$$T(\phi)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$

Here the matrix $T$ is parametrized by $\phi$ and $\theta$=some constant angle. Can I find out the generators of this orthogonal transformation parametrized by the angle $\phi$ ? If my approach is wrong, how do I find the generators of this matrix and exponentiate it?

I have derived an infinitesimal transformation which leads to $$T(\delta\phi)=\bigg[I+t\delta\phi\bigg]$$ where $t$ is, $$t= \begin{bmatrix} 0 &0 & 0 \\ 0 & 0 & 1\\ \sin(\theta) & -\cos(\theta) &0 \end{bmatrix}.$$

Here $\theta$ is some fixed angle, say $120$ degrees or $69$ degrees or anything, but it remains constant. Can I exponentiate this matrix to get $$e^{-\hat{t}\phi}$$ Is it correct? Where am I going wrong if I am completely incorrect?

Edit: if $\theta$ is a fixed constant, there is no way I could get the identity element, so what if $T$ is parametrized by both $\theta$ and $\phi$ ? I will surely get the identity element. Now how do I proceed from here?

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    $\begingroup$ This is wrong, starting with the zeroth-order term. If you set $\phi=0$ you don't get the identity, in contrast to the identity you've written. $\endgroup$ – Emilio Pisanty Nov 26 '19 at 10:33
  • $\begingroup$ [Formatting Note] Use \sin and \cos in math expressions. It looks better, especially when things multiply the trig functions. $A sin(\omega t)$ vs. $A \sin(\omega t)$. $\endgroup$ – ja72 Nov 26 '19 at 13:03
  • $\begingroup$ @EmilioPisanty I have edited the question as single parameter does not get me to identity element and I assume $T$ to be having two parameters $\endgroup$ – user135580 Nov 26 '19 at 13:08
  • $\begingroup$ This is not the most general parametrisation of a rotation - you can't parametrise rotations with only two angles. $\endgroup$ – jacob1729 Nov 26 '19 at 13:37
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    $\begingroup$ If nothing else, if you took that at fist glance, then the exponential form would imply that the rotation axis lies along the $x,z$ plane. However, this rotation axis can be obtained explicitly as $(\cot(\theta/2)\tan(\varphi/2), \tan(\varphi/2), 1)$ (unnormalized) via Mathematica's symbolic Eigensystem. This again indicates that the matrix isn't in exponential form. $\endgroup$ – Emilio Pisanty Nov 26 '19 at 17:13

Your orthogonal matrix $$R(\phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$ must have antisymmetric generators.

To find them, you must expand around the origin, $\phi=\pi, \theta=0$. To allay confusion, define $\Phi\equiv = \pi-\phi$, so the origin is at $\Phi=\theta=0$. $$R(\Phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ - \sin(\theta)\cos(\Phi) & \cos(\theta)\cos(\Phi) & \sin(\Phi)\\ \sin(\theta)\sin(\Phi) & -\cos(\theta)\sin(\Phi) &\cos(\Phi) \end{bmatrix}$$

Evaluate $R(\delta\Phi, 0)=\bigg[I+t\delta\Phi\bigg]$, so $$t= \begin{bmatrix} 0 &0 & 0 \\ 0 & 0 & 1\\ 0 & - 1 &0 \end{bmatrix}.$$

Can you also evaluate $R(0,\delta \theta)$?

Note added as per comments.

The above rotation matrix R then, in the conventions of WP, is but $$ e^{-\Phi L_x} e^{-\theta L_z} , $$ which you might choose to compose by BCH, $$\exp (-\Phi L_x -\theta L_z+ \Phi \theta [L_x,L_z]/2+... ), $$ or the Gibbs finite rotation formula, etc. if you were so inclined. For orthogonal axes like yours, Gibbs' formula all but collapses: the effective axis of rotation is just parallel to $\hat z \tan (\theta/2) +\hat x \tan (\Phi/2) +\hat y \tan (\theta/2) \tan (\Phi/2)$ ! (Can you see that this is precisely the invariant vector of R?)

In any case, the limiting procedure at the origin yielding the generators from your finite rotation matrix has sacrificed information: convince yourself that several different rotation matrices may share this identical behavior at the origin, of course--think of reversing the order of the two factors above; so you should not expect to reconstitute this specific rotation matrix from the tangent space behavior at the origin, in general. (Here you already factored your finite rotations in advance. What Lie's 3rd theorem guarantees is essentially Euler's theorem: the two component rotations will combine to a single rotation about a new axis.)

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    $\begingroup$ Thank you so much. After seeing your comments, I found one thing that this $T$ matrix seems to be a product of $R_{x}(\pi - \phi) R_{z}(\theta)$. $\endgroup$ – user135580 Nov 26 '19 at 16:50
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    $\begingroup$ @CosmasZachos The product exponentials of angular momenta isn't the exponential of the sum of the generators, though. Saying "this matrix has antisymmetric generators" implies not just that the derivatives are antisymmetric, but that the matrix equals the exponential of the derivatives, which isn't true. $\endgroup$ – Emilio Pisanty Nov 26 '19 at 17:15
  • $\begingroup$ @Emilio The BCH composition formula will yield an expression in the Lie algebra of traceless antisymmetric matrices, which, for small angles, will be the leading term in the BCH expansion. I am not sure the axis of rotation was asked for... but the SO(3) composition law may easily produce it for perpendicular component axes. $\endgroup$ – Cosmas Zachos Nov 26 '19 at 17:44
  • $\begingroup$ @Cosmas That is ultimately up to OP to prioritise, given that the result they originally wanted is impossible, but here I find it important that we be crystal clear with what does and does not hold. $\endgroup$ – Emilio Pisanty Nov 26 '19 at 17:48
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    $\begingroup$ @Emilio... I never understood the OP asking if the logarithm of that Rotation matrix is linear in 𝜃 and Φ. We are in full agreement it isn't! If he wishes a deduction of the effective axis of rotation, it is straightforward, but also trivial by looking for that matrix's identity vector. $\endgroup$ – Cosmas Zachos Nov 26 '19 at 19:54

The parametrization you've given just isn't of the exponential form you're after.

The matrix you've written down is parametrized by the location of the $x$ axis after the rotation, which has polar angle $\theta$ and azimuthal angle $\phi$ in polar spherical coordinates drawn around the old $x$ axis. It is not a rotation by angle $\phi$ about an axis at angle $\theta$, nor is it a rotation by angle $\theta$ about any clean axis, nor is it a combination of rotations by angles $\theta$ and $\phi$. It can of course be expressed as a rotation by a certain angle about a certain axis, but that angle is neither $\theta$ nor $\phi$.

As such, there is no useful way in which finding a low-order expansion in terms of $\theta$ or $\phi$ will give you a useful generator that will re-create your matrix upon exponentiation.


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