<p>The answer to this question is obviously affirmative, though the change of shape will be minuscule.</p>
<p>Let <span class="math-container">$\boldsymbol{\omega}$</span> be the angular velocity vector, and <span class="math-container">$\bf r$</span> be a point on the Earth surface.  Employing the vector formula
<span class="math-container">$$ 
\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = \boldsymbol{\omega}~(\boldsymbol{\omega}\cdot{\bf r}) - {\bf r}~\boldsymbol{\omega}^{\,2} =
 \nabla\left[\,\frac{1}{2}\,\left(\boldsymbol{\omega}\cdot{\bf r}\right)^2-\frac{1}{2}\,\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\,\right]\;,
$$</span>
and introducing the colatitude <span class="math-container">$\varphi$</span> through
<span class="math-container">$$
\cos\varphi=\frac{\boldsymbol{\omega}}{\,|\boldsymbol{\omega}|\,}\cdot\frac{\bf r}{\,|{\bf r}|\,}\;,
$$</span>
we split the centrifugal force into a second-harmonic part and a purely radial part:
<span class="math-container">$$
-\rho~\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = -\nabla\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}
 \,P_2\left(\cos\varphi\right)\right]
 +\nabla\;\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\right]~,\qquad(*)
$$</span>
where we assumed the body homogeneous, and used the expression
<span class="math-container">$P_2(\cos\varphi)=\frac{\textstyle 1}{\textstyle 2}\left(3\,\cos^2\varphi-1\right)$</span> for the degree-2 Legendre polynomial.</p>
<p>Interested in oscillating deformation solely, we now must change, in equation (*), <span class="math-container">$\boldsymbol{\omega}^{\,2}~$</span> to <span class="math-container">$~\boldsymbol{\omega}^{\,2}-\langle \boldsymbol{\omega}^{\,2}\rangle $</span>, where <span class="math-container">$\langle ... \rangle$</span> denotes time averaging.</p>
<p>From expression (*), we observe that the forcing splits into two parts -- quadrupole and purely radial.</p>
<p>Spherical asymmetry aside, the radial forcing causes a radial deformation. For terrestrial bodies, this is a teeny-tiny effect, because under weak forcing these bodies are virtually incompressible.</p>
<p>The quadrupole term mimics the quadrupole component of tides, wherefore the resulting deformation will be easy to calculate using the standard tidal machinery. I am not expert on ocean tides, but can say that the deformation of the solid part of the Earth will be expressed via the Love number <span class="math-container">$h_2$</span>.</p>

CC BY-SA 4.0
2024-04-09T06:36:47.180Z
It's possible, but a bit painful. See https://math.meta.stackexchange.com/q/29967/207316 I suppose you could make a macro using `\newcommand`
PM 2Ring
CC BY-SA 4.0
2024-04-09T12:25:45.680Z
@PM2Ring Thank you. I didn't realise that here we can make macros. But I also see that `planetmaker' has kindly fixed my answer using $\backslash$boldsymbol
Michael_1812
CC BY-SA 4.0
2024-04-09T06:20:04.057Z
2024-04-09T06:57:45.800Z
2024-04-09T06:57:45.800Z
Michael_1812
<p>The Earth is oblate because it spins and is not a rigid body. Since its spin axis changes relative to its crust, then the meridian of maximum diameter changes too, right?</p>
<p>I understand that polar motion is very small (just a few meters), but the tidal bulge is larger (over 20 kilometers), so I wonder if the side effects of polar motion are more dramatic than I had initially thought.</p>
<p>To be clear: I am talking here about polar motion, not precession or nutation.</p>

CC BY-SA 4.0
2022-09-25T09:45:38.257Z
Reading the wiki article, I wonder whether the definition of North (usually averaged over a year) doesn't follow the polar wander. https://en.m.wikipedia.org/wiki/Polar_motion. As tides play a large role, it matters whether it is spring tide or not (this sun-moon relative alignment). Hard to guesstimate 
planetmaker
CC BY-SA 4.0
2022-09-25T10:43:15.743Z
20km is more than I have read before,  I've heard of crustal tides in the order of 30cm.
James K
CC BY-SA 4.0
2022-09-25T15:29:24.733Z
The way they are modeled in the Explanatory Supplement to the Astronomical Almanac, the answer is no.  The algorithm for computing an observer's geocentric position does not account for polar motion.  The observer's lat, lon and altitude may change, but this is modeled as the observer moving, but the reference ellipsoid doesn't change because of it.  Put another way: of course it does, but to simplify things, different effects are classified into different models and accounted for separately.
Greg Miller
CC BY-SA 4.0
2023-10-20T01:18:37.613Z
You can check Earth SE. See if you find anything related.
Nilay Ghosh
<earth><planetary-science>
Does polar motion cause the Earth ellipsoid to change?
prideout
# Does polar motion cause the Earth ellipsoid to change?

