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Given an affine manifold $(M,\nabla)$, the geodesic equation $\ddot{x}^j+\dot{x}^k \dot{x}^l\Gamma_{kl}^j=0$ completely characterizes the geodesics on the manifold. This is often called the Euler-Lagrange equation. I was wondering what the connection between geodesics and Lagrangian mechanics are. Given a mechanical system, its solutions can be found by solving the resulting Euler-Lagrange equation. Is there some affine connection $\nabla$ we can equip the configuration space $C$ so that all solutions are hence geodesics on the "manifold" $(C,\nabla)$? Reading this paper, it seems one can construct a Riemannian metric from the kinetic energy. However, it seems this really only works when there's no potential in the system, which I find kinda odd since what about the case where there is a potential.

Qmechanic
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Spencer Kraisler
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  • As for references, you could read the first chapter or two of Lanczos' "The Variational Principles of Mechanics". I remember that he does touch on this topic, briefly, in the introductory chapter(s). And I think he does mention how the potential changes/breaks the interpretation of a system as following geodesics in a metric space. Another book is Classical Mechanics by Goldstein, which might contain similar information. – Myridium Feb 07 '23 at 01:47
  • A geometric perspective that does work out for generic mechanical systems (including ones with a potential) is the symplectic geometry of Hamiltonian mechanics, which you might find interesting: https://physics.stackexchange.com/q/564834/ – Andrew Feb 07 '23 at 02:43
  • Comments to the post (v4): 1. Non-Levi-Civita connections and potentials seem to be 2 separate issues. Consider to only ask 1 question per post. 2. Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Feb 07 '23 at 03:03
  • Possible duplicates: Can Lagrangian be thought of as a metric? , Lagrangian mechanics and geodesics in configuration space? , Auto-parallel Transport or Principle of Extremum Action? and links therein. – Qmechanic Feb 07 '23 at 03:34
  • The equations of motion are $\nabla_{\dot{\gamma}}\dot{\gamma}=-X\circ \gamma$, where $X=\text{grad}_g(V)$ is the metric-gradient of the potential $V$ defined on $M$. By writing this out in coordinates, you can see that for non-trivial $V$, it cannot be interpreted as the geodesic equation of a modified connection on $M$ (the LHS depends on $\dot{x}$ and $\ddot{x}$, while the RHS does not). However, what you can do is consider the manifold $N=\Bbb{R}\times M$, and you can “lift” the connection $\nabla^M$ on $M$ to a connection $\nabla^N$ by defining… – peek-a-boo Feb 07 '23 at 04:09
  • $(\Gamma^N)^{\alpha}{\beta\delta}(t,x)=(\Gamma^M)^{\alpha}{\beta\delta}(x)$, and $(\Gamma^N)^{\alpha}{00}(t,x)=X^{\alpha}(x)$, and setting all other ‘unrelated’ $\Gamma^N$’s to be zero. Here, the indices $\alpha,\beta,\delta\in{1,\dots, n}$. I just checked manually (as you should) that this does indeed have the right transformation laws, so this gives a well-defined torsion-free connection on $N$. Now, a curve $t\mapsto (t,\gamma(t))$ satisfies the geodesic equation for $\nabla^N$ if and only if $\gamma$ satisfies $\nabla^M{\dot{\gamma}}\dot{\gamma}=-X\circ \gamma$ on $M$. – peek-a-boo Feb 07 '23 at 04:09
  • but note that this isn’t a fruitful point of view in general. In GR the reason why we do this (to combine space and time into spacetime) is because of the incompatibility of Newton’s description of gravity as a vector field on $\Bbb{R}^3$ with the postulates of SR, and also in order to incorporate the idea that the gravitational and inertial masses are the same for particles. So, combining things yields the notion that freely-falling particles traverse geodesics in spacetime, which is a conceptual leap. But, if you’re trying to combine merely for the sake of it, then you’ll be disappointed. – peek-a-boo Feb 07 '23 at 04:15
  • btw @Qmechanic I don’t think this is a dupe. From a brief read through, the question there is whether we can combine space and time and always interpret the old equations as a purely kinetic part arising from a metric tensor, but here OP’s question seems to be whether we can interpret it as a ‘geodesic equation’ arising from some arbitrary (not-necessarily Levi-Civita) connection. Of course the answer is no in general for trivial reasons, but these weren’t mentioned in the links; also the answer becomes yes if we allow ourselves to combine space and time (but we don’t get a metric-connection). – peek-a-boo Feb 07 '23 at 04:31
  • It seems OP conflates the autoparallel equation $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ and the geodesic equation $\nabla^{LC}_{\dot{\gamma}}\dot{\gamma}=0$. OP should clarify. – Qmechanic Feb 07 '23 at 04:43
  • @Qmechanic I never heard of an autoparallel equation before. From my understanding, given an affine manifold $(M,\nabla)$ geodesics are simply solutions to $\ddot{x}i + \dot{x}^i \dot{x}^j \Gamma{ij}^k$. – Spencer Kraisler Feb 07 '23 at 18:12
  • @Andrew Thank you for that link. I'll check it out, along with Lanczos' book suggested by Myridium. – Spencer Kraisler Feb 07 '23 at 18:13
  • Is this question mainly about non-Levi-Civita connections? Then there is no standard variational principle and no standard EL equations for the autoparallel equation. – Qmechanic Feb 07 '23 at 18:52

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