Conservation Law from Metric

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Main Question or Discussion Point

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

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

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?
 

Answers and Replies

  • #2
haushofer
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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
pervect
Staff Emeritus
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Consider the following metric for a 2D spacetime:
##g_{tt} = -x ##
##g_{tx} = g_{xt} = 3##
##g_{xx} = 0##

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

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?
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).
 

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