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Wikipedia Article re: Hamiltonians: Could it have been better stated?

  1. Nov 19, 2011 #1
    See:

    http://en.wikipedia.org/wiki/Hamiltonian_mechanics

    I refer specifically to this passage:

    "Basic physical interpretation

    The simplest interpretation of the Hamilton equations is as follows, applying them to a one-dimensional system consisting of one particle of mass m under time-independent boundary conditions: The Hamiltonian represents the energy of the system (provided that there are NO external forces, or additional energy added to the system), which is the sum of kinetic and potential energy, traditionally denoted T and V, respectively. Here q is the x coordinate and p is the momentum, mv. Then

    H = T + V, T = p^2/2m, V = V(q) = V(x)

    Note that T is a function of p alone, while V is a function of x (or q) alone.

    Now the time-derivative of the momentum p equals the Newtonian force, and so here the first Hamilton equation means that the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location. (Force equals the negative gradient of potential energy.)

    The time-derivative of q here means the velocity: the second Hamilton equation here means that the particle’s velocity equals the derivative of its kinetic energy with respect to its momentum. (Because the derivative with respect to p of p2/2m equals p/m = mv/m = v.)"

    ***

    My problem here is the statement to the effect that "the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location." I feel that a more correct statement would be that the force on the particle equals the rate at which its potential energy is being converted into kinetic energy (assuming it's falling within a vacuum, eliminating the need to account for frictional atmospheric resistance and consequent draining of gravitational potential energy into heat-generastion through the process of overcoming aerodynamic drag). Furthermore, seeing as we're only speaking of mono-dimemsional motion here, why use the phrase "changes in x, its location", as opposed to the standard term "translational displacement"?

    :confused::confused:
     
  2. jcsd
  3. Nov 19, 2011 #2

    D H

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    You are ignoring an important part of that statement in the article. The statement in full is "the first Hamilton equation means that the force on the particle equals the rate at which it loses potential energy with respect to changes in x, its location. (Force equals the negative gradient of potential energy.)"

    This "first Hamilton equation" to which the article is referring is
    [tex]\dot p = -\frac {\partial \mathcal H}{\partial q}[/tex]
    When the Hamiltonian can be separated into a form H(p,q,t)=T(p,t)+U(q,t) the above reduces to
    [tex]\dot p = -\frac {\partial U}{\partial q}[/tex]
    In words, the time derivative of the generalized momentum is the negative gradient of the potential energy with respect to generalized position. Specializing to ordinary momentum and position leads to force being the negative gradient of potential energy.

    Could the wikipedia article have been better written? Of course. But your improvement misses the key point that relates force with gradient of potential. The wikipedia article should have hammered this home rather than making it a short parenthetic remark.
     
  4. Nov 19, 2011 #3
    OK. So, here, (http://en.wikipedia.org/wiki/Gradient) Wikipedia says:

    "In vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change."

    ***

    I guess what I'm not understanding here is: why is the vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change in the same, not proportional to the rate of conversion of gravitational potential energy into gravitational kinetic energy, assuming the particle is moving within a vaccuum? (I here recognize my earlier confusion of the proportionality of the rate of energy conversion along this vector with identity of such rate of conversion.)

    EDIT:

    I guess what I'm really saying is, is there redistribution (obviously not loss, but redistribution) of gravitational potential energy of the particle along vectors other than the negative gradient sufficient to prevent an increase of mass in the particle as it moves directly into the negative gradient?
     
    Last edited: Nov 19, 2011
  5. Nov 19, 2011 #4

    D H

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    That's wikipedia for ya. A typical problem with wiki articles is not that they are wrong; they are just poorly written. That is an abuse of the term "rate" and it is going to lead to confusion. "Rate" typically means with respect to time.

    And right here is where that confusion bites you. That vector field you speak of has units of energy over length, or force. On the other hand, your "rate of conversion of gravitational potential energy into gravitational kinetic energy" has units of energy over time, or power. You have incommensurate units!

    You are missing something very profound in those seemingly simple Hamilton's equations.
    [tex]\begin{aligned}
    \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\
    -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}
    \end{aligned}[/tex]
    These equations relate the total derivatives with respect to time of the generalized position / generalized momentum with the partial derivatives of energy with respect to the conjugate variable. This is incredibly profound. It is the basis of quantum mechanics; it would have been very hard to develop quantum mechanics without Hamilton's reformulation of Newtonian mechanics.


    Increase of mass?? :confused:
     
  6. Nov 19, 2011 #5
    I see it now. Thanks!

    I've never seen the equations in quite those forms before.



    I was looking for lost energy, and, so far as I could see, lost energy meant increased mass. I wasn't expecting to actually find an increase in mass, that was just my form of asking where did the lost energy go?

    I now see that no energy is lost.

    Thanks!
     
  7. Nov 19, 2011 #6

    D H

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    That's the same form as the wiki article, and many texts. I just changed the Newtonian fluxion into Liebniz d/dt form. As a recap, the wiki article says "The Hamilton equations are generally written as follows:"

    [tex]\begin{aligned}
    \dot p = -\frac{\partial \mathcal H}{\partial q} \\ \\
    \dot q = \phantom{-}\frac{\partial \mathcal H}{\partial p}
    \end{aligned}[/tex]

    And here's how I wrote Hamilton's equations:

    [tex]\begin{aligned}
    \phantom{-}\frac{dq}{dt} = \frac{\partial \mathcal H}{\partial p} \\ \\
    -\frac{dp}{dt} = \frac{\partial \mathcal H}{\partial q}
    \end{aligned}[/tex]

    Obviously the same two equations.

    I like the Leibniz d/dt form as opposed to the Newtonian dot form because expressing it the way I did makes it painstakingly clear that you have a time derivative on one side and a gradient on the other.


    Glad I could help.
     
  8. Nov 19, 2011 #7

    The Leibnizian form is how I was taught calculus: to be honest with you, I've never heard of "Newtonian dot form" in my life. Honestly never knew it existed. Obviously the source of my confusion.

    Thanks again!!
     
  9. Nov 19, 2011 #8
    Anyways, I posted this little piece on another website about how well this discussion has gone for me:

    "For whatever it's worth (probably very little), I got into an argument on the Physics Forum tonight, and here's about how it went for me:




    :blushing:
     
    Last edited by a moderator: Sep 25, 2014
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