Christoffel Symbol of the Second Kind

Dropping the parallel postulate in the Euclidean space axiom will give a curved space. The Euclidean space without the parallel postulate is called the Riemann space.

Derivative of Basis Set

For a general set of coordinate defined using the constant basis set $(x,y)$ $$\begin{align} \xi &= \xi(x,y) \\ \eta &= \eta(x,y). \end{align}$$ Since the general coordinate set is not neccesary constant, $$\begin{align} d\xi &= \frac{\partial \xi}{\partial x}dx + \frac{\partial \xi}{\partial y}dy \\ d\eta &= \frac{\partial \eta}{\partial x}dx + \frac{\partial \eta}{\partial y}dy, \end{align}$$ or in matrix form, $$\begin{pmatrix} d\xi \\ d\eta \end{pmatrix} = \begin{pmatrix} \frac{\partial \xi}{\partial x} &\frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial x} &\frac{\partial \eta}{\partial y} \end{pmatrix} \begin{pmatrix} dx \\ dy \end{pmatrix}.$$ To have a one-to-one mapping, the matrix should be invertible and gives 1 unique solution. So, its deteminent $$del\begin{pmatrix} \frac{\partial \xi}{\partial x} &\frac{\partial \xi}{\partial y} \\ \frac{\partial x}{\partial y} &\frac{\partial x}{\partial y} \end{pmatrix} \neq 0,$$ must be non-zero. Denote the matrix as $$\Lambda^{\alpha'}_{\beta}\equiv \begin{pmatrix} \frac{\partial \xi}{\partial x} &\frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial y} &\frac{\partial \eta}{\partial y}. \end{pmatrix}$$ Let $\phi$ be a scalar field. A one-form is defined as $$\tilde{d\phi} \to \left(\frac{\partial \phi}{\partial \xi},\frac{\partial \phi}{\partial \eta}\right).$$ Then, by chain rule, $$\frac{\partial \phi}{\partial \xi} = \frac{\partial x}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial y}{\partial \xi}\frac{\partial \phi}{\partial y}.$$ Similarly for $\eta$, we have $$\begin{align} \begin{pmatrix}\frac{\partial \phi}{\partial \xi} &\frac{\partial \phi}{\partial \eta}\end{pmatrix} = \begin{pmatrix}\frac{\partial \phi}{\partial x} &\frac{\partial \phi}{\partial y}\end{pmatrix} \begin{pmatrix} \frac{\partial \xi}{\partial x} &\frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial y} &\frac{\partial \eta}{\partial y}. \end{pmatrix} \end{align}$$ The components are given by $$(\tilde{d\phi})_{\beta'}=\Lambda^{\alpha}_{\beta'}(\tilde{d\phi})_{\alpha}.$$ Note that $\Lambda^{\alpha}_{\beta'}$ is the inverse of $\Lambda^{\alpha'}_{\beta}$.

Christoffel Symbol of the Second Kind

For an arbitrary vector, $\vec{V}$, in any vector space, its derivative in general is $$\begin{align} \frac{\partial \vec{V}}{x^\beta} &= \frac{\partial}{\partial x^{\beta}}\left(V^{\alpha}\vec{e}_\alpha\right)\\ &= \frac{\partial V^{\alpha}}{\partial x^{\beta}}\vec{e}_{\alpha} + V^{\alpha}\frac{\partial \vec{e}_\alpha}{\partial x^\beta}. \end{align}$$ As the derivative of a vector is also a vector, the term $$\frac{\partial \vec{e}_\alpha}{\partial x^\beta}$$ is also a vector and can be expressed in terms of the linear combination of the basis $$\bbox[5px,border:2px solid #666]{ \frac{\partial \vec{e}_\alpha}{\partial x^\beta} = \Gamma^{\mu}_{\alpha\beta}\vec{e}_\mu, }$$ where the $\Gamma^{\mu}_{\alpha\beta}$ is called the Christoffel symbol of the second kind. It is the $\mu$th component of $\frac{\partial \vec{e}_\alpha}{\partial x^\beta}$.

