Angular velocity in terms of Euler angles

[aliens123]

In Chapter 4, derivation 15 of Goldstein reads:
"Show that the components of the angular velocity along the space set of axes are given in terms of the Euler angles by
$$\omega_x = \dot{\theta} \cos \phi + \dot{\psi} \sin \theta \sin \phi,

\omega_y = \dot{\theta} \sin \phi - \dot{\psi} \sin \theta \cos \phi,

\omega_z = \dot{\psi} \cos \theta + \dot{\phi}$$"

I want to know what is wrong with the following line of reasoning:
A vector which has been rotated by the Euler angles can be given by
$$\vec{r'} = A \vec{r}$$ where $$A$$ is the matrix and has been given in the text. Then $$\Delta \vec{r} = \vec{r'} - \vec{r} = (A-I)\vec{r}.$$ So
$$\frac{d\vec{r}}{dt} = \frac{dA}{dt}\vec{r}.$$
We also know that a vector which has been rotated has the defining condition
$$\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}.$$
So by comparing $$\frac{dA}{dt}\vec{r}$$ to $$\vec{\omega} \times $$ we should be done. But this does not work. In fact, $$\frac{dA}{dt}\vec{r}$$ does not even have the same form as $$\vec{\omega} \times. $$

 

[vanhees71]

You have a space-fixed basis (inertial system) with the Cartesian righthanded basis vectors $\vec{e}_i$ which are time-independent and a body-fixed Cartesian right-handed basis $\vec{e}_j'$ which is time-dependent. Then there's a time-dependent rotation matrix $D_{jk}$ such that (Einstein summation convention used)
$$\vec{e}_j'=D_{ij} \vec{e}_i.$$
Then for any vector $\vec{V}$ you have
$$\vec{V}=V_j' \vec{e}_j' = D_{ij} V_j' \vec{e}_i.$$
Now you want to calculate the time derivative
$$\dot{\vec{V}}=(D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') \vec{e}_i.$$
But you want to have this in the components of the body-fixed frame, which you get, using $\hat{D}^{-1} = \hat{D}^{\text{T}}$) from
$$\vec{e}_i=D_{ik} \vec{e}_k'$$
from which
$$\dot{\vec{V}} = (D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') D_{ik} \vec{e}_k' = (\dot{V}_k' + \Omega_{kj} V_j') \vec{e}_k'.$$
Now [EDIT: corrected typos leading to index confusion ;-))]
$$\Omega_{kj} = D_{ik} \dot{D}_{ij}=(\hat{D}^{\text{T}} \dot{D})_{kj}.$$
Since
$$\hat{D}^{\text{T}} \hat{D}=\hat{1}$$
you have
$$\hat{D}^{\text{T}} \dot{\hat{D}} + \dot{\hat{D}}^{\text{T}} \hat{D}=\hat{0}$$
or
$$\hat{D}^{\text{T}} \dot{\hat{D}}=\hat{\Omega}' = -\dot{\hat{D}}^{\text{T}} \hat{D}=-\hat{\Omega}^{\prime \text{T}}.$$
In components this means that
$$\Omega_{kj}'=\epsilon_{klj}\omega_l'$$
with three components $\omega_l'$, which are the angular-velocity components wrt. the body-fixed frame.

From this you get
$$\dot{\vec{V}}=(\mathrm{D}_t V')_k \vec{e}_k'$$
with a kind of "covariant time derivative" for vector components in the body-fixed frame
$$\mathrm{D}_t \underline{v}'=\dot{\underline{v}}'+\underline{\omega}' \times \underline{V}'.$$
Here and in the following I denote the column vectors built out of the components wrt. the body-fixed and space-fixed basis as $\underline{V}'$ and $\underline{V}$.

