# Geodesic on a plane

I guess that each geodesic on a plane is a straight line. Is it right? What can I use to prove it? I guess I have to use somehow Levi-Civita connection.

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## 3 Answers

This just adds a few words and references, but maybe it will help you organize your thoughts. If you use the standard coordinates for $\mathbf R^2$, then the Euclidean metric has constant matrix $$(g_{ij}) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ Therefore, the Christoffel symbols of the Levi-Civita connection are all zero. If you trace through the definition of the covariant derivative along a curve $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ applied to the tangent vector field of $\gamma$, then you end up with the geodesic equation, which in this case requires that $$\frac{\partial^2\gamma_1}{\partial t^2} = \frac{\partial^2\gamma_2}{\partial t^2} = 0.$$

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I'm very willing to expand upon the proofs, but I didn't want to write a chapter of Riemannian geometry out. Let me know! – Dylan Moreland Jan 15 '12 at 18:37

I think it is quite simple, it is a classical example of calculus of variations. You find a function that minimizes a Lagrangian that represents the function's length. You use Euler-Lagrange equation to find that the function's second derivative is zero, concluding that it must be a straight line.

For a full proof, see here.

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Similar to what Dylan said, let $\mathbf{u}$ be a tangent vector to a path $\gamma$, then $\nabla_{u}u=0$ describes geodesic motion. Then noting that the Christoffel symbols vanish, we have $\partial_{A}u^{B}=0$. If we parameterize in $\tau$ we have $$\frac{d u^{x}(\tau)}{d\tau}=0$$ and $$\frac{du^y(\tau)}{d\tau}=0$$ which imples $u^x=v_{0}^x$ and $u^y=v_{0}^y$ and $x(\tau)=v_{0}^x\tau+x_0$ and $y(\tau)=v_{0}^y\tau +y_0$. These are lines.

I hope this helps.

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