\begin{align} \left( \begin{array}{rrrr} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \end{align}

i.e. the inner product

(4)
\begin{align} \left( v_t , v_x , v_y , v_z \right) \left( \begin{array}{rrrr} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} w_t \\ w_x \\ w_y \\ w_z \end{array} \right) \end{align}

It is because 3 defines an inner product giving Lorentz-invariant lengths that it is called the metric for special relativity. The space $\mathbb{R}^4$ endowed with the metric 3 is called Minkowski space after Hermann Minkowski, and 3 is called the Minkowski metric.

Now, even though Minkowski space does not have a Euclidean metric (that would be all 1's down the main diagonal and 0's elsewhere, yielding the standard dot product), it is still a flat space because the components of its metric do not vary over spacetime (they aren't functions of the coordinates $t , x , y, z$). So although special relativity allows us to think of spacetime as a manifold with a particular group of symmetries, it considers the geometry of the manifold (Minkowski space) to be unaffected by the presence of matter or energy.

## General relativity

This is what changes in general relativity (GR). The big idea is that the metric is actually a dynamical object; i.e. that the metric components should be allowed to vary as functions of the space and time coordinates, and that they are dynamically determined by the presence of matter and energy. Thus GR takes place not on a flat but a curved manifold.

One thing from special relativity will carry over into general relativity, and that is what's called the "signature" of the metric. The Minkowski metric 3 is written in diagonal form, so we easily see that its eigenvalues are -1, 1, 1, 1, with -1 being the time eigenvalue. In GR the metrics might not be diagonal (they will always be symmetric, but not necessarily diagonal), but it will always be the case that when we find the eigenvalues, there will be one negative eigenvalue for time and three positive ones for the spatial directions. We say that the spacetimes of GR have "Lorentzian signature" by which we mean metrics with eigenvalues of the signs (-+++).

Note that once we upgrade to considering curved spacetime, we can no longer think of vectors as pointing from one point to another (they wouldn't lie flat!). So instead we switch to thinking about vectors tangent to the manifold at a given point. For example, a 2-d sphere $S^2$ is a curved manifold with two angles $\theta \in \left[0, 2 \pi \right]$ and $\phi \in \left[ 0 , \pi \right]$ as its coordinates. At any given latitude and longitude (value of $\theta$ and $\phi$), there is a plane tangent to the sphere, and the tangent vectors are vectors in this plane.

However, thinking of a tangent space in this literal way is only convenient for 2-dimensional surfaces, and we are actually interested in 4-d manifolds! The more abstract way of thinking about tangent vectors is to note that they are in one-to-one correspondence with directional derivatives of functions defined on the manifold. For example in the case of the sphere $S^2$, we can define functions $f: S^2 \to \mathbb{R}$ in terms of the coordinates as $f = f \left( \theta , \phi \right)$. At a particular point $\left( \theta_0 , \phi_0 \right)$, the directional derivative of $f$ corresponding to the vector $\mathbf{v} = \left( v^1 , v^2 \right)$ in the tangent plane to the sphere is:

(5)
\begin{align} \left( v^1 \partial_{\theta} f + v^2 \partial_{\phi} f \right) \bigg|_{\theta = \theta_0 , \phi = \phi_0}, \end{align}

so we can think of the tangent vector $\mathbf v$ as represented by the differential operator $v^1 \partial_{\theta} + v^2 \partial_{\phi}$ evaluated at the point $\left( \theta_0 , \phi_0 \right)$. This also makes it easy to think of vector fields defined over the sphere, because we just let the components $v^1, v^2$ be functions of $\theta$ and $\phi$ (of course, we need $v^i \left( 0 , \phi \right) = v^i \left( 2 \pi , \phi \right)$ and $v^i \left( \theta , 0 \right) = v^i \left( \theta , \pi \right)$).

For a manifold $M$, denote the space of tangent vectors (linear first-order differential operators) at the point $p$ by $T_p M$. Denote vector fields on $M$ by $TM$. As is evident in the $S^2$ example, the coordinate derivatives $\partial_{\theta}, \partial_{\phi}$ (and in any other example, however many more coordinate derivatives there are) form a basis for the tangent vectors. For tangent vector fields, of course, the coordinate derivatives form a basis not over the real numbers but over infinitely differentiable functions $C^{\infty} \left( M \right)$ - i.e. in the $S^2$ example, the tangent vector fields are of the form $v^1 \left( \theta , \phi \right) \partial_{\theta} + v^2 \left( \theta , \phi \right) \partial_{\phi}$.

The metric of general relativity is an inner product on these tangent vectors. In order to give you the full effect, we also need not only this new abstract definition of tangent vectors, but also the definition of forms.