• $\alpha$ is the $\alpha$th component of the vector being differentiated
• $\beta$ is the $\beta$th component of the vector differentiated with respect to
• $\mu$ is the $\mu$th component of the newly differentiated vector

Example

For polar coordinate basis in 2D Euclidean space $$\begin{align} &\frac{\partial\vec{e}_r}{\partial r} = 0 \imply \Gamma^{\mu}_{rr} = 0, \prime \mu \\ &\frac{\partial vec{e}_r}{\partial \theta} = \frac{1}{r}\vec{e}_\theta \imply \Gamma^{r}_{r\theta} = 0, \,\Gamma^\theta_{r\theta}=\frac{1}{r}\\ &\frac{\partial \vec{e}_\theta}{\partial r} = \frac{1}{r}\vec{e}_\theta \imply \Gamma^r_{\theta r}=0,\,\Gamma^\theta_{\theta r} = \frac{1}{r}\\ &= \frac{\partial \vec{e}_\theta}{\partial \theta} = -r\vec{e}_r \imply \Gamma^r_{\theta\theta} = -r,\,\Gamma^\theta_{\tehta\theta}=0 \end{align}$$

Covariant Derivative

With the defintion of Christoffel symbol, the derivative of the vector $\vec{V}$ is $$\frac{\partial\vec{V}}{\partial x^\beta}=\frac{\partial V^\alpha}{\partial x^\beta}\vec{e}_\alpha+V^\alpha\Gamma^\mu_{\alpha\beta}\vec{e}_\mu.$$ The second term is sum $\alpha$ and sum $\mu$. So, the dummy variables can be relabelled $$\begin{align} \frac{\partial\vec{V}}{\partial x^\beta} &= \frac{\partial V^\alpha}{\partial x^\beta}\vec{e}_\alpha + V^\mu\Gamma^\alpha_{\mu\beta}\vec{e}_\alpha \\ &= \left(\frac{\partial V^\alpha}{\partial x^\beta}+V^\mu\Gamma^\alpha_{\mu\beta}\right)\vec{e}_\alpha. \end{align}$$ Define the covariant derivative of a vector, $\vec{V}$, as $$\bbox[5px,border:2px solid #666] { V^\alpha_{;\beta} := V^\alpha_{,\beta} + V^\mu\Gamma^\alpha_{\mu\beta}. }$$ Then, $$\frac{\partial \vec{V}}{\partial x^\beta}=V^\alpha_{;\beta}\vec{e}_\alpha.$$ Consider a scalar $$\phi = p_\alpha V^\alpha,$$ where $p_\alpha$ is the component of a one-form $\tilde{p}$ and $V^\alpha$ is the component of a vector $\vec{V}$. Then, the derivative of $\phi$, by product rule, is $$\begin{align} \nabla_\beta\phi &= \phi_{,\beta} = \frac{\partial p_\alpha}{\partial x^\beta}V^\alpha + p_\alpha\frac{\partial V^\alpha}{\partial x^\beta} \\ &= \frac{\partial p_\alpha}{\partial x^\beta}V^\alpha + p_\alpha V^\alpha_{;\beta} - p_\alpha V^\mu\Gamma^\alpha_{\mu\beta} \\ &= \left(\frac{\partial p_\alpha}{\partial x^\beta}-p_\mu\Gamma^\mu_{\alpha\beta}\right)V^\alpha + p_\alpha V^\alpha_{;\beta}. \end{align}$$ Define the covariant derivative of a one-form, $\tilde{p}$, as $$\bbox[5px,border:2px solid #666] { p_{\alpha ;\beta} := (\nabla_\beta\tilde{p})_\alpha := (\nabla \tilde{p})_{\alpha\beta} = p_{\alpha,\beta}-p_\mu\Gamma^\mu_{\alpha_beta}. }$$ For a (0, 2) tensor, similar derivation can be done with two vectors $$\nabla_\beta T_{\mu\nu} = T_{\mu\nu,\beta} - T_{\alpha\nu}\Gamma^\alpha_{\mu\beta} - T_{\mu\alpha}\Gamma^\alpha_{\nu\beta}.$$ In general, $$\begin{align} \nabla_\beta T_{\mu\nu} &= T_{\mu\nu,\beta}-T_{\alpha\nu}\Gamma^\alpha_{\mu\beta}-T_{\mu\alpha}\Gamma^\alpha_{\nu\beta}\\ \nabla_\beta A^{\mu\nu} &= A^{\mu\nu}_{,\beta} + A^{\alpha\nu}\Gamma^\mu_{\alpha\beta} + A^{\mu\alpha}\Gamma^\nu_{\alpha\beta}\\ \nabla_\beta B^\mu_\nu &= B^\mu_{\nu,\beta} + B^\alpha_nu\Gamma^\mu_{\alpha\beta} - B^\mu_\alpha \Gamma^\alpha_{\nu\beta}. \end{align}$$

