Tensor #16. The Metric Tensor Field텐서 #16. 메트릭 텐서장

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 16 of the playlist, The metric tensor field. In the previous post we saw how the basis changes when the coordinates change, so this time I sort out the language for calling the spaces at all points one body and set up the tensor field that puts an inner product at every point at once.

The hand-waving part

The story of putting one vector space at each point was in fact a hand-waving argument. Strictly speaking, spacetime is a manifold, and the structure that gathers the points together with the vector spaces attached to them is a fiber bundle, and more precisely the tangent bundle. A coordinate system cannot be laid on just any object, and manifold theory has its own conditions for when we may lay one and its own rules for changing coordinates. This series does not go deep into that formalism. The lecture recommends climbing up from topology if you want to know it properly.

All the spaces as one body

The tangent space at a point P is $T_P$, and if we call the whole of spacetime $\M$, then seeing the spaces at all points as one body gives $T\M$. Pulling one vector from each point gives a vector field, and a vector field is an element of $T\M$.

So from now on, the tangent space by itself means this whole body rather than the space at one point, and an element of it means one vector field. We almost never fix our eyes on one point. This convention matters because the connection, coming soon, is a map that starts from the set of all vector fields.

Put in fields and a function comes out

At one point, a covector $\omega$ takes a vector $v$ and returns a real number $\map{\omega}{v}$. What happens when we put in a covector field $\lambda$ and a vector field $X$?

Pick a point x, find the values of the two fields at that point, and compute the pairing of those values to get a real number. We do this at every point, so the result is not one real number but a function that gives a real number at each point of spacetime, a smooth real function on $\M$. From now on we read the pairing of the dual space as a map that takes fields and returns a function. The inner product is the same. Put in two vector fields instead of two vectors and a function comes out instead of a real number.

Q. Why is it computed point by point when two fields go in?
A. Because the components of a field are functions of the coordinates, and only after fixing a point do they become numbers. With the point x fixed, the vector field shrinks to a single vector of $T_P$ and the covector field to a single covector of the cotangent space, and then the old pairing holds as it was. Spreading that out over all x gives the function.

Coordinate transformations of fields

When the coordinates change, the basis and the components of a tensor field change together. For example, to write the tensor field $Q = Q^{\alpha\beta}{}_{\delta}\,\partial_\alpha \otimes \partial_\beta \otimes dx^\delta$ in the new coordinates, we put the rule of the previous post into the basis.

The matrices of partial derivatives attached to each basis element move over to the components, and the components change like this.

Each upper index gets one $\partial x'/\partial x$, and each lower index gets one $\partial x/\partial x'$. If we put this product back together with the basis, the factors of the old and new coordinates cancel each other and the same tensor comes out. A tensor is an object that does not care what we call the coordinates.

There is no metric yet

So far we have treated the tangent spaces as pure vector spaces. But what we want are inner product spaces. With an inner product we can leave one slot empty as in $\map{v}{\,\cdot\,}$ and make an element of the dual space, and this correspondence is the natural correspondence between the vector space and the dual space. It exists only once an inner product is given.

But right now there are only vector spaces at the points and no inner product. So there is no way to pair vector fields with covector fields, and no way to move between the tensor product spaces. Putting an inner product at only one point is no use either. It has to be put at all points at once, connected between neighboring points.

The metric tensor field

We solve this problem in one stroke with a single tensor field.

We often omit the tensor product symbol, but it is there. This is a (0, 2) tensor field that picks one element of the (0, 2) tensor product space at every point. Give it a point and sixteen real numbers $g_{\mu\nu}$ come out. Putting in the coordinates of a point P and writing $g_{\mu\nu}\big(x^\delta(P)\big)$ is the formal way, and we usually just write $g_{\mu\nu}(x)$. Once this field is laid at all points, we get the correspondence between the tangent space and the cotangent space, and an inner product, at every point. Without tools to compare, we cannot even begin physics, so this is the core of what we wanted from this structure.

Q. Is any (0, 2) tensor field a metric?
A. No. There are many (0, 2) tensor fields and the metric is one of them. To behave like an inner product it must be symmetric, $g_{\mu\nu} = g_{\nu\mu}$, and the only vector whose inner product with every vector is zero must be the zero vector. (This explanation takes the definition of the inner product from earlier posts and was not covered in the lecture.)

Lengths and angles

With a metric we can measure the length of a vector and the angle between two vectors at a point.

These are the length and angle fixed by this formula, not those of the picture drawn on the screen. (This formula carries over the definition of the inner product and was not covered in the lecture.) In the toy below, pick a metric and drag the point P and the tips of the two vectors. The endpoints of the vectors of length 1 are drawn as an ellipse that differs from point to point, and two vectors that are perpendicular in the picture can fail to be perpendicular under the metric, or the other way round. When the metric differs from point to point, the size and shape of the ellipses change across the grid.

Q. Can I not write $g_{\mu\nu}(P)$?
A. P is a point of spacetime and the component function takes four coordinates, so formally we have to go through the coordinate functions and write $g_{\mu\nu}\big(x^\delta(P)\big)$. Abbreviating loosely to $g_{\mu\nu}(x)$ as in this post means the same thing.

Spacetime itself

The metric is a field that gives a value at every point, so at points P, Q, and R we compute $g_{\mu\nu}(P),\ g_{\mu\nu}(Q),\ g_{\mu\nu}(R)$ respectively. But there are as many (0, 2) tensor fields as we like, and exactly one of them is the metric. Which one it has to be for our model to resemble the real world is decided by physics. Whether a mass is attracted to another mass or just passes by depends on how spacetime is shaped, and what fixes that is the Einstein equation. A Schwarzschild spacetime or a curved spacetime refers to a metric that solves the Einstein equation in a certain situation.

So the metric is so important that we sometimes just call it spacetime. The coordinates are in it, and so are the ways to measure tangent vectors and to pair vectors with covectors. The place this series aims at is fixing that metric by the equation. The question of what a tensor is ends here, and now we look at what tensors do for physics. In the next post we cover the covariant derivative, which connects the spaces that stand apart from each other.