Since for any vector
$$\vec{V}=V_j' \vec{e}_j' = V_i \vec{e}_i = D_{ij} V_j' \vec{e}_i \; \Rightarrow \; V_i=D_{ij} V_j' \; \Rightarrow \; \underline{V}=\hat{D} \underline{V}'$$
and thus
$$\underline{V}'=\hat{D}^{\text{T}} \underline{V}$$
you get the body-fixed components of the angular velocity as
$$\hat{\Omega}'=\hat{D}^{\text{T}} \dot{\hat{D}}=\begin{pmatrix}
0 &-\omega_3' & \omega_2' \\
\omega_3' & 0 &-\omega_1' \\
-\omega_2 & \omega_1 0
\end{pmatrix}.$$
From this you read off $\underline{\omega}'$. The angular-velocity components wrt. the space-fixed frame you then get from
$$\underline{\omega}= \hat{D} \underline{\omega}'.$$

 

[aliens123]

Vanhees71, thank you very much for your reply. It seems like you understand this well, which makes me glad, because I would like to follow up with more questions if that is alright. Before we start though, I would like to clarify something. In your post you have a line which says

Now
$$\Omega_{kj} D_{ik} \dot{D}_{ij}=(\hat{D}^{\text{T}} \dot{D})_{ik}.$$

but was this a typo? Is what you meant
$$\Omega_{kj} =D_{ik} \dot{D}_{ij}=(\hat{D}^{\text{T}} \dot{D})_{ik}?$$

Other than that, I follow your derivation, and I don't struggle with the machinery you are using nor do I struggle with the properties of rotation matrices you are using (transpose is the inverse, etc.). What I am confused about though is why this particular route was chosen, if that makes sense.

I don't see why we need to worry about a space fixed and a body fixed basis. In other words, I want to think about this as being an active rotation. We have some initial vector and we apply a time dependent rotation to it to get a time dependent vector. As a concrete example, let's suppose that we have the following:

$$\vec{r}(t)= \begin{pmatrix}
\cos (\omega t) &-\sin (\omega t) & 0 \\
\sin(\omega t) & \cos (\omega t) & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}
=Dv\vec{r_0}$$
In other words, our vector starts out in the xz plane and then rotates around the z axis. The angular velocity vector should just clearly be $$(0,0,\omega).$$ The time derivative of our vector is then just
$$\dot{\vec{r}} = \dot{D} \vec{r_0} = \dot{D}D^T \vec{r}.$$
In my initial post I was confused, and didn't realize until after I had read your post and once I was typing that I also needed to include that second equality $$\dot{D} \vec{r_0} = \dot{D}D^T \vec{r}$$
if we want to compare it to the $$\omega \times $$ version. Is this also a valid way to proceed?

 

[vanhees71]

The typo is now corrected. It's of course
$$\Omega_{kj} = (\hat{D}^{\text{T}} \dot{D})_{kj}.$$
Also I don't understand your notation. I want to keep everything cleanly in one frame, i.e., you have to work either with components with respect to the space-fixed or the body-fixed frame. An at this point imho it is important to distinguish between frame-independent vectors (which I indicate with a vector arrow over the symbol) and component-column vectors wrt. the different bases of the frames, which I indicate by an underlined symbol without a prime (space-fixed inertial system) or with a prime (body-fixed non-inertial/rotating system). I always had big problems to understand the textbooks' treatment, because they don't make clearly this distinction.

The best treatment I've found so far is (no surprise) Sommerfeld, Lectures on Theoretical Physics, vol. 1.

 

[etotheipi]

The best treatment I've found so far is (no surprise) Sommerfeld, Lectures on Theoretical Physics, vol. 1

There's also some notes that I think are quite nice here:
https://www.damtp.cam.ac.uk/user/tong/dynamics/three.pdf
and section 3.5 is relevant to the OP.

 

[vanhees71]

Yes, very nice. This is pretty much the same approach, I posted above.

 

[aliens123]

Also I don't understand your notation. I want to keep everything cleanly in one frame, i.e., you have to work either with components with respect to the space-fixed or the body-fixed frame

I am confused by your confusion. I am describing a setup where there is only one (space fixed) frame which does not change with time. I am not describing a body which moves at all. I am just describing a vector which is rotating with time according to the equation:

$$\vec{r}(t)=
\begin{pmatrix}
x(t) \\
y(t) \\
z(t)
\end{pmatrix} =
\begin{pmatrix}
\cos (\omega t) &-\sin (\omega t) & 0 \\
\sin(\omega t) & \cos (\omega t) & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 \\
0 \\
1
\end{pmatrix}
$$

