Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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 \[ \frac{\partial^2\gamma_1}{\partial t^2} = \frac{\partial^2\gamma_2}{\partial t^2} = 0. \]

share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.