## Friday, October 7, 2022

### Solution To Helioseismology Challenge Problem

Explain the appearance of the spherical surface shown below if it describes m= 5 and ℓ = 5. Thence, find the associated Legendre function: P ℓm (q ) and also Y ℓm (q,j) assuming q = p/2: (Assume  n nℓm =  1 /s)

Solution:

Since ℓ = 5 = m then there are no lines of latitude because these are defined: (ℓ  - m), so (5 – 5) = 0.  Since m = 5 we have an azimuthal order of 5 which indicates 5 nodes (each with embedded contours) around the equator on the three dimensional spherical surface. Three of these are shown and the remaining two are on the distal side of the sphere.

Finally we have the angular degree ℓ or the number of nodes along longitude or meridian lines. Again 3 of these (“doubled” longitudes each enfolding a node m) are shown, and 2 others are on the other side of the sphere.

The associated Legendre polynomials are given by:

P ℓm (q )  =  (1 – z2) m/2 / ℓ! 2   d (ℓ+m) / dz(ℓ+m)  (z2   -1)

But   = m = 5 ,  so:

P ℓm (q )  =  (1 – z2) 0 / 5! 25   d(5+5)  / dz(5+5)   (z2   -1) 5

For q = p/2 ,  d(10)  / dz(10)   (z2   -1) 5= 0  (since z = cos  q)

Therefore: P ℓm (q )  =  0

Y ℓm (q,j) =

(1) m [2 ℓ +1 (ℓ - m)!/ 4n ((ℓ +m)!] ½ P m (z ) exp (i m j),

Since ℓ = 5 = m then there are no lines of latitude because these are defined: (ℓ  - m), so (5 – 5) = 0.  Since m = 5 we have an azimuthal order of 5 which indicates 5 nodes (each with embedded contours) around the equator on the three dimensional spherical surface. Three of these are shown and the remaining two are on the distal side of the sphere.

Finally we have the angular degree ℓ or the number of nodes along longitude or meridian lines. Again 3 of these (“doubled” longitudes each enfolding a node m) are shown, and 2 others are on the other side of the sphere.     The associated Legendre polynomials are given by:

P ℓm (q)  =  (1 – z2) m/2 / ℓ! 2   d (ℓ+m) / dz(ℓ+m)  (z2   -1)

But   = m = 5 ,  so:

P ℓm (q)  =  (1 – z2) 0 / 5! 25   d(5+5)  / dz(5+5)   (z2   -1) 5

For q = p/2 ,  d(10)  / dz(10)   (z2   -1) 5= 0  (since z = cos  q)

Therefore: P ℓm (q )  =  0

Y ℓm (q,j) =

(1) m [2 ℓ +1 (ℓ - m)!/ 4n ((ℓ +m)!] ½ P m (z ) exp (i m j),

But since P ℓm (q)  =  0  is a factor then: Y ℓm (q,j) =  0