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Have a little question regarding infinitesimal rotations. In the Cohen Book, volumen 1, Complement B-VI, it says that the transform of a vector $\textbf{OM}$ under an infinitesimal rotation can be written, to first order in $d\alpha$ is

\begin{equation} \Re_{\textbf{u}}(d\alpha)\textbf{OM}=\textbf{OM}+d\alpha \textbf{u}\times \textbf{OM} \end{equation}

what I don't understand is the about the "first orden in $d\alpha$". Is a taylor expansion? Also I don't know how to deduce the that equation. Considering the differential that appears there, I think is this type of expansion, but I can't see why there's a vectorial product also.

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The rotation group of three dimensional space has three generators $T^a$ given by

$$ T^3 = \left(\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right) \quad T^2 = \left(\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{matrix}\right) \quad T^1= \left(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{matrix}\right)$$

which yield rotations associated to a vector $\vec \phi \in \mathbb{R}^3$ by (summation over repeated indices implied in the following)

$$ R_{\vec \phi} = \mathrm{e}^{\phi^aT^a} = 1 + \phi^aT^a + \mathcal{O}(\phi^2)$$

and, by inspection, we see that they fulfill (for $e^i$ the $i$-th standard basis vector)

$$ (T^a v)^c = (T^a (v^b e^b))^c = (v^b T^a e^b)^c = v^b \epsilon^{abc} = (e^a \times \vec v)^c$$

since the cross product is given as

$$ (\vec v \times \vec w)^k = \epsilon^{ijk}v^i w^j$$

with $\epsilon$ the Levi-Civita symbol. Therefore,

$$ R_{\vec \phi} v = (1 + \phi^a T^a + \mathcal{O}(\phi^2))\vec v = \vec v + \vec\phi\times \vec v + \mathcal{O}(\phi^2)$$

and writing $\vec \phi = \phi \hat{\phi}$ with $\hat{\phi}$ being a unit vector and saying $\phi$ is infinitesimal, i.e. neglecting higher orders in $\phi$, yields your relation.

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  • $\begingroup$ Group theory is beautiful, but sometimes is better to go deep in the bones of the groups, for Lie groups, in analytical way with our equally beautiful coordinate systems. Especially, for new ones that don't understand what is happens and confuses group theory with magic. $\endgroup$ – Nogueira Jun 9 '15 at 2:02
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It is easy to verify that every proper rotation in space, which is the subgroup of the full group of rotations of $\mathbb R^3$ that is connected to the identity $I$, can be expressed in the form $$R_u = e^{A_u},$$ where $A$ is a $3\times3$ skew-symmetric matrix. Indeed $(e^A)^Te^A = e^{-A}e^A = I$ and $\det(e^A) = e^{Tr(A)} = 1$ and therefore $e^A$ is a proper rotation. Since skew-symmetric $3\times3$ matrices are in one-to-one correspondence with vectors in $\mathbb R^3$ (by Hodge duality one has an isomorphism between $\mathbb R^3$ and $\bigwedge^2\mathbb R^3$), one can identify any skew-symmetric matrix $A$ with a vector $u_A$ representing the axis of rotation, with the magnitude of $u_A$ giving the idea of the extent of this rotation around $u_A$. For an infinitesimal rotation $\delta\alpha$ around the direction of the unit vector $u$ one then gets the rotation $$e^{\delta\alpha A_u},$$ where $A_u$ is the skew-symmetric matrix corresponding to the vector $u$ through the isomorphism mentioned earlier. A series expansion around the identity then gives $$e^{\delta\alpha A_u} = I + \delta\alpha A_u + o(\delta\alpha^2).$$ The way any $A_u$ acts on a vector $x$ of $\mathbb R^3$ can be seen to be $$A_u x = u\times x,\qquad\forall x\in\mathbb R^3$$ and therefore the infinitesimal rotation of a vector $x\in\mathbb R^3$ has the approximation $$e^{\delta\alpha A_u}x = x + \delta\alpha u\times x+o(\delta\alpha^2).$$

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  • $\begingroup$ I changed your $o(\delta a)$ to $o(\delta a^2)$. If that's not what you meant, feel free to revert my edit. $\endgroup$ – David Hammen Jun 8 '15 at 16:30
  • $\begingroup$ As I'm employing the little-o notation I think it was right in the original form. $\endgroup$ – Phoenix87 Jun 8 '15 at 19:06
  • $\begingroup$ Why anyone in this question are spitting group theory approach to rotations without answering the OP's question. $\endgroup$ – Nogueira Jun 9 '15 at 16:42
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    $\begingroup$ 1) I don't see an explicit reference to group theory in my answer; 2) I don't really see the problem with using group theory. IMHO it gives a different perspective on this question. For instance, what is your matrix that you get when you compute that derivative in your answer? Is it an accident that it is skew-symmetric? What happens if one now wants to consider the third dimension as well? How do the formulae generalise? I don't think this quite apparent from your answer... $\endgroup$ – Phoenix87 Jun 9 '15 at 16:46
  • $\begingroup$ I understand your point, but my answering is in terms of matrices in $3d$. I'm consider the three dimensions, but as anyone know, rotations in $3d$ always preserve one axis. I use this feature of rotation, and suitable rectangular coordinates, for uses a short notation. My matrix is the usual rotation that anyone that can't derive the $1+\delta \theta \vec{u} \times$ knows. I'm not saying that you are wrong, but saying that this answering is not quiet compatible with OP. $\endgroup$ – Nogueira Jun 10 '15 at 3:28
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It is a Taylor series of the rotation matrix, but the idea is that we don't know the form of the matrix when we write the expansion. Instead we use the definition of rotations, as linear transformations which preserve the length of vectors, to determine the form of the expansion coefficients.

Its not hard to prove that every rotation matrix is orthogonal, i.e., $R^T=R^{-1}$.

We can write infinitesimal rotations, which are very close to the identity, as,

$$ R= (\mathbb{I} + \epsilon \ \rho + O(\epsilon^2)) \qquad (\epsilon << 1).$$

The orthogonality of $R$ requires that,

$$ \mathbb{I} = R^T R= (\mathbb{I}+\epsilon \ \rho^T + O(\epsilon^2))(1 + \epsilon \ \rho +O(\epsilon^2)),$$

$$ \mathbb{I} = \mathbb{I}+\epsilon \ ( \rho^T+\rho) + O(\epsilon^2),$$

if we require that coefficient of each power of $\epsilon$ vanish independently then we get,

$$ \boxed{ \rho^T + \rho =0}.$$

We now have that $\rho$ must be an antisymmetric matrix. In three dimensions there are only three linearly independent antisymmetric matrices,

$$ \rho_1 = \left( \begin{array}{ccc} \ 0 & 1 & 0 \\ \ -1 & 0& 0 \\ \ 0 & 0 & 0 \\ \end{array}\right); \ \rho_2 = \left( \begin{array}{ccc} \ 0 & 0 & 1 \\ \ 0 & 0& 0 \\ \ -1 & 0 & 0 \\ \end{array}\right); \ \rho_3 = \left( \begin{array}{ccc} \ 0 & 0 & 0 \\ \ 0 & 0& 1 \\ \ 0 & -1 & 0 \\ \end{array}\right). $$

It turns out that we can write the cross product of two vectors in terms of these matrices,

$$ \vec{\omega} \times \vec{v} = \left( \begin{array} \ 0 & -\omega_3 & \omega_2 \\ \ \omega_3 & 0& -\omega_1 \\ \ -\omega_2 & \omega_1 & 0 \\ \end{array}\right) \left( \begin{array} \ v_1 \\ v_2 \\ v_3 \end{array} \right)$$

Since a rotation about the axis $u$ is supposed to leave any vector parallel to that axis unchanged it is natural to identify $R_u(\epsilon)$ with the corresponding "cross product matrix",

$$ R_u(\epsilon) = \mathbb{I} + \epsilon \ \vec{u} \times, $$

if we want a finite rotation about this axis we can use the formula,

$$ R_u(\theta) = \lim_{N\rightarrow \infty} (\mathbb{I} + \frac{\theta}{N} \ \vec{u} \times)^N = \exp(\theta \vec{u} \times ).$$

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  • $\begingroup$ I think you forgot to take a limit in your last equation $\endgroup$ – Phoenix87 Jun 8 '15 at 19:05
  • $\begingroup$ Wrong: by the specification of $\vec{u}$ and $\delta \theta$ we already know the rotation. $\endgroup$ – Nogueira Jun 9 '15 at 2:16
  • $\begingroup$ @Nogueira, only if you already know the form of the rotation matrix, not if you are trying to derive it. The technique I outlined here applies to more general situations, which Physicists need to deal with, than just rotations in three dimensions. $\endgroup$ – Spencer Jun 9 '15 at 7:18
  • $\begingroup$ Yes, you are dealing with SO(3) symmetries. But this is not the OP's question. $\endgroup$ – Nogueira Jun 9 '15 at 16:39
  • $\begingroup$ @Nogueira, The OP literally said "I don't know how to deduce the that equation", my answer shows him how to derive the equation. $\endgroup$ – Spencer Jun 10 '15 at 1:46
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Yes, is a Taylor expansion! Take the rotational direction and define this direction as $z$ of some cartesian coordinate system (you have freedom to do that, you are a free physicist). Ignore the $z$ component of the vector, because don't change with the rotation.

The rotation take the form of:

$$ \vec{v}(\theta)=\left( \begin{array} \ v_x(\theta) \\ v_y(\theta) \end{array} \right)=\left( \begin{array} \ cos(\theta) & -sin(\theta)\\ \ sin(\theta) & cos(\theta)\\ \end{array}\right) \left( \begin{array} \ v_x \\ v_y \end{array} \right) $$

By taylor expansion in $\theta=0$ we have: $$ \vec{v}(\theta)=\vec{v}(0)+\theta\frac{d}{d\theta}\left( \begin{array} \ cos(\theta) & -sin(\theta) \\ \ sin(\theta) & cos(\theta) \\ \end{array}\right)_{\theta=0} \left( \begin{array} \ v_x \\ v_y \end{array} \right)+...=\vec{v}(0)+\theta\left( \begin{array} \ 0 & -1 \\ \ 1& 0 \\ \end{array}\right) \left( \begin{array} \ v_x \\ v_y \end{array} \right)+... $$

Then, we have $\vec{v}(\theta)=\vec{v}(0)+(\vec{z}\times\vec{v}(0))\theta+...$, and, without loss of generality, for small rotations $\delta \theta$ in any direction in $3d$ space $\vec{u}$ we have:

$$ \vec{v}(\delta \theta)=\vec{v}+(\vec{u}\times\vec{v})\delta \theta. $$

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  • $\begingroup$ Why the down vote? $\endgroup$ – Nogueira Jun 10 '15 at 3:14

protected by Qmechanic Jun 8 '15 at 17:41

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