Visual walkthrough — Streamlines, pathlines, streaklines

2.2.10 · D2 · Physics › Fluid Mechanics › Streamlines, pathlines, streaklines

Step 1 — Blank stage banao: space aur clock

KYA. Koi bhi curve exist karne se pehle, humein ek jagah (plane) aur ek time (ek clock) chahiye. Jo kuch bhi hum draw karte hain woh plane par hota hai; jo bhi rule hum likhte hain woh clock ki reading se refer karta hai.

KYUN. Teeno curves ek hi velocity field se aati hain, lekin woh time se alag-alag tarike se behave karwati hain. Ek streamline clock ko freeze karti hai; ek pathline use aage chalati hai; ek streakline puri history yaad rakhti hai. Isliye hume space aur time ko line one se clearly alag rakhna hoga.

PICTURE. Neeche: ek horizontal axis $x$ (rightward position), ek vertical axis $y$ (upward position), aur ek chhota clock face side mein jo hum spin ya freeze karenge is hisaab se ki hum kaun sa curve bana rahe hain.

Step 2 — Har point par ek arrow lagao: velocity field

KYA. Har point par hum ek aisa arrow lagate hain jiska direction aur length $\vec{v}=(u,v)=(1,t)$ hai.

KYUN. Ek velocity field teeno constructions ke liye raw material hai. Arrow kehta hai "agar fluid ka ek kana abhi yahan baitha hai, toh use itni tez aur iss direction mein carry kiya ja raha hai." Dhyan do ki arrow kahan ho is par depend nahi karta (andar koi $x$ ya $y$ nahi), sirf kab par — yahi iss flow ko spatially uniform lekin temporally changing banata hai.

PICTURE. Arrow-forest ke do snapshots side by side:

$\underset{\text{rightward speed}}{\underbrace{u}}=1,\underset{\text{upward speed}}{\underbrace{v}}=\underset{\text{clock ke saath badhta hai}}{\underbrace{t}}$

• $t=0$ par: har arrow horizontal hai ($v=0$), pure-right point kar raha hai.
• $t=2$ par: har arrow steeply upar tilt hai ($v=2$), kyunki upward speed ne rightward speed ko pakad liya aur paar kar gaya.

Step 3 — Streamline: clock freeze karo, arrows follow karo

KYA. Ek streamline ek aisi curve hai jisko thread kiya gaya hai taki iska tangent har jagah ek frozen instant par arrows ke saath line up kare.

KYUN. Hum snapshot answer chahte hain: "sab kuch abhi kis taraf point kar raha hai?" Toh hum clock ko kisi value par, jaise $t=2$, jam kar dete hain, aur woh curve draw karte hain jo kabhi kisi arrow ko cross nahi karti balki hamesha uske saath chalti hai.

KAISE — equation, term by term. "Tangent arrow ke parallel hai" ka matlab hai ki curve ke saath tiny step $d\vec{r}=(dx,dy)$ usi direction mein point karta hai jisme $\vec{v}=(u,v)$ hai. Parallel vectors ke component-ratios match karte hain:

$\underset{\begin{array}{c}\text{tiny rightward step}\\ \text{÷ rightward speed}\end{array}}{\underbrace{\frac{dx}{u}}}=\underset{\begin{array}{c}\text{tiny upward step}\\ \text{÷ upward speed}\end{array}}{\underbrace{\frac{dy}{v}}}(\text{clock frozen at }t)$

$u=1$, $v=t$ plug in karo jahan $t$ ek constant number ki tarah hold hai:

$\frac{dx}{1}=\frac{dy}{t}⟹dy=tdx⟹\underset{\text{slope }t\text{ ki ek straight line}}{\underbrace{y=tx+C}}$

PICTURE. Frozen time $t=2$ par streamlines slope $2$ ki straight lines hain. Yeh straight hain kyunki is ek instant par har arrow ka tilt same hai (field spatially uniform hai).

Step 4 — Pathline: clock release karo, ek particle follow karo

KYA. Ek pathline fluid ke ek kane ka actual trail hai jab clock sach mein chalta hai.

KYUN. Ab hum tracking question poochh rahe hain: "YEH particle kahan jaata hai?" Particle ki velocity wahi field hai jise particle jahan-aur-jab bhi ho wahan evaluate kiya gaya hai. Toh $t$ freeze karne ki bajaye, hum use flow karne dete hain aur time mein integrate karte hain.

KAISE — equations, term by term.

$\underset{\text{har second mein }x\text{ kitna badlta hai}}{\underbrace{\frac{dx}{dt}}}=u=1,\underset{\text{har second mein }y\text{ kitna badlta hai}}{\underbrace{\frac{dy}{dt}}}=v=t$

Particle ko $t_0=0$ par origin se release karo (toh $x_0=y_0=0$):

$x={\int }_0^{t}1d{t}^{\prime }=\underset{\text{speed }1\text{ se right drift karta hai}}{\underbrace{t}},y={\int }_0^t{t}^{\prime }d{t}^{\prime }=\underset{\text{tezi se tezi se climb karta hai}}{\underbrace{\frac{1}{2}{t}^2}}$

Clock eliminate karo: since $x=t$ hai, $t=x$ substitute karo:

$\boxed{y=\frac{1}{2}{x}^2}\text{— ek parabola upar ki taraf khuulti hui}$

PICTURE. Ek particle origin se nikalti hai, steadily right drift karti hai lekin ever-faster climb karti hai (kyunki $v=t$ badhta rehta hai), ek parabola banati hai.

