Invariance of Lagrangian under rotations in a constant magnetic field 
The Lagrangian for the motion of a particle with mass $m$ and charge $q$ in a constant magnetic field $B$ is given by $$\mathcal{L}(x,v)=\frac{m}{2}\left|v\right|^2-\frac{q}{2c}\left(v\cdot[x\times B]\right).$$
Show that rotations around the $B$-axis leave the Lagrangian invariant, where each rotation is given by $O_{\eta}:=\exp(\eta\,[B\,\times \,.]),\,\eta\in\mathbb{R}$.


I can see that $\left|O_{\eta}(v)\right|^2=\left|v\right|^2$, since rotations are supposed to leave the "length" unchanged but that's about as far as I've gotten with this. I'm guessing that one needs to apply some certain identities here regarding the cross product and the $\exp$ function, which I haven't been able to find on Wikipedia or other websites so far.
 A: you have to show that $\mathcal{L}'=\mathcal{L}$
with:
$$\mathcal{L}=\frac{m}{2}\vec{v}^T\,\vec{v}-\frac{q}{2c}\vec{v}^T\,\vec{\omega}$$
where $\vec{\omega}=\vec{x}\times\vec{B}$
$$\mathcal{L}'=\frac{m}{2}\vec{v}'^T\,\vec{v}'-\frac{q}{2c}\vec{v}'^T\,\vec{\omega}'$$
with:

*

*$\vec{v}'=O_{\eta}\,\vec{v}$

*$\vec{\omega}'=O_{\eta}\vec{\omega}$

*$O_{\eta}=\exp(\eta\,[\hat{\vec{B}}\,\times ])$
and
$O_{\eta}^T=\exp(-\eta\,[\hat{\vec{B}}\,\times  ])\quad $ thus $O_{\eta}^T\,O_{\eta}=I_3$
you obtain :
$$\vec{v}'^T\,\vec{v}'=\left(O_{\eta}\,\vec{v}\right)^T\,O_{\eta}\,\vec{v}=\vec{v}^T\,\vec{v}$$
$$\vec{v}'^T\,\vec{\omega}'=\left(O_{\eta}\,\vec{v}\right)^T\,O_{\eta}\,\vec{\omega}=
\vec{v}^T\,\vec{\omega}$$
thus:
$$\mathcal{L}'=\mathcal{L}$$
Edit:
$$\exp(\eta\,A)=I_3+\eta\,A+\eta^2\frac{1}{2}\,A\,A+\ldots+$$
$$\left[\exp(\eta\,A)\right]^T=I_3+\eta\,A^T+\eta^2\frac{1}{2}\,A^T\,A^T+\ldots+$$
with:
$$A=[{\vec{B}}\,\times  ]=\left[ \begin {array}{ccc} 0&-B_{{z}}&B_{{y}}\\ B_{
{z}}&0&-B_{{x}}\\ -B_{{y}}&B_{{x}}&0\end {array}
 \right]
$$
$$A^T=-A 
$$
A: The conservation of the second term seems intuitive, as simply a preservation of volume, formed by vectors $v, x, B$. Nevertheless, I also present a way to deduce this for infinitesimal transformation, which will imply for a finite rotation. Under a small rotation with $\eta \ll 1$, the variation of vector $a$ is $\delta a = \eta [a, B]$. Then:
$$
\delta (v \cdot [x \times B]) = ( \delta v \cdot [x \times B]) + (v \cdot [\delta  x \times B]) = \eta [v \times B] \cdot [x \times B] + \eta  \ v \cdot [[x \times B] \times B] = 
$$
$$
= \eta (v \cdot x \ B \cdot B - v \cdot B \ x \cdot B) + 
\eta (v \cdot (-x (B \cdot B) + B (B \cdot x))) = 0
$$
Here we have used following identitites:
$$
(a \times b) \times c = - (c \cdot b) a + (c \cdot a) b \qquad
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) \equiv
  (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})
$$
A: Write your Lagrangian in cylindrical coordinates. You will see that the Lagrangian doesn't depend on $\theta$, where $\theta$ is the angle that measures the rotatión about the z-axis.
