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Hamiltonian for an unknown dissipative system

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the following Hamiltonian

    [itex]H=\frac{p^2}{2m}e^{\frac{-q}{a}}[/itex]

    a: constant
    m: mass of the particle

    q corresponds to the coordinate, and p its momentum.
    note: q' stands for the derivative of q.

    a) Prove that for p(t) > 0 this system seems to describe a particle subject to a force proportional to [itex]q'^2[/itex]
    b) Solve q(t) and p(t) using the initial conditions q(0)=0 and p(t)=mv , v>0
    c) Find the asymptotic conditions for b), i.e. find q(t), p(t) and q'(t) for t-> infinity
    d) What is the relation between this hamiltonian and the kinetic energy [itex]T=\frac{1}{2}mq'^2[/itex]
    e) Determine the ratio of kinetic energy dissipation, i.e. dT/dt



    I would like to ask for a little bit of help in order to understand this problem better. As far as I can tell, it corresponds to a dissipative system (hence question e), but I'm not familiarized with it. Is this a well known Hamiltonian for a common problem?

    For a) I tried calculating the hamiltonian for a particle subject to a force given by [itex]F=\alpha q'^2[/itex]
    I integrated the force to get a potential, thus constructing a Lagrangian with kinetic energy equal to [itex]T=\frac{1}{2}mq'^2[/itex] and potential energy [itex]V=-\frac{\alpha}{3}q'^3[/itex]
    After calculating hamiltonian equations of motion, I'm kind of confused about how to get the exponential term around so I can relate it to the problem. Maybe I should start from another point?


    Thanks for your help and time!
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2

    Mute

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    Part (a) is asking you to use the Hamiltonian you are given to show that it describes a force proportional to ##(q')^2##. You are approaching it backwards (starting with the force and trying to derive a Hamiltonian).

    In this case, you want to use Hamilton's equations to derive an equation for the force which, hopefully, will come out to be proportional to ##(q')^2##. (I believe it may also have some q-dependence due to the exponential term).

    The Hamiltonian is not a standard Hamiltonian, but that won't prevent you from using Hamilton's equations to derive equations of motion, etc.
     
  4. Jun 9, 2013 #3
    Thanks for your reply.

    The Hamilton equations of motion read:

    [itex]\dot{q}=\frac{∂H}{∂p} [/itex]
    [itex]\dot{p}=-\frac{∂H}{∂q}[/itex]

    From which I obtained the following equations:

    [itex]\dot{q}=\frac{p}{m}e^{-\frac{q}{a}}[/itex] (1)
    [itex]\dot{p}=\frac{p^2}{2am}e^{-\frac{q}{a}}[/itex] (2)

    From (1), I obtained an expression for p, and p-dot. I replaced those terms in eq. (2). After some manipulation, I ended up with the following equation:

    [itex]m\ddot{q}+m\dot{q}-\frac{{m^2}\dot{q^2}}{2am}=0[/itex] (3)

    If I hadn't gotten the second term of eq. (3) I'd have been more satisfied. Can we draw any conclusion from what I got here?


    Thanks for your time again.
     
  5. Jun 9, 2013 #4

    dextercioby

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    You must check your algebra. From (1) [itex] p= m\dot{q} e^{q/a} [/itex]. Now square this term to substitute it in (2).
     
  6. Jun 9, 2013 #5
    I would suggest taking the derivative of your equation (1), giving
    [itex]\ddot{q}=\frac{\dot{p}}{m}e^{-\frac{q}{a}} - \frac{p}{ma}\dot{q}e^{-\frac{q}{a}} [/itex]
    Now plugging in for [itex] \dot{p} [/itex] gives:
    [itex] \ddot{q} = \frac{p^2}{2am^2}e^{-\frac{2q}{a}} - \frac{p}{ma}\dot{q}e^{-\frac{q}{a}} [/itex]
    And from (1) we know [itex] \dot{q}^2 = \frac{p^2}{m^2}e^{-\frac{2q}{a}} [/itex].
    We also have [itex] F = m\ddot{q} [/itex]. I think this should give the solution to part (a), but I didn't entirely work it out, so let me know.
     
  7. Jun 9, 2013 #6

    Mute

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    Hm, are you sure that the force is not ##F = \dot{p}##?

    Also, Siberion, to get some physical intuition for this system, consider the limit ##a \rightarrow \infty##. This corresponds to the limit of no dissipation, and the Hamiltonian reduces to a familiar form. This may help you conceptualize the physics described by this Hamiltonian.
     
  8. Jun 9, 2013 #7
    Hello, thanks for your replies.


    Yes this is exactly what I did. You are right, I commited a manipulation failure, but as far as I can tell it doesn't affect the equation that much. This is the procedure I followed:

    From eq. (1), we have
    [itex] p= m\dot{q} e^{q/a} [/itex].

    Deriving this equation,
    [itex] \dot{p}= m\ddot{q} e^{q/a}+\frac{\dot{q}m}{a}e^{\frac{q}{a}} [/itex].

    Moreover,
    [itex] p^2= m^2\dot{q}^2 e^{2q/a} [/itex].

    Plugging these results into eq. (2) yields
    [itex]m\ddot{q} e^{\frac{q}{a}}+\frac{\dot{q}m}{a}e^{\frac{q}{a}} = \dot{q}^2m e^{\frac{2q}{a}}e^{\frac{-q}{a}} [/itex].
    [itex]\ddot{q}+\frac{\dot{q}}{a}-\frac{\dot{q}^2}{2a}=0[/itex]

    If you want to express the equations in terms of [itex]\dot{p}[/itex], then I can omit the derivative part and I can just replace p² into eq. (2):

    [itex]\dot{p}=\frac{m\dot{q}^2}{2a}e^{\frac{q}{a}}[/itex]

    In both cases, I would have expected to end up with an equation [itex]F=α\dot{q}^2[/itex], but I'm getting these extra terms...

    Also, Mute, I believe p corresponds to the conjugate momenta, which isn't neccesarily mv. Would it still be ok to consider it as equal to the force in Newton law?
     
    Last edited: Jun 9, 2013
  9. Jun 9, 2013 #8

    Mute

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    It is definitely the case that ##p \neq m\dot{q}##. I admit, however, that I don't recall if the force should be interpreted as the time derivative of the canonical momentum or ##m\ddot{q}## in Hamiltonian mechanics. I asked the question to prompt you to double-check this. I suspect the force is determined from the canonical momentum, because generally we regard ##F = \dot{p}## as more fundamental than ##F = ma##, but I'm not 100% sure. An example using an electromagnetic Hamiltonian on the wikipedia page for Hamiltonian mechanics calls the ##\dot{p}## the force. However, being a wikipedia article on an advanced topic, you can't trust it too much, so you should double-check with a textbook.
     
  10. Jun 9, 2013 #9

    Hmm, I found a similar example of electromagnetic interaction in order to find the Lorentz Force. Here [itex]\dot{p}\neq mv[/itex], and the force F actually corresponded to the derivative of the linear momentum, which differed form [itex]\dot{p}[/itex].

    At this point I'm kind of lost, I was thinking about a generalized force in the Lagrangian formulation (D'alembert principle I believe) but then again I didn't expect this question to be that troublesome.

    Edit: Well it looks like I've been drowning in a glass of milk. I made a mistake when taking the total derivative of q, I actually missed a q dot, so the correct equation and replacements yield:

    [itex]\ddot{q}=-\frac{\dot{q}²}{2a}[/itex]

    hence it seems like the particle is subject to a force proportional to [itex]\dot{q}²[/itex]

    I'll move on and solve the rest.

    Thanks for your help!
     
    Last edited: Jun 9, 2013
  11. Jun 11, 2013 #10

    Mute

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    Oh good, you found that. I just checked the calculation of ##\ddot{q}## this morning and got that same result. I'm still not sure why ##m\ddot{q}## appears to be the appropriate measure of the force, but hopefully the other Homework Helpers and I can figure that bit out and explain it too.
     
  12. Jun 14, 2013 #11
    Perhaps it would be easiest to write out the Lagrangian [itex] L = pq - H [/itex] and use the formula for the generalized force [itex] F = \frac{\partial{L}}{\partial{q}} [/itex]; however, I haven't worked it out explicitly. I only suggest this because you're right, I'm not sure why [itex] \ddot{q} [/itex] rather than [itex] \dot{p} [/itex] gives the desired answer.
     
  13. Jun 14, 2013 #12
    Hello Siberion! A couple of comments on this interesting problem...

    I think when they say "force" here they're just after [itex]m{\ddot q}[/itex], rather than the "generalized force", and that's why they put in the word "seems". You're supposed to judge the "force" acting on the object by looking at its acceleration.

    Anyway when you do that, it looks like you've been given a Hamiltonian that successfully describes motion with a dissipative force slowing down our object with a force proportional to the square of the object's speed. Which is probably fairly reasonable for say a meteor speeding through the earth's atmosphere.

    The strange thing is that Hamiltonian mechanics isn't supposed to be able to describe dissipative forces. That's because Hamiltonian mechanics conserves entropy, and entropy is supposed to increase during a dissipative process. So we can expect bad things to happen!

    And indeed bad things do happen - namely, the solutions in (b) do weird things for negative values of t. For sufficiently far back in the past, they're undefined. That's the kind of thing that happens when you try to model dissipative processes within the framework of classical mechanics.

    However note that the Hamiltionian for a charged particle in an EM field mentioned above is perfectly OK from this point of view because the magnetic force on the particle isn't dissipative. The equations of motions resulting from that Hamiltonian admit solutions that are valid for all t.
     
  14. Jun 14, 2013 #13

    Mute

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    This sounds like the most reasonable interpretation to me. The others that I talked to had the same intuition that ##\dot{p}## was in some sense the more fundamental quantity, but I agree that it probably encapsulates more than just the "force" on the particle, hence why it doesn't boil down to ##m\ddot{q}## in this system. Thanks for chiming in!
     
  15. Jun 14, 2013 #14
    Such a beautiful and elegant explanation! This was indeed a very interesting problem. It had me thinking for a long time, dealing with logarithmic solutions and trying to understand what they really meant... By the due date of the homework (yesterday), my professor commented on how this Hamiltonian described a particle subject to friction. Now it is very satisfying to open this forum and read more replies from all of you. All this conversation was really helpful and, at the end of this day, it makes me smile.

    Thank you all once again for your replies! :biggrin:
     
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