Step 5 — Streakline: clock chalta rakho, har release yaad rakho

KYA. Ek streakline dye plume hai: ek observation time par, un sab particles ko mark karo jo kabhi bhi ek fixed source se release hue the.

KYUN. Ek smoke chimney ya dye needle fluid continuously inject karta hai. Har puff apne time $\tau$ par release hua tha aur tab se drift kar raha hai. Visible streak ek photograph abhi hai bahut saare alag particles ka, har ek ke paas alag amount ka travel time hai.

KAISE — equations, term by term. Time $\tau$ par origin se release hua particle Step 4 ke same rules follow karta hai, lekin uska clock $\tau$ par start hua:

$x={\int }_{\tau }^{t}1d{t}^{\prime }=\underset{\text{kitne time se drift kar raha hai}}{\underbrace{(t-\tau )}},y={\int }_{\tau }^t{t}^{\prime }d{t}^{\prime }=\underset{\text{uski accumulated climb}}{\underbrace{\frac{1}{2}(t^2-{\tau }^2)}}$

• $t$ ::: fixed observation instant (shutter click) — har puff ke liye same.
• $\tau$ ::: release time — yeh woh knob hai jise hum sweep karte hain, kaafi pehle se ($\tau \to -\infty$) abhi tak ($\tau =t$).

Shutter ko $t=2$ par fix karo aur $\tau$ ko $2$ se negative values tak sweep karo:

$x=2-\tau ,y=\frac{1}{2}(4-{\tau }^2)$

$\tau =2-x$ use karke $\tau$ eliminate karo:

$y=\frac{1}{2}(4-(2-x{)}^2)=\frac{1}{2}(4x-x^2)=2x-\frac{1}{2}{x}^2$

$\boxed{y=2x-\frac{1}{2}{x}^2}\text{— ek *alag* parabola}$

PICTURE. $t=2$ par, dots $\tau =2,1,0,-1,\dots$ par release hue particles mark karti hain. Unhe connect karo: ek downward-bending arc jo clearly Step 4 ki pathline nahi hai.

Step 6 — Overlay: teeno ko alag hote dekho

KYA. Teeno ko ek grid par $t=2$ par, sab origin se, rakh do.

KYUN. Unhe simultaneously alag hote dekhne se yeh cement ho jaata hai ki woh unsteady flow mein teeno genuinely alag cheezein hain.

PICTURE.

• Streamline (straight, slope $2$): $y=2x$.
• Pathline (upar khuulti): $y=\frac{1}{2}{x}^2$.
• Streakline (neeche khuulti): $y=2x-\frac{1}{2}{x}^2$.

Teeno origin chhod ke jaati hain, teeno ka wahan same starting slope hai — check karo: streamline slope $=2$; pathline slope $\frac{dy}{dx}=x=0$ origin par... ruko, dhyan se dekho — woh sirf source par agree karti hain aur turant fan out ho jaati hain.

Step 7 — Degenerate case: flow ko steady banao aur unhe fuse hote dekho

KYA. Clock-dependent field $v=t$ ko ek steady field, $v=1$ (constant), se replace karo. Teeno redo karo.

KYUN. Parent ka "golden rule" kehta hai ki steady flow mein teeno coincide karte hain. Hume yeh dekhna hai, sirf bataya nahi jaana. Is limiting case ko cover karna hi derivation ko complete banata hai.

KAISE. $u=1,v=1$ ke saath:

• Streamline ($t$ freeze karo): $\frac{dx}{1}=\frac{dy}{1}\Rightarrow y=x+C$.
• Pathline ($t_0=0$ par origin se release): $x=t,y=t\Rightarrow y=x$.
• Streakline ($\tau$ par release, $t$ par observe): $x=t-\tau ,y=t-\tau \Rightarrow y=x$.

Teeno ek hi line $y=x$ par collapse ho jaati hain. Arrows ko tilt karne ke liye koi clock-turning nahi hai, toh koi history effect nahi, toh koi splitting nahi.

PICTURE. Teeno overlaid curves bilkul ek doosre ke upar hain — ek single line.

Ek-picture summary

Yeh final panel poori kahaani stack karta hai: arrow-field time ke saath upar turn karta hua (upar), aur teeno curves ek source se unsteady case mein fan out hote hue alongside steady case ki single fused line (neeche). Ek nazar = puri derivation.

Recall checkpoints

Connections

• Velocity field and material derivative
• Steady vs unsteady flow
• Lagrangian vs Eulerian description
• Stream function ψ
• Continuity equation
• Flow visualization techniques (dye, smoke, PIV)

Concept Map