(Asked by: username_0 on 2022-09-25)

The Earth is oblate because it spins and is not a rigid body. Since its spin axis changes relative to its crust, then the meridian of maximum diameter changes too, right?

I understand that polar motion is very small (just a few meters), but the tidal bulge is larger (over 20 kilometers), so I wonder if the side effects of polar motion are more dramatic than I had initially thought.

To be clear: I am talking here about polar motion, not precession or nutation.

**Tags:** earth, planetary-science

> **Comments:**
> **username_1**: Reading the wiki article, I wonder whether the definition of North (usually averaged over a year) doesn't follow the polar wander. https://en.m.wikipedia.org/wiki/Polar_motion. As tides play a large role, it matters whether it is spring tide or not (this sun-moon relative alignment). Hard to guesstimate   
> **username_2**: 20km is more than I have read before,  I've heard of crustal tides in the order of 30cm.  
> **username_3**: The way they are modeled in the Explanatory Supplement to the Astronomical Almanac, the answer is no.  The algorithm for computing an observer's geocentric position does not account for polar motion.  The observer's lat, lon and altitude may change, but this is modeled as the observer moving, but the reference ellipsoid doesn't change because of it.  Put another way: of course it does, but to simplify things, different effects are classified into different models and accounted for separately.  
> **username_4**: You can check Earth SE. See if you find anything related.  

---

## 1 answers

### Answer by username_5 on 2024-04-09. Score: 4 | Accepted answer

The answer to this question is obviously affirmative, though the change of shape will be minuscule.

Let $\boldsymbol{\omega}$ be the angular velocity vector, and $\bf r$ be a point on the Earth surface. Employing the vector formula
$$
\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = \boldsymbol{\omega}~(\boldsymbol{\omega}\cdot{\bf r}) - {\bf r}~\boldsymbol{\omega}^{\,2} =
\nabla\left[\,\frac{1}{2}\,\left(\boldsymbol{\omega}\cdot{\bf r}\right)^2-\frac{1}{2}\,\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\,\right]\;,
$$
and introducing the colatitude $\varphi$ through
$$
\cos\varphi=\frac{\boldsymbol{\omega}}{\,|\boldsymbol{\omega}|\,}\cdot\frac{\bf r}{\,|{\bf r}|\,}\;,
$$
we split the centrifugal force into a second-harmonic part and a purely radial part:
$$
-\rho~\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times{\bf r}\right) = -\nabla\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}
\,P\_2\left(\cos\varphi\right)\right]
+\nabla\;\left[\frac{\rho}{3}~\boldsymbol{\omega}^{\,2}\,{\bf r}^{\,2}\right]~,\qquad(\*)
$$
where we assumed the body homogeneous, and used the expression
$P\_2(\cos\varphi)=\frac{\textstyle 1}{\textstyle 2}\left(3\,\cos^2\varphi-1\right)$ for the degree-2 Legendre polynomial.

Interested in oscillating deformation solely, we now must change, in equation (\*), $\boldsymbol{\omega}^{\,2}~$ to $~\boldsymbol{\omega}^{\,2}-\langle \boldsymbol{\omega}^{\,2}\rangle $, where $\langle ... \rangle$ denotes time averaging.

From expression (\*), we observe that the forcing splits into two parts -- quadrupole and purely radial.

Spherical asymmetry aside, the radial forcing causes a radial deformation. For terrestrial bodies, this is a teeny-tiny effect, because under weak forcing these bodies are virtually incompressible.

The quadrupole term mimics the quadrupole component of tides, wherefore the resulting deformation will be easy to calculate using the standard tidal machinery. I am not expert on ocean tides, but can say that the deformation of the solid part of the Earth will be expressed via the Love number $h\_2$.

> **Comments:**
> **username_6**: It's possible, but a bit painful. See https://math.meta.stackexchange.com/q/29967/207316 I suppose you could make a macro using `\newcommand`  
> **username_5**: @username_6 Thank you. I didn't realise that here we can make macros. But I also see that `username_1' has kindly fixed my answer using $\backslash$boldsymbol  

---