 

[vanhees71]

? What has this then to do with Euler angles? In this case it's much easier, because then you can show by just differentiating wrt. $t$ that (modulo a sign, perhaps ;-))
$$\dot{\vec{r}}=\omega \vec{e}_3 \times \vec{r}.$$

 

[aliens123]

? What has this then to do with Euler angles? In this case it's much easier, because then you can show by just differentiating wrt. $t$ that (modulo a sign, perhaps ;-))
$$\dot{\vec{r}}=\omega \vec{e}_3 \times \vec{r}.$$

I want to do the same thing, but replace my matrix with the Euler angle matrix shown in the text. Then I want to differentiate that matrix to find an expression for $$\dot{\vec{r}}$$ in terms of the matrix $$D, $$ and then compare it to the equation $$\dot{\vec{r}} = \omega \times \vec{r}$$ to determine $$\omega$$ in terms of $$D.$$

 

[aliens123]

Let me try to make this question better. I have a matrix which rotates a vector, and this matrix is written in terms of the Euler angles. This matrix can be found on page 153 of Goldstein (equation 4.46);
$$
A = \left(
\begin{array}{ccc}
\cos (\psi ) \cos (\phi )-\cos (\theta ) \sin (\psi ) \sin (\phi ) & \cos (\theta ) \sin (\psi ) \cos (\phi )+\cos (\psi ) \sin (\phi ) & \sin (\theta ) \sin (\psi ) \\
-\cos (\theta ) \cos (\psi ) \sin (\phi )-\sin (\psi ) \cos (\phi ) & \cos (\theta ) \cos (\psi ) \cos (\phi )-\sin (\psi ) \sin (\phi ) & \sin (\theta ) \cos (\psi ) \\
\sin (\theta ) \sin (\phi ) & \sin (\theta ) (-\cos (\phi )) & \cos (\theta ) \\
\end{array}
\right)$$
This was all done in terms of passive rotations though, so I just flip the sign of the angles (or take the transpose) to get:
$$A=
\left(
\begin{array}{ccc}
\cos (\psi (t)) \cos (\phi (t))-\cos (\theta (t)) \sin (\psi (t)) \sin (\phi (t)) & -\cos (\theta (t)) \sin (\psi (t)) \cos (\phi (t))-\cos (\psi (t)) \sin (\phi (t)) & \sin (\theta (t)) \sin (\psi (t)) \\
\cos (\theta (t)) \cos (\psi (t)) \sin (\phi (t))+\sin (\psi (t)) \cos (\phi (t)) & \cos (\theta (t)) \cos (\psi (t)) \cos (\phi (t))-\sin (\psi (t)) \sin (\phi (t)) & \sin (\theta (t)) (-\cos (\psi (t))) \\
\sin (\theta (t)) \sin (\phi (t)) & \sin (\theta (t)) \cos (\phi (t)) & \cos (\theta (t)) \\
\end{array}
\right)$$
So, this matrix rotates a vector:
$$\vec{r}(t) = A(t)\vec{r_0}$$
$$A(t)^{T} \vec{r}(t) = \vec{r_0}$$
Now I want to find how this changes with time:
$$\frac{ d \vec{r}(t)}{dt} = \frac{ d A(t)}{dt} \vec{r_0} =\dot{A(t)} A(t)^{T} \vec{r}(t) $$
Next I compare this to
$$\frac{ d \vec{r}(t)}{dt} = \vec{\omega} \times \vec{r} = Q \vec{r} $$
So by comparing
$$\dot{A(t)} A(t)^{T} = Q = \vec{\omega} \times$$
I should be able to read off the components of the angular velocity vector. When I actually do this in Mathematica though, I get an incorrect expression. The correct expression is to use
$$ A(t)^{T} \dot{A(t)}$$
Which is, incidentally, what you found as well. So what is incorrect about my derivation? Notably, according to the text, what I have found is $$\omega_{x'}, \omega_{y'}, \omega_{z'}$$ which is easy to see because
$$A^T (\dot{A} A^{T})) A = A^{T} \dot{A}.$$
I think this is because the equation
$$\frac{ d \vec{r}(t)}{dt} = \vec{\omega} \times \vec{r} $$
Assumes $$ \vec{\omega}$$ is in the body system, but I am not sure.
Edit: I have taken a closer look at your original post, and I think you essentially include enough information there for me to answer my own question. I think you would agree that the equation I wrote above does assume this is in the prime system? Or maybe that question is ill-posed.

