13. ISOGONAL NETS

Explanation 114 : We can represent a curve on the surface x(u₁,u₂) by the equation φ(u₁,u₂) = λ ; for example (u cos(v), u sin(v), u³) and u² + v = 4 give the curve with parameter representation (u cos(4-u²), u sin(4-u²), u³).
When a family of curves is such that through every point of the surface there goes exactly one of these curves, and it is given by the equations φ(u₁,u₂) = λ, λ ∈ I (where φ is a fixed function and I an interval on the real axis), we call it a system of curves on the surface.
Now we can ask, for instance, for another system of curves on the surface such that through every point of the surface there goes exactly one curve of both systems, whilst these two curves intersect under a fixed angle.
Then we call the pair of systems an isogonal net. In the special case that this fixed angle is 90 degrees, we call it an orthogonal net and the two systems each other's orthogonal trajectories.

Explanation 115 : Let the curve x(u₁(t),u₂(t)) on the surface also be given by an equation φ(u₁,u₂) = λ.
Then we have for all t : φ(u₁(t),u₂(t)) = λ, and hence φ_u1du₁/dt + φ_u2du₂/dt = 0.
So the tangent vector x_u1du₁/dt + x_u2du₂/dt has the same or opposite direction as -φ_u2x_u1 + φ_u1x_u2.

Explanation 116 : We find the isogonal trajectories of a system φ(u₁,u₂) = λ on the surface x(u₁,u₂) as follows:
let x(u₁(t),u₂(t)) be such a trajectory, and α the fixed angle; then we must have (see 107 ii))

cos(α) = (-φ_u2du₁ a_{1 1} -φ_u2du₂ a_{1 2} +φ_u1du₁ a_{1 2} +φ_u1du₂ a_{2 2})/(√(a_{1 1}φ_u2² - 2a_{1 2}φ_u1φ_u2 + a_{2 2}φ_u1²).√(a_{1 1}du₁² + 2 a_{1 2}du₁ du₂ + a_{2 2}du₂²)).

When we solve this differential equation, we get the isogonal trajectories with angle α in the form ψ(u₁,u₂) = μ.
In the special case that α is ninety degrees, the differential equation becomes

du₁/du₂ = (a_{2 2}φ_u1 - a_{1 2}φ_u2)/(a_{1 1}φ_u2 - a_{1 2}φ_u1).

Problem 117 : Determine the orthogonal trajectories of the v-lines u=λ of (u+v,u-v,uv).

Problem 118 : Determine the curves on the sphere that make a fixed angle α with the parallels of latitude (we call them the loxodromes of the sphere).

Explanation 119 : A quadratic equation in du₁ and du₂, say c_{1 1}du₁² + 2 c_{1 2}du₁ du₂ + c_{2 2}du₂² = 0, determines in each point (u₁,u₂) two directions (if the roots of the equation are real).
Each of both solutions du₁/du₂ = ψ_ui(u₁,u₂) (i=1,2) determines a system of curves on the surface, so the quadratic form determines a net . For example, du₁ du₂ = 0 determines the net of the parameter lines.

Proposition 120 : The net c_{1 1}du₁² + 2 c_{1 2}du₁ du₂ + c_{2 2}du₂² = 0 on a surface with first fundamental form a_{1 1}du₁² + 2 a_{1 2}du₁ du₂ + a_{2 2}du₂² is orthogonal if and only if a_{1 1}c_{2 2} - 2 a_{1 2}c_{1 2} + a_{2 2}c_{1 1} = 0.

Proof : In 'each' point the quadratic equation c_{1 1} + 2 c_{1 2}du₂/du₁ + c_{2 2}(du₂/du₁)² = 0 has two solutions du₂/du₁ = ψ₁ and du₂/du₁ = ψ₂, where ψ₁+ψ₂ = -2c_{1 2}/c_{2 2} and ψ₁ψ₂ = c_{1 1}/c_{2 2}.
The requirement of orthogonality becomes (x_u1 + x_u2ψ₁).(x_u1 + x_u2ψ₂) = 0, so a_{1 1} - a_{1 2}(ψ₁+ψ₂) + a_{2 2}ψ₁ψ₂ = 0, so a_{1 1}c_{2 2} - 2 a_{1 2}c_{1 2} + a_{2 2}c_{1 1} = 0.

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