Intuition The big picture (WHY this exists)A 2D incompressible flow has a velocity field V ⃗ = ( u , v ) \vec{V} = (u,v) V = ( u , v ) two functions of position. But these two functions are not independent — they are tied together by physics:
Incompressibility (mass conservation, ∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0 ψ \psi ψ Irrotationality (no spin, ∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 ϕ \phi ϕ
So instead of solving for two velocity components, we solve for one scalar field . That's the whole point: fewer unknowns, easier math.
Definition Velocity potentialIf a flow is irrotational (∇ × V ⃗ = 0 \nabla\times\vec V = 0 ∇ × V = 0 ϕ \phi ϕ V ⃗ = ∇ ϕ ⇒ u = ∂ ϕ ∂ x , v = ∂ ϕ ∂ y . \vec V = \nabla\phi \qquad\Rightarrow\qquad u=\frac{\partial\phi}{\partial x},\quad v=\frac{\partial\phi}{\partial y}. V = ∇ ϕ ⇒ u = ∂ x ∂ ϕ , v = ∂ y ∂ ϕ . potential flow .
WHY does ϕ \phi ϕ From vector calculus: the curl of a gradient is always zero, ∇ × ( ∇ ϕ ) = 0 \nabla\times(\nabla\phi)=0 ∇ × ( ∇ ϕ ) = 0 if ∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 then V ⃗ \vec V V ∇ ϕ \nabla\phi ∇ ϕ
Intuition Why is it like the electric potential?Same math as E ⃗ = − ∇ V \vec E=-\nabla V E = − ∇ V ϕ \phi ϕ + + + increasing ϕ \phi ϕ
Definition Stream function (2D incompressible)If a flow is incompressible (∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0 ψ \psi ψ u = ∂ ψ ∂ y , v = − ∂ ψ ∂ x . u=\frac{\partial\psi}{\partial y},\qquad v=-\frac{\partial\psi}{\partial x}. u = ∂ y ∂ ψ , v = − ∂ x ∂ ψ .
WHY this works (DERIVE it): Continuity in 2D is
∂ u ∂ x + ∂ v ∂ y = 0. \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0. ∂ x ∂ u + ∂ y ∂ v = 0. u , v u,v u , v automatically satisfied. Try u = ∂ ψ / ∂ y u=\partial\psi/\partial y u = ∂ ψ / ∂ y v = − ∂ ψ / ∂ x v=-\partial\psi/\partial x v = − ∂ ψ / ∂ x ∂ ∂ x ( ∂ ψ ∂ y ) + ∂ ∂ y ( − ∂ ψ ∂ x ) = ψ x y − ψ y x = 0. \frac{\partial}{\partial x}\!\left(\frac{\partial\psi}{\partial y}\right)+\frac{\partial}{\partial y}\!\left(-\frac{\partial\psi}{\partial x}\right)=\psi_{xy}-\psi_{yx}=0. ∂ x ∂ ( ∂ y ∂ ψ ) + ∂ y ∂ ( − ∂ x ∂ ψ ) = ψ x y − ψ y x = 0. So by construction, any ψ \psi ψ That is why the minus sign sits on v v v
Intuition Why streamlines = lines of constantψ \psi ψ d y d x = v u \frac{dy}{dx}=\frac{v}{u} d x d y = u v u d y − v d x = 0 u\,dy - v\,dx = 0 u d y − v d x = 0 ψ \psi ψ ( d x , d y ) (dx,dy) ( d x , d y ) d ψ = ψ x d x + ψ y d y = − v d x + u d y . d\psi=\psi_x\,dx+\psi_y\,dy = -v\,dx + u\,dy. d ψ = ψ x d x + ψ y d y = − v d x + u d y . u d y − v d x u\,dy - v\,dx u d y − v d x d ψ = 0 d\psi=0 d ψ = 0 streamlines are curves of constant ψ \psi ψ Beautiful.
