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How is the geodesic equation obtained from the Euler-Lagrange equation?

The geodesic equation is obtained by applying the Euler-Lagrange equation to the Lagrangian L=gijx˙ix˙j\mathscr{L} = g_{ij}\dot{x}^i\dot{x}^j. The resulting differential equation for the curve components is d2xkdλ2+Γijkdxidλdxjdλ=0\frac{d^2x^k}{d\lambda^2} + \Gamma^k_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda} = 0, where Γijk\Gamma^k_{ij} are the connection coefficients.

Conditions

  • λ\lambda is an affine parameter.
  • Γijk\Gamma^k_{ij} are connection coefficients.
  • The equation is presented as the general solution of the Euler-Lagrange step for F.

Reasoning, step by step

  1. Start with the Euler-Lagrange equation: ddλ(∂L∂x˙k)=∂L∂xk\frac{d}{d\lambda}\left(\frac{\partial\mathscr{L}}{\partial\dot{x}^k}\right) = \frac{\partial\mathscr{L}}{\partial x^k}.
  2. Substitute the Lagrangian L=gijx˙ix˙j\mathscr{L} = g_{ij}\dot{x}^i\dot{x}^j.
  3. Perform the differentiation and algebraic manipulation (omitted in the video).
  4. Arrive at the geodesic equation: d2xkdλ2+Γijkdxidλdxjdλ=0\frac{d^2x^k}{d\lambda^2} + \Gamma^k_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda} = 0.

Example

The video states: "To minimise this integral, we can use the Euler-Lagrange equation ... The general solution to this equation is this ... This equation is called the geodesic equation".

Common misconceptions

  • Assuming the video shows the full algebraic derivation; it skips the intermediate steps.
  • Thinking the connection coefficients Γijk\Gamma^k_{ij} are defined in this clip; the narrator explicitly says they will not be treated in detail.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.