A 1-form, defined at a point on a manifold, is a linear function from the vectors in the tangent space at that point to the real numbers $\omega : T_p M \to \mathbb{R}$. So in the example of $S^2$, the 1-forms at the point $\left( \theta_0 , \phi_0 \right)$ are linear functions $\omega : T_{\left( \theta_0 , \phi_0 \right)} S^2 \to \mathbb{R}$ (i.e. functions of the form $\omega \left( \mathbf{v} \right) = \omega_1 v^1 + \omega_2 v^2$).

Differential forms are linear functions from vector fields on a manifold $M$ to the real numbers, $\omega : TM \to \mathbb{R}$. (In this context, linearity is over infinitely differentiable functions on the manifold $C^{\infty} \left( M \right)$, so that for instance in the $S^2$ example, the differential forms are (evaluated on a vector field $\mathbf{v} = v^1 \left( \theta , \phi \right) \partial_{\theta} + v^2 \left( \theta , \phi \right) \partial_{\phi}$) like $\omega \left( \mathbf{v} \right) = \omega_1 \left( \theta , \phi \right) v^1 \left( \theta , \phi \right) + \omega_2 \left( \theta , \phi \right) v^2 \left( \theta , \phi \right)$).

Just as the coordinate partial derivatives form a basis of the tangent vectors (and a basis of the vector fields over $C^{\infty}$ functions on M), there is a basis of differential 1-forms coming from the coordinates. These are denoted as (in the $S^2$ example) $d \theta, d \phi$ and defined by (on a vector field $\mathbf{v} = v^1 \left( \theta , \phi \right) \partial_{\theta} + v^2 \left( \theta , \phi \right) \partial_{\phi}$) $d \theta \left( \mathbf{v} \right) = v^1 \left( \theta , \phi \right)$ and $d \phi \left( \mathbf{v} \right) = v^2 \left( \theta , \phi \right)$ (i.e. the basis differential form corresponding to a given coordinate picks out the component of the vector field corresponding to that coordinate).

The differential 1-forms can be multiplied together (this is called a tensor product) to form bilinear forms on two vectors. In terms of the basis 1-forms in the $S^2$ example, the bilinear forms can be written as $\omega_{11} \left( \theta , \phi \right) d \theta ^2 + \omega_{12} d \theta d \phi + \omega_{21} d \phi d \theta + v_{22} d \phi^2$. These 2-forms act on two vector fields $\mathbf{v} = v^1 \left( \theta , \phi \right) \partial_{\theta} + v^2 \left( \theta , \phi \right) \partial_{\phi}$ and $\mathbf{x} = x^1 \left( \theta , \phi \right) \partial_{\theta} + x^2 \left( \theta , \phi \right) \partial_{\phi}$ as $\omega \left( \mathbf{v} , \mathbf{x} \right) = \omega_{11} v^1 x^1 + \omega_{12} v^1 x^2 + \omega_{21} v^2 x^1 + \omega_{22} v^2 x^2$. Here the $\omega_{ij}$ are functions of $\theta$ and $\phi$ (as are the components of $\mathbf{v}$ and $\mathbf{x}$)

Note that this is exactly the formal structure we need to define a metric on a curved space. Imposing the stipulation of symmetry (i.e. $g_{\mu \nu} = g_{\nu \mu}$), we get a symmetric bilinear form $g$ on vector fields. In terms of coordinates on a given manifold $M$, the vector fields can be expressed as $v^{\mu} \partial_{\mu}$. Note two new notations here: one is that we use Greek sub/superscripts to index the coordinates on $M$, which is a standard convention in physics when working on a manifold endowed with a metric of Lorentzian signature (as before, having eigenvalues of signs (-+++)). Also, in the expression for the tangent vector field $v^{\mu} \partial_{\mu}$ there is a summation over the index $\mu$, but we have dropped this and agree to view summation over repeated indices as understood. This is called the Einstein summation convention. So now the inner product of vector fields $\mathbf{v} = v^{\mu} \partial_{\mu}$ and $\mathbf{x} = x^{\nu} \partial_{\nu}$ is $g \left( \mathbf{v} , \mathbf{x} \right) = g_{\mu \nu} v^{\mu} x^{\nu}$. * Exercise: Use the Einstein summation convention and the fact that $g_{\mu \nu} = g_{\nu \mu}$ to show that $g \left( \mathbf{v} , \mathbf{x} \right) = g \left( \mathbf{x} , \mathbf{v} \right)$.

The metrics used in general relativity are of the form $g \left( \mathbf{v} , \mathbf{x} \right) = g_{\mu \nu} v^{\mu} x^{\nu}$, with $\mu$ and $\nu$ running from 0 to 3 ($x^0$ is the timelike coordinate, and $x^i, i=1,2,3$ are the spacelike coordinates). It is symmetric, $g_{\mu \nu} = g_{\nu \mu}$, and has Lorentzian signature so as to distinguish the timelike direction from the spacelike directions.

page revision: 11, last edited: 05 Apr 2014 15:48