- #31
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The discussion seems to overcomplicate things. What you want to solve is the Killing equation
$$\partial_{\mu} V_{\nu} + \partial_{\nu} V_{\mu}=0.$$
The trick to find the general solutions is to first take one more derivative of this equation:
$$\partial_{\rho} \partial_{\mu} V_{\nu} + \partial_{\rho} \partial_{\nu} V_{\mu}=0.$$
Now write down also the other two equations obtained from this by simply cyclically permute the indices:
$$\partial_{\mu} \partial_{\nu} V_{\rho} + \partial_{\mu} \partial_{\rho} V_{\nu}=0,\\
\partial_{\nu} \partial_{\rho} V_{\mu} + \partial_{\nu} \partial_{\mu} V_{\rho}=0.$$
Now add the first two and subtract the last equation. Using that partial derivatives commute, this leads to
$$\partial_{\rho} \partial_{\mu} V_{\nu}=0.$$
This means all 2nd deviatives of ##V_{\nu}## vanish, so that it must be a linear equation of ##x^{\mu}##:
$$V_{\nu}=T_{\nu} + \omega_{\nu \mu} x^{\mu}.$$
However, the 1st-order equation is more restrictive than the 2nd-order equation. So we must check this result with the first-order equation. Indeed
$$\partial_{\rho} V_{\nu} + \partial_{\nu} V_{\rho}=\omega_{\nu \rho} + \omega_{\rho \nu} \stackrel{!}{=} 0.$$
This implies that
$$\omega_{\mu \nu}=-\omega_{\nu \mu}.$$
Now remember that ##V_{\mu}## describes infinitesimal transformations of the coordinates that are symmetries. It's obvious that the ##T_{\nu}## define spacetime translations and the ##\omega_{\mu \nu}## Lorentz transformations (i.e., boosts and rotations and any combination thereof), and these are the well-known symmetries of Minkowski spacetime.
$$\partial_{\mu} V_{\nu} + \partial_{\nu} V_{\mu}=0.$$
The trick to find the general solutions is to first take one more derivative of this equation:
$$\partial_{\rho} \partial_{\mu} V_{\nu} + \partial_{\rho} \partial_{\nu} V_{\mu}=0.$$
Now write down also the other two equations obtained from this by simply cyclically permute the indices:
$$\partial_{\mu} \partial_{\nu} V_{\rho} + \partial_{\mu} \partial_{\rho} V_{\nu}=0,\\
\partial_{\nu} \partial_{\rho} V_{\mu} + \partial_{\nu} \partial_{\mu} V_{\rho}=0.$$
Now add the first two and subtract the last equation. Using that partial derivatives commute, this leads to
$$\partial_{\rho} \partial_{\mu} V_{\nu}=0.$$
This means all 2nd deviatives of ##V_{\nu}## vanish, so that it must be a linear equation of ##x^{\mu}##:
$$V_{\nu}=T_{\nu} + \omega_{\nu \mu} x^{\mu}.$$
However, the 1st-order equation is more restrictive than the 2nd-order equation. So we must check this result with the first-order equation. Indeed
$$\partial_{\rho} V_{\nu} + \partial_{\nu} V_{\rho}=\omega_{\nu \rho} + \omega_{\rho \nu} \stackrel{!}{=} 0.$$
This implies that
$$\omega_{\mu \nu}=-\omega_{\nu \mu}.$$
Now remember that ##V_{\mu}## describes infinitesimal transformations of the coordinates that are symmetries. It's obvious that the ##T_{\nu}## define spacetime translations and the ##\omega_{\mu \nu}## Lorentz transformations (i.e., boosts and rotations and any combination thereof), and these are the well-known symmetries of Minkowski spacetime.
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