# What is the physical significance/interpretation of a vanishing Lie Derivative?

Imagine you have a vector field $\xi^\mu$. Imagine you have a coordinate system $x^\mu$ and want to define a new coordinate system

$$ x'^\mu = x^\mu + \varepsilon \xi^\mu $$ which we can also write as $$ x^\mu = x'^\mu - \varepsilon \xi^\mu $$ where $\varepsilon$ is tiny. What is the metric in the $x'^\mu$ frame? Well, $$ g'(x')_{\mu \nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu}g(x)_{\alpha \beta} $$ and $$ \frac{\partial x^\alpha}{\partial x'^\mu} = \delta^\alpha_\mu - \varepsilon \partial_\mu\xi^\alpha $$ so, only keeping terms to the first order in $\varepsilon$, \begin{align*} g'(x + \varepsilon \xi)_{\mu \nu} &= (\delta^\alpha_\mu - \varepsilon \partial_\mu\xi^\alpha)(\delta^\beta_\nu - \varepsilon \partial_\nu\xi^\beta)g(x)_{\alpha \beta} \\ g'(x)_{\mu\nu} + \varepsilon \xi^\rho \partial_\rho g_{\mu \nu}(x)&= g(x)_{\mu \nu} - \varepsilon \partial_\mu\xi^\alpha g(x)_{\alpha \nu} - \varepsilon \partial_\nu\xi^\beta g(x)_{\mu \beta} \end{align*}

Therefore, under this infinitesimal diffeomorphism of translation along the vector field, we can see that the change in the metric at point $x$ is \begin{align*} \delta g_{\mu \nu} &= \tfrac{1}{\varepsilon}(g'(x)_{\mu \nu} - g(x)_{\mu \nu}) \\ &= -\xi^\rho \partial_\rho g_{\mu \nu} - \partial_\mu\xi^\alpha g_{\alpha \nu} - \partial_\nu\xi^\beta g_{\mu \beta}. \end{align*}

Next, note that the Lie derivative of the metric along $\xi$ is \begin{align*} \mathcal{L}_\xi g_{\mu \nu} &= \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu \\&= g_{\nu \alpha} \nabla_\mu \xi^\alpha + g_{\beta \mu} \nabla_\nu \xi^\beta \\ &= \big( g_{\nu \alpha} \partial_\mu \xi^\alpha + \Gamma_{\nu \mu \rho} \xi^\rho \big) +\big( g_{\beta \mu} \partial_\nu \xi^\beta + \Gamma_{\mu \nu \rho }\xi^\rho\big) \\ &= g_{\nu \alpha} \partial_\mu \xi^\alpha + g_{\beta \mu} \partial_\nu \xi^\beta + \big( \Gamma_{\nu \mu \rho} + \Gamma_{\mu \nu \rho } \big) \xi^\rho \\ &= g_{\nu \alpha} \partial_\mu \xi^\alpha + g_{\beta \mu} \partial_\nu \xi^\beta + \partial_\rho g_{\mu \nu} \xi^\rho \\ &= -\delta g_{\mu \nu} \end{align*}

So we have now seen the interpretation: the Lie derivative of the metric along the vector field is the infinitesimal change of the metric under the tiny diffeomorphism of a translation along that vector field.

In fact, this is in general intepretation of the Lie derivative of any tensor, it's just the change in the tensor under the tiny diffeomorphism $x^\mu \mapsto x^\mu + \varepsilon \xi^\mu$.

For a concise introduction, consult section 1.4 of Eric Poisson's book "A Relativist's Toolkit." Lie differentiation is actually a more "primative" operation than covariant differentiation. You don't need a metric or Christoffel symbols to define it. However, it can also be written in terms of covariant derivatives, which is fortunate. When $\xi$ is a "symmetry" of our metric, then the metric doesn't change under this transformation.