 

[aliens123]

From this you get
$$\dot{\vec{V}}=(\mathrm{D}_t V')_k \vec{e}_k'$$
with a kind of "covariant time derivative" for vector components in the body-fixed frame
$$\mathrm{D}_t \underline{v}'=\dot{\underline{v}}'+\underline{\omega}' \times \underline{V}'.$$

Is this a typo as well? Should it say:
$$\mathrm{D}_t \underline{v}=\dot{\underline{v}}'+\underline{\omega}' \times \underline{V}'.$$

 

[aliens123]

I just want to compare and contrast the approach taken by Goldstein and the approach taken by you. On page 172, Goldstein writes:

"The component of G along the ith space axis is related to the components along the body axes:
$$G_i = \tilde{a}_{ij} G'_j = a_{ji} G'_j.$$
As the body moves in time, the components G'_j will change as will the elements a_ij of the transformation matrix. Hence, the change in G_i in a differential time element dt is
$$dG_i = a_{ji} dG'_j + da_{ji}G'_j.$$
It is no loss of generality to take the space and body axes as instantaneously coincident at the time t. Components in the two systems will then be the same instantaneously, but differential will not be the same, since the two systems are moving relative to each other. Thus, G'_j = G_j but a_ji dG'_j = dG'_i, the prime emphasizing the differential is measured in the body axis system. The change in the matrix A in the time dt is thus a change from the unit matrix and therefore corresponds to the matrix ε of the infinitesimal rotation. Hence,
$$da_{ji} = (\tilde{\epsilon})_{ij} = -\epsilon_{ij},$$
using the antisymmetry property of ε. In terms of the permutation symbol ε_ijk, the elements of ε are such that
$$-\epsilon_{ij} = -\epsilon_{ijk} d\Omega_k = \epsilon_{ikj} d\Omega_k.$$
Equation (4.84) can now be written
$$dG_i = dG'_i + \epsilon_{ikj} d \Omega_{k} G_j.$$
...
$$dG_i = dG'_i + ( d \Omega_{k} \times )_i.$$
...
an operator equation acting on some given vector:
$$\left( \frac{d}{dt} \right)_s = \left( \frac{d}{dt} \right)_r + \omega \times .$$"
Now compare this to what you have

You have a space-fixed basis (inertial system) with the Cartesian righthanded basis vectors $\vec{e}_i$ which are time-independent and a body-fixed Cartesian right-handed basis $\vec{e}_j'$ which is time-dependent. Then there's a time-dependent rotation matrix $D_{jk}$ such that (Einstein summation convention used)
$$\vec{e}_j'=D_{ij} \vec{e}_i.$$
Then for any vector $\vec{V}$ you have
$$\vec{V}=V_j' \vec{e}_j' = D_{ij} V_j' \vec{e}_i.$$
Now you want to calculate the time derivative
$$\dot{\vec{V}}=(D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') \vec{e}_i.$$
But you want to have this in the components of the body-fixed frame, which you get, using $\hat{D}^{-1} = \hat{D}^{\text{T}}$) from
$$\vec{e}_i=D_{ik} \vec{e}_k'$$
from which
$$\dot{\vec{V}} = (D_{ij} \dot{V}_j' + \dot{D}_{ij} V_j') D_{ik} \vec{e}_k' = (\dot{V}_k' + \Omega_{kj} V_j') \vec{e}_k'.$$

I like the way you have done this much better, however can anybody shed some light on what Goldstein means? The textbook's treatment is leaving me confused, so maybe it is just best to ignore it.

 

[vanhees71]

I've always been in the minority among those who don't like Goldstein's classic textbook on mechanics. I find his notation utterly confusing. In the case of rigid-body dynamics it's precisely the vagueness about the question whether he refers to components in the space- or body-fixed system. Maybe somebody else can help, who understands Goldstein's notation.