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:

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

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    $\begingroup$ What's your definition of metric tensor? It sounds like you're studying tensors in a very coordinate-heavy way. $\endgroup$ Jun 26, 2013 at 2:23
  • $\begingroup$ @jesseMadnick I'm studying a book that begins with Linear algebra and ends in tensors, metric tensors, and Christoffherr Symbols to describe R3 vector spaces, I'm wanting to move to differential geometry, but for now trying to understand most of the math that pertains to beginning theoretical physics, the book uses physics II examples for the Lin Alg portion and Electro Mag in second part I believe. $\endgroup$ Jun 26, 2013 at 4:50

2 Answers 2


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.

  • $\begingroup$ So, if you have two points in $\mathbb{R}^3$, the metric tensor -- defined as a certain array $g_{ij}$ satisfying various properties -- will (somehow, magically) determine a notion of distance between them. Will this notion of distance change if we switch between Cartesian coordinates and spherical coordinates? I could be wrong, but I think that's the OP's question here. $\endgroup$ Jun 26, 2013 at 2:26
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    $\begingroup$ @JesseMadnick I see; I've added a brief paragraph about that. $\endgroup$
    – anon
    Jun 26, 2013 at 2:52
  • $\begingroup$ One of the definitions for metric tensor was given as -> ds^2 = gsubi ds + gsupj ds. A comment that was made is that gsubisupj gives the space of non orthogonal but linearly independant basis vectors a usable dot product. This is the background I'm commingled from. As a side note what is the scripting language that this site uses so I can format my comments correctly? $\endgroup$ Jun 26, 2013 at 4:59
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    $\begingroup$ @mathacka We use $\LaTeX$; see meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – anon
    Jun 26, 2013 at 5:00
  • $\begingroup$ Using g as a way to keep covarient and contravarient measurements accurate was the reason for introducing the metric tensor. So for instance covarient measurements of the x and y axis of non orthagonal basis vectors would give an inaccurate, if you will, distance from the origin, yet, say the sines of the contravarient vectors wouldn't be "accurate" in that sender either, but both of them have their purposes, such as the gradient of the covarient axis, but when used in conjunction give accurate measurements. $\endgroup$ Jun 26, 2013 at 5:08

@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$.

  • $\begingroup$ I'm using "A Student's Guide to Vectors and Tensors". The misunderstanding is all mine. I'm just not getting how to 1. find the metric tensor for a given change of basis, and then 2. what's it's uses are beyond finding a distance formula. $\endgroup$ Jul 9, 2013 at 4:34
  • $\begingroup$ After I understand this, I plan on trying to get an idea what Christhoffer (sp?) symbols mean and are used for. $\endgroup$ Jul 9, 2013 at 4:37

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