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I Conservation Law from Metric

  1. May 22, 2016 #1
    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?
     
  2. jcsd
  3. May 22, 2016 #2

    haushofer

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    Science Advisor

    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.
     
  4. May 22, 2016 #3

    pervect

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    Staff Emeritus
    Science Advisor

    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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