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The Classical Path, QM Path Integrals and Paths in Curved Spacetime

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  1. Nov 24, 2013 #1
    "The" Classical Path, QM Path Integrals and Paths in Curved Spacetime

    Hey Guys!

    I've got an exciting question! It's been burning on my mind for years, but I think I can formulate it now. It's not so much a specific question, but rather a physical story which perhaps this thread can uncover.

    It starts in Chapter 6 of Thornton & Marion, Classical Dynamics of Particles and Systems, the chapter on calculus of variations. This is like the "mathematics you're gonna need" for the following chapter on Lagrangian and Hamiltonian dynamics. In this chapter, they use Fermat's principle as an example: Light travels by the path which takes the least amount of time.

    1) I was thinking about the propagation of a photon on a curved spacetime background, (general relativity) and, how at every location in the spacetime which is on the particle's (the photon's) classical path, (its worldline) the particle "chooses" which, possibly new, direction is, for the following infinitesimal bit of its motion *only,* the direction which will contribute the least amount of elapsed time if the particle has in its "mind" that it is trying to get to some "final" location of the spacetime. If there are no humans present to specify this "final" point, I presume the photon will merely follow the curves of the spacetime, that is, a null geodesic, at the rate of 299,792,458 m/s for all eternity, thereby taking the most time-efficient path to its "final location," which, in this case, is at infinity. (How can there be a most time-efficient path to infinity… )

    2) Non-Uniqueness: I understand the classical path is not necessarily unique. In quantum mechanics, I believe these extra classical paths simply get summed over in some way in the overall path integral, and everything is all right because we have Heisenberg's uncertainty relation anyway, but what does it mean in classical mechanics? In classical mechanics, a definite path must be realized. How is this path selected amongst the solution set?

    3) Path Integration: In quantum mechanics, one begins with an initial point, a final point and an elapsed time T. One then evaluates a path integral to obtain the amplitude for a particle to begin at the initial point and end at the final point a time T later. You modulus-square this (complex) amplitude to obtain the probability for the aforementioned "motion." (however one interprets what "motion" is in quantum mechanics, given the Heisenberg uncertainty relation)

    I'm trying to make 1), 2) and 3) make sense simultaneously in my poor head. I'm trying to bring all these ideas together. I guess what this is driving at is a path integral formulation of general relativity which, in the classical limit and in the particular solution for a single photon, produces a single, realized path which minimizes the physical quantity, time. Again, there is this confusion in 2) about the non-uniqueness of the classical path.
     
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  3. Nov 24, 2013 #2
    I think in 1) I am wondering about what happens in the limit as your "final point" becomes infinitesimally close to your "initial point," as you effectively "shrink" the length of the classical path q_{cl}(t) by considering smaller and smaller motions of the particle.
     
  4. Nov 24, 2013 #3

    Simon Bridge

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    You hit quantum mechanical limits.
     
  5. Nov 24, 2013 #4
    Regarding (1) and (2), you need to realize that the particle does not actually know where it is going to end up. It just follows the equations of motion at each instant of time, without any pre-planned destination or goal. Now, if you for some reason happen to know where the particle is going to end up, you can note the interesting fact that the equations of motion look like the Euler-Lagrange equations which are the conditions for a path to be a stationary point of a certain action functional. Then you can say that of all paths between the initial and final point [where the "final point" can be *any* future point along the trajectory], the actual path is a stationary point of the action. But this is just an interesting fact about the path; it is not how the particle actually decides how to move. The particle's motion is controlled by the equations of motion, not by the requirement that it minimize the action while traveling to some specific foreordained destination.

    Furthermore if there are multiple paths between the initial and final point that are stationary points of the action, this is not a problem. No one ever promised that there wouldn't be *other* stationary points of the action, besides the actual path followed. But those other paths have different initial velocities.

    I'm not sure what curved spacetime has to do with any of this though. All the things you are talking about occur in flat spacetime too.
     
  6. Nov 24, 2013 #5
    Yeah, I know photons (particles) don't have brains :) but it brings out a certain point of view.

    What happens when you take a "final point" which is off the particle's classical trajectory? Like you say, if you happen to know of a future point which the particle will travel through, (or you simply get lucky with a guess) then you go through your analysis and it is essentially a standard derivation of Euler-Lagrange, which would be in any classical mechanics textbook. In that case, all kinds of cool things come out which you already explained. (Thanks!) But to me that sounds like a special case of something.

    So this "final point" is not part of the classical path, but perhaps one could regard it as being on a different path which at some point intersects the classical path. That path would contribute something (possibly naught I suppose) to the quantum mechanical path integral for our new "final point," which does lie on the classical path. I don't know... I'm trying to find something here.

    Ah, that makes sense! Aren't there some 2nd-order differential equations in here somewhere? Like the equations of motion themselves, referring to acc. (time) = whatever, and the way you fix the two integration constants (were you to solve said equations by direct integration) is by plugging in boundary value data on both the position and velocity. So this is supposed to uniquely determine q_{cl}(t) for whatever particular dynamical problem you are trying to solve?
     
  7. Nov 24, 2013 #6

    Simon Bridge

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    It would be helpful if you tried to describe what you are trying to find.

    An endpoint no on the classical trajectory?
    Then it is not an endpoint is it?

    You mean " quantum mechanical path integral" as in the Feynman integral?
    In QM you have to be very careful to set up the situation ... the source and the detector make the endpoints of the particle's path. We can then talk about the probability that a particle created at the source will be detected at the detector.
    It sounds like you are grasping at a situation where the detector is deliberately positioned off the classically determined path the particle should follow. Indeed you may get particles detected there ... the classical trajectory is followed only on average.

    If you try to use thought experiments instead of general terms, you'll find you'll make more headway.

    You mean like ##F=m\ddot{q}## ?
    Or something more like this?
     
  8. Dec 18, 2013 #7
    I think you are correct, but let me pose my confusion as a slightly different question.

    Sticking with the "Newtonian" formalism, the governing dynamical equation (Which one solves for particular physics problems in order to obtain equation(s) of motion) is

    [itex] \dot{\vec{p}} = - \nabla V(x,y,z). [/itex]

    So one begins with some potential "landscape," [itex]V[/itex]. (That is, hills and valleys which are smooth) Imagine a particle sitting somewhere initially at rest. (We have chosen an initial condition of [itex] \dot{q} = 0 [/itex]). Immediately, it will fall down the potential landscape in the direction which is steepest. As long as neither this direction nor its rate of descent changes, the particle will continue to accumulate velocity in a steady fashion. (Uniform acceleration in a straight line) If the rate increases, for instance, (Steeper slope) its straight-line acceleration will increase. If the "steepest direction" changes, its trajectory will arc such that if the aforementioned change is uniform, uniform circular motion will eventually result.

    This approach to generating [itex] q_{cl}(t) [/itex] amongst the hills and valleys of the potential is local. The particle responds locally to changes in the potential landscape.

    Conjecture: A path which is generated in this fashion, responding to the local potential landscape, when taken through from some initial point A to some other final point B *must* also minimize the "action?"

    1) Is this true?
    2) I'd be interested to see the explicit proof if anyone has it.

    The connection seems reasonable. In trying to get at smaller and smaller values of potential energy, over the course of the particle's motion, [itex] \int_A^B \mathrm{d}t\,T - V [/itex] *should* also, somehow, get minimized amongst all possible paths from A to B through the potential landscape.

    I understand the Lagrangian and Hamiltonian formalisms are related by a Legendre transformation. Is the above how the Lagrangian (Global approach) and Newtonian (Local approach) formalisms are related?
     
  9. Dec 19, 2013 #8

    Simon Bridge

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    I think you want to revisit that last part - to get circular motion, the particle must return to a previous position ... to do that while always going downhill would require some work done on it from somewhere.

    Circular motion requires a very specific set of conditions. The description you gave is too general.

    You can get periodic motion about a local minima - but not if the particle approaches the local minima from beyond the barrier ... i.e. if you have a local maxima next to a local minima, a particle can have periodic motion in the minima but not if it started out on the maxima.

    If the direction of the steepest gradient changes, then the direction of ##\dot{p}## changes.

    Fair enough.

    Restricting ourselves to classical physics - yes.

    The kind of trajectory you described was generated by application of Newton's Laws.
    It is possible to get Newton's Laws from the principle of least action ... showing that the one is another way of writing the other.
    That good enough?

    Something like this perhaps:
    http://www.eftaylor.com/pub/NewtonFromAction.pdf
     
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