Circular waves and the Bessel function

Ole Witt-Hansen 2018

Contents 1. The wave equation ...... 1 2. Polar coordinates in the plane and space ...... 1 3. Solving the wave equation for plane waves...... 4 4. The drum head ...... 7

Circular waves and the Bessel function 1

1. The wave equation In this article, we shall look into the physics of plane circular waves based on the wave equation:

1 2u 2u 2u 1 2u (1.1) 2u     v2 t 2 x2 y2 v2 t 2 where u is a scalar field.

2. Polar coordinates in the plane and space The polar coordinates in the plane are usually chosen as (r, ) as shown in the figure: We have:

(2.1) x  r cos and y  r sin

The infinitesimal distance vector is seen geometrically to be:  ds  (rd,dr) Since the direction of rd and dr are orthogonal the square of the distance element is.

(2.2) ds2  r 2d 2  dr 2

And the area element dA is:

(2.3) dA  rddr

These results obtained geometrically, may also be formally proven, by making a coordinate transformation.

x x dx  d  dr  r sin d  cosdr  r y y dy  d  dr  r cosd  sindr  r

ds2  dx2  dy 2  (r sind  cosdr)2  (r cosd  sindr)2 

(2.4) ds2  r 2d 2  dr 2

Since we shall consider a rotational symmetric wave, we shall begin by rewriting the Laplace operator in polar coordinates. For general orthogonal curvilinear coordinates in the plane ( p , p ) or in space ( p , p , p ) , the  1 2  1 2 3 distance element is ds  (g1dp1, g2dp2 ) in the plane, and in space is ds  (g1dp1, g2dp2 , g3dp3 ) - Here the g’s are functions of the coordinates (p1, p2) or (p1, p2, p3). If the curvilinear coordinates are orthogonal we have:

Circular waves and the Bessel function 2

2 2 2 2 2 2 2 2 2 2 2 2 (2.5) ds  g1 dp1  g2 dp2 and ds  g1 dp1  g2 dp2  g3 dp3

In space a point P(x, y, z) has the polar coordinates (r, θ, φ). See figure below. θ is the polar angle and φ is the azimuth angle. The coordinates are, when expressed by the two angles.

(5.1) x  r sin cos, y  r sin sin, z  r cos

To come from one point P (r,θ,φ) to another Q(r+dr, θ + dθ, φ + dφ), you first go “horizontally” rsinθ dφ then “vertically” the distance rdθ, and finally the radial distance dr. The infinitesimal distance vector therefore becomes

 (5.2) ds  (r sind, rd, dr)

Since the three “steps” are perpendicular to each other, the square of the distance is:

(5.3) ds2  r 2d 2  r 2 sin 2 d 2  dr 2

Likewise the infinitesimal volume element is the product of the three orthogonal steps.

(2.6) dV  r 2 sindddr

The gradient in the plane is defined as:

 u u (2.7) u  ( , ) x y u u  The significance of the gradient is: du  dx  dy  u (dx, dy) x y That is, the infinitesimal increase in u resulting from the displacement (dx,dy) .

 The infinitesimal displacement along the curvilinear coordinates is in the plane ds  (g dp , g dp )  1 1 2 2 and in space: ds  (g1dp1, g2dp2 , g3dp3 ) , dx is replaced by g1dp1 and so on, and therefore the gradient of a scalar field U in plane curvilinear becomes.

 U U  1 U 1 U U  ( , )  U  ( , ) x y g1 p1 g2 p2 And in space:

 U U U  1 U 1 U 1 U (2.8) U  ( , , )  U  ( , , ) x y z g1 p1 g2 p2 g3 p3 Circular waves and the Bessel function 3

 Since ds  (rd,dr) in polar coordinates in the plane we have:

 1 U U (2.9) U  ( , ) r  r

 And in polar coordinates in space: ds  (r sind,rd,dr)

 1 U 1 U U (2.10) U  ( , , ) r sin  r d r

Concerning the divergence, things become somewhat more complicated: This comes about because the lines inclosing the “rectangular” square or box in curvilinear coordinates the area of the front end and the back end are not necessarily equal.

