What does this infinitesimal Eulerian change describe? This is a question I originally posted in math.se which received an answer that was far too mathematically sophisticated for what I wanted; given that basic multivariable calculus was used through out the paper.
I'm reading a paper where the orientation of a coordinate system is specified by y-convention Euler angles (eqns 30-47)$(\phi_0, \theta_0, \psi_0)$ and rotation matrix $\xi$. It then goes on to say let $$d\xi = (d\psi_0\sin\theta_0\cos\phi_0 - d\theta_0\sin\phi_0,~d\psi_0\sin\theta_0\sin\phi_0 + d\theta_0\cos\phi_0,~d\phi_0 + d\psi_0\cos\theta_0)$$
What sort of infinitesimal change is it describing?
This is equation 3.22 of http://148.216.10.84/archivoshistoricosMQ/ModernaHist/Thomas1927.pdf
 A: This is a hard question, because Thomas is using terrible language. Most of the same calculation is done easily in my answer to this question: How did L.H. Thomas derive his 1927 expressions for an electron with an axis?
The right way to do it is to just choose the x axis to be in the direction of the velocity at one time, and the x-y plane to include the acceleration and velocity, and the origin to be the position of the electron. Then all of Thomas's manipulations can be quickly done with no vector notation. His results are needlessly obfuscated, especially in this follow-up paper, where the introduction of Eulerian angles so deeply irrelevant to the underlying essentially trivial calculation, that I suspect that Thomas was on a mission to make his paper unreadable.
If you insist on an answer
The Eulerian angles introduced by Thomas are not used later, they just serve to parametrize the rotation matrix $\zeta_0$. The quantity $d\zeta_0$ is confusing, because it is not the infinitesimal change in the $\zeta_0$ matrix, which would be a tensor of second rank, but a vector, specifying an axis and a magnitude of a rotation. This vector is given by a well known procedure for specifying an infinitesimal rotation.
When you have two rotation matrices which differ infinitesimally, $R$ and $R+dR$, then $dR$ is a bad quantity, because adding rotation matrices does not produce a rotation matrix. The correct way to measure the infinitesimal difference is to consider $R^{-1}(R+dR) = I + R^{-1}dR$, and this tells you that $R^{-1}dR$ is an element of the Lie algebra (by definition--- the product of two rotations is a rotation, and a rotation which is infinitesimally different from the identity is the identity plus an element of the Lie algebra). There is also $dR R^{-1}$ which works too, they two differ by a rotation.
The Lie algebra in this case consists of all infinitesimal rotations, which are rotations about some axis with an infinitesimal angle. The axis unit vector times the infinitesimal angle is a vector, not a tensor, and this vector tells you what the infinitesimal difference between two nearby rotations is. The interpretation is that R+dR is R followed by a rotation specified by the vector of $dR R^{-1}$. The interpretation of the other order $R^{-1} dR$ is that a rotation by this, followed by a rotation by R, gives R+dR.
The quantity $d\xi_0$ is this infinitesimal rotation vector for the rotation between the two nearby frames he is considering. Rotation vectors are well known from undergraduate mechanics courses--- they are the angular velocities--- the things you cross with the position to find the rate of change of the position under rotation.
It is difficult to see this, because he did not say what he was doing. The only purpose of the Euler angles is to specify that he is talking about the infinitesimal rotation vector, in the most obscure possible way.
Once you know that there is an explicit angular velocity vector which is doing a rotation in addition to the Thomas precession rotation calculated in the linked answer, you just add the rotation vector of this additional rotation to the rotation vector of the Thomas precession to get the full rotation of the frame. This gives his final formula. Nowhere do you need to use his intermediate steps.
If you want his Euler angle conventions...
His convention for Euler angles is as follows:
$$ \zeta_0 = A(\phi) B(\theta) C(\psi) $$
Where $A$ is a rotation about the z axis by angle $\phi$, $B$ is a rotation about the y axis by angle $\theta$, and $C$ is a rotation about the z axis by angle $\psi$. From this Euler angle convention, you can find the formula for the infinitesimal rotation vector by the procedure above. Done abstractly, you get:
$$ d\zeta_0 \,\, \zeta_0^{-1} = dA\,A^{-1}  + A\,\, dB\,B^{-1} \,\, A^{-1} + AB \,\,dC C^{-1}\,\, B^{-1}A^{-1} $$
The infinitesimal rotation M(\omega) corresponding to the vector $\omega$ has the property that
$$ R M(\omega) R^{-1} = M(R\omega) $$
This is easy to prove--- it is transforming between the the two different orders $R^{-1}dR$ and $dR R^{-1}$, which just changes the axis of rotation by R. Mathematicians say that the Lie algebra is the adjoint representation, which transforms by matrix-on-one-side, inverse-matrix-on-the-other. For 3d rotations, the adjoint representation is the vector representation, and the equivalence between vectors and adjoints is the formula above. But you can ignore all this jawboning--- the above is something you can understand and check directly.
When you turn it into vectors, the first quantity is a rotation purely along the z axis by an amount d\phi, so it's omega vector is purely along the z axis:
$$ \omega(dA A^{-1}) = (0,0,1) d\phi$$
the second starts off as a pure rotation about the y-axis by an amount $d\theta$, but it gets rotated about the z axis by $\phi$ (because of the $A$ and $A^{-1}$), so its omega vector is the y-axis rotated by $\phi$ around the z axis.
$$ \omega(A\,\,dB\,B^{-1}\,\,A^{-1}) = (-\sin(\phi),\cos(\phi),0) d\theta $$
The third would be a rotation about the z axis, or (0,0,1) but it gets rotated around the y axis by $\theta$, so into $(\sin\theta, 0, \cos\theta)$, then it gets rotated around the z axis by $\phi$, so into $(\sin\theta \cos\phi, \sin\theta \sin\phi , \cos\theta)$. This is the rotation vector of the third term.
$$ \omega(AB\,\,dC\,C^{-1}\,\,A^{-1}B^{-1} ) = (\sin\theta \cos\phi , \sin\theta \sin\phi, \cos\theta) d\psi$$
Adding the three angular velocities together is Thomas's formula. Again, it is never used in the equations that follow.
