Lecture 19 Spherical Polar Coordinates Remember Phils Problems and your notes = everything Come to see me before the end of term I’ve put more sample questions and answers in Phils Problems Past exam papers Have a look at homework 2 (due in on 15/12/08)

Hints for Homework 2 You know that and that So if you hadyou must write In questions that ask you to prove an expression is a solution of an equation, simply stick the expression into the equation starting at the heart of the equation working outwards and show that the LHS = RHS. (see Q25 of tutorial questions and corresponding answer at Phils Problems)

Polar Coordinate Systems 1. Spherical Polar Coordinates Spherical polars are the coordinate system of choice in almost all 3D problems. This is because most 3D objects are shaped more like spheres than cubes, e.g. atoms, nuclei, planets, etc. And many potentials (Coulomb, gravitational, etc.) depend on radius. Physicists define r,  as shown in the figure. They are related to Cartesian coordinates by:. 2. 3D Integrals in Spherical Polars The volume element is (given on data sheet). To cover over all space, we take Example 1 Show that a sphere of radius R has volume 4  R 3 /3. So

Polar Coordinate Systems Example 2 Find the Fourier transform of a screened Coulomb potential, As before we have the 3D Fourier transform In this case f(r) is a function only of the magnitude of r and not its direction and so has perfect radial symmetry. Again the volume element is We therefore have There is a standard ‘trick’ which is to chose the direction of k to be parallel to the polar (z) axis for the integral. Then k.r becomes. Now clearly the whole integral is a function only of the magnitude of k, not its direction, i.e. F(k) becomes F(k):

Polar Coordinate Systems We therefore write The integral over  is trivial: it just gives a factor of 2 . But note that the factor involves r and . Which integral should we do next? so The presence of the together with the makes integration by substitution  the obvious choice: let Rewrite So

Polar Coordinate Systems We are then left with the integral over r: From previous page: This type of integral was met earlier in the tutorial question exercises on Fourier transforms. It’s best to write the sine in terms of complex exponentials: This gives the final result:

Polar Coordinate Systems 3.  2 in Spherical Polars: Spherical Solutions As given on the data sheet, (Spherically symmetric’ means that V is a function of r but not of  or .) Example 3 Find spherically symmetric solutions of Laplace’s Equation  2 V(r) = 0. Therefore we can say Really useful bit!!!! If (as in the homework) we were given an expression for V(r) and had to prove that it was a solution to the Laplace equation, then we’d just stick it here and start working outwards until we found the LHS was zero. If on the other hand we have to find V(r) then we have to integrate out the expression.

Polar Coordinate Systems Multiplying both sides by r 2 gives Integrating both sides gives where A is a constant. This rearranges to and so …. Integrating we get the general solution:. We’ve just done Q3(i) of the homework backwards!!! (see earlier note in red)

Polar Coordinate Systems 4. The Wave Equation in polar coordinates Let’s only look for spherically symmetric solutions  (r,t), so the equation can be written The wave equation is As usual we look for solutions of the form As usual substitute this back in As usual separate the variables We equate both sides to a constant and since we expect LHO solutions this is -ve To make maths easier let this be

Polar Coordinate Systems The equation for T(t) is easy to solve giving Now we need to solve There is a standard trick which is to define, solve for u(r) and thus find R(r). This is tricky to solve ….. Start by differentiating R(r) with respect to r using the product rule. Multiply both sides of the expression above by r 2 gives Now differentiate again. (*)

Polar Coordinate Systems Therefore Remember we originally needed to solve So equation (*) becomes: (*) so Thus we have solutions of the form: Remember and so And from the start of this example

Polar Coordinate Systems Yet again I’ve made a mistake in the notes !!!!

Polar Coordinate Systems These are spherical waves moving in and out from the origin. Note the factor of 1/r. Intensity is related to amplitude squared. Our solution gives intensity as For waves moving out from the central point (origin). This is the well known inverse square law.