Then (20) can be replaced by
{ λ₃
{ ∫δω dλ = 0
(20a) { λ₁
{
{ ω² = gμν(dxμ/dλ)(dxν/dλ)
But
δω = (1/ω){½(∂gμν/∂xσ) · (dxμ/dλ) · (dxν/dλ) · δxσ
+ gμν(dxμ/dλ)δ(dxν/dλ)}
So we get by the substitution of δω in (20a), remembering that
Then (20) can be replaced by
{ λ₃
{ ∫δω dλ = 0
(20a) { λ₁
{
{ ω² = gμν(dxμ/dλ)(dxν/dλ)
But
δω = (1/ω){½(∂gμν/∂xσ) · (dxμ/dλ) · (dxν/dλ) · δxσ
+ gμν(dxμ/dλ)δ(dxν/dλ)}
So we get by the substitution of δω in (20a), remembering that