Do I understand metric tensor correctly?

So I've been studying vectors and tensors, and I'm trying to understand metric tensors.

As I understand them, besides a vast array of explanations, they provide an invariant distance between vectors regardless of whether their basis has changed.

So if you had a set of vectors in say a Cartesian coordinate space, using a metric tensor to describe the distances between the vectors even if the primed basis describes them in Spherical coordinates, are he same distances? This isn't becoming quite intuitive to me yet.

This equation might clarify where I understand the definition of a metric tensor:

$ds^2= g_{11} dx^2 + g_{22} dy^2$

Solution 1:

Suppose $U\subseteq{\bf R}^2$ is some open region and $f:U\to{\bf R}^3$ defines some smooth surface in space. Now let $\gamma:I\to U$ be a path in $U$ that corresponds to a path $f\circ\gamma:I\to{\cal S}=f(U)\subset {\bf R}^3$ on the surface which we denote by $\cal S$ (here $I$ is an open interval in $\bf R$). How do we find the length of this curve?

The length is computed as

$${\rm length}=\int_I\underbrace{\left\|\frac{d}{dt}(f\circ\gamma)\right\|}_{\large\rm ``speed"}dt=\int_I\left\|J_f(\gamma)\gamma'(t)\right\|dt $$

where $J_F$ is the Jacobian, or matrix of partials, of $f$ with respect to $\gamma$. Note that an expression of the form $\| Jv\|$ may be rewritten as $\sqrt{Jv\cdot Jv}=\sqrt{v^T(J^TJ)v}$. Note that every matrix $A$ determines a bilinear form as $Q_A(u,v)= u^TAv$.

In general, then, suppose we have a manifold $\cal M$ (the device intended to axiomatize curved space), which is essentially a collection of points with some topology; there is no a priori notion of distance between points. To create such a notion of distance, form a collection of bilinear forms $g_p$, one associated to every point $p$ in space, so that any path $\gamma:I\to\cal M$ can be "measured" via

$$\int_I \sqrt{g_\gamma(\gamma',\gamma')}dt.$$

(I am sort of fudging up usual notational conventions for the sake of clarity.) Generally, given charts aka coordinate patches, we represent $g$ as a matrix, and so put two indices under it.

This is the motivation of the metric tensor in differential geometry. Vectors are usually understood to be tangent vectors; there is an abstract object called a "vector space" whose elements are by fiat called "vectors," and to each point in space we attach a vector space called a "tangent space." The derivative $\gamma'$ of a nice path $\gamma:I\to\cal M$ doesn't reside in just one space, it moves along the tangent spaces, which means that $\gamma'(t)$ is always in the tangent space of the point $\gamma(t)\in\cal M$, and furthermore the bilinear form $g_p$ at the point $p\in\cal M$ is always defined on $p$'s tangent space.

There are different, purely algebraic ways of understanding vectors and tensors in the context of abstract algebra (you'll see $\otimes$ symbols everywhere and no Einstein summation notation, for example): these purely algebraic ways are ultimately married to the differential-geometric ways of understanding tensors at the more advanced levels of geometry.

The distance between two points will not change if you change your coordinates. Also the metric tensor itself, as a function, will not change. But the matrix which represents the metric tensor depends on what coordinate system is being imposed on all of the tangent spaces, so the matrix representation of the metric tensor will in fact change with changes of coordinates.

Solution 2:

@Anon's answer focuses on the distance function, but perhaps one could relate to the question as stated, in the context of the metric tensor.

@mathacka: your notation for the metric tensor is highly unusual. Which book are you using? Usually what one writes down is an expression of the sort $ds^2= g_{11} dx^2 + g_{22} dy^2$ (in the simplest diagonal case). To be more fancy, one could set $du^1=dx$ and $du^2=dy$, and write $ds^2 = g_{ij} du^i du^j$ with the implied summation over both indices (here $g_{12}=0$ in the diagonal case). This last expression looks somewhat reminiscient of what you wrote. More details can be found in my course notes at http://u.cs.biu.ac.il/~katzmik/88-526.html

To answer the follow-up question, one key use of the $\Gamma_{ij}^k$ symbols is in writing down the geodesic equation: ${\alpha^k}^{\prime\prime}+\Gamma_{ij}^k$ $ {\alpha^i}^{\prime}{\alpha^j}^{\prime}=0$.