In space the areas of a box with sides (1-2) (1-3) and (2-3) are

A23  g2 g3p2p3 , A13  g1g3p1p3 , and A12  g1g2p1p2 , so in the calculation of the u “change of flux”, along one of the coordinates in a Cartesian coordinate system: x xyz must x 1   be replaced by (g2 g3u1)g1p1p2p3  (g2 g3u1)p1p2p3 , because g2 g3 may depend g1 p1 p1 on p1.   The divergence of a vector field, however, is defined as the flux out of the volume dV:   udV The flux in curvilinear coordinates calculated in this manner from a “rectangular” box is, however:

   ( (g g u )  (g g u )  (g g u ))dp dp dp   p 2 3 1 p 1 3 2 p 1 2 3 1 2 3 (2.11) 1 2 3 1    ( (g g u )  (g g u )  (g g u ))g dp g dp g dp  2 3 1 1 3 2 1 2 3 1 1 2 2 3 3 g1g2 g3 p1 p2 p3

Since dV  g1dp1g2dp2 g3dp3 is the correct expression for the volume element, the divergence becomes.

  1        (2.12)  u   (g2 g3u1)  (g1g3u2 )  (g1g2u3 ) g1g2 g3  p1 p2 p 3 

In the plane we get similarly:

  1       (2.13)  u   (g2u1)  (g1u2 ) g1g2  p1 p2 

 In polar coordinates in the plane ds  (rd,dr) , we find:

Circular waves and the Bessel function 4

  1  u1   1 u1 1 (ru2 ) (2.14)  u    (ru2 )   r   r  r  r r

And for the Laplace operator in the plane we find:

   1 U U 1  1 U 1  U 2U   U    ( , )  ( )  (r ) r  r r  r  r r r

1 2U 1  U (2.15) 2U   (r ) r 2  2 r r r

1 2U The wave equation 2U  , where v is the speed of propagation is in polar coordinates: v2 dt 2

1 2U 1  U 1 2U (2.16)  (r )  r 2  2 r r r v2 dt 2

3. Solving the wave equation for plane waves. Assuming that the time dependence is independent of r and θ, we write: U(r,,t)  u(r, ) (t) . Inserting in (2.16) and subsequently dividing by u , we have:

1  2u 1  u 1  2   (r )  u  r 2  2 r r r v 2 dt 2 (3.1) 1 1  2u 1  u 1 1  2 (  (r ))  u r 2  2 r r r v 2  dt 2

In this equation the left side depends only on r and θ, whereas the right side depends only on t. But that means that they are both equal to the same constant, which we for physical reasons put to  –k2, where k is the wave number: k  . So we have the two equations: v

1 1 2 2  k 2   k 2v2  (3.2) v2  dt 2 dt 2   Aeikvt    Aeit

1  2u 1  u 1  2u 1  u (3.3)  (r )  k 2u   (r )  k 2u  0 r 2  2 r r r r 2  2 r r r

u If we further assume that u = u(r, θ) does not depend explicitly on θ, such that:  0 , we have  the equation: Circular waves and the Bessel function 5

1  u  2u 1 u (3.4) (r )  k 2u  0    k 2u  0 r r r r 2 r r

If we put   kr we can eliminate the constant k.

 2u 1 u k 2  k 2  k 2u  0   2 r  (3.5)  2u 1 u   u  0  2 r 

The solution to this differential equation cannot be expressed with already known functions, but the solution is called the Bessel function of zero order and is denoted J 0 () or J 0 (x) .

We can see that if J 0 () is a solution to (3.5) , then J 0 (kr) is a solution to (3.4).

Shifting the independent variable from ρ to x, and the dependent variable from u to f , we can making a power series expansion of f:

 k (3.6) f (x)   ak x . k0  2 f 1 f Inserting this expression in the differential equation:   f  0 , collecting terms with the x2 x x same power of x, we find in the following steps:

2 3 4 5 6 f (x)  a0  a1x  a2 x  a3 x  a4 x  a5 x  a6 x  ...

2 3 4 5 f '(x)  a1  2a2 x  3a3 x  4a4 x  5a5 x  6a6 x  ...

f '(x) a  1  2a  3a x  4a x2  5a x3  6a x4  ... x x 2 3 4 5 6 2 3 4 f ''(x)  2a2  6a3 x 12a4 x  20a5 x  30a6 x  ...

