Solutions to Solar Oscillation Problems

1)  Solution: The acoustic cutoff frequency is:

w _ac   = g g/ 2c

Then:

w _ac   =  (5/3) (273 ms^-2)/ 2 (900 ms^-1 ) 

w _ac   =    0.25 s^-1

2) Solution:

The spherical surface would appear as shown below:


The critical parameter is the difference  (ℓ  - m) given it yields the lines corresponding to parallels of latitude.  In this case, (ℓ  - m) =   (2 – 0) = 2, hence two parallels of latitude which are shown bounding the white region. (Since m = 0 there are no longitudinal lines)

The associated Legendre polynomial function given by:

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

But ℓ  =2  and  m = 0 ,  so:

P _ℓm (q )  =  (1 – z²) ⁰ / 2! 2²   d⁽²⁾  / dz⁽²⁾   (z²   -1) ²

(1 – z²) ⁰   = 1  so:

P _ℓm (q )  =  (1)/ 2! 2²   d⁽²⁾  / dz⁽²⁾   (z²   -1) ²

=   1/ 8   [d² / dz²   (z²   -1) ²]

3) Solution:

The depth of reflection is r, and the Lamb frequency is:

L   =  2p/ T  = 2p/ 340.61 s =   0.0184 s^-1

Where: L   =    Ö( ℓ(ℓ + 1) c² / r² )

Solving for r:     r=  Ö[(ℓ(ℓ + 1)]  c / L

r=   Ö[ (20(20 +1)] 900 m/s/ 0.0184 s^-1  

r =   10 ⁶   m

4) The sketch denotes a vibration for which m= 1 and ℓ = 2.  Then (ℓ  - m) =   (2 – 1) = 1 so a single parallel of latitude must appear in the sketch. Also since m= 1 we have at least one meriodonal or longitude line. 

Then : 

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

Since ℓ  =2  and  m = 1 :

P _ℓm (q )  =  (1 – z²) ^1/2   / 2! 2²   d⁽³⁾  / dz⁽³⁾   (z²   -1) ²

z =  cos q  =    cos  p/4  =    Ö2 / 2

But: d⁽³⁾  / dz⁽³⁾   (z²   -1) ²   =  16.97

Then:

 

P _ℓm (q )  =  (Ö2 / 2) / 2! 2 ²   [16.97]  =  Ö2/ 16  [16.97]