Tensor #17. The Covariant Derivative (Elementary Pedagogy)텐서 #17. 공변 미분 (초보적인 방법)

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 17 of the playlist, The covariant derivative (elementary pedagogy). In the previous post we put a metric at every point, so this time I find the connection that links the separated spaces and use it to differentiate a vector field. This is the elementary method leaning on pictures of arrows, and the method without coordinates is done again in the next lecture.

The metric alone does not connect them

The metric gives an inner product at each point and pairs vectors with the dual space, but it does not tear down the wall between the spaces at different points. The tangent spaces are independent from point to point, so adding or comparing vectors at different points is mathematically meaningless. The device that tears down this wall is the connection. It is a deep concept that starts in general relativity and appears in a different form in quantum field theory too.

The reason we have to tear down the wall is differentiation. Physics is written in differential equations, and if we build models from vector fields, covector fields, and tensor fields, we must be able to differentiate them on the manifold.

The parallel vector

Let the coordinates of a point P be $x$ and those of a nearby point Q be $x + \Delta x$. Both are shorthand for four coordinates. The components of a vector field X are $X^\alpha(x)$ at P and $X^\alpha(x+\Delta x)$ at Q, and the two vectors live in $T_P$ and $T_Q$ respectively.

We want to carry the vector at P to the space at Q in parallel. Parallel means the vector does not change while it is carried, and in ordinary flat space it is an arrow pointing in the same direction. The problem is that we have no tool yet that tells us which vector is parallel. A vector field does not put mutually parallel vectors at all points either. If it did, every vector field would always point in one direction, but a vector field that draws streamlines keeps changing direction.

Still, if we assume we know the parallel vector $X^\alpha_\parallel$, we gain a lot. At Q, both the vector the field placed and the parallel vector carried from P are in $T_Q$, so we can take their difference. If the difference is zero, the field has not changed in that direction, and if we divide the difference by the coordinate difference and let $\Delta x \to 0$, it becomes the derivative of the vector field.

The change in the field’s value

First let us sort out the side we know. The components of a vector field are smooth functions of the coordinates, so we use a Taylor expansion and look only at points close enough that the square terms in $\Delta x$ can be ignored.

We know the field, so this change can be computed entirely.

Stuffing the unknown into one symbol

We write the components of the parallel vector as the components at P plus some difference, and we pack that difference into a symbol $\bar{\Delta}X^\alpha$.

Then the difference of the two vectors at Q is $\Delta X^\alpha - \bar{\Delta}X^\alpha$, and the limit after dividing by $\Delta x^c$ is the following.

The unknown part, parallelism, is entirely inside $\bar{\Delta}X^\alpha$. We have only crammed the unknown into a small symbol, and the problem is not solved.

The connection Γ

We do not know what $\bar{\Delta}X^\alpha$ looks like, but we can expect a few things. It must get smaller as Q approaches P, so it must be linear in the displacement $\Delta x^c$. If the components of the field are zero the correction must be zero too, so it must be linear in the components $X^\beta$ as well. It has to be linear in every combination of components on both sides, so we bundle the coefficients with an index sum.

$\Gamma$ is a function of the coordinates, and we take its value at P. The correction that looks for the parallel vector has to be tied to the point P where the vector lives. This function has three indices, so it is $4 \times 4 \times 4$ functions. Everything we know about parallelism is in these functions, and they are called the affine connection.

Q. Why put a minus sign in front of $\Gamma$?
A. It is a convention. With this definition, the minus sign is multiplied by the minus sign of $\bar{\Delta}X$ in the derivative, and a plus sign remains. When we differentiate covector fields or other tensors the minus sign comes back, so it is not a great convenience, but it has hardened into the standard form.

Q. It has three indices, up and down, so is $\Gamma$ not a tensor too?
A. No. $\Gamma$ is only a bundle of functions gathered as the coefficients of an index sum, so writing it attached to a basis like $e_\alpha \otimes e^\beta \otimes e^c$ makes no sense. A short reason is that the partial derivative $\partial_c X^\alpha$ alone does not change like a tensor when the coordinates change, and $\Gamma$ corrects that mismatched part. $\Gamma$ itself does not obey the tensor rule either. (This explanation is added to help understanding and was not covered in the lecture.)

The covariant derivative

Put $\Gamma$ into the limit above. The two minus signs multiply to a plus, and $\Delta x^c$ cancels. What we get is the covariant derivative.

Here $c$ is simply an index pointing to the direction of differentiation, and it does not mean covariant. The $\Gamma$ term is added to the partial derivative as a correction, and it becomes a derivative that measures the difference from the parallel vector.

Looking at the indices, there is one up and one down. A vector field is a (1, 0) tensor and this derivative is the components of a (1, 1) tensor, so one differentiation raises the rank by one. With the basis written out, it is as follows.

The most common notation uses a semicolon, $X^\alpha{}_{;c}$, and there is another with two bars. They all mean the same, and usually only the components are written. A scalar field s is neither a vector nor a covector, so it needs no connection, and the partial derivative is the covariant derivative. The rank goes from (0, 0) to (0, 1).

Touching it in polar coordinates

Even in a flat plane, $\Gamma$ is not zero when we use curved coordinates. In polar coordinates $(r,\theta)$ there are only these three nonzero values. (These values come from the metric of the plane and were not covered in the lecture.)

In the toy below, pick a field and drag the point P. It draws the vector at P and the vector the field places at a nearby point Q, and overlays the parallel vector made by correcting with $\Gamma$. As the gap $\Delta x$ shrinks, the difference between the two vectors shrinks, and that difference divided by the gap is the number the covariant derivative points to. If a field is constant in Cartesian coordinates its covariant derivative must be zero, but in polar coordinates the partial derivative alone does not give zero, and only with the $\Gamma$ term added does it become zero.

The connection has to be given

$\Gamma$ is mathematically arbitrary. Any bundle of functions will do as long as they are differentiable in the coordinates, and having a manifold does not by itself fix a natural connection that belongs to it. We have to place a connection on the manifold separately before we can speak of parallelism, and only after speaking of parallelism can we define the covariant derivative. In pure mathematics we declare what parallelism is and measure all derivatives against it. In physics, physics tells us. Physics gives the metric and the metric gives the connection, and that is the road we are heading down.

This derivation leans on coordinates from start to finish. In the next lecture the same derivative is derived again without coordinates. There the symmetry of $\Gamma^\alpha{}_{\beta c}$ in its two lower indices is pointed out, and with this symmetry we say there is no torsion, which is the case general relativity deals with. In the next post we look at the same covariant derivative again without coordinates.