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Lagrangian of object with air resistance

  1. May