← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Vortex motion; circulation · Mechanics & Fluid Dynamics · asked 6× in 13 yrs · Read the full method →

Question

Prove that the necessary and sufficient condition that the vortex lines may be at right angles to the stream lines are

u,v,w=μ ⁣(∂ϕ∂x,∂ϕ∂y,∂ϕ∂z),u,v,w=\mu\!\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right),

where μ\mu and ϕ\phi are functions of x,y,z,tx,y,z,t.

Technique

Sufficiency by direct curl computation + scalar triple product; necessity via Frobenius integrability.

Solution

Setup. Velocity V⃗=(u,v,w)\vec V=(u,v,w). Stream lines are tangent to V⃗\vec V; vortex lines are tangent to ω⃗=∇×V⃗\vec\omega=\nabla\times\vec V (vorticity).

Stream and vortex lines are perpendicular at each point iff V⃗⋅ω⃗=0\vec V\cdot\vec\omega=0 at each point (and time).

We prove V⃗⋅ω⃗=0  ⟺  V⃗=μ∇ϕ\vec V\cdot\vec\omega=0\iff\vec V=\mu\nabla\phi for some scalar functions μ,ϕ\mu,\phi of (x,y,z,t)(x,y,z,t).

Sufficiency (⇐\Leftarrow): V⃗=μ∇ϕ\vec V=\mu\nabla\phi implies V⃗⋅ω⃗=0\vec V\cdot\vec\omega=0

Compute the curl using the product rule:

ω⃗=∇×(μ∇ϕ)=(∇μ)×(∇ϕ)+μ (∇×∇ϕ)=(∇μ)×(∇ϕ),\vec\omega=\nabla\times(\mu\nabla\phi)=(\nabla\mu)\times(\nabla\phi)+\mu\,(\nabla\times\nabla\phi)=(\nabla\mu)\times(\nabla\phi),

since ∇×∇ϕ=0\nabla\times\nabla\phi=0 for any scalar ϕ\phi.

Now compute V⃗⋅ω⃗\vec V\cdot\vec\omega:

V⃗⋅ω⃗=(μ∇ϕ)⋅[(∇μ)×(∇ϕ)]=μ[∇ϕ⋅(∇μ×∇ϕ)]=μ⋅0=0,\vec V\cdot\vec\omega=(\mu\nabla\phi)\cdot[(\nabla\mu)\times(\nabla\phi)]=\mu\bigl[\nabla\phi\cdot(\nabla\mu\times\nabla\phi)\bigr]=\mu\cdot 0=0,

since the scalar triple product with a repeated factor (∇ϕ\nabla\phi appears twice) vanishes.

So V⃗⊥ω⃗\vec V\perp\vec\omega everywhere ✓.

Necessity (⇒\Rightarrow): V⃗⋅ω⃗=0\vec V\cdot\vec\omega=0 implies V⃗=μ∇ϕ\vec V=\mu\nabla\phi

Strategy. This is the Frobenius integrability theorem applied to the 1-form dual to V⃗\vec V. The condition V⃗⋅(∇×V⃗)=0\vec V\cdot(\nabla\times\vec V)=0 is the integrability condition for the distribution of planes perpendicular to V⃗\vec V.

Argument. Consider the 1-form α=u dx+v dy+w dz\alpha=u\,dx+v\,dy+w\,dz. Its exterior derivative is

dα=(wy−vz) dy∧dz+(uz−wx) dz∧dx+(vx−uy) dx∧dy.d\alpha=(w_y-v_z)\,dy\wedge dz+(u_z-w_x)\,dz\wedge dx+(v_x-u_y)\,dx\wedge dy.

The Frobenius integrability condition for the distribution {α=0}\{\alpha=0\} (tangent planes perpendicular to V⃗\vec V) is α∧dα=0\alpha\wedge d\alpha=0, which expands to

u(wy−vz)+v(uz−wx)+w(vx−uy)=0.u(w_y-v_z)+v(u_z-w_x)+w(v_x-u_y)=0.

This is precisely V⃗⋅ω⃗=0\vec V\cdot\vec\omega=0.

When Frobenius holds, there exist scalar functions ϕ\phi and μ\mu such that α=μ dϕ\alpha=\mu\,d\phi — equivalently (u,v,w)=μ(ϕx,ϕy,ϕz)=μ∇ϕ(u,v,w)=\mu(\phi_x,\phi_y,\phi_z)=\mu\nabla\phi. The level surfaces ϕ=\phi= const are tangent everywhere to the distribution {α=0}\{\alpha=0\}, and V⃗\vec V is normal to these surfaces (scaled by μ\mu).

So V⃗⋅ω⃗=0\vec V\cdot\vec\omega=0 implies V⃗=μ∇ϕ\vec V=\mu\nabla\phi.

Combining

Answer

  Vortex lines⊥stream lines  ⟺  V⃗=μ∇ϕ.  \boxed{\;\text{Vortex lines}\perp\text{stream lines}\iff\vec V=\mu\nabla\phi.\;}
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