WHAT we need first: how a vector's time-derivative looks different to an observer rotating with the frame versus a fixed observer.
WHY (derive it). Imagine A written in the rotating basis {e^1,e^2,e^3}:
A=A1e^1+A2e^2+A3e^3.
The components Ai are numbers; the basis vectors themselves rotate. Differentiating in the inertial frame using the product rule:
(dtdA)in=(dA/dt)roti∑A˙ie^i+∑iAidtde^i.
A unit vector rigidly rotating with ω obeys dtde^i=ω×e^i (its tip traces a circle of radius sinθ). So the second sum is ω×∑iAie^i=ω×A. Why this step? Because rotation only reorients basis vectors — that reorientation rate is precisely ω×. ■
Apply the transport theorem to the position vectorr to get velocity, then again to get acceleration.
Step 1 — velocity. Set A=r:
vin=vrot+ω×r.Why? The inertial velocity = velocity you measure inside the frame + the velocity the frame drags you with.
Step 2 — acceleration. Apply the operator (dtd)in=(dtd)rot+ω× to vin (taking ω constant):
ain=(dtd)rot(vrot+ω×r)+ω×(vrot+ω×r).
Expand term by term:
(dtdvrot)rot=arot
(dtd)rot(ω×r)=ω×vrot (since ω˙=0)
+ω×vrot
+ω×(ω×r)
Collecting:
ain=arot+2ω×vrot+ω×(ω×r)
Step 3 — put Newton in. True forces obey Freal=main. Solve for what the rotating observer sees, marot:
marot=FrealCoriolis−2mω×vrotcentrifugal−mω×(ω×r)
Recall Forecast-then-Verify: predict before peeking
Q: A ball is dropped from a tall tower at the equator. Which way does Coriolis deflect it?
Forecast… then verify: v is downward (toward axis-ish), ω along Earth's axis → −2ω×v points east. The ball lands slightly east of the plumb line. (Confirmed experimentally — Hall's drop experiments.)
Recall Feynman: explain to a 12-year-old
Imagine you're on a spinning merry-go-round. When the floor spins under you, your body wants to keep going straight (that's just being lazy/inertia). But to you, spinning along, it looks like something is shoving you outward — that "shove" is the centrifugal make-believe force. Now roll a marble across the spinning floor: it curves sideways, like a ghost hand bending its path — that's the Coriolis make-believe force. Nobody is really pushing! The floor is turning under the marble, so a straight path looks bent to you. We invent these pretend forces just so our usual "force = mass × acceleration" rule still works while we're dizzy and spinning.
Dekho, Newton ka F=ma sirf inertial frame me sidha kaam karta hai — yaani jo frame na accelerate kar raha ho na ghoom raha ho. Lekin hum to ghoomti hui Earth pe, ya merry-go-round ke andar baithte hain. Aise rotating frame me cheezein "bina wajah" ghoomti dikhti hain. To hum Newton ko chhodte nahi — balki do fictitious (nakli) forces add kar dete hain taaki hisaab match ho jaye: centrifugal aur Coriolis.
Centrifugal force hamesha axis se bahar ki taraf lagti hai, magnitude mω2r⊥. Yeh sirf position pe depend karti hai — chahe tum hilo ya na hilo, lagti rahegi. Jab tum car me turn lete ho aur door se chipakte ho, woh feel centrifugal ka hai (par asli force to door ka andar ka push hai, tumhari inertia seedha jaana chahti thi).
Coriolis force tab aati hai jab tum frame ke andar chal rahe ho: FCor=−2mω×v. Yeh velocity ke perpendicular hoti hai, isliye sirf raasta ghumaati hai, speed nahi badhaati (work zero!). Aur factor 2 mat bhulna — woh derivation me do baar ω×v aane se aata hai. Ek mast yaad rakhne wali baat: Northern hemisphere me Coriolis cheezon ko unke right taraf modti hai — isi wajah se cyclone counter-clockwise ghoomte hain.
80/20 funda: locally Coriolis bahut chhoti hoti hai (ω⊕ bahut tiny hai), par badi distance aur lambe time pe iska effect jama hota jaata hai — isiliye weather aur ocean currents pe yeh raaj karti hai. Bas yaad rakho: centrifugal = bahar bhaago (position), Coriolis = mod do (velocity, 2, no work).