Derivation of Lorentz algebra commutation relation

[spaghetti3451]

Homework Statement

1. Show that the Lorentz algebra generator ##J^{\mu \nu} = i(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})## lead to the commutation relation ##[J^{\mu \nu}, J^{\rho \sigma}] = i(g^{\nu \rho}J^{\mu \sigma} - g^{\mu \rho}J^{\nu \sigma}-g^{\nu \sigma}J^{\mu \rho}+g^{\mu \sigma}J^{\nu \rho})##.

2. Show that one particular representation of the the Lorentz group given by ##(J^{\mu \nu})_{\alpha \beta} = i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})##.

Homework Equations

The Attempt at a Solution

##[J^{\mu \nu}, J^{\rho \sigma}]##

##= J^{\mu \nu}J^{\rho \sigma}-J^{\rho \sigma}J^{\mu \nu}##

##= -(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})+(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})##.

How do I get the metric tensor from here?

 

[fzero]

When you distribute terms you will find various factors like ##\partial^\nu x^\rho##.

 

[spaghetti3451]

How are those factors related to the metric tensor?

As far as I know, a term like ##\partial_{\nu}x^{\rho}## gives the Kronecker delta ##\delta_{\nu}^{\rho}##, not the metric tensor.

 

[fzero]

Now raise the index.

 

[spaghetti3451]

So, you say ##\delta_{\nu}^{\rho}## equals ##g^{\nu \rho}##?

 

[spaghetti3451]

Or, do I use ##\delta_{\nu}^{\rho} = g^{\nu \rho}g_{\nu \rho}##?

 

[fzero]

So, you say ##\delta_{\nu}^{\rho}## equals ##g^{\nu \rho}##?

Not exactly, ##\partial^\nu x^\rho = g^{\nu\mu} \partial_\mu x^\rho##. So use the result for ##\partial_\mu x^\rho## here.