1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hamiltonian for elastic collision

  1. Nov 5, 2013 #1
    What would the Hamiltonian for a system of two classical point particles, with no interaction except for an elastic collision between them at a point look like?

    My gut says it's the usual T + V, with T = p12/2m1 + p22/2m2 and

    V = Kδ(r1-r2)

    With K approaching infinity -- each particle acts to the other like an infinite potential barrier.

    I am bothered by the infinity multiplying the (infinite) delta function...it doesn't seem absolutely necessary that it's infinite; if I'm not mistaken, it should only need to be later than the kinetic energy , but I wanted to avoid momentum-dependent forces if possible: I'm reminded of the "normal force" of intro mechanics, which is exerted by an immovable barrier on anything acting on it.

    I'm also bothered that I cannot find this Hamiltonian anywhere, since it 'a such a fundamental system...but maybe my search terms are just no good ( Hamiltonian "elastic collision" "interaction term")
    Last edited: Nov 5, 2013
  2. jcsd
  3. Nov 5, 2013 #2


    User Avatar
    2017 Award

    Staff: Mentor

    What about a very large K?

    Two classical point particles will never meet in a general setup - and if you force them to collide, you have no way to evaluate how they behave there.
  4. Nov 5, 2013 #3
    Any finite constant (eg not momentum-dependent) K would allow fast-enough particles to "push through" one another...

    I don't think the problem is that they're points -- I have the same problem coming up with a Hamiltonian for finite-sized particles (with the δ(r1 - r2) function replaced by a step function θ((r1 - r2)2 - R2), where the radius is R)...the points just seemed easier to work with.

    And as for them never meeting, I'd have to disagree there, too: in a 1D system, if their momenta are pointing towards each other, they MUST collide...and at this point, I'd be happy with a 1D Hamiltonian.
    Last edited: Nov 5, 2013
  5. Nov 5, 2013 #4
    I'm inclined to think that if you calculated the scattering cross section for that system, you'd find it to be zero. Two point particles should have zero probability of colliding.

    If, instead, you represent the interaction as a sharp Gaussian function, or some other approximation to the delta function, then I think you'd have something you can work with.
  6. Nov 5, 2013 #5

    Ok, let's forget the point thing, because that seems to be causing more problems than simplifications.

    What would the interaction Hamiltonian for perfect elastic scattering look like for two beads constrained to move frictionlessly along a wire?

    To reiterate, this is a classical system .

    I found a similar question, and the response is *almost* what I need -- http://physics.stackexchange.com/questions/39679/why-cant-collisions-be-elastic -- e.g. The responder suggests that the potential is "something like" an infinite step function...but for such a basic, simple problem with a known analytic solution, it really seems like there should be an equally simple (and unique) answer...but so far, that doesn't seem to be forthcoming...and even that unique answer *is* the infinite step function, it's not clear how that would integrate to an elastic collision.

    Last edited: Nov 5, 2013
  7. Nov 6, 2013 #6
    Intuitively, I aggree with dipole's opinion. Since you want to describe an elastic collision, adjust the parameters of the sharp function that you chosed, so that the Hamiltonian and the momentum have vanishing Poisson brackets.
  8. Nov 6, 2013 #7


    User Avatar
    2017 Award

    Staff: Mentor

    Okay, 1D motion....

    What about an 1/r^10-potential (or something similar) defined for a very small region around r1=r2, with an arbitrary value for r1=r2 as they will never meet exactly anyway?
  9. Nov 6, 2013 #8
    My problem with all of these finite, non-delta function interaction potentials is that, while they presumably reproduce the same qualitative behavior, the do not reproduce the exact (trivial) Newtonian equations of motion (e.g.where the momenta are swapped in the center-of-mass system).

    To simplify it even further, what would the Hamiltonian be for a *single* particle (or sphere) moving in 1D, elastically colliding with a wall at x=0? From the Newtonian equations, the momentum remains constant "p" until x=0, and changes to "-p"; so the impulse, the integral of the force dp/dt over time, is the total change in momentum: -2p, so the force would be -2pδ(x), right?

    So how would you reproduce that in Hamiltonian (or Lagrangian) form?
  10. Nov 6, 2013 #9
    I'm not sure about the coefficient, but that seems like a reasonable force for this situation (note that it's unphysical though - there's no such thing as a perfectly rigid wall/particle).

    To find the potential, you need to ask yourself what sort of function, when differentiated, will give me a delta function? The delta function isn't actually a function (very bad name for it), but what is called a distribution - and you need to be careful working with them. However, the potential,

    [itex] V(x) = \begin{cases} \infty & x \leq 0 \\ 0 & x > 0 \end{cases} [/itex]

    Is what you need (think of a particle in a box). Try integrating the Hamiltonian and see what you get.
  11. Nov 7, 2013 #10
    So the derivative of the unit step/sign function is the delta function....∂H/∂x = ∂V/∂x = -dp/dt in your case is ∞δ(x(t)), so the impulse (Δp) over any finite amount of time around a zero of x(t) will be -∞...or am I missing something? It seems like to get the right Δp -- which IIRC is -2p(t) in this case -- the correct V(x) would need to be

    [itex] V(x) = \begin{cases} -2p(t) & x \leq 0 \\ 0 & x > 0 \end{cases} [/itex]

    Or something similar...but this not only contradicts the scant sources i've seen on this, but also makes the force momentum-dependent (which isn't the end of the world, but it sure doesn't make integrating the equations of motion any easier for someone who's never had a diff-eq class)...