Symmetry of Christoffel Symbol of 2nd Kind in Euclidean space

For any arbitrary scalar field $\phi$, its gradient $\nabla_\beta\phi$ is a one-form with components $\phi_{,\beta}$. The second derivative of its gradient is a tensor with components $$\phi_{,\beta;\alpha} = \phi_{,\beta,\alpha}-\phi_{,\mu}\Gamma^\mu_{\beta\alpha}.$$ In Cartiesian coordinate of Euclidean space, the covariant derivative is the same as directional derivative, and since $$\phi_{,\beta,\alpha} = \frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial x^\beta}\phi= \frac{\partial}{\partial x^\beta}\frac{\partial}{\partial x^\alpha}\phi,$$ the tensor is symmetric. As a tensor is a function and changing basis will not alter the output value, if it is symmetric in Cartesian coordinate, it is symmetric in other coordinate system as well. Therefore, $$\frac{d}{dx^\alpha}\frac{d}{dx^\beta}\phi = \frac{d}{dx^\beta}\frac{d}{dx^\alpha}\phi$$ Since $$\frac{d}{dx^\alpha}\frac{d}{dx^\beta}\phi = \frac{d}{dx^\beta}\frac{d}{dx^\alpha}\phi,$$ we have $$\begin{align} \phi_{,\beta,\alpha}-\phi_{,\mu}\Gamma^\mu_{\beta\alpha} &= \phi_{,\alpha,\beta}-\phi_{,\mu}\Gamma^\mu_{\alpha\beta} \\ \Gamma^\mu_{\alpha\beta}\phi_{,\mu} &= \Gamma^\mu_{\beta\alpha}\phi_{,\mu} \Gamma^\mu_{\alpha\beta} &= \Gamma^\mu_{\beta\alpha}. \end{align}$$ So, the Christoffel symbol of the 2nd kind is symmetric in lower index.

Finding Christoffel Symbol

If the Christoffel symbol is symmetric in a vector space, we can make use of the symmetry property to find its expression in terms of the metric tensor. Substituting the metric tensor into the definition of covariant derivative of (0,2) tensor, we have $$\begin{align} g_{\alpha\beta,\mu} &= \Gamma^\mu_{\alpha\mu}g_{\nu\beta} + \Gamma^\nu_{\beta\mu}g_{\alpha\nu} \\ g_{\alpha\mu,\beta} &= \Gamma^\nu_{\alpha\beta}g_{\nu\mu}+\Gamma^nu_{\mu\beta}g_{\alpha\nu} \\ g_{\beta\mu,\alpha} &= -\Gamma^nu_{\beta\alpha}g_{\nu\mu} - \Gamma^\nu_{\mu\alpha}g_{\beta\nu}. \end{align}$$ If the metric tensor is symmetric, that is $$g_{\beta\nu}=g_{\nu\beta}.$$ Then, $$\begin{align} g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha}\\ &= \left(\Gamma^\nu_{\alpha\mu}-\Gamma^\nu_{\mu\alpha}\right)g_{\nu\beta} + \left(\Gamma^\nu_{\alpha\beta}-\Gamma^\nu_{\beta\alpha}\right)g_{\nu\mu} + \left(\Gamma^\nu_{\beta\mu}+\Gamma^\nu_{\mu\beta}\right)g_{\alpha\nu}\\. \end{align}$$ Because of the symmtry of lower index of the Christoffel symbol, the first two terms vanish and the two Christoffel symbols of the third term are identical. Therefore, we have $$g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha}= 2g_{\alpha\nu}\Gamma^\nu_{\beta\mu}.$$ Since $$g^{\alpha\gamma}g_{\alpha\nu}=\delta^\gamma_\nu,$$ the Christoffel symbol is given by $$\bbox[5px,border:2px solid #666] { \Gamma^\gamma_{\beta\mu} = \frac{1}{2}g^{\alpha\gamma}\left(g_{\alpha\beta,\mu}+g_{\alpha\mu,\beta}-g_{\beta\mu,\alpha}\right) }$$