Right Invariant Randers metrics I'm hoping to determine the geodesic equation for a right invariant Randers metric $F(x) = \sqrt{a(x,x)} + b(x)$ on $SU(N)$. In my special case the navigation data $(h,W)$ for the Randers metric are such that $h$ is the biinvariant metric and $W$ is right invariant. 
The geodesic spray coefficients induced by a Randers metric are known and can be found in "Finsler Geometry, An Approach via Randers Spaces" as formula (2.30) and in many other places. However, formulating the problem on a Lie group seems to encounter the problem that no coordinates are holonomic coordinates for any basis for the tangent space in which either the metric $h$ or $W$ take a practical form.
Is there a coordinate free way to solve this problem which results in a first order equation for the tangent vector to a geodesic involving only objects in the Lie algebra $\mathfrak{su}(N)$?
 A: You are asking about a particular case of the general right invariant Lagrangian for curves on a Lie group.  This is a well-known story, but I can summarize it here:
Let $G$ be a Lie group with Lie algebra ${\frak{g}}=T_eG$ and dual ${\frak g}^\ast$, with the canonical pairing $\langle,\rangle:{\frak g}^\ast\times{\frak g} \to \mathbb{R}$.  Let $\mathrm{ad}$ and $\mathrm{ad}^\ast$ be the adjoint and co-adjoint representations, respectively, so that, for example
$$
\langle \mathrm{ad}^\ast(x)\xi,y\rangle = -\langle\xi,\mathrm{ad}(x),y\rangle = -\langle\xi,[x,y]\rangle.
$$
(Some people often forget about this minus sign, which is why I am reminding you of it now.)
Now, let $F:{\frak g}\to\mathbb{R}$ be a function that is smooth away from $0\in{\frak g}$ and has the property that $L = F^2$ is strictly convex on $\frak g$.  Then we want to know the geodesics of the right-invariant functional
$$
\lambda(\gamma) = \int_a^b F\bigl(\rho(\dot\gamma(t))\bigr)\ dt
$$
where $\gamma:[a,b]\to G$ is a differentiable curve and $\rho:TG\to{\frak g}$ is the canonical right-invariant $1$-form on $G$.  Usually, to get a convex functional (and fix the parametrization), we instead consider the energy functional
$$
E(\gamma) = \int_a^b \bigl(F\bigl(\rho(\dot\gamma(t))\bigr)\bigr)^2\ dt
=\int_a^b L\bigl(\rho(\dot\gamma(t))\bigr)\ dt.
$$
Here is the standard formula:  Let $L':{\frak g}\to {\frak g}^\ast$ be the Legendre transform of $L$, i.e., $d L = \langle L'(p), d p\rangle$.
Then a curve $\gamma:[a,b]\to G$
satisfies the Euler-Lagrange equations if and only if $p(t)=\rho\bigl(\dot\gamma(t)\bigr)$
satisfies the Euler equation
$$
\frac{d\ }{dt}\bigl(L'(p(t))\bigr) 
= -\mathrm{ad}^\ast\bigl(p(t)\bigr)\bigl(L'(p(t))\bigr).
$$
You should have no difficulty specializing this to your case.
