In the last post, Geodesics of left invariant metrics on matrix Lie groups – Part 1,we have derived Arnold’s equation – that is a half of the problem of finding geodesics on a Lie group endowed with left-invariant metric.
Suppose is a Lie group, and is a scalar product (i.e. a nondegenerate bilinear form) on its Lie algebra . Then, using left translations defines a left invariant (Riemannian or pseudo-Riemannian) metric on the whole group If is a path in we use left translations to define the image of the tangent vector in
Usually it is written more carefully, using instead of , but I am using a simplified notation, well adapted to dealing with problems for matrix groups.
If is a geodesic for the metric , then satisfies the Arnold’s equation
is defined as
and are the structure constants of , that is
where form a basis for and
Equation (2) is a system of nonlinear ordinary differential equations with constant coefficients. In this form it can be found in Arnold’s 1966 paper “Sur la g\’eom\’etrie differentielle des groupes de Lie de dimension inﬁnie et ses applications \`a l’hydrodynamique des ﬂuides parfaits”, Ann. Inst.
Fourier (Grenoble) 16 (1966), 319-361.
In his blog post “The Euler-Arnold equation” Terrence Tao mentions that some people call this equation the “Euler-Arnold” equation, while some other prefer to skip Arnold’s name completely and call it “Euler-Poincare equation”. Go-figure!
Solving equations (2) is just one half of the whole problem of finding a geodesic. Once are known, we need to solve the linear differential equation with variable coefficients that results from the definition of (??):
For the case of the rotation group and free rigid body we did it in Taming the T-handle continued.
B. Kolev in his paper “Lie groups and mechanics. An introduction” shows that for the rotation group O(n) the equations of motion for the free rigid body are completely integrable. We will not need this result, but is useful to know that in general case there are always two quadratic constants of motion corresponding to “kinetic energy” and “square of the angular momentum”. We were discussing these constants of motion for the group O(3) in
“Asymmetric Spinning Top – The Hardest Concept To Grasp In Physics” – they are used in so-called Poinsot construction. We will discuss a version of it for the group O(2,1) in the future.
The first observation is that the “kinetic energy” is, in fact, a constant, independent of To see that this is the case, we differentiate:
In the previous post we wrote Eq. (2) as
Substituting into Eq. (7) we get
The result is zero because is antisymmetric in while is symmetric. That is an often used property: if and then the contraction Indeed where in the last equality we have exchanged dummy indices names . If a number is equal to its negative, it must be zero.
To get the formula for the second quadratic invariant we need to return to the Ad-invariant scalar product that we have denoted in Killing vectors, geodesics, and “Noether’s theorem”:
The fact that is Ad-invariant implies an important relation between the matrix and the structure constants
that we are going to use. Ad-invariance means that:
for all and
Differentiating at we get
Setting we get
Multiplying both sides by we obtain
We can now derive the second conservation law. The angular momentum is defined as
Notice that is a covector, a one-form on , it is in the dual of It is the metric that connects the space to its dual. While vectors in play an active role, they generate transformations, elements in the dual, one-forms from , are “passive”, they evaluate vectors to numbers. It is the metric that is the third element here, that allows the active principle to connect to the passive principle. The metric depends on the mass distribution. In application to rigid bodies the inertia tensor is encoded in the metric on the rotation group.
The second conservation law states that the square of the angular momentum evaluated with the Ad-invariant metric is constant:
To verify we differentiate and use Eq. (8) rewritten as
Now, according to Eq. (14) is antisymmetric in , while the product is symmetric, therefore we get zero:
In the following posts we will first return to the case of the rotation group in three dimensions and the rigid body, and then try to apply a similar reasoning to the case of the Lorentz group O(2,1) in 2+1 dimensions.