# How does a two-tensor transform under an infinitesimal shift?

This is a follow-up to this question I posted yesterday: How does a vector field transform under an infinitesimal coordinate transformation?

If I have an infinitesimal coordinate shift of the form $x^{\mu} \to x^{\mu} + v^{\mu}(x)$, The the vector field $A^{\mu}(x)$ will transform like: $$A^{\mu}(x) \to A^{\mu}(x) + \mathcal{L}_{v}(A^{\mu})(x)$$ where $\mathcal{L}_{v}(A^{\mu}) = v^{\nu} \partial_{\nu}A^{\mu} - A^{\nu} \partial_{\nu} v^{\mu}$ is the Lie derivative of $A$ wrt $v$.

If I build a field strength tensor $F^{\mu\nu} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$, is it true that this tensor transforms to the following object? $$\to \partial^{\mu} \left( A^{\nu} + v^{\rho} \partial_{\rho}A^{\nu} - A^{\rho} \partial_{\rho} v^{\nu} \right) - \partial^{\nu} \left( A^{\mu} + v^{\rho} \partial_{\rho}A^{\mu} - A^{\rho} \partial_{\rho} v^{\mu} \right)$$

I think this is the case, but I'd like to confirm this. It's pretty messy.

• Would Mathematics be a better home for this question? Feb 7 '17 at 21:04

In components, the Lie derivative of a doubly contravariant tensor such as $F^{\mu\nu}$ is given by $$({\mathcal L}_X F)^{\mu \nu}= X^\lambda \partial_\lambda F^{\mu\nu}- F^{\lambda\nu} \partial_\lambda X^\mu - F^{\mu\lambda}\partial_\lambda X^{\nu}.$$ For doubly covariant tensor we have $$({\mathcal L}_X g)_{\mu \nu}= X^\lambda \partial_\lambda g_{\mu\nu}+ g_{\lambda\nu} \partial_\mu X^\lambda+g_{\mu\lambda}\partial_\nu X^{\lambda}.$$ The genral pattern should now be clear.