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Integral curves and one-parameter groups of diffeomorphisms

  1. Sep 28, 2009 #1

    Fredrik

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    I think I understand why a vector field must have a unique set of integral curves, but I don't see why they must define a one-parameter group of diffeomorphisms.

    Let X be a vector field on a manifold M, and p a point in M. A smooth curve C through p is said to be an integral curve of X if C(0)=p and

    [tex]X_{C(t)}=\dot C_{C(t)}[/tex]

    for all t in some interval (-a,a). (I hope it's obvious that what I mean by "dot C" is the tangent vector of C). I'm confused about how this defines a one-parameter group of diffeomorphisms. Supposedly, if we write the integral curve at q as Cq, we can define [itex]\phi_t(q)=C^q(t)[/itex] for all t and q, and now this [itex]\phi[/itex] is a one-parameter group of diffeomorphisms. In particular, this would imply that

    [tex]\phi_t(\phi_s(p))=\phi_{t+s}(p)[/tex]

    which is equivalent to

    [tex]C^{C^p(s)}(t)=C^p(t+s)[/tex]

    Let's simplify the notation by calling the integral curve at p C, and the integral curve at C(s) B. The equation becomes

    [tex]B(t)=C(t+s)[/tex]

    How do we prove that this equation holds?

    Also, why does the definition of an integral curve require that [tex]X=\dot C[/tex] holds over an interval rather than just at a single point? I suspect that the answer is that as long as the curve is smooth, it doesn't matter.
     
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  3. Sep 29, 2009 #2

    quasar987

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    Since integrality of a curve is a local thing, we may assume that M=R^n.
    To see that the equality
    [tex]
    C^{C^p(s)}(t)=C^p(t+s)
    [/tex]
    holds, fix s. Then notice that when evaluated at t=0, both sides are equal (to C^p(s)). And, if you differentiate both sides with respect to t, you will see that both side satisfy the differential equation
    [tex]X(\gamma(t))=\dot{\gamma}(t)[/tex]
    By the fundamental theorem on existence and unicity of the solutions to inital value (Cauchy) problems, it must be that both sides agree for all t where they are defined. Now since s was arbitrary, both sides coincide for all s and t where they are defined.

    As to your second question... given a vector field X, we say that a curve is an integral curve of X at p if it passes through p, and if X coincides with the its tangent vector in a nbh of p. The idea that this definition tries to capture is that if X is thought of as a speed field for a particle of unit mass, then an integral curve at p is one that, locally near p, corresponds to the motion of a unit mass particle caught in that speed field. This is why integral curves are also called "flow lines": because if X is thought of as the speed of a fluid on M, then an integral curve at p follows the motion of a particle of this fluids passing through p.

    If we only require that [tex]X\circ C=\dot C[/tex] hold at t=0 in the definition of an integral curve, then cleary this would not capture the idea we want for there is usually curves other that the one following the "flow lines" satisfying [tex]X\circ C=\dot C[/tex]
     
  4. Sep 29, 2009 #3

    Fredrik

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    Thanks quasar. That helped. I think I get it now.
     
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