Tensor #21. The Lie Derivative텐서 #21. 리 미분

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 21 of the playlist, The Lie derivative. Following the previous post, I collect the Lie derivative, the second of three kinds of derivatives. The first is the covariant derivative, and the third is the exterior derivative, which comes later.

One definition, and the reason

Pick a reference vector field $V$ and write $\Lie{V}X$ for the derivative of another vector field $X$ in that direction. In components it reads as follows.

We could simply declare this formula to be the definition, but the formula alone does not show what it measures. So we start by understanding vector fields again through curves. We take this road because the Killing vector fields, which we use later to dig into the structure of spacetime, are built on the Lie derivative.

Curves and tangent vectors

In a spacetime with coordinates $x^\mu$, a curve is a map from an interval of real numbers into spacetime. With parameter $\lambda$ the curve is four functions $x^\mu(\lambda)$, and another curve $x^\mu(\delta)$ with parameter $\delta$ is a different function even though we use the same symbol. Restricting a function $f$ on spacetime to the curve gives $f(x^0(\lambda), \dots, x^3(\lambda))$, and its rate of change in $\lambda$ comes from the chain rule.

Since $x^\mu$ is a function of the single variable $\lambda$, we use ordinary derivatives rather than partial derivatives. Stripping away $f$ leaves an operator.

Computing the components $dx^\mu/d\lambda$ at a point $P$ gives four real numbers, and they sit in front of the coordinate basis $\partial_\mu$ as a vector. One curve hands out one tangent vector at each point on it. Collecting the tangent vectors of all curves through a point gives the vector space of that point. Addition is the sum of operators, the inverse comes from parametrizing the curve backwards, and the zero vector is the curve that stays at the point. This rebuilds, on firm ground, the earlier way of placing a coordinate basis at every point.

A congruence of curves makes a vector field

A congruence of curves is a collection of curves that fills spacetime without meeting each other. Think of the streamlines of a fluid. All curves of the congruence use the same parameter $\lambda$, and another value tells which curve we are on. Each point $P$ has only one curve through it, so one tangent vector is fixed.

One vector sits at each point, so this is a vector field. A different congruence gives a different vector field, and the vector field of a congruence with parameter $\delta$ is written $d/d\delta$.

A vector field makes a congruence

The other direction works too. Given a smooth vector field $V$, we get curves by solving the equation below.

This is a first-order differential equation, a theorem says a solution exists and is unique, and one curve is fixed by its initial condition. Collecting the solutions gives a congruence. In the simplest example $V = \partial_0$, we have $V^0 = 1$ and the other components are 0, so this happens.

The three constants choose the curve, and the congruence is the straight lines parallel to the $x^0$ axis. For $V = \partial_1$ we get the straight lines parallel to the $x^1$ axis. A general vector field gives four coupled equations, but a solution exists just the same.

Q. Which comes first, the congruence or the vector field?
A. Neither comes first. Collecting tangent vectors from a congruence gives a vector field, and solving the equation from a vector field gives a congruence. That we can go both ways is decisive for the argument that follows. Even for curves of the same shape, running the parameter twice as fast doubles the tangent vector, so a vector field fixes not only the shape of the curves but also the speed of the parameter. (This explanation follows from the definition of the tangent vector and was not covered in the lecture.)

Moving by ε along a curve

The coordinates of the point reached from the point $\lambda_0$ by a parameter amount $\varepsilon$ can be written with a Taylor series. We assume $x^\mu$ is analytic.

Q. What does the exponential notation mean?
A. It is the Taylor series of an operator written all at once. The formula is exact even when $\varepsilon$ is a finite move. For an infinitesimal move we drop $\varepsilon^2$ and higher and keep only the linear term.

Lie dragging

Pick a curve, set up so that every point on it has the parameter value $\lambda_0$ in the congruence of $V$. Sending every point of this curve along its own curve by $\varepsilon$ makes a new curve. This act of pushing a curve along the reference vector field $V$ is called Lie dragging. Repeating it while changing $\varepsilon$ makes a new congruence. Call its parameter $\delta$. Then the new curves are curves of constant $\lambda$, and the original curves are curves of constant $\delta$. The two congruences form a net.

On this net, moving $\varepsilon$ in the $\lambda$ direction and $\varepsilon$ in the $\delta$ direction reaches the same point in either order.

Expanding both sides in series and subtracting, all the lower orders cancel, and the first surviving term is $\varepsilon^2$ times the commutator. This must vanish, so we get the following.

This commutation holds for pairs of vector fields made by Lie dragging, and for two arbitrary vector fields it generally fails. For example, $[\partial_0, x^0 \partial_1] = \partial_1$, so the two flows reach different points when the order is swapped. (This example is added for understanding and was not covered in the lecture.)

The definition of the Lie derivative

Now take the reference vector field $V$ (parameter $\lambda$) and an arbitrary vector field $X$ (parameter $\eta$) that was not made by Lie dragging. The curve of $X$ through the point $P$ and the curve of $X$ through the point reached from $P$ along $V$ by $\Delta\lambda$ are not connected by Lie dragging. We measure exactly that difference.

1. Drag the curve of $X$ through the point $\Delta\lambda$ ahead of $P$ back along $V$ by $\Delta\lambda$. This makes a new curve through $P$. Call its parameter $\eta^*$.
2. Now there are two curves at $P$, so there are two tangent vectors: $\frac{dx^\mu(\eta)}{d\eta} \partial_\mu$ of the original curve and $\frac{dx^\mu(\eta^*)}{d\eta^*} \partial_\mu$ of the dragged curve. They are in the same vector space at the same point, so we can subtract them.
3. Divide the difference by $\Delta\lambda$ and send it to 0.

The lecture skips the computation of this limit, saying it is in the textbooks, and shows only the result. The result equals the formula we wrote at the start.

This commutator is also called the Lie bracket. A definition that was one line came out of a picture of curves in this way. In the toy below, pick the reference $V$ and the $X$ to compare, and drag the point $P$. The difference between the dragged vector and $X(P)$, divided by $\varepsilon$, approaches the Lie bracket as you shrink $\varepsilon$. (The toy computes this value numerically by sending the point along the same flow by $\varepsilon$ and dragging the vector back.)

Q. Why does the Lie derivative need no rule of parallelism?
A. To compare vectors at different points, we must carry the vector at one point to the other. The covariant derivative carries it by fixing the meaning of parallel with connection coefficients. For the Lie derivative the flow of the reference vector field $V$ already gives the road from point to point, so there is nothing more to agree on. Instead we must choose one $V$.

Properties and uses

The covariant derivative adds one covariant index to the result, so differentiating a vector gives a $(1, 1)$ tensor. The Lie derivative of a vector field is a vector field, and no index is added. Also, just as every value of the covariant derivative changes when we change the meaning of parallel, every value of the Lie derivative changes when we change the reference vector field $V$. So $V$ has to be chosen for a reason.

On functions, the vector field acts as an operator.

If $X$ is a vector field Lie dragged by $V$, the dragged-back curve lies exactly on the original curve and the difference of the two tangent vectors is 0. Then the Lie derivative is 0 and the two vector fields commute. The condition that the Lie derivative is zero can be read as “unchanged under the flow of $V$”, and this connects to how Killing vectors are found. The lecture previews that in general relativity a vector field that leaves the metric unchanged under Lie dragging is a Killing vector, and says it will come later. In the same way we can build the Lie derivative of tensor fields and covector fields, and the lecture says the formula for covector fields may be listed in the next lecture.

Q. What does it mean when the Lie derivative is 0?
A. It means $X$ does not change when dragged along the reference vector field $V$, and since $[V, X] = 0$, the two flows reach the same point in either order. In the toy above, choosing $X$ to be $V$ itself gives this case.

In the next post we meet p-forms, the objects the exterior derivative acts on.