In general relativity The Christoffel symbols find frequent use in Einstein's theory of
general relativity, where
spacetime is represented by a curved 4-dimensional
Lorentz manifold with a
Levi-Civita connection. The
Einstein field equations—which determine the geometry of spacetime in the presence of matter—contain the
Ricci tensor, and so calculating the Christoffel symbols is essential. Once the geometry is determined, the paths of particles and light beams are calculated by solving the
geodesic equations in which the Christoffel symbols explicitly appear.
In classical (non-relativistic) mechanics Let x^i be the generalized coordinates and \dot{x}^i be the generalized velocities, then the kinetic energy for a unit mass is given by T = \tfrac{1}{2} g_{ik}\dot{x}^i \dot{x}^k, where g_{ik} is the
metric tensor. If V\left(x^i\right), the potential function, exists then the contravariant components of the generalized force per unit mass are F_i = \partial V/\partial x^i. The metric (here in a purely spatial domain) can be obtained from the line element ds^2 = 2T dt^2. Substituting the Lagrangian L = T - V into the
Euler-Lagrange equation, we get g_{ik}\ddot{x}^k + \frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^l} + \frac{\partial g_{il}}{\partial x^k} - \frac{\partial g_{lk}}{\partial x^i}\right) \dot{x}^l \dot{x}^k = F_i. Now multiplying by g^{ij}, we get \ddot{x}^j + {\Gamma^j}_{lk} \dot{x}^l \dot{x}^k = F^j. When Cartesian coordinates can be adopted (as in inertial frames of reference), we have an Euclidean metrics, the Christoffel symbol vanishes, and the equation reduces to
Newton's second law of motion. In curvilinear coordinates (forcedly in non-inertial frames, where the metrics is non-Euclidean and not flat), fictitious forces like the
Centrifugal force and
Coriolis force originate from the Christoffel symbols, so from the purely spatial curvilinear coordinates.
In Earth surface coordinates Given a
spherical coordinate system, which describes points on the Earth surface (approximated as an ideal sphere). \begin{align} x(R, \theta, \varphi) &= \begin{pmatrix} R\cos\theta\cos\varphi & R\cos\theta\sin\varphi & R\sin\theta \end{pmatrix} \\ \end{align} For a point x, is the distance to the Earth core (usually approximately the
Earth radius). and are the
latitude and
longitude. Positive is the northern hemisphere. To simplify the derivatives, the angles are given in
radians (where d sin(x)/dx = cos(x), the degree values introduce an additional factor of 360 / 2 pi). At any location, the tangent directions are e_{R} (up), e_{\theta} (north) and e_{\varphi} (east) - you can also use indices 1,2,3. \begin{align} e_{R} &= \begin{pmatrix} \cos\theta\cos\varphi & \cos\theta\sin\varphi & \sin\theta \end{pmatrix} \\ e_{\theta} &= R \cdot \begin{pmatrix} -\sin\theta\cos\varphi & - \sin\theta\sin\varphi & \cos\theta \end{pmatrix} \\ e_{\varphi} &= R\cos\theta \cdot \begin{pmatrix} -\sin\varphi & \cos\varphi & 0 \end{pmatrix} \\ \end{align} The related
metric tensor has only diagonal elements (the squared vector lengths). This is an advantage of the coordinate system and not generally true. \begin{align} g_{RR} = 1 \qquad & g_{\theta\theta} = R^2 \qquad & g_{\varphi\varphi} = R^2\cos^2\theta \qquad & g_{ij} = 0 \quad \mathrm{else} \\ g^{RR} = 1 \qquad & g^{\theta\theta} = 1/R^2 \qquad & g^{\varphi\varphi} = 1/(R^2\cos^2\theta) \qquad & g^{ij} = 0 \quad \mathrm{else} \\ \end{align} Now the necessary quantities can be calculated. Examples: \begin{align} e^{R} = e_{R} g^{RR} = 1 \cdot e_{R} &= \begin{pmatrix} \cos\theta\cos\varphi & \cos\theta\sin\varphi & \sin\theta \end{pmatrix} \\ {\Gamma^R}_{\varphi \varphi} = e^{R} \cdot \frac{\partial}{\partial \varphi} e_\varphi &= e^{R} \cdot \begin{pmatrix} -R\cos\theta\cos\varphi & -R\cos\theta\sin\varphi & 0 \end{pmatrix} = -R\cos^2\theta \\ \end{align} The resulting Christoffel symbols of the second kind {\Gamma^k}_{ji} = e^k \cdot \frac{\partial e_j}{\partial x^i} then are (organized by the "derivative" index in a matrix): \begin{align} \begin{pmatrix} {\Gamma^R}_{RR} & {\Gamma^R}_{\theta R} & {\Gamma^R}_{\varphi R} \\ {\Gamma^\theta }_{RR} & {\Gamma^\theta }_{\theta R} & {\Gamma^\theta }_{\varphi R} \\ {\Gamma^\varphi }_{RR} & {\Gamma^\varphi }_{\theta R} & {\Gamma^\varphi }_{\varphi R} \\ \end{pmatrix} &= \quad \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1/R & 0 \\ 0 & 0 & 1/R \end{pmatrix} \\ \begin{pmatrix} {\Gamma^R}_{R\theta } & {\Gamma^R}_{\theta \theta } & {\Gamma^R}_{\varphi \theta } \\ {\Gamma^\theta }_{R\theta } & {\Gamma^\theta }_{\theta \theta } & {\Gamma^\theta }_{\varphi \theta } \\ {\Gamma^\varphi }_{R\theta } & {\Gamma^\varphi }_{\theta \theta } & {\Gamma^\varphi }_{\varphi \theta } \\ \end{pmatrix} \quad &= \begin{pmatrix} 0 & -R & 0 \\ 1/R & 0 & 0 \\ 0 & 0 & -\tan\theta \end{pmatrix} \\ \begin{pmatrix} {\Gamma^R}_{R\varphi } & {\Gamma^R}_{\theta \varphi } & {\Gamma^R}_{\varphi \varphi } \\ {\Gamma^\theta }_{R\varphi } & {\Gamma^\theta }_{\theta \varphi } & {\Gamma^\theta }_{\varphi \varphi } \\ {\Gamma^\varphi }_{R\varphi } & {\Gamma^\varphi }_{\theta \varphi } & {\Gamma^\varphi }_{\varphi \varphi } \\ \end{pmatrix} &= \quad \begin{pmatrix} 0 & 0 & -R\cos^2\theta \\ 0 & 0 & \cos\theta\sin\theta \\ 1/R & -\tan\theta & 0 \end{pmatrix} \\ \end{align} These values show how the tangent directions (columns: e_{R}, e_{\theta}, e_{\varphi}) change, seen from an outside perspective (e.g. from space), but given in the tangent directions of the actual location (rows: , , ). As an example, take the nonzero derivatives by in {\Gamma^k}_{j\ \theta}, which corresponds to a movement towards north (positive dθ): • The new north direction e_{\theta} changes by -R dθ in the up (R) direction. So the north direction will rotate downwards towards the center of the Earth. • Similarly, the up direction e_{R} will be adjusted towards the north. The different lengths of e_{R} and e_{\theta} lead to a factor of 1/R . • Moving north, the east tangent vector e_{\varphi} changes its length (-tan(θ) on the diagonal), it will shrink (-tan(θ) dθ 0) on the southern hemisphere. These effects are maybe not apparent during the movement, because they are the adjustments that keep the measurements in the coordinates , , . Nevertheless, it can affect distances, physics equations, etc. So if e.g. you need the exact change of a
magnetic field pointing approximately "south", it can be necessary to also
correct your measurement by the change of the north direction using the Christoffel symbols to get the "true" (
tensor) value. The Christoffel symbols of the first kind {\Gamma_l}_{ji} = g_{lk} {\Gamma^k}_{ji} show the same change using metric-corrected coordinates, e.g. for derivative by : \begin{align} \begin{pmatrix} {\Gamma_R}_{R\varphi } & {\Gamma_R}_{\theta \varphi } & {\Gamma_R}_{\varphi \varphi } \\ {\Gamma_\theta }_{R\varphi } & {\Gamma_\theta }_{\theta \varphi } & {\Gamma_\theta }_{\varphi \varphi } \\ {\Gamma_\varphi }_{R\varphi } & {\Gamma_\varphi }_{\theta \varphi } & {\Gamma_\varphi }_{\varphi \varphi } \\ \end{pmatrix} &= R\cos\theta \begin{pmatrix} 0 & 0 & -\cos\theta \\ 0 & 0 & R\sin\theta \\ \cos\theta & -R\sin\theta & 0 \end{pmatrix} \\ \end{align}
