Geometric interpretation of Christoffel symbols and some alternative approaches to calculating them

In a classic paper in 1869, Elwin Bruno Christoffel (1829-1900) introduced his famous Christoffel symbols \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} to represent an array of numbers describing a metric connection. They are also known as connection coefficients (and sometimes less respectfully as `Christ-awful symbols’). In differential geometry one usually first encounters them when studying covariant derivatives of tensors in tensor calculus. For example, suppose we try to differentiate the contravariant vector A = A^{\alpha} e_{\alpha}, where e_{\alpha} denotes a coordinate basis vector (and we are using the Einstein summation convention). We get

\frac{\partial A}{\partial x^{\beta}} = \frac{\partial A^{\alpha}}{\partial x^{\beta}} e_{\alpha} + A^{\alpha} \frac{\partial e_{\alpha}}{\partial x^{\beta}}

In general, the partial derivative in the second term on the right will result in another vector which we can write in terms of its coordinate basis as

\frac{\partial e_{\alpha}}{\partial x^{\beta}} \equiv \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma}

This defines the Christoffel symbol \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta}. The downstairs indices refer to the rate of change of the basis components e_{\alpha} with respect to the coordinate variable x^{\beta} in the direction of the coordinate basis vector e_{\gamma} (\gamma being the upstairs index). Substituting the second equation into the first we get

\frac{\partial A}{\partial x^{\beta}} = \frac{\partial A^{\alpha}}{\partial x^{\beta}} e_{\alpha} + A^{\alpha} \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma}

To enable us to factor out the coordinate basis vector we can exchange the symbols \alpha and \gamma in the second term on the right to get

\frac{\partial A}{\partial x^{\beta}} = \frac{\partial A^{\alpha}}{\partial x^{\beta}} e_{\alpha} + A^{\gamma} \Gamma^{\alpha}_{\hphantom{\alpha} \gamma \beta} \ e_{\alpha}

= \big( \frac{\partial A^{\alpha}}{\partial x^{\beta}} + A^{\gamma} \Gamma^{\alpha}_{\hphantom{\alpha} \gamma \beta}\big) \ e_{\alpha}

The expression in the bracket is called the covariant derivative of the contravariant vector A, i.e., the rate of change of A^{\alpha} in each of the directions \beta of the coordinate system x^{\beta}. It has the important property that it is itself tensorial (unlike the ordinary partial derivative of the tensor on its own). This covariant derivative is often written using the notation

\nabla_{\beta} \ A^{\alpha} = \partial_{\beta} \ A^{\alpha} + A^{\gamma} \Gamma^{\alpha}_{\hphantom{\alpha} \gamma \beta}

Having thus established the meaning of the Christoffel symbols, one then goes on to work out that the covariant derivative of a one-form is

\nabla_{\beta} \ A_{\alpha} = \partial_{\beta} \ A_{\alpha} - A_{\gamma} \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta}

and that the covariant derivatives of higher rank tensors are constructed from the building blocks of \nabla_{\beta} \ A^{\alpha} and \nabla_{\beta} \ A_{\alpha} by adding a \Gamma^{\alpha}_{\hphantom{\alpha} \gamma \beta} term for each upper index \gamma and a \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} term for each lower index \gamma. For example, the covariant derivative of the (1, 1) rank-2 tensor X^{\mu}_{\hphantom{\mu} \sigma} is

\nabla_{\beta} \ X^{\mu}_{\hphantom{\mu} \sigma} = \partial_{\beta} \ X^{\mu}_{\hphantom{\mu} \sigma} + X^{\alpha}_{\hphantom{\mu} \sigma} \ \Gamma^{\mu}_{\hphantom{\mu} \alpha \beta} - X^{\mu}_{\hphantom{\mu} \alpha} \ \Gamma^{\alpha}_{\hphantom{\alpha} \sigma \beta}

Christoffel symbols then go on to play vital roles in other areas of differential geometry, perhaps most notably as key components in the definition of the Riemann curvature tensor.

It is possible to have a working knowledge of all of this without truly understanding at a deep level, say geometrically, what Christoffel symbols really mean. In the present note I want to delve a bit more deeply into how one might calculate and interpret Christoffel symbols geometrically. I also want to explore some alternative ways of calculating them in the context of a simple plane polar coordinate system (r, \theta) which is related to the usual Cartesian (x, y) coordinate system via the conversion equations

x = r \cos \theta

y = r \sin \theta

In an n-dimensional manifold there are potentially n^3 Christoffel symbols to be calculated, though this number is usually reduced by symmetries. In the present plane polar coordinate case, we will need to calculate 2^3 = 8 Christoffel symbols. These are

\Gamma^{r}_{\hphantom{r} \theta \theta}

\Gamma^{\theta}_{\hphantom{\theta} \theta \theta}

\Gamma^{\theta}_{\hphantom{\theta} \theta r}

\Gamma^{r}_{\hphantom{r} \theta r}

\Gamma^{r}_{\hphantom{r} r r}

\Gamma^{\theta}_{\hphantom{\theta} r r}

\Gamma^{\theta}_{\hphantom{\theta} r \theta}

\Gamma^{r}_{\hphantom{r} r \theta}

Geometric approach
Consider the situation shown in the diagram below where two vectors (e_{\theta})_P and (e_{\theta})_S of the basis vector field e_{\theta} are drawn emanating from points P and S respectively:

If we parallel transport the vector (e_{\theta})_P from P to S we end up with the situation shown in the next diagram:

Now, in plane polar coordinates the magnitude of e_{\theta} is

|e_{\theta}| = r

Therefore the length of the arc L in the diagram is

L = r \Delta \theta

If \Delta \theta is small, we have

L \approx |\Delta_{\theta} e_{\theta}|

where \Delta_{\theta} e_{\theta} is the vector connecting the endpoints of (e_{\theta})_P and (e_{\theta})_S, i.e., \Delta_{\theta} e_{\theta} = (e_{\theta})_S - (e_{\theta})_P.

Therefore

 |\Delta_{\theta} e_{\theta}| \approx r \Delta \theta

Passing to the differential limit as \Delta \theta \rightarrow 0 we get

|d_{\theta} e_{\theta}| = r d \theta

From the diagram we see that d_{\theta} e_{\theta} points in the opposite direction of e_r. Therefore we have

d_{\theta} e_{\theta} = - r d \theta e_r

(note that in plane polar coordinates e_r is of unit length). From this equation we have

\frac{d_{\theta} e_{\theta}}{d \theta} \equiv \frac{\partial e_{\theta}}{\partial \theta} = -r e_r

But from the definition of Christoffel symbols we have

\frac{\partial e_{\theta}}{\partial \theta} = \Gamma^{r}_{\hphantom{r} \theta \theta} e_r + \Gamma^{\theta}_{\hphantom{\theta} \theta \theta} e_{\theta}

Therefore we conclude

\Gamma^{r}_{\hphantom{r} \theta \theta} = -r

\Gamma^{\theta}_{\hphantom{\theta} \theta \theta} = 0

We have obtained the first two Christoffel symbols on our list from the geometric setup and the nice thing about this approach is that we can see what the underlying changes in the coordinate basis vectors looked like.

To obtain the next two Christoffel symbols on our list, we consider a change in the vector field e_{\theta} due to a displacement in the radial direction from P to Q in the following diagram:

We have moved outwards by a small amount \Delta r and as a result the length of the vectors in the vector field e_{\theta} has increased by a small amount |\Delta_r e_{\theta}| shown in the diagram. From the diagram we see that the proportions of the two increases must be same, so we have

\frac{|\Delta_r e_{\theta}|}{|e_{\theta}|} = \frac{\Delta r}{r}

or

|\Delta_r e_{\theta}| = \Delta r \frac{1}{r} |e_{\theta}|

Passing to the differential limit as \Delta r \rightarrow 0 we get

|d_r e_{\theta}| = dr \frac{1}{r} |e_{\theta}|

Since d_r e_{\theta} is directed along the vector e_{\theta} we can write the vector equation

d_r e_{\theta} = dr \frac{1}{r} e_{\theta}

so

\frac{d_r e_{\theta}}{dr} \equiv \frac{\partial e_{\theta}}{\partial r} = \frac{1}{r} e_{\theta}

But

\frac{\partial e_{\theta}}{\partial r} = \Gamma^{\theta}_{\hphantom{\theta} \theta r} e_{\theta} + \Gamma^{r}_{\hphantom{r} \theta r} e_r

from which we conclude

\Gamma^{\theta}_{\hphantom{\theta} \theta r} = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} \theta r} = 0

We have thus found two more Christoffel symbols from the geometrical setup. To get the next two Christoffel symbols on our list we observe that the basis vector field e_r does not change as we move in the radial direction (either in magnitude or direction) so we must have

\frac{\partial e_r}{\partial r} = 0

where the right hand side here denotes a zero vector. But we know that

\frac{\partial e_r}{\partial r} = \Gamma^{r}_{\hphantom{r} r r} e_r + \Gamma^{\theta}_{\hphantom{\theta} r r} e_{\theta}

so we conclude

\Gamma^{r}_{\hphantom{r} r r} = 0

\Gamma^{\theta}_{\hphantom{\theta} r r} = 0

Finally, to get the last two remaining Christoffel symbols on our list, we consider a change in the vector field e_r due to an angular displacement. In the diagram below two vectors (e_r)_P and (e_r)_S of the basis vector field e_r are drawn emanating from points P and S respectively:

If we parallel transport the vector (e_r)_P from P to S we end up with the situation shown in the next diagram:

The arc length L is

L = |e_r| \Delta \theta = \Delta \theta

(since the magnitude of the coordinate basis vector e_r is |e_r| = 1). But for small \Delta \theta we also have

L \approx |\Delta_{\theta} e_r|

where \Delta_{\theta} e_r is the vector connecting the endpoints of (e_r)_P and (e_r)_S, i.e., \Delta_{\theta} e_r = (e_r)_S - (e_r)_P. Therefore

|\Delta_{\theta} e_r| = \Delta \theta

Passing to the differential limit as \Delta \theta \rightarrow 0 we have

|d_{\theta} e_r| = d \theta

But d_{\theta} e_r has the same direction as e_{\theta}. Therefore

d_{\theta} e_r = \frac{1}{r} d \theta e_{\theta}

where the factor \frac{1}{r} is needed to correct for the magnitude r of e_{\theta} (we only want the direction of e_{\theta} here). Therefore we see that

\frac{d_{\theta} e_r}{d \theta} \equiv \frac{\partial e_r}{\partial \theta} = \frac{1}{r} e_{\theta}

But

\frac{\partial e_r}{\partial \theta} = \Gamma^{\theta}_{\hphantom{\theta} r \theta} e_{\theta} + \Gamma^{r}_{\hphantom{r} r \theta} e_r

from which we conclude

\Gamma^{\theta}_{\hphantom{\theta} r \theta} = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} r \theta} = 0

This completes the geometric calculation of all the Christoffel symbols for plane polar coordinates.

Algebraic approach

It is possible to calculate the eight Christoffel symbols quite easily for plane polar coordinates by first expressing the basis components e_r and e_{\theta} in terms of the Cartesian components e_x and e_y. Note that these basis components are one-forms, so they transform as

e^{\prime}_{\alpha} = \frac{\partial x^{\beta}}{\partial x^{\prime \alpha}} e_{\beta}