If a flow is both incompressible and irrotational ("ideal flow"), both ϕ \phi ϕ ψ \psi ψ
Each satisfies Laplace's equation. WHY:
Incompressible + V ⃗ = ∇ ϕ \vec V=\nabla\phi V = ∇ ϕ ∇ ⋅ ∇ ϕ = 0 ⇒ ∇ 2 ϕ = 0 \nabla\cdot\nabla\phi=0\Rightarrow \nabla^2\phi=0 ∇ ⋅ ∇ ϕ = 0 ⇒ ∇ 2 ϕ = 0
Irrotational + ψ \psi ψ ∂ v ∂ x − ∂ u ∂ y = 0 ⇒ − ψ x x − ψ y y = 0 ⇒ ∇ 2 ψ = 0 \dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}=0 \Rightarrow -\psi_{xx}-\psi_{yy}=0\Rightarrow \nabla^2\psi=0 ∂ x ∂ v − ∂ y ∂ u = 0 ⇒ − ψ xx − ψ y y = 0 ⇒ ∇ 2 ψ = 0
∇ 2 ϕ = 0 , ∇ 2 ψ = 0 \boxed{\nabla^2\phi=0,\qquad \nabla^2\psi=0} ∇ 2 ϕ = 0 , ∇ 2 ψ = 0
ϕ \phi ϕ ψ \psi ψ ∇ ϕ = ( ϕ x , ϕ y ) = ( u , v ) \nabla\phi=(\phi_x,\phi_y)=(u,v) ∇ ϕ = ( ϕ x , ϕ y ) = ( u , v ) ∇ ψ = ( ψ x , ψ y ) = ( − v , u ) \nabla\psi=(\psi_x,\psi_y)=(-v,u) ∇ ψ = ( ψ x , ψ y ) = ( − v , u ) ( u ) ( − v ) + ( v ) ( u ) = 0 (u)(-v)+(v)(u)=0 ( u ) ( − v ) + ( v ) ( u ) = 0 So equipotential lines and streamlines cross at right angles — they form an orthogonal grid (a "flow net").
Worked example Uniform flowTake ϕ = U x \phi = U x ϕ = U x u = ϕ x = U u=\phi_x=U u = ϕ x = U v = ϕ y = 0 v=\phi_y=0 v = ϕ y = 0 Why this step? Differentiating the potential gives velocity directly.
Stream function: u = ψ y = U ⇒ ψ = U y u=\psi_y=U\Rightarrow\psi=Uy u = ψ y = U ⇒ ψ = U y v = − ψ x = 0 v=-\psi_x=0 v = − ψ x = 0 ψ = U y = \psi=Uy= ψ = U y = Why? Constant y y y
Worked example Source flow (strengthm m m u r = m 2 π r u_r=\dfrac{m}{2\pi r} u r = 2 π r m u θ = 0 u_\theta=0 u θ = 0 ϕ \phi ϕ u r = ∂ ϕ / ∂ r u_r=\partial\phi/\partial r u r = ∂ ϕ / ∂ r ϕ = m 2 π ln r \phi=\dfrac{m}{2\pi}\ln r ϕ = 2 π m ln r Why? Because ∂ ∂ r m 2 π ln r = m 2 π r = u r \frac{\partial}{\partial r}\frac{m}{2\pi}\ln r=\frac{m}{2\pi r}=u_r ∂ r ∂ 2 π m ln r = 2 π r m = u r u r = 1 r ∂ ψ ∂ θ = m 2 π r ⇒ ψ = m 2 π θ u_r=\frac1r\frac{\partial\psi}{\partial\theta}=\frac{m}{2\pi r}\Rightarrow \psi=\dfrac{m}{2\pi}\theta u r = r 1 ∂ θ ∂ ψ = 2 π r m ⇒ ψ = 2 π m θ ψ = \psi= ψ = θ = \theta= θ = Why? A source pushes fluid straight outward, so streamlines are spokes.
Worked example Check orthogonality for the sourceEquipotentials ϕ = m 2 π ln r = \phi=\frac{m}{2\pi}\ln r= ϕ = 2 π m ln r = r = r= r = 90 ° 90° 90° Why this matters: confirms the flow-net theorem visually.
Worked example Flux between two streamlinesBetween ψ 1 \psi_1 ψ 1 ψ 2 \psi_2 ψ 2 Q = ψ 2 − ψ 1 Q=\psi_2-\psi_1 Q = ψ 2 − ψ 1 θ = 0 \theta=0 θ = 0 θ = 2 π \theta=2\pi θ = 2 π Q = m 2 π ( 2 π ) − 0 = m Q=\frac{m}{2\pi}(2\pi)-0=m Q = 2 π m ( 2 π ) − 0 = m Why? Total outflow from a source of strength m m m m m m
ϕ \phi ϕ Why it feels right: Velocity is a vector field, and we love potentials.
The fix: ϕ \phi ϕ only if the flow is irrotational (∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 ϕ \phi ϕ
ψ \psi ψ Why it feels right: Streamlines always exist, so surely ψ \psi ψ The fix: The scalar ψ \psi ψ 2D incompressible flow. For compressible flow you need a modified (density-weighted) stream function.
Common mistake Forgetting the minus sign:v = − ψ x v=-\psi_x v = − ψ x Why it feels right: Symmetry tempts you to write v = ψ x v=\psi_x v = ψ x The fix: The minus sign is required so that continuity cancels (ψ x y − ψ y x = 0 \psi_{xy}-\psi_{yx}=0 ψ x y − ψ y x = 0
Common mistake Mixing the roles: "ϕ \phi ϕ Why it feels right: Both are level curves of a scalar.
The fix: ψ = \psi= ψ = streamlines (along the flow). ϕ = \phi= ϕ = equipotentials (perpendicular to the flow).
Recall Feynman: explain to a 12-year-oldImagine a river. Instead of tracking how fast water moves in the left–right and up–down directions separately (two things), we use ONE clever number for each spot:
The stream function ψ \psi ψ ψ \psi ψ
The velocity potential ϕ \phi ϕ ϕ \phi ϕ
Mnemonic Remember the definitions"ψ \psi ψ ϕ \phi ϕ
ψ \psi ψ u = ψ y u=\psi_y u = ψ y v = − ψ x v=-\psi_x v = − ψ x cross and a minus appears.ϕ \phi ϕ u = ϕ x u=\phi_x u = ϕ x v = ϕ y v=\phi_y v = ϕ y match , no minus.
And "STeam = STreamlines" (ψ \psi ψ "Potential = Perpendicular Pressure-like" (ϕ \phi ϕ
Condition for velocity potential ϕ \phi ϕ Flow must be irrotational,
∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 (since
∇ × ∇ ϕ = 0 \nabla\times\nabla\phi=0 ∇ × ∇ ϕ = 0 ).
Condition for stream function ψ \psi ψ 2D incompressible flow,
∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0 .
Define u , v u,v u , v ϕ \phi ϕ u = ∂ ϕ / ∂ x , v = ∂ ϕ / ∂ y u=\partial\phi/\partial x,\; v=\partial\phi/\partial y u = ∂ ϕ / ∂ x , v = ∂ ϕ / ∂ y .
Define u , v u,v u , v ψ \psi ψ u = ∂ ψ / ∂ y , v = − ∂ ψ / ∂ x u=\partial\psi/\partial y,\; v=-\partial\psi/\partial x u = ∂ ψ / ∂ y , v = − ∂ ψ / ∂ x .
Why is the minus sign needed in v = − ψ x v=-\psi_x v = − ψ x So continuity
ψ x y − ψ y x = 0 \psi_{xy}-\psi_{yx}=0 ψ x y − ψ y x = 0 is satisfied automatically.
What does ψ = \psi= ψ = A streamline (curve tangent to the velocity).
What does ϕ = \phi= ϕ = An equipotential line, perpendicular to streamlines.
Flow rate between two streamlines Q = ψ 2 − ψ 1 Q=\psi_2-\psi_1 Q = ψ 2 − ψ 1 (per unit depth).
Equation satisfied by ϕ \phi ϕ ψ \psi ψ Laplace's equation,
∇ 2 ϕ = 0 , ∇ 2 ψ = 0 \nabla^2\phi=0,\ \nabla^2\psi=0 ∇ 2 ϕ = 0 , ∇ 2 ψ = 0 .
Why are equipotentials ⟂ streamlines? ∇ ϕ ⋅ ∇ ψ = ( u ) ( − v ) + ( v ) ( u ) = 0 \nabla\phi\cdot\nabla\psi=(u)(-v)+(v)(u)=0 ∇ ϕ ⋅ ∇ ψ = ( u ) ( − v ) + ( v ) ( u ) = 0 .
Cauchy–Riemann (fluid) relations ϕ x = ψ y \phi_x=\psi_y ϕ x = ψ y and
ϕ y = − ψ x \phi_y=-\psi_x ϕ y = − ψ x .
ϕ \phi ϕ ψ \psi ψ U U U ϕ = U x , ψ = U y \phi=Ux,\ \psi=Uy ϕ = U x , ψ = U y .
ϕ \phi ϕ ψ \psi ψ m m m ϕ = m 2 π ln r , ψ = m 2 π θ \phi=\frac{m}{2\pi}\ln r,\ \psi=\frac{m}{2\pi}\theta ϕ = 2 π m ln r , ψ = 2 π m θ .
Continuity Equation — source of the stream function (∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0 Vorticity and Irrotational Flow — source of the velocity potential (∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 Laplace Equation — both ϕ \phi ϕ ψ \psi ψ Bernoulli Equation — used once ϕ \phi ϕ Cauchy-Riemann Equations — complex potential w = ϕ + i ψ w=\phi+i\psi w = ϕ + i ψ Electrostatic Potential — direct mathematical analogue (E ⃗ = − ∇ V \vec E=-\nabla V E = − ∇ V Flow Nets — orthogonal grid of ϕ \phi ϕ ψ \psi ψ
auto-satisfies continuity
combined with incompressible
combined with irrotational
Incompressibility div V=0
Intuition Hinglish mein samjhoDekho, 2D flow mein velocity ke do components hote hain, u u u v v v incompressible hai (mass conserve, ∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0 stream function ψ \psi ψ u = ψ y u=\psi_y u = ψ y v = − ψ x v=-\psi_x v = − ψ x
Stream function ka sabse pyaara fact: ψ = \psi= ψ = streamlines hain, yaani fluid actually inhi raston pe behta hai. Aur do streamlines ke beech ka difference ψ 2 − ψ 1 \psi_2-\psi_1 ψ 2 − ψ 1 Q Q Q ψ \psi ψ
Dusri taraf, agar flow irrotational hai (koi spinning nahi, ∇ × V ⃗ = 0 \nabla\times\vec V=0 ∇ × V = 0 velocity potential ϕ \phi ϕ V ⃗ = ∇ ϕ \vec V=\nabla\phi V = ∇ ϕ u = ϕ x u=\phi_x u = ϕ x v = ϕ y v=\phi_y v = ϕ y ϕ \phi ϕ
Jab flow dono ho — incompressible aur irrotational (ideal flow) — tab dono ϕ \phi ϕ ψ \psi ψ ∇ 2 ϕ = 0 \nabla^2\phi=0 ∇ 2 ϕ = 0 ∇ 2 ψ = 0 \nabla^2\psi=0 ∇ 2 ψ = 0 90 degree pe milti hain (flow net). Exam mein bas yaad rakhna: ψ \psi ψ ϕ \phi ϕ