Lagrangian approach at finding a solution In cylindrical coordinates, Cartesian and cylindrical polar coordinates exist as: \begin{cases}x = r \cos\varphi \\ y = r \sin\varphi \\ z = h \end{cases} and \begin{cases} r = \sqrt{x^2 + y^2}\\ \varphi = \arctan\left(\frac{y}{x}\right) \\ h = z \end{cases} Cartesian points exist and Christoffel Symbols vanish as time passes, therefore, in cylindrical coordinates: \Gamma^r_{rr} = \Gamma^r_{\varphi r} = \frac{\partial^2 x}{\partial r^2} \frac{\partial r}{\partial x} + \frac{\partial^2 y}{\partial r^2} \frac{\partial r}{\partial y} + \frac{\partial^2 z}{\partial r^2} \frac{\partial r}{\partial z} = 0 \Gamma^r_{r \varphi} = \Gamma^r_{\varphi r} = \frac{\partial^2 x}{\partial r \partial \varphi} \frac{\partial r}{\partial x} + \frac{\partial^2 y}{\partial r \partial \varphi} \frac{\partial r}{\partial y} + \frac{\partial^2 z}{\partial r \partial \varphi} \frac{\partial r}{\partial z} = - \sin \varphi \cos \varphi + \sin \varphi \cos \varphi = 0 \Gamma^r_{\varphi \varphi} = \frac{\partial^2 x}{\partial \varphi^2} \frac{\partial r}{\partial x} + \frac{\partial^2 y}{\partial \varphi^2} \frac{\partial r}{\partial y} + \frac{\partial^2 z}{\partial \varphi^2} \frac{\partial r}{\partial z} = -\frac{x}{r} - \frac{y}{r} = -r \Gamma^\varphi_{rr} = \Gamma^\varphi_{\varphi r} = \frac{\partial^2 x}{\partial r^2} \frac{\partial \varphi}{\partial x} + \frac{\partial^2 y}{\partial r^2} \frac{\partial \varphi}{\partial y} + \frac{\partial^2 z}{\partial r^2} \frac{\partial \varphi}{\partial z} = 0 \Gamma^\varphi_{r \varphi} = \Gamma^\varphi_{\varphi r} = \frac{\partial^2 x}{\partial r \partial \varphi} \frac{\partial \varphi}{\partial x} + \frac{\partial^2 y}{\partial r \partial \varphi} \frac{\partial \varphi}{\partial y} + \frac{\partial^2 z}{\partial r \partial \varphi} \frac{\partial \varphi}{\partial z} = -\frac{y}{r^2} + \cos \varphi \frac{x}{r^2} = \frac{1}{r} \Gamma^\varphi_{\varphi \varphi} = \frac{\partial^2 x}{\partial \varphi^2} \frac{\partial \varphi}{\partial x} + \frac{\partial^2 y}{\partial \varphi^2} \frac{\partial \varphi}{\partial y} + \frac{\partial^2 z}{\partial \varphi^2} \frac{\partial \varphi}{\partial z} = -\frac{x}{r^2} - \frac{y}{r^2} = 0
Spherical coordinates (using Lagrangian 2x2x2) ds^2 = d\theta^2 + \sin^2\theta d\phi^2 The Lagrangian can be evaluated as: L = \dot{\theta}^2 + \sin^2\theta\dot{\phi}^2 Hence, \begin{cases}\ddot{\phi} + 2\frac{\cos\theta}{\sin\theta}\dot{\theta}\dot{\phi} = 0 \\ \ddot{\theta} - \sin\theta\cos\theta\dot{\phi}^2 = 0 \\ \frac{d^2x^k}{d\lambda^2} + \Gamma^k_{ij} \frac{dx^i}{d\lambda} \frac{dx^j}{d\lambda} = 0 \\ \frac{\partial L}{\partial\ddot{\theta}} = 0\end{cases} can be rearranged to \begin{cases}\ddot{\phi} + 2\frac{\cos\theta}{\sin\theta}\dot{\theta}\dot{\phi} = 0 \\ \ddot{\theta} - \sin\theta\cos\theta\dot{\phi}^2 = 0\end{cases} By using the following geodesic equation: \frac{d^2x^k}{d\lambda^2} + \Gamma^k_{ij} \frac{dx^i}{d\lambda} \frac{dx^j}{d\lambda} = 0 The following can be obtained: \Gamma_{22}^1 = -\sin\theta\cos\theta(\Gamma_{12}^2) = \Gamma_{21}^2\frac{\cos\theta}{\sin\theta} == Lagrangian mechanics in geodesics (principles of least action in Christoffel symbols) ==