# Lie derivative of a vector along itself

The Lie derivative for a covariant and contravariant vector is:

$$\mathcal{L}_U V^\mu=U^\nu\nabla_\nu V^\mu- V^\nu\nabla_\nu U^\mu$$

$$\mathcal{L}_U V_\mu=U^\nu\nabla_\nu V_\mu+ V_\nu\nabla_\mu U^\nu$$

Therefore, $$\mathcal{L}_V V^\mu=0$$ using the above definition. My question is: what is the intuition behind this result?

Moreover, when foliating a spacetime in Cauchy surfaces as

$$h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$$

where $$n^\mu$$ is a spacelike normal vector, we get an extra identity of the form

$$n^\alpha \nabla_\beta n_\alpha=0$$

derivating the definition $$n^\alpha n_\alpha=1$$. With this in mind, the computation of $$\mathcal{L}_n n_\alpha$$ gives

$$\mathcal{L}_n n_\alpha=n^\beta\nabla_\beta n_\alpha$$

that is different from zero. Once again, what's the intuition behind this result?

• You should double check your second equation. It’s seems to me that the indices are inconsistent, and that you introduced a vector $n$ without defining it. Jul 13, 2018 at 14:00

When you take the Lie derivative of a vector, you are looking at how it changes as you move along integral curves. Now if you look at $L_UU$ you are asking how does $U$ change along its integral curves. But the point of an integral curve is that it’s tangent is always $U$. So $U$ does not change as you travel along the curve (its always pointing ahead)

• Thanks for the intuition it was very useful. What about the second part of the question? Jul 12, 2018 at 21:34
• @P.G.A. Maybe I’m missing something, but if you raise the index in the LHS of your last equation, you get the Lie derivative of $n^\alpha$ by itself and thus 0, is the right hand side is also 0. Jul 13, 2018 at 13:59

The first kind of Lie derivative is anticommutative, which implies the desired result. In fact, identifying $V^\mu$ with $V^\mu\nabla_\mu$, the Lie derivative is a commutator.

The Lie derivative is a more primitive notion than the covariant derivative $$\nabla$$, since it does not require specification of a connection (although it does require a vector field $$U$$):

$$\mathcal{L}_U V^\mu=U^\nu \partial_\nu V^\mu- V^\nu \partial_\nu U^\mu \\ \mathcal{L}_U V_\mu=U^\nu\partial_\nu V_\mu+ V_\nu\partial_\mu U^\nu$$

However, it is sometimes useful to write it in terms of the covariant derivative, see Why can I use the Covariant Derivative in the Lie Derivative?.

Consider the case where $$U=u$$, $$V^\mu =u^\mu$$ and $$u_\mu u^\mu = 1$$. Hence, since $$u_\mu u^\mu$$ is a scalar (that is also a constant field),

$$\partial_\beta ( u_\mu u^\mu) = \nabla_\beta ( u_\mu u^\mu) = u_\mu \nabla_\beta u^\mu = 0 ,$$

where we used the fact that the covariant derivative is defined to obey the Leibnitz rule. Now, you have:

$$\mathcal{L}_u u^\alpha = u^\nu \partial_\nu u^\alpha - u^\nu \partial_\nu u^\alpha = 0$$

and

$$\mathcal{L}_u u_\alpha = u^\nu\partial_\nu u_\alpha + u_\nu \partial_\alpha u^\nu = u^\nu \nabla_\nu u_\alpha + u_\nu \nabla_\alpha u^\nu = u^\nu\nabla_\nu u_\alpha$$

In the second passage of the above equation we used the fact that for scalar fields, vector fields and 1-forms (i.e. covectors), the use of standard $$\partial$$ or covariant $$\nabla$$ derivative is equivalent in computing a Lie derivative. The point is that, while it is intuitive that $$\mathcal{L}_u u^\alpha=0$$, this does not need to be the case for the corresponding 1-form because the Lie-derivative of the metric tensor does not need to be zero.