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.

  • 3
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 7, 2017 at 21:04

1 Answer 1


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.


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