We shall assume that J 0 (0) 1 and J 0 '(0)  0

Inserting in the differential equation: a 2a  6a x 12a x2  20a x3  30a x4 ... 1  2a  3a x  4a x2  5a x3  6a x4 ... 2 3 4 5 6 x 2 3 4 5 6 2 3 4 5 6 a0  a1x  a2 x  a3 x  a4 x  a5 x  a6 x  ...= 0

We shall multiply this equation with x.

2 3 4 5 2 3 4 5 2a2 x  6a3 x 12a4 x  20a5 x  30a6 x  ... a1  2a2 x  3a3 x  4a4 x  5a5 x  6a6 x  ... 2 3 4 5 6 7 a0 x  a1x  a2 x  a3 x  a4 x  a5 x  a6 x  ...= 0 Circular waves and the Bessel function 6

The condition J 0 (0) 1 gives a0 = 1, and from above we see that a1 = 0. We then collect the terms with x.

1 ( 2a2 + 2a2 + a0 )x = 0  a2   4

2 ( 6a3  3a3  a1)x = 0  9a3  0  a3  0

3 1 (12a4 + 4a4 + a2 )x  0  a4  64

Continuing we get the first 4 terms in the series expansion of the Bessel function of zero order.

(3.7) J (x) 1 ( x )2  1 ( x )4  1 ( x )6 ... 0 2 2!2 2 3!2 2

On the other hand, if u(r, ) is not independent of θ, we have the differential equation:

1  2u 1  u (3.8)  (r )  k 2u  0 r 2  2 r r r

in We replace u(r, ) with un (r)e , finding:

1 1  u u (n2 )ein  ein (r )  ein k 2u  0  r 2 n r r r

 2u 1 u n2 (3.9) n  n  (k 2  )u  0 r 2 r r r 2 n

Or when putting x  kr :

 2u 1 u n2 (3.10) n  n  (1 )u  0 x2 x x x2 n

Let J n (x) denote the solution of this differential equation. J n (x) is then denoted the Bessel function of order n. The series expansions for J n (x) turn out to be:

1 x n 1 x n2 1 x n4 (3.11) J n (x)  n! ( 2 )  1!(n1)! ( 2 )  2!(n2)! ( 2 ) ...

We shall only apply the zero order Bessel function in the following, but below we state some theorems for the Bessel functions, (without proof).

d (3.12) J (x)   J (x) (Can be verified, by applying (3.11) 1 dx 0

Circular waves and the Bessel function 7

d (3.13) xJ (x)  (xJ (x)) (Can be verified, by applying (3.11) 0 dx 1

This relation integrated to an arbitrary upper limit, yields the following formula.

x0 xJ (x)dx  x J (x )  0 0 1 0 0

Below are shown the graphs of the first three Bessel functions

4. The drum head Assuming that we have a circular drum head having radius R, and the drum skin has the elastic velocity of propagation v, and assuming that the solution has no explicit dependence on θ, then the solution, (as demonstrated above), will obey the following differential equations.

1 1 2 2  k 2   k 2v2  v2  dt 2 dt 2   Aeikvt    Aeit And  2u 1 u   k 2u  0 r 2 r r

The last equation having the solution J 0 (kr) .  The wave number is k  , so that the wave number is proportional to ω. v

Any solution must obey the boundary condition J0 (kR)  0, since the oscillations must vanish, at the edge. Circular waves and the Bessel function 8

The first three zero points of J0 (r) occur at: r = 2.40, r = 5.52 and r = 8.65.

If R = 0.10 m , then it gives: kR = 2.40, kR = 4.52, and kR = 8.65, resulting in the values for the wave number: k = 24.0 m-1 , k = 45.2 m-1 and k = 86.5 m-1.

The first three modes are shown below in a 3-dim mapping. (Borrowed from Wikipedia: “Vibrations of a circular membrane”)

1 3 5 Assuming that the modes correspond to standing waves having: R  4  , R  4  and R  4  , the corresponding wavelengths are 0.40 m, 0.133 m and 0.80 m.

If the speed of propagation in the drum skin is 100 m/s, we find for the frequencies: v   = 250 Hz, 750 Hz and 1250 Hz, which does not seem unreasonable, but we have really no  way of estimating the speed of propagation, without measuring the frequency.