Translating Between Worlds
We need to be able to move from \((x, y)\) to \((r, \theta)\) and back. The bridge between them is basic trigonometry.
- Polar to Cartesian: \(x = r \cos\theta, y = r \sin\theta\)
- Cartesian to Polar: \(r^2 = x^2 + y^2, \tan\theta = y/x\)
Worked Examples
Example 1: To Cartesian
Convert \((4, \pi/6)\) to Cartesian.
- \(x = 4 \cos(\pi/6) = 4(\sqrt{3}/2) = 2\sqrt{3} \approx 3.46\).
- \(y = 4 \sin(\pi/6) = 4(1/2) = 2\).
- Result: \((3.46, 2)\)
Example 2: To Polar
Convert \((1, 1)\) to polar.
- \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\).
- \(\tan\theta = 1/1 = 1 \to \theta = 45^\circ\).
- Result: \((\sqrt{2}, 45^\circ)\)
The Bridge to Quantum Mechanics
In Quantum Mechanics, the "Position Operator" in 3D is a vector \(\vec{r}\). We often define a "Potential" \(V(r)\) that only depends on the distance from the center, not the angle. This is called a Central Potential. To solve for the behavior of a particle in this environment, we use these conversion equations to turn the Schrödinger Equation into its polar form. This allows us to separate the "radial" part of the problem from the "angular" part, which is the only way to solve for the orbitals of an atom.