<p>Citing from <a href="https://en.wikipedia.org/wiki/Spherical_harmonics" rel="noreferrer">Wikipedia</a>: <span class="math-container">$Y_l^m$</span> is called a spherical harmonic function of degree <span class="math-container">$l$</span> and order <span class="math-container">$m$</span>. If we take the real part of the spherical harmonics only, there is a nice visual explanation for order and degree.</p>
<p>The order <span class="math-container">$m$</span> 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: <span class="math-container">$l-|m|$</span> is the number of zero crossings if we walk from pole to pole.</p>
<p>Below you can see an example for <span class="math-container">$l=3$</span> and <span class="math-container">$0\leq m \leq 3$</span> (keep in mind that in order to count the number of zonal waves you also have to walk around the &quot;back&quot; of the sphere which you can not see in the picture):<a href="https://i.sstatic.net/1sGCD.png" rel="noreferrer"><img src="https://i.sstatic.net/1sGCD.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-06-23T07:05:03.250Z
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)
Edlothiad
CC BY-SA 4.0
2021-06-23T07:18:03.997Z
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.
Joscha Fregin
CC BY-SA 4.0
2021-06-23T07:21:35.197Z
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?
Edlothiad
CC BY-SA 4.0
2021-06-23T07:33:46.513Z
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.
Joscha Fregin
CC BY-SA 4.0
2021-06-23T08:04:27.223Z
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?
Edlothiad
CC BY-SA 4.0
2021-06-23T10:06:46.587Z
I think you are correct for $l=3$, $m=3$ if you walk around the whole sphere (not just pole to pole), but I like to emphasize again that, to my knowledge, it is hard to connect order and degree to the meridional wavenumber. That is why we use the number of zero crossings between poles instead.
Joscha Fregin
CC BY-SA 4.0
2021-06-23T10:59:14.877Z
Alright, thank you very much for your help and explanation!
Edlothiad
CC BY-SA 4.0
2021-06-22T13:09:14.683Z
2021-06-22T13:09:14.683Z
Joscha Fregin
<p>It's Legendre polynomials, not Lagrangian.</p>

<p>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.</p>

<p>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.)</p>

<p>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.</p>

CC BY-SA 3.0
2017-05-02T13:18:29.923Z
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
Edlothiad
CC BY-SA 3.0
2017-05-02T12:51:34.047Z
2017-05-02T12:51:34.047Z
David Hammen
<p>I've been reading up about Gravitational Models and Spherical Harmonics.</p>

<p>Where I first read about them is in this <a href="https://arxiv.org/pdf/1609.05225.pdf" rel="nofollow noreferrer">scholarly article</a> on Autonomous navigation with Gravity Gradients. It talks about Gravity Models being of order <em>x</em> and degree <em>y</em>.</p>

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

<p>Another example is found <a href="http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/egm08_wgs84.html" rel="nofollow noreferrer">here</a> which states that EGM2008 is complete to spherical harmonic degree and order 2159</p>

CC BY-SA 4.0
2021-06-21T08:21:04.307Z
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 ).
Aditya sahu
<models><gravity>
What does the degree and order of Spherical Harmonics mean?
Edlothiad
# What does the degree and order of Spherical Harmonics mean?

(Asked by: username_0 on 2017-05-02)

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

Where I first read about them is in this [scholarly article](https://arxiv.org/pdf/1609.05225.pdf) 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](http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/egm08_wgs84.html) which states that EGM2008 is complete to spherical harmonic degree and order 2159

**Tags:** models, gravity

> **Comments:**
> **username_1**: 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 ).  

---

## 2 answers

### Answer by username_2 on 2021-06-22. Score: 5 | Accepted answer

Citing from [Wikipedia](https://en.wikipedia.org/wiki/Spherical_harmonics): $Y\_l^m$ 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](https://i.sstatic.net/1sGCD.png)](https://i.sstatic.net/1sGCD.png)

> **Comments:**
> **username_0**: 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)  
> **username_2**: 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.  
> **username_0**: 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?  
> **username_2**: 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.  
> **username_0**: 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?  
> **username_2**: I think you are correct for $l=3$, $m=3$ if you walk around the whole sphere (not just pole to pole), but I like to emphasize again that, to my knowledge, it is hard to connect order and degree to the meridional wavenumber. That is why we use the number of zero crossings between poles instead.  
> **username_0**: Alright, thank you very much for your help and explanation!  

---

### Answer by username_3 on 2017-05-02. Score: 4

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.

> **Comments:**
> **username_0**: 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  

---

