Skip to content
← All questions

Why does the video minimize F instead of L?

Minimizing the length functional LL is described as tricky. When λ\lambda is an affine parameter, minimizing the simpler functional F=∫ABgijx˙ix˙j dλF = \int_A^B g_{ij}\dot{x}^i\dot{x}^j\,d\lambda is equivalent to minimizing LL and is much easier.

Conditions

  • λ\lambda must be an affine parameter.
  • FF is used as an easier surrogate for LL under that condition.

Reasoning, step by step

  1. Recognize that minimizing LL involves a square root, making it complex.
  2. Introduce the functional FF without the square root.
  3. Assert that for affine λ\lambda, minimizing FF is equivalent to minimizing LL.
  4. Proceed to minimize FF using the Euler-Lagrange equation.

Example

The narrator says: "However, this is tricky, but in the case that lambda is an affine parameter, minimising F given here is equivalent to minimising L and is much easier."

Common misconceptions

  • Assuming FF and LL are always equivalent; the equivalence holds only for affine parameters.
  • Thinking the square root in LL can be ignored without justification.

Watch the explanation

Connected concepts

Explore next

Related questions

Find a method

↗
Find a method

↗
Meet the concept

↗
Meet the concept

↗
Meet the concept

↗

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