# Let’s differentiate tensors!

Posted by on Jan 20, 2013 in Science | 0 comments

This is a nasty and highly mathematical topic, but I am going to try to keep this presentation of it as absolutely simple as possible. The question is, can we differentiate tensors and get tensors? Jumping ahead, the answer is no, not unless the process is done in a very special way. This will take us to the notion of covariant derivates and the Christoffel symbol, but I’ll go in simple steps.

First, imagine that we have a scalar field defined as a constant everywhere. This could be a uniform temperature in some region of space; maybe a uniform humidity or electric potential. It really doesn’t matter much. Let’s take the gradient of this scalar field:

$\frac{\partial \phi }{\partial x^m}=0$

It should be almost intuitively obvious that (a) this gradient is a vector, and (b) it’s being 0 must be true in any coordinate system. We don’t even have to consider coordinate transformation machinery. If this scalar field is everywhere constant, then its derivatives must vanish in any coordinates. However, we can still write

$\frac{\partial x^m}{\partial x^{m'}}\frac{\partial \phi }{\partial x^m}=0$

because whatever the relationship between the primed coordinates and our original coordinates, the gradient is 0 everywhere in the original coordinates. The partials for coordinate conversion are total “don’t cares” in this situation, assuming they’re not infinite or singular in other ways.

Let’s go up one level and consider a constant vector field, $V_m$ now. Suppose we differentiate this vector to get the candidate tensor:

$\frac{\partial V_m}{\partial x^n}$

Whether this object is a tensor depends on how it transforms through a change of coordinates. It certainly appears to be of rank 2, given the appearance of both $m$ and $n$ indices since we are differentiating each of the components of $V_m$ with respect to each of the spatial coordinates, $x^n$. We should have a $DxD$ matrix, where $D$ are the number of dimensions of the space that we’re working in.

The question is then, what is the relationship between this derivative in the unprimed coordinates and the same derivative in new, primed coordinates?

$\frac{\partial V_{m'}}{\partial x^{n'}}$

Nominally, we would go from the unprimed to the primed coordinates in the following manner:

$\frac{\partial x^n}{\partial x^{n'}}\frac{\partial x^m}{\partial x^{m'}}\frac{\partial V_m}{\partial x^n}$

by multiplying through by the appropriate partial derivates for the coordinates themselves. So, is this last construction equal to the derivative in the primed coordinates? We can collapse two of these derivates to get a sort of mixed expression as follows:

$\frac{\partial x^m}{\partial x^{m'}}\frac{\partial V_m}{\partial x^{n'}}$

which includes the derivative of the vector expressed in the unprimed coordinates relative to the primed coordinates. We can go back to our expression for the derivative completely in the primed coordinates and convert it back towards the unprimed coordinates as well for comparison and contrast.

$\frac{\partial V_{m'}}{\partial x^{n'}}= \frac{\partial }{\partial x^{n'}}\left(\frac{\partial x^s}{\partial x^{m'}}V_s\right)$

Working through the derivatives in this last expression gives us

$\frac{\partial V_{m'}}{\partial x^{n'}}=\frac{\partial x^m}{\partial x^{m'}}\frac{\partial V_m}{\partial x^{n'}}+ \frac{\partial }{\partial x^{n'}}\frac{\partial x^m}{\partial x^{m'}}V_m$

and we see that we have picked up an extra term that looks like a differential operator on the original vector in the unprimed coordinates. We have one part on the RHS that corresponds to actual changes in the vector as a function of coordinates and another part that corresponds only to changes in the coordinates themselves.

It is now customary to substitute the form $\Gamma _{m'n'}^m$ for the last combination of partial derivatives on the RHS:

$\frac{\partial V_{m'}}{\partial x^{n'}}=\frac{\partial x^m}{\partial x^{m'}}\frac{\partial V_m}{\partial x^{n'}}+ \Gamma _{m'n'}^m V_m$

Now, recall that we began with the vector $V_m$ and set about trying to create a tensor by differentiating with respect to its components. We have found that the resulting mathematical object cannot be a tensor because in going through a coordinate change from unprimed to primed coordinates, we have picked up a term that doesn’t have to do with variations in the vector, but rather, with the coordinate systems themselves.

This suggests that we redefine the derivative of a tensor to include a part that cancels out this coordinate-sensitive term; viz., $\Gamma _{m'n'}^m$. The common name for this operation is “covariant derivative” and it is defined as follows:

$\nabla _n V_m \equiv \frac{\partial V_m}{\partial x^n} + \Gamma _{\text{mn}}^r V_r$

By including the extra $\Gamma _{mn}^r$ term in the definition, we guarantee that the resulting mathematical object from this form of vector (or more generally, tensor) derivative transforms as a tensor.

In the example I just ran through, I was assuming Cartesian coordinates. Hence, the relatively simple expression for $\Gamma _{mn}^r$ that arose is not the most general form on curved spaces and with curvilinear coordinates.

First, let’s show the expression for the covariant derivative of a tensor of rank 2:

$\nabla _pT_{\text{mn}}= \frac{\partial T_{\text{mn}}}{\partial x^p} - \Gamma _{\text{pn}}^rT_{\text{mr}} - \Gamma _{\text{pn}}^rT_{\text{rn}} = U_{\text{pmn}}$

For each index of the tensor, the $\Gamma _{mn}^r$ occurs once. In the case above, you can see how the covariant derivative increases the rank by 1.

The questions are now, what is this $\Gamma _{mn}^r$ thingy and how do we find a general form for it? First, let’s give our friend here its official name: it is a Christoffel symbol of the second kind. Since the Christoffel symbol has rank 3, it has DxDxD components, where D is the number of dimensions of the manifold under consideration. Next, we’ll see that the Christoffel symbol depends only on the metric tensor, or more specifically, on derivatives of the metric tensor. In this respect, and further down the road we’re headed on towards General Relativity, we’ll find that the metric tensor is a representation of a gravitational potential and the Christoffel symbol, as a form of derivative of the metric tensor qua potential, is a measure of gravitational force.

But I’m getting ahead of myself…

Consider the covariant derivative of the metric tensor for Cartesian coordinates on a Euclidean space. Then the metric tensor is just the Kroenecker delta, $g_{mn} = \delta_{mn}$, everywhere. Since this is a constant tensor, it’s covariant derivative should have the desirable property of vanishing everywhere; that is, if this covariant derivative thing is worth its salt. And since this is a tensor, it should vanish no matter what other set of curvilinear coordinates we might introduce into this Euclidean space.

So, for the metric tensor, we can set

$\nabla _pg_{\text{mn}}= \frac{\partial g_{\text{mn}}}{\partial x^p}-g_{\text{rn}}\Gamma _{\text{mp}}^r- g_{\text{mr}}\Gamma _{\text{np}}^r\text{ }= 0$

This is a tensor equation for the Christoffel symbol in terms of the metric tensor and its derivatives. Using the notation that a comma in the sequence of indices on a tensor implies a partial derivative with respect to that index, this can be rewritten more compactly as

$g_{mn,p} = g_{\text{rn}}\Gamma _{\text{mp}}^r + g_{\text{mr}}\Gamma _{\text{np}}^r$

It is straightforward to generate two new equations from this one simply by permuting the indices of $g_{mn,p}$ around; that is, we map $m,n,p$ to $n,p,m$ and then to $p,m,n$ as follows:

$g_{np,m} = g_{\text{rp}}\Gamma _{\text{nm}}^r +g_{\text{nr}}\Gamma _{\text{pm}}^r$

$g_{pm,n} = g_{\text{rm}}\Gamma _{\text{pn}}^r +g_{\text{pr}}\Gamma _{\text{mn}}^r$

We now recall that the metric tensor is symmetric, $g_{mn} = g_{nm}$ since its components are just the dot products of partial derivatives of the $m$th and $n$th spatial directions. And since the Christoffel symbols depend only upon the metric tensor, it must also be symmetric in the same indices. Using this, it is possible to identify pairs of equal elements in each of these three equations.

These are just

$A = g_{\text{rn}}\Gamma _{\text{mp}}^r = g_{\text{nr}}\Gamma _{\text{pm}}^r$

$B = g_{\text{mr}}\Gamma _{\text{np}}^r = g_{\text{rm}}\Gamma _{\text{pn}}^r$

$C = g_{\text{rp}}\Gamma _{\text{nm}}^r = g_{\text{pr}}\Gamma _{\text{mn}}^r$

Recognizing these symmetries, we can rewrite our three equations as

$g_{mn,p} = A + B$

$g_{np,m} = C + A$

$g_{pm,n} = B + C$

If we add the second and third equations and subtract the first, we have

$2C = 2g_{\text{rp}}\Gamma _{\text{nm}}^r = g_{np,m} + g_{pm,n} - g_{mn,p}$

Now, recognizing that we can easily write the inverse of the metric tensor $g_{mn}$ as $g^{mn}$, we can use the inverse of the metric tensor to multiply through on both sides. This yields the following expression for the Christoffel symbol in terms of the metric tensor and its derivatives (after reorganizing the indices):

$\Gamma _{\text{np}}^m= \frac{1}{2}g^{\text{mr}}\left(g_{\text{rn},p}+g_{\text{rp},n}-g_{\text{np},r}\right)$

Whooppeee! Now that wasn’t that difficult, was it?

Just for the record, there’s also an object called the Christoffel symbol of the first kind too. It is just the object we’ve been considering here, but with the upper index lowered by the metric tensor; viz.,

$\Gamma_{mnp} = g_{mr} \Gamma_{np}^r$

It is worth observing that the Christoffel symbols are not themselves tensors, in the sense that they do not transform as tensors with coordinate changes. Nonetheless, they are written in a tensor notation and act on tensors, especially as elements of the covariant derivative.

So far, so good… We have discovered just the right correction factors to include along with the simple derivative of a vector or tensor (the Christoffel symbols) that will ensure that the covariant derivative transforms as a tensor. Recall that we started down this path (as it were) through a consideration of parallel transport on curved surfaces. In a very real sense, the notion that some line segment or vector is parallel to another at a point in space constitutes information about the relationship between these two vectors. In considering this relationship at one point, and then transporting that information to another point (parallel transport), we are considering the connection of information between points in space. This is as simple as considering whether up at the North Pole is the same thing as up at the South Pole. Up is the opposite of down and down is the direction in which things fall, in which they are accelerated by gravity.

The Christoffel symbols represent one approach to a solution of this “connection” problem, using a differential operator. There are others. If you look at the Christoffel symbol’s components, you will see that they depend on the metric tensor and upon derivatives of the metric tensor. The metric tensor is itself a set of dot products of the derivatives of tangent vectors to a manifold with respect to coordinates in that manifold. The Christoffel symbol then adds second derivatives into this picture. We’ll find that Christoffel symbols connect the idea of down at the North Pole with down at the South Pole through a smoothly varying set of differentials over any path between these two points.

Stay tuned.