Energy Momentum Tensor and Conserved Current

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.

$$\nabla_\mu e_\nu+ \nabla_\nu e_\mu=0,$$ 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 $T^{\mu\nu}=T^{\nu\mu}$.
• 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 $g_{mu\nu}$ – mike stone Aug 30 '18 at 16:48