# How to best calculate Christoffel symbols of metric $du^2 +g^2(u) dv^2$?

Suppose that the first fundamental form is $du^2+g^2(u)dv^2$. Calculate $\Gamma_{11}^1, \Gamma_{11}^2, \Gamma_{12}^1, \Gamma_{12}^2, \Gamma_{22}^1, \Gamma_{22}^2$.

In lectures we've been given 6 formulas for the Christoffel symbols, all of this style: $$\Gamma_{11}^1=\frac{GE_u-2FF_u+FE_v}{2(EG-F^2)}$$ but all slightly different.

We've also been given 6 equations like this: $$\Gamma_{11}^1 \cdot E + \Gamma_{11}^2 \cdot F=\frac{1}{2}E_u$$ and $$\Gamma_{11}^1 \cdot F + \Gamma_{11}^2 \cdot G=F_u-\frac{1}{2}E_v$$ Again, they're all the same style but all slightly different.

What is the best way to calculate the Christoffel symbols? Is it using either of these sets of equations or is there another way?

I ask if there is another way because these aren't the easiest of equations to memorise, especially when there are 6 (or 12) of them so another method would be appreciated.

• To avoid confusion, some authors would, in this concrete setting, index the symbols as $\Gamma^{u}_{vv}$ and similar, since your coordinates are $u,v$ and not $x^1,x^2$. – Arthur Mar 30 '16 at 15:24

The way that I calculate Christoffel symbols (when I must) is using this formula: $$\Gamma^{i}_{jk}=\sum_{m=1}^2\frac{1}{2}(g^{-1})^{im}\left(\frac{\partial g_{mj}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^m}\right),$$ where $g$ is the matrix of your first fundamental form, and $x^1=u$ and $x^2=v$ (in comparison with your coordinates). So here, each of the indices $i$, $j$, and $k$ can take the value $1$ or $2$, $g_{11}=E$, $g_{12}=g_{21}=F$, and $g_{22}=G$. Using this formula, we can calculate $\Gamma^1_{11}$ as

\begin{align*} \Gamma^1_{11}&=\frac{1}{2}(g^{-1})^{11}\left(\frac{\partial g_{11}}{\partial x^1}+\frac{\partial g_{11}}{\partial x^1}-\frac{\partial g_{11}}{\partial x^1}\right)+\frac{1}{2}(g^{-1})^{12}\left(\frac{\partial g_{21}}{\partial x^1}+\frac{\partial g_{21}}{\partial x^1}-\frac{\partial g_{11}}{\partial x^2}\right)\\ &=\frac{1}{2}\frac{G}{EG-F^2}\left(E_u+E_u-E_u\right)+\frac{1}{2}\frac{-F}{EG-F^2}\left(F_u+F_u-E_v\right)\\ &=\frac{GE_u-2FF_u+FE_v}{2(EG-F^2)}. \end{align*} In this same way, you can recover all of the formulas you received in lecture. This one way of calculating the Christoffel symbols is a little more complicated than the formulas you were given, but at least you only have to remember one equation.

Finally, notice that $\Gamma^i_{jk}=\Gamma^i_{kj}$, so you only have to memorize four formulas at most.

• I think I understand how to use most of the formula, the only thing I'm unclear on is the $(g^{-1})^{im}$ bit. Am I correct in saying that $\frac{1}{EG-F^2}=\frac{1}{\text{det}(g)}$? But then how do I know what the numerator is? i.e. where does the $G$ and then the $-F$ come from? – user320757 Mar 31 '16 at 8:14
• In fact, I think I've got it now. $(g^{-1})^{im}$ refers to the $im$ entry of the matrix $g^{-1}$ doesn't it? – user320757 Mar 31 '16 at 8:17
• Also, this might be a stupid question, but in my example, where $E=1$, say I want to calculate $\Gamma_{11}^1$, I need $E_v$. Does $E_v=0$? Just because this means lots of the Christoffel symbols in this case will then be $0$ by similar logic. – user320757 Mar 31 '16 at 8:39
• Yes, $(g^{-1})^{im}$ refers to the $im$ entry of $g^{-1}$. When $E$ is constant, it does indeed mean that $E_v=0$. And this will mean that many of the Christoffel symbols are zero, an those that are not zero will likely be pretty simple. – Alex S Mar 31 '16 at 14:09

Use the Euler-Lagrange equations. If $L = \frac{1}{2} (u'(t)^2 + g(u(t))^2 v'(t)^2)$, the equations $$\frac{\partial L}{\partial u} -\frac{d}{dt}\left(\frac{\partial L}{\partial u'}\right) = 0,\qquad \frac{\partial L}{\partial v} -\frac{d}{dt}\left(\frac{\partial L}{\partial v'}\right) = 0$$will produce the geodesic equations, from which you can read the Christoffel symbols easily. I'll do the first one, you do the other. $$g(u(t))g'(u(t))v'(t)^2 - \frac{d}{dt}\left(u'(t)\right) = 0 \implies u''(t) - g(u(t))g'(u(t)) v'(t)^2 = 0,$$so $\Gamma_{11}^1 = \Gamma_{12}^1 = 0$ and $\Gamma_{22}^1 =- g(u)g'(u)$.

In general, if $ds^2 = \sum g_{ij} du^i du^j$, the energy of a curve is given by $E = \frac{1}{2}\int_I ds^2_{\gamma(t)}(\gamma'(t),\gamma'(t))\,dt$, and every geodesic is a critical point of the energy functional. The Lagrangian $L = \frac{1}{2} ds^2_{\gamma(t)}(\gamma'(t),\gamma'(t))$ must then satisfy $$\frac{\partial L}{\partial u^k} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{u}^k}\right) =0,$$where $(u^1(t),\cdots,u^n(t))$ are the geodesic's coordinates. These equations, after some algebraic juggling show up as $\ddot{u}^k + \sum \Gamma_{ij}^k\dot{u}^i\dot{u}^j = 0$ .