We use the conversion equations

x = r \cos \theta

y = r \sin \theta

to calculate the coefficients. We get

e_r = \frac{\partial x}{\partial r} e_x + \frac{\partial y}{\partial r} e_y

e_{\theta} = \frac{\partial x}{\partial \theta} e_x + \frac{\partial y}{\partial \theta} e_{\theta}

and therefore

e_r = \cos \theta e_x + \sin \theta e_y

e_{\theta} = -r \sin \theta e_x + r \cos \theta e_y

Then we calculate the Christoffel symbols as follows. First,

\frac{\partial e_r}{\partial r} = 0

so

\frac{\partial e_r}{\partial r} = \Gamma^{r}_{\hphantom{r} r r} e_r + \Gamma^{\theta}_{\hphantom{\theta} r r} e_{\theta} = 0

and we conclude

\Gamma^{r}_{\hphantom{r} r r} = 0

\Gamma^{\theta}_{\hphantom{\theta} r r} = 0

Next,

\frac{\partial e_r}{\partial \theta} = - \sin \theta e_x + \cos \theta e_y = \frac{1}{r} e_{\theta}

so

\frac{\partial e_r}{\partial \theta} = \Gamma^{\theta}_{\hphantom{\theta} r \theta} e_{\theta} + \Gamma^{r}_{\hphantom{r} r \theta} e_r = \frac{1}{r} e_{\theta}

from which we conclude

\Gamma^{\theta}_{\hphantom{\theta} r \theta} = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} r \theta} = 0

Next,

\frac{\partial e_{\theta}}{\partial \theta} = -r \cos \theta e_x - r \sin \theta e_y = -r e_r

so

\frac{\partial e_{\theta}}{\partial \theta} = \Gamma^{r}_{\hphantom{r} \theta \theta} e_r + \Gamma^{\theta}_{\hphantom{\theta} \theta \theta} e_{\theta} = -r e_r

Therefore we conclude

\Gamma^{r}_{\hphantom{r} \theta \theta} = -r

\Gamma^{\theta}_{\hphantom{\theta} \theta \theta} = 0

Finally,

\frac{\partial e_{\theta}}{\partial r} = -\sin \theta e_x + \cos \theta e_y = \frac{1}{r} e_{\theta}

so

\frac{\partial e_{\theta}}{\partial r} = \Gamma^{\theta}_{\hphantom{\theta} \theta r} e_{\theta} + \Gamma^{r}_{\hphantom{r} \theta r} e_r = \frac{1}{r} e_{\theta}

from which we conclude

\Gamma^{\theta}_{\hphantom{\theta} \theta r} = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} \theta r} = 0

Metric tensor approach

The previous approach relied on knowing the functional relationship between the Cartesian coordinates (x, y) and the plane polar coordinates (r, \theta). There is another more generally useful method of calculating the Christoffel symbols from the components of the metric tensor, using the formula

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} = \frac{1}{2} g^{\gamma \mu} [\partial_{\beta} \ g_{\alpha \mu} + \partial_{\alpha} \ g_{\mu \beta} - \partial_{\mu} \ g_{\alpha \beta}]

I will first derive this formula from first principles, then use it to find the Christoffel symbols for the plane polar coordinates case.

The first step is to show that Christoffel symbols are symmetric in their lower indices, i.e.,

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} = \Gamma^{\gamma}_{\hphantom{\gamma} \beta \alpha}

as this property will be needed in the derivation of the formula. To prove the symmetry property we start from the defining equation for Christoffel symbols,

\frac{\partial e_{\alpha}}{\partial x^{\beta}} \equiv \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma}

Suppose we now decompose the basis vectors e_{\alpha} in a local Cartesian coordinate system. Then using the transformation rule for one-forms we have

e_{\alpha} = \frac{\partial x^m}{\partial x^{\alpha}} e_m

where the x^m are the Cartesian coordinates and the e_m are the coordinate basis vectors (which are constant in both magnitude and direction in the Cartesian system). Differentiating gives

\frac{\partial e_{\alpha}}{\partial x^{\beta}} = \frac{\partial^2 x^m}{\partial x^{\alpha} \partial x^{\beta}} e_m

Equating the expressions for \frac{\partial e_{\alpha}}{\partial x^{\beta}} we get

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma} = \frac{\partial^2 x^m}{\partial x^{\alpha} \partial x^{\beta}} e_m

But then

\Gamma^{\gamma}_{\hphantom{\gamma} \beta \alpha} \ e_{\gamma} = \frac{\partial^2 x^m}{\partial x^{\beta} \partial x^{\alpha}} e_m

so it follows from Young’s Theorem (equality of cross-partials) that

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} = \Gamma^{\gamma}_{\hphantom{\gamma} \beta \alpha}

We conclude that Christoffel symbols are symmetric in their lower indices, as claimed.

Note too that the components g_{\mu \gamma} of the general metric tensor are also symmetric with respect to their indices. This follows from the defining equation of the metric tensor components in terms of the basis vector fields e_{\gamma}, namely

g_{\mu \gamma} \equiv e_{\mu} \cdot e_{\gamma}

Since e_{\mu} \cdot e_{\gamma} = e_{\gamma} \cdot e_{\mu}

the metric is symmetric, i.e.,

g_{\mu \gamma} = g_{\gamma \mu}

To derive the formula for the Christoffel symbols in terms of the metric tensor components, we begin again with the defining equation for Christoffel symbols,

\frac{\partial e_{\alpha}}{\partial x^{\beta}} \equiv \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma}

Taking the scalar product with another basis vector on both sides we get

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ e_{\gamma} \cdot e_{\mu} = \frac{\partial e_{\alpha}}{\partial x^{\beta}} \cdot e_{\mu}

= \frac{\partial (e_{\alpha} \cdot \  e_{\mu})}{\partial x^{\beta}} - e_{\alpha} \cdot \frac{\partial e_{\mu}}{\partial x^{\beta}}

= \frac{\partial g_{\alpha \mu}}{\partial x^{\beta}} - \Gamma^{\rho}_{\hphantom{\rho} \mu \beta} \ e_{\alpha} \cdot \ e_{\rho}

= \partial_{\beta} \ g_{\alpha \mu} - \Gamma^{\rho}_{\hphantom{\rho} \mu \beta} \ g_{\alpha \rho}

Therefore we have

\partial_{\beta} \ g_{\alpha \mu} = \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ g_{\gamma \mu} + \Gamma^{\rho}_{\hphantom{\rho} \mu \beta} \ g_{\alpha \rho}

In the second term on the right hand side we can rename \rho \rightarrow \gamma and use the fact that the metric is symmetric to reverse the indices. We get

\partial_{\beta} \ g_{\alpha \mu} = \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ g_{\gamma \mu} + \Gamma^{\gamma}_{\hphantom{\gamma} \mu \beta} \ g_{\gamma \alpha}

By cyclically renaming the indices \beta, \alpha, and \mu we can generate two more similar equations. From the cyclic permutation \beta, \alpha, \mu \rightarrow \alpha, \mu, \beta we get

\partial_{\alpha} \ g_{\mu \beta} = \Gamma^{\gamma}_{\hphantom{\gamma} \mu \alpha} \ g_{\gamma \beta} + \Gamma^{\gamma}_{\hphantom{\gamma} \beta \alpha} \ g_{\gamma \mu}

and from the cyclic permutation \alpha, \mu, \beta \rightarrow \mu, \beta, \alpha we get

\partial_{\mu} \ g_{\beta \alpha} = \Gamma^{\gamma}_{\hphantom{\gamma} \beta \mu} \ g_{\gamma \alpha} + \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \mu} \ g_{\gamma \beta}

Now we add the first two equations and subtract the third to get

\partial_{\beta} \ g_{\alpha \mu} + \partial_{\alpha} \ g_{\mu \beta} - \partial_{\mu} \ g_{\beta \alpha} = 2 \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} \ g_{\gamma \mu}

where we have taken advantage of the symmetry in the lower indices of the Christoffel symbols to cancel some terms. Using the fact that

g^{\mu \gamma} \ g_{\gamma \mu} = 1

we multiply both sides by \frac{1}{2}g^{\mu \gamma} to get the final formula:

