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I'm a mathematician studying Arnold's Mathematical Methods of Classical Mechanics.

On p. 83 the following definition is given.

Let $M$ be a differentiable manifold, $TM$ its tangent bundle, and $L:TM \to \mathbf{R}$ a differentiable function. A map $\gamma: \mathbf{R} \to M$ is called a motion in the lagrangian system with configuration manifold $M$ and lagrangian function $L$ if $\gamma$ is an extremal of the functional $\Phi(\gamma)=\int_{t_0}^{t_1}L(\gamma')dt$, where $\gamma'$ is the velocity vector $\gamma'(t) \in TM_{\gamma(t)}$.

(1) Is it correct then that geodesics in $M$ (if $M$ is Riemannian) are just a special case of this construction?

(2) Homogeneous spaces, i.e. manifolds of the form $G/K$, where $G$ is a Lie group and $K$ a closed subgroup, provide interesting examples of manifolds. What would a physical system look like which has $G/K$ as a configuration space? I remember hearing something along the lines "global $G$-invariance, local $K$-invariance" but I'm not sure.

And what interpretation do the geodesics then have in this model?

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2 Answers

  1. Yes, the geodesics are obtained in this way if $L(\gamma')$ is the proper length of the vector. However, the Arnold construction is not meant to describe geodesics only - but any motion of particles with any configuration space (with velocities added). The Lagrangian doesn't have to be identified with a proper length in general, and for most systems in mechanics, it is not.

  2. These quotients are omnipresent in supergravity theories, among others. For example, 11-dimensional supergravity compactified on a 7-torus and dimensionally reduced is a 4-dimensional theory known as $N=8$, $d=4$ supergravity with 133-63 = 70 scalars spanning the $E_{7(7)}/SU(8)$ coset. The global and local interpretation of the groups is exactly as you suggest: the $E_{7(7)}$ exceptional group in the numerator is a global symmetry while the denominator, $SU(8)$, is a local symmetry. It's because you may define an action of $E_{7(7)}$ on the whole $E_{7(7)}/SU(8)$ quotient (it's a symmetry of the rest of the theory, too). However, most of the generators are spontaneously broken. Only the generators of $SU(8)$ are unbroken: the vacuum state is invariant under them which is why you don't get new vacua by $SU(8)$ transformations, so there are no new scalars that would label these non-existent new vacua. That's why it's quotient.

The remaining generators of $E_{7(7)}(R)$ change the vacuum in supergravity. They change it to a "different vacuum" whose physics is, however, completely equivalent: it's a global symmetry. I added the label $R$ to emphasize it is a continuous group. In quantum gravity, however, electric and magnetic charges have to be quantized because of the Dirac quantization rule. They have to belong to a lattice and the lattice is changed under almost any $E_{7(7)}(R)$ transformation: so these transformations map a vacuum to a physically distinct vacuum and $E_{7(7)}$ is no longer a global symmetry. Only its discrete subgroup of elements that preserve the lattice, the $E_{7(7)}(Z)$ group, remains an exact symmetry of M-theory on $T^7$. It's also known as the U-duality group.

Well, maybe, I should have avoided supergravity. It would be enough to mention the Goldstone bosons. Scalars span a $G/K$ space is $G$ is the symmetry group of the action while $K$ is the unbroken subgroup. For each broken generator, you obtain one boson - a coordinate parameterizing the quotient. The supergravity story was an unnecessarily advanced variation of the same topic.

Let me mention a simple nontrivial example. Imagine that you describe the orientation of an object by an element of $SO(3)$. Now, imagine that the object is an axially symmetric gyroscope that preserves the $SO(2)$ symmetry. The actual "shape" of the gyroscope in space won't be affected by this $SO(2)$ symmetry, so the orientation will only be given by an element of $SO(3)/SO(2)$ which is isomorphic to a two-sphere, $S^2$.

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(1) Is it correct then that geodesics in M (if M is Riemannian) are just a special case of this construction?

Yes, it is correct to say this. Give a Riemannian manifold $M$ with a metric $g_{ab}$ and an affine connection $\Gamma^a_{bc}$, one can construct various geometric invariants on $M$ such as the Riemann and Ricci tensors. Now for a point particle propagating on this static manifold, i.e without taking General Relativity into account, one could write various Lagrangians of the form:

$$ L = \sqrt{|g|} ( K_m - V_m) $$

where $|g|$ is the absolute value of the determinant of $g_{ab}$ and $K_m$ and $V_m$ are kinetic and potential terms respectively which depending on the position and velocity of the particle or more generically is energy-momentum n-vector $j^a$. A generic example of the kinetic term in this case is:

$$ K_m := \frac{1}{2} \left( \partial^a j_a \right)^2 + \frac{1}{2}(\partial^a j^b \partial_a j_b) $$

and $V_m$ is some polynomial in $j^a$.

If we wish to incorporate the dynamics of the manifold itself (i.e. move into the domain of GR) then our Lagrangians must contain "kinetic" and "potential" terms for the various geometric invariants. These quantities however lack a simple expression for a general manifold and are subsumed into the Ricci scalar:

$$ L = \sqrt{|g|}(K_m - V_m + \mathcal{R}) $$

where the Ricci scalar $\mathcal{R} = g^{ab}\mathcal{R}_{ab} = g^{ab}g^{cd} \mathcal{R}_{acdb} $ is the trace of the Ricci tensor which in turn is a partial trace of the Riemann tensor.

Now, geodesics on a Riemannian manifold $M$ with a metric $g_{ab}$ and an affine connection $\Gamma^a_{bc}$ are given by solutions (integral curves) of the of the geodesic equation:

$$ v^a\nabla_a v^b = 0 $$

where $v := v^a \partial/\partial_a $ is a vector field on $M$, the covariant derivative is given in terms of the affine connection $\nabla_a v_b = \partial_a v_b + \Gamma^c_{ab}v_c $ and indices are raised and lowered with the metric $ v^a = g^{ab} v_b $. Your question amount to asking if there exists a Lagrangian $L(v^a)$ whose equation of motion is this geodesic equation. The answer is 'yes'. The following Lagrangian

$$ L = \sqrt{g^{ab} v_a v_b} $$

which is nothing more than the proper distance along the particle's wordline, yields the geodesic equation on variation w.r.t $\delta v_a$.

(2) Homogeneous spaces, i.e. manifolds of the form G/K, where G is a Lie group and K a closed subgroup, provide interesting examples of manifolds. What would a physical system look like which has G/K as a configuration space? I remember hearing something along the lines "global G-invariance, local K-invariance" but I'm not sure.

An example of a physical system with this configuration space is MacDowell-Mansouri gravity. A beautifully written reference for this is the review paper MacDowell-Mansouri gravity and Cartan geometry by Derek Wise, a former student of John Baez.

There one starts with a five-dimensional spacetime with symmetry group $G$ which can be deSitter, Anti-deSitter or Minkowski with a lie-algebra valued antisymmetric tensor $B$ and the curvature of the gauge connection $F$. The action is that of a topological BF theory given by:

$$ S_{BF} = \int tr B \wedge F $$

We can identify $B:= B_{\mu\nu}^{IJ}$ with the wedge product of two fermion fields: $ B^{IJ} = \psi^I \wedge \psi^J $ which transform in the fundamental representation of the gauge group. Furthermore, as in the case of the BCS mechanism, let us assume that the dynamics of the system contains a four-fermion term which acquires a vev (vacuum expectation value) due to the formation of a condensate of these fermions. Such a term results in the action:

$$ S'_{BF} = \int tr B \wedge F - \frac{G\Lambda}{6} B \wedge \star B $$

where $\star B$ is the Hodge dual of $B$. $G$ and $\Lambda$ are Newton's constant and the cosmological constant respectively. The formation of the condensate can be physically described by writing the five-dimensional gauge connection in the following form:

$$ {}^5 A = \left( \begin{array}{cc} {}^4 A && \frac{1}{l}\{e^0,e^{i} \} \\ \frac{1}{l} \{ e^0, \epsilon e^i \} && 0 \end{array} \right) $$

where ${}^4 A$ is a four-dimensional connection and $\{e^0,e^{i}\}$ (where $ i \in {1,2,3}$) is a vier-bien (tetrad). $\epsilon = \{-1,0,1\}$ for $G = \{ SO(4,1)$ (deSitter), $ISO(3,1)$ (Minkowski), $SO(3,2)$ (Anti-deSitter) $\}$ respectively. The group $H$ in each case is $SO(3,1)$ and the resulting theory has gauge group $\{SO(4,1)/SO(3,1), ISO(3,1)/SO(3,1),SO(3,2)/SO(3,1)\}$ respectively.

The resulting theory describes general relativity in four dimensions with the addition of topological terms such as the Nieh-Yan and Pontyargin terms in the action. For more details see Wise's excellent paper reference above and also the seminal paper (1) by Friedel and Starodubstev who first proposed this formulation of the MacDowell-Mansouri mechanism.

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Nice answer. But for (2), could you be more explicit about $G$, $K$ and $G/K$? I suppose you have e.g. $M^4 = (SO(1,3) \ltimes {\mathbb R}^4) / SO(1,3)$? –  Marek Feb 14 '11 at 11:21
    
Thanks @Marek. Good point. I'll edit that in. –  user346 Feb 14 '11 at 11:45
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