Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is a configuration space?

  1. Feb 22, 2010 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I keep reading that the configuration space of a given system is the "space of all it's posible positions". Along with this is the inevitable example that the configuration space of a double pendulum is the 2-torus, S^1 x S^1. This makes sense: the possible positions of the first bob is a circle around the "fixed point" , while the possible positions of the second bob is a circle around the first bob. So it is clear how one may identify all the possible positions of the system with the points of S^1 x S^1. Additionally, the motion of the system in time is clearly a continuous path on S^1 x S^1 with the its usual (product) topology.

    So as I said, this is all intuitively clear to me. But when I try to find a formal definition of configuration space in order to make the above reasoning rigourous, I fail.

    So, does anyone know the formal definition of configuration space?? Thanks!
     
    Last edited: Feb 23, 2010
  2. jcsd
  3. Feb 24, 2010 #2
    I guess the most rigorous way to do mechanics is to define your phase space first, i.e. a cotangent bundle. The base space of this bundle is your configuration space. That defines all the degrees of freedom of your physical system.

    After that you need to define the Hamiltonian, in order to define the time evolution of the mechanical system.

    Maybe look around here:
    http://math.ucr.edu/home/baez/classical/

    Torquil
     
  4. Feb 24, 2010 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In

    http://math.ucr.edu/home/baez/classical/classical7.pdf

    it is said that the configuration space of a rigid body with its center of mass pined down to the origin in R³ is SO(3). How do you make sense of that? Why isn't it S²: take any point in the body other that the COM. Then the position of this point determines the position of the body entirely and this point is free to move on a sphere around the origin, "hence" Q=S².

    Thanks for these notes by the way, they look great.
     
  5. Feb 24, 2010 #4
    Say that you mark a point on a sphere. After you have transported this point to any (phi,theta) solid angle point, the rigid body can still rotate about the axis connecting this point and the origin. So there is one more angular variable in addition to the S2 you mention, thus making up SO(3). SO(3) is by definition all these transformations that move the 2-sphere into inequivalent configurations.

    Torquil
     
  6. Feb 24, 2010 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hm, yes, I had overlooked that rotation once the point is in place. Again, this makes perfect sense at an intuitive level...

    So... could it be that physicists rely only on their instinct & experience & understanding of mechanics to establish what the proper configuration manifold Q is for a given mechanical system?
     
  7. Feb 24, 2010 #6
    Yes, physicists very much use their intuition/experience/understanding to get to the interesting parts (for the physicist!) of the problem quicker. Physicists will usually go on and try to calculate results even if they know that a rigorous mathematical foundation doesn't exist. This is because we can compare with experiment, which is the most important thing for us. The perfect example here is quantum field theory. This is a bit more complicated than classical mechanics and configuration manifolds, though.

    Classical mechanics has a very well-defined and mathematically elegant rigorous foundation, so I don't think there are any such problems there. Often, mathematicians have made rigorous definitions of physics afterwards, e.g. for mechanics and also certain simple quantum field theories, but lots of other stuff as well. This is good, because they can discover fundamental problems in the formalism that way.

    In the rigid object example I don't think there is much more to prove, since group manifold of SO(3) is by definition all the inequivalent configurations of the sphere. I'm no expert on mechanics, so I don't really know if the definition of the configuration space can be a problem or not in practice. Perhaps some people work with complicated systems where the configuration space is difficult to determine. Myself, I'm only used to simple configuration spaces, e.g. like your SO(3) or R^{3N} for N point particles moving in R^3.

    Torquil
     
  8. Feb 24, 2010 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Apparently, people interested in stability questions for satellites (ex: will doing such and such to a given satellite will make it spin out of control or will we be able to restore its stability?) work with symplectic geometry to answer their complicated questions. The relevant symplectic manifold here is just the cotangent bundle of the configuration space of the satelite (its phase space, as you said).

    This manifold is often 100+ dimensional and very complicated (again, so I read). These guys must have a pretty stressful job, knowing that their misjudging what the proper configuration space for the satellite is might result in losing the satellite as a result of it crashing down on our heads!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook