PHYS 419: Classical Mechanics Lecture Notes. POLAR COORDINATES. A
vector in two dimensions can be written in Cartesian coordinates as r = xˆx + y ˆy.
(1).
PHYS 419: Classical Mechanics Lecture Notes POLAR COORDINATES A vector in two dimensions can be written in Cartesian coordinates as ˆ + y yˆ r = xx
(1)
ˆ and yˆ are unit vectors in the direction of Cartesian axes and x and y are the where x components of the vector, see also the figure. It is often convenient to use coordinate systems other than the Cartesian system, in particular we will often use polar coordinates. ˆ see the figure. These coordinates are specified by r = |r| and the angle φ between r and x, The relations between the polar and Cartesian coordinates are very simple: x = r cos φ and r=
y = r sin φ
p
y φ = arctan . x
x2 + y 2
ˆ The former one is The unit vectors of polar coordinate system are denoted by rˆ and φ. defined accordingly as rˆ =
r r
(2)
Since ˆ + r sin φ y, ˆ r = r cos φ x ˆ + sin φ y. ˆ rˆ = cos φ x ˆ is to require it to be orthogonal to rˆ, i.e., to have rˆ · φ ˆ = 0. The simplest way to define φ This gives the condition cos φ φx + sin φ φy = 0. 1
The simplest solution is φx = − sin φ and φy = cos φ or a solution with signs reversed. This gives ˆ = − sin φ x ˆ + cos φ y. ˆ φ This vector has unit length ˆ·φ ˆ = sin2 φ + cos2 φ = 1. φ ˆ points in the direcThe unit vectors are marked on the figure. With our choice of sign, φ ˆ are drawn from the position of the point tion of increasing angle φ. Notice that rˆ and φ ˆ is considered. Notice also that due to Eq. (2), the expression for r in terms of rˆ and φ r = rrˆ. An expression analogous to Eq. (1) is wrong ˆ r 6= rrˆ + φφ. We will need also the derivatives of vector r expressed in polar coordinates. We have r˙ =
dr = r˙ rˆ + rrˆ˙ dt
and dˆ r ˆ ˆ + φ˙ cos φ yˆ = φ˙ (− sin φ x ˆ + cos φ y) ˆ = φ˙ φ rˆ˙ = = −φ˙ sin φ x dt (notice that in contrast to Cartesian coordinate system, derivatives of unit vectors of the polar system are not zero) so that ˆ r˙ = r˙ rˆ + rφ˙ φ. Now get the second derivative ˆ + rφ¨φ ˆ + rφ˙ φ, ˆ˙ r¨ = r¨rˆ + r˙ rˆ˙ + r˙ φ˙ φ ˆ so that the only new derivative is that of φ: ˆ˙ = φ˙ (− cos φ x ˙ r. ˆ − sin φ y) ˆ = −φˆ φ Grouping terms together, we finally get: ˙ φ. ˆ r¨ = (¨ r − rφ˙ 2 )ˆ r + (rφ¨ + 2r˙ φ)
