# I Conservation Law from Metric

1. May 22, 2016

### Rococo

Consider the following metric for a 2D spacetime:
$g_{tt} = -x$
$g_{tx} = g_{xt} = 3$
$g_{xx} = 0$

i.e.
$$g_{\mu \nu} = \left( \begin{array}{cc} -x & 3\\ 3 & 0 \end{array} \right)$$

Now, since the metric is independent of time (t), there is supposedly a conservation law containing $\frac{dx(\tau)}{d\tau}$ and $\frac{dt(\tau)}{d\tau}$. What is this conservation law, and why does the time independence of the metric imply it?

2. May 22, 2016

### haushofer

To find the isometries of such a spacetime, you have to solve for the Killing equations. The tt-component of this equation should answer your question.

A note: GR in two dimensions is a bit problematic; the Einstein-Hilbert action is a topological term, meaning that the Einstein equations are an identity.

3. May 22, 2016

### pervect

Staff Emeritus
Consider a curve $u^0(\tau) = t(\tau), u^1(\tau) = x(\tau)$. The tangent to this curve is a vector, i.e.

$$\left( \frac{dt}{d\tau}, \frac{dx}{d\tau} \right)$$

is a vector.

Because none of the components of the metric is a function of time, the vector (in the above sense) $\xi^a = (1,0) = dt/d\tau$ is a Killing vector.

The dot product of a tangent vector of a geodesic curve and a Killing vector is a conserved quantity - it is the same everywhere along the geodesic curve. If we let $u^a = (u^0(\tau), u^1(\tau) ) = (t(\tau), x(\tau))$ be a geodesic curve then we can say that the dot product is given by $g_{ab} \xi^a u^b = \xi_b u^b$ is constant. Because $\xi^a = (1,0)$ and $\xi_a = (-x,3)$ we can say that $-x (dt/d\tau) + 3 (dx/d\tau)$ is constant along the curve, i.e. is independent of $\tau$.

As a check, we can consider $(d/d\tau) (-x (dt/d\tau) + 3 (dx/d\tau)) = 0$, we know $d/d\tau$ of a constant is zero. Expanding this out using the chain rule, we get:
$$-\frac{dx}{d\tau} \frac{dt}{d\tau} + -x \frac{dt^2}{d\tau^2} + 3 \frac{dx^2}{d\tau^2} = 0$$

We can derive this from the geodesic equation. We compute the christoffel symbols

$$\Gamma^t{}_{tt} = \frac{1}{6} \quad \Gamma^x{}_{tt} = \frac{x}{18} \quad \Gamma^x{}_{tx} = -\frac{1}{6}$$

and confirm that we can derive the above from the geodesic equations, which in tensor notation are:

$$\frac{d^2 u^i}{d\tau^2} + \Gamma^i{}_{jk} \frac{du^j}{d\tau}\frac{du^k}{d\tau} = 0$$

Tensor notaton is very compact - the non-repeated index i generates two equations, one where we substitute i=0, the other where we substitute i=1. For each of these equations, j and k are repeated indices, so we sum over the repeated indicies, i.e we sum , for all four combinations of j,k, (0,0), (0,1), (1,0), (1,1).