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} = \frac{1}{2}g^{\mu \gamma}[\partial_{\beta} \ g_{\alpha \mu} + \partial_{\alpha} \ g_{\mu \beta} - \partial_{\mu} \ g_{\beta \alpha}]

= \frac{1}{2}g^{\gamma \mu}[\partial_{\beta} \ g_{\alpha \mu} + \partial_{\alpha} \ g_{\mu \beta} - \partial_{\mu} \ g_{\alpha \beta}]

This is made easier to remember by noting the following facts. A factor of the inverse metric generates the Christoffel symbol’s upper index. The negative term has the symbol’s lower indices as the indices of the metric. The other two terms in the bracket are cyclic permutations of this last term.

Having derived the formula we can now employ it to calculate the eight Christoffel symbols for plane polar coordinates. We can work out the metric tensor using the distance formula

ds^2 = dx^2 + dy^2

with the conversion equations

x = r \cos \theta

y = r \sin \theta

Then

dx = \cos \theta dr - r \sin \theta d \theta

dy = \sin \theta dr + r \cos \theta d \theta

so

dx^2 = \cos^2 \theta dr^2 + r^2 \sin^2 \theta d \theta^2 - 2 r \sin \theta \cos \theta dr d \theta

dy^2 = \sin^2 \theta dr^2 + r^2 \cos^2 \theta d \theta^2 + 2 r \sin \theta \cos \theta dr d \theta

Therefore the metric in plane polar coordinates is

ds^2 = dx^2 + dy^2 = dr^2 + r^2 d \theta^2

The metric tensor is therefore

[g_{\alpha \beta}] = \begin{pmatrix} 1 & 0 \\ \ \\ 0 & r^2 \end{pmatrix}

and the inverse metric is

[g^{\alpha \beta}] = \begin{pmatrix} 1 & 0 \\ \ \\ 0 & \frac{1}{r^2} \end{pmatrix}

Now, in the formula for \Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} the indices \alpha, \beta, \gamma and \mu represent the polar coordinates r and \theta in various permutations. Inspection of [g_{\alpha \beta}] shows that the only partial derivative terms which do not equal zero are

\partial_r \ g_{\theta \theta} = \partial_r (r^2) = 2r

Inspection of [g^{\alpha \beta}] shows that this equals zero except when

g^{rr} = 1

and

g^{\theta \theta} = \frac{1}{r^2}

Substituting these values of the metric tensor components into the formula

\Gamma^{\gamma}_{\hphantom{\gamma} \alpha \beta} = \frac{1}{2} g^{\gamma \mu} [\partial_{\beta} \ g_{\alpha \mu} + \partial_{\alpha} \ g_{\mu \beta} - \partial_{\mu} \ g_{\alpha \beta}]

we get

\Gamma^{r}_{\hphantom{r} \theta \theta} = \frac{1}{2} g^{r r} \big( - \partial_r \  g_{\theta \theta}\big) = \frac{1}{2} (-2r) = -r

\Gamma^{\theta}_{\hphantom{\theta} \theta \theta} = 0

\Gamma^{\theta}_{\hphantom{\theta} \theta r} = \frac{1}{2} g^{\theta \theta} \big( \partial_r \  g_{\theta \theta}\big) = \frac{1}{2} \frac{1}{r^2} 2r = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} \theta r} = 0

\Gamma^{r}_{\hphantom{r} r r} = 0

\Gamma^{\theta}_{\hphantom{\theta} r r} = 0

\Gamma^{\theta}_{\hphantom{\theta} r \theta} = \frac{1}{2} g^{\theta \theta} \big( \partial_r \  g_{\theta \theta}\big) = \frac{1}{2} \frac{1}{r^2} 2r = \frac{1}{r}

\Gamma^{r}_{\hphantom{r} r \theta} = 0