4
$\begingroup$

In my lectures, an isometry of the metric is introduced as follows:

A flow on a manifold $M$ is a one-parameter family of differomorphisms $\sigma_t:M \to M$. The flow is said to be an isometry if the metric looks the same at each point along a given flow line.

This definition is very intuitive and makes quite a lot of sense but the mathematical definition is less intuitive:

Mathematically, this means that an isometry satisfies $$ \mathcal{L}_K (g) = 0 \iff \nabla_\mu K_\nu + \nabla_\nu K_\mu = 0 $$ where $ K^\mu = \frac{dx^\mu}{dt}$ is the vector field on $M$ that is tangent to the flow at every point and $\mathcal{L}_K$ is the Lie Derivative of the metric.

I understand how the LHS expression is mathematically equivalent to the RHS expression, but I am not so sure about the physical significant of either expression. In particular, I am curious if there is any physical significance or interpretation to the vanishing of a Lie Derivative (of tensors or vectors)?

$\endgroup$
  • $\begingroup$ The vector field $K$ acts as an infinitesimal isometry. If $K$ is restricted to a geodesic $\gamma$, $K_{\gamma}$ then $<\gamma^{'},K>$ is constant along $\gamma$ - and $K_{\gamma}$ is also a Jacobi field. $\endgroup$ – Cinaed Simson Nov 22 '19 at 23:09
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you very much for the insightful answer! My lecturer did not introduce nor interpret the Lie Derivative in the way you have described, which is way more intuitive than the way I have been taught. $\endgroup$ – alfred Nov 23 '19 at 20:11
0
$\begingroup$

The geometric/physical meaning is that the components of the metric tensor do not change along the direction of the field $K$. Such a field is called a Killing vector.

An isometry of manifolds is something else and there is no preferred direction.

I do not know why your lecturer uses this non standard terminology

$\endgroup$
  • $\begingroup$ A vector field being Killing, implies that the diffeomorphisms descirbed by its flow are isometries in the sense meant in the Wikipedia link. $\endgroup$ – mmeent Nov 22 '19 at 23:38
  • $\begingroup$ And there's no error the instructors description of the Killing vector field. $\endgroup$ – Cinaed Simson Nov 23 '19 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.