4
$\begingroup$

I've been reading up about Gravitational Models and Spherical Harmonics.

Where I first read about them is in this scholarly article on Autonomous navigation with Gravity Gradients. It talks about Gravity Models being of order x and degree y.

What does order and degree mean in this circumstance? Is it something to do with the Lagrangian Polynomials used in the models?

Another example is found here which states that EGM2008 is complete to spherical harmonic degree and order 2159

$\endgroup$
1
  • $\begingroup$ I am also a bit confused as far as I know in Spherical harmonics:- 1. Zonal Terms has something to do with latitudes only Zn = f(Jn) where n>= 2 2. Tesseral Terms has something to do with orbital axis symmetry Tnm = f(Cn, Sm) where n>=2, n>=m>=1 , (The higher number of terms we consider better the Geopotential function of earth becomes? Is this statement right ). $\endgroup$ Commented Jun 21, 2021 at 8:21

2 Answers 2

5
$\begingroup$

Citing from Wikipedia: $Y_l^m$ is called a spherical harmonic function of degree $l$ and order $m$. If we take the real part of the spherical harmonics only, there is a nice visual explanation for order and degree.

The order $m$ is the zonal wave number, that is how many waves we count walking around the sphere at constant latitude. The degree is a little more difficult to interpret, because we need to take the order into account: $l-|m|$ is the number of zero crossings if we walk from pole to pole.

Below you can see an example for $l=3$ and $0\leq m \leq 3$ (keep in mind that in order to count the number of zonal waves you also have to walk around the "back" of the sphere which you can not see in the picture):enter image description here

$\endgroup$
7
  • $\begingroup$ This is an exceptional answer! Although I still have a few questions. In the l=3 m=0 case, if we walk along a constant latitude would we not see any of the zonal waves? Because the 3 we see are each along a unique latitude. We can only count the three waves when looking vertically. The degree makes sense. that the number of "vertical" waves (let's say) is equal to m and the number of "horizontal" is equal to l-m. (apologies if my terminology is all wrong, It has been a long time since I undertook this project and my work has changed rather significantly) $\endgroup$
    – Edlothiad
    Commented Jun 23, 2021 at 7:05
  • $\begingroup$ Yes for $l=3$, $m=0$ you don't see any zonal waves "because" the zonal wave number $m$ is zero. For the vertical case: don't confuse zero crossings with waves. Take for example $l=3$, $m=1$. there are $3-1 = 2$ zero crossings from pole to pole (2 horizontal white lines in the picture), but we count 3 waves walking around the sphere at constant longitude. $\endgroup$ Commented Jun 23, 2021 at 7:18
  • $\begingroup$ Ah of course, I misread l and m! That makes a lot of sense! I have accepted this answer as it provides exactly the explanation I had hoped for! For the vertical case, would the three waves be what is coloured the blue, red and blue sections? Or am I missing how the waves are counted? $\endgroup$
    – Edlothiad
    Commented Jun 23, 2021 at 7:21
  • $\begingroup$ You passed one wave if you walk over both colors. so going vertical and having something like this is one wave: Start at white horizontal line, pass red color, pass white horizontal line, pass blue color, arrive at white horizontal line. This is one wave. Sorry I should have added some sort of legend. $\endgroup$ Commented Jun 23, 2021 at 7:33
  • $\begingroup$ Does this mean for the l=3, m=3 case, walking vertically we only pass one wave? Or are all three waves counted at the junction point at the pole? $\endgroup$
    – Edlothiad
    Commented Jun 23, 2021 at 8:04
4
$\begingroup$

It's Legendre polynomials, not Lagrangian.

The zonal harmonics (the $Y^0_l$ terms) depend only on latitude. These zonal harmonics are closely associated with the Legendre polynomials $P_l(x)$, where $x$ is the sine of the geocentric latitude.

The tesseral and sectoral harmonics (the $Y^m_l$ terms, where $m\ne 0$) are closely associated with the associated Legendre functions $P_{lm}(x)$. (Note: These are sometimes called associated Legendre polynomials, but they aren't polynomials.)

Spherical harmonics are widely used in physics, so the presentations readily found on the internet generally reflect how physicists use spherical harmonics. There are some key differences between these easily found references on the internet and the forms used to represent gravitation. One is that physicists typically use colatitude, and hence use $P_l(\cos\theta)$ as opposed to $P_l(\sin\phi)$. Another is that physicists tend to use the complex form, while gravity models are expressed in terms of real sine and cosine forms. Finally, a set of gravity coefficients is inevitably going to be fully normalized coefficients, while physicists use a form with a different normalization.

$\endgroup$
1
  • 1
    $\begingroup$ So what does the order and degree of the harmonic dictate? The order and degree to which the Legendre Polynomials are taken? If that's even correct speaking $\endgroup$
    – Edlothiad
    Commented May 2, 2017 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.