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I have $$j_{\epsilon} = T_{\mu\nu}\epsilon^{\nu}.$$ I need to show $$\nabla_{\mu}j^{\mu} = 0,$$ which I am told is possible via taking into account the Killing equation $$\partial_{\mu}\epsilon_{\nu} + \partial_{\nu}\epsilon_{\mu} = \partial_{\rho}\epsilon^{\rho}g_{\mu\nu}.$$

Since

$$\partial^{\mu}T_{\mu\nu}\epsilon^{\nu} = (\partial^{\mu}T_{\mu\nu})\epsilon^{\nu} + T_{\mu\nu}(\partial^{\mu}\epsilon^{\nu}),$$

my problem is to show $\partial^{\mu}\epsilon^{\nu}$ because $\partial^{\mu}T_{\mu\nu}$ is already zero because the energy-momentum tensor is constant along motions.

My Work:

  1. Multiplied both sides of equation by inverse metric

    $$(\partial_{\mu}\epsilon_{\nu} + \partial_{\nu}\epsilon_{\mu}) g^{\mu\nu}= (\partial_{\rho}\epsilon^{\rho}g_{\mu\nu}) g^{\mu\nu}$$

    which gave

    $$\partial_{\mu}\epsilon_{\nu}g^{\mu\nu} + \partial_{\nu}\epsilon_{\mu} g^{\mu\nu}= \partial_{\rho}\epsilon^{\rho}g_{\mu\nu} g^{\mu\nu}$$

  2. Contracting indices gave

    $$\partial^{\nu}\epsilon_{\nu} + \partial^{\mu}\epsilon_{\mu} = \partial_{\rho}\epsilon^{\rho}*2$$

    where $g_{\mu\nu} g^{\mu\nu}=2$ because we are in 2-D.

  3. Moving term to one side

    $$\partial^{\nu}\epsilon_{\nu} = 2*\partial_{\rho}\epsilon^{\rho} - \partial^{\mu}\epsilon_{\mu} = 2\partial_{\rho}\epsilon^{\rho}- \partial^{\mu}\epsilon_{\mu} = 2\partial_{\rho}\epsilon^{\rho}- \partial_{\mu}\epsilon^{\mu}$$

    where in the last step I flipped the indices.

How do I continue? Please point out any of my mistakes in reasoning or understanding.

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You need covariant derivatives rather than partials. The Killing equation can be written
$$ \nabla_\mu e_\nu+ \nabla_\nu e_\mu=0, $$ This follows from your form of Kiiling equation either by some tedious manipulation with the explicit form of the christoffel symbols, or more quicky by using geodesic coordinates. From my form of Killing, we have $$ \nabla_\mu (T^{\mu\nu}e_\nu) = (\nabla_\mu T^{\mu\nu})e_\nu + T^{\mu\nu}(\nabla_\mu e_\nu) $$ The first term in the last line is zero because

$T^{\mu\nu}$ is conserved, and the last term is zero because $$ T^{\mu\nu}(\nabla_\mu e_\nu)= \frac 12 T^{\mu\nu}(\nabla_\mu e_\nu+ \nabla_\nu e_\mu) $$ where we have taken into accout that $T^{\mu\nu}=T^{\nu\mu}$.

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  • $\begingroup$ I am not sure what you mean by taking it into account that the energy-momentum tensor is symmetric. How does that imply the final term is zero? Plus, I was given the third equation in my question. The right side of the equation is not zero, and they contained partials. What would the equation I was given mean, then? How did you know to use the form above for the Killing equation to obtain the solution? $\endgroup$ – user570877 Aug 30 '18 at 15:14
  • $\begingroup$ I'll amend my answer to make this explicit. $\endgroup$ – mike stone Aug 30 '18 at 15:16
  • $\begingroup$ Thanks Mike, makes sense. What does the form of the Killing equation I was given mean? I was given this with partials and with the non-zero right-hand side. $\endgroup$ – user570877 Aug 30 '18 at 16:01
  • $\begingroup$ It looks like the the Lie derivative of $g$ with respect to the Killing field ${\bf e}$. The Lie derivative is given $({\mathcal L}_{\bf e} g)_{\mu\nu} = e^\alpha\partial_\alpha g_{\mu\nu}+ g_{\alpha \nu}\partial_\mu e^\alpha+g_{\mu \alpha} \partial_\nu e^\alpha$ which has the same tensor character as $g_{mu\nu}$ $\endgroup$ – mike stone Aug 30 '18 at 16:48

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