    I know what the equations of motion *should* look like, for 1D elastic collision against a wall -- x(t) should be a properly shifted/stretched |t|, and p(t) should be p(0)sign(t - t0), where the collision happens at t = t0...but I'm having trouble even getting that far.
  12. Apr 1, 2014 #11
    jjustinn, have a look at my recent question on PSE http://physics.stackexchange.com/questions/105318 . The solution for a particle hitting a wall is to use a [itex]\varepsilon[/itex]-parameterised potential which goes to hard wall abrupt potential with [itex]\varepsilon \to 0^+[/itex].

    I agree that it seems to be an overkill to regularise potential because it makes both the problem statement and its solution more complex. All I want is an approach that would combine the elegance of generality of Hamiltonian formalism with the underlying simplicity of the hard wall model. But it is nowhere where I could find it.
    Last edited: Apr 1, 2014
  13. Apr 1, 2014 #12
    Very interesting: thanks for sharing...after suggestions from one or another of my iterations of the question, I had looked at some other similar limit-based solutions with nascent delta functions, but my PDE-disability kept me from getting anything meaningful (similar to your problem in the original question...it seemed like it was always either over- or under-determined and hopelessly self-referential, ala dx/dt = f(x(t)) for nontrivial f). At a glance it seems I'd have had the same problem with the "elastic" potential, but at least now knowing it's correct (e.g. doesnt lead to a final velocity proportional to 1/ε), I might give it another shot.

    Anyway, thanks again. I'll keep an eye on that PSE thread.
  14. Apr 1, 2014 #13


    User Avatar
    Science Advisor
    Gold Member

    The "problem" here is that you are trying to use a method based on [kinetic - potential] energy in a situation where there is no potential energy!

    Thus you are searching for a work-around. Since it is an "elastic" collision, why not make the balls elastic? That is, use a Hook's law potential which is zero everywhere away from the balls, but increases from U=0 at contact to an increasing amount as the balls "penetrate" each other.

    BTW, you should start by writing the Lagrangian, then transform it. There may be something to learn there!

    PS: Hard spheres are treated here: http://en.wikipedia.org/wiki/Dynamical_billiards

    PPS: Sometimes the Hamiltonian method is easier, sometimes the Newtonian vector approach is easier ...
    Last edited: Apr 1, 2014
  15. Apr 1, 2014 #14
    That may technically be true, but that seems to be a minority viewpoint: virtually everyplace discussing Hamiltonian mechanics w/ hard spheres (including the dynamical billiards Wikipedia page you cited) state plainly that both a "hard sphere" and a barrier (e.g. "An [infinite] potential well") can be treated with a step-function potential...and, as discussed here and elsewhere, the conservation of momentum/energy treatment (including the "non-existetent" step potential function) trivially yields the correct equations of motion...the problem only arises when trying to arrive at the same solution via the more general Hamiltonian framework.
  16. May 31, 2014 #15
    Maybe changing coordinates could help. Well, forget about two particles, as simplest model collision of a single particle with a wall at q=0 should be taken. If you imagine the phase space (q,p) the collision in it takes a form of a teleportation from (0,p) to (0,-p). What you are trying to do by modifying the potential energy is to find such U that would instantly shoot the particle from one point in phase space to another distant point, that is that would provide such a non-infinitesimal vector.

    An alternative approach could be gluing two points together into one to see if things will be better there. Practically it can be achieved by changing the coordinates, for example p -> q*p , q -> q, though this is not a canonical transformation. You may like to find a better (canonical) transformation. Don't know yet if it will work out, but I guess we have not so many variants to choose from.
  17. Jun 9, 2014 #16
    jjustinn, I think I have solved the problem for a particle colliding with a hard wall --- http://physics.stackexchange.com/a/118415/4020

    It could be done by modifying a space phase topology to equate points (0,p) and (0, -p). Formally it is achieved via a coordinate transform. Then Hamiltonian approach works as a charm. While I was doing this I also studied a bit of a field of dynamical billiards (http://en.wikipedia.org/wiki/Dynamical_billiards). Here they also resort to this trick, but they do not make much applications of it, at least they do not follow with Hamiltonian equations.

    Though there is a problem with such an approach. To find this coordinate transform I resorted to visualising the phase space of the problem. It is simple for a single particle in 1D. Unfortunately I'm afraid it is much harder to do the same even for a two colliding particles in 1D, not to say for several hard spheres in 3D. Yes, the problem can be tackled within Hamiltonian framework by glueing phase space along certain lines/surfaces, but it is extremely hard to visualise it and to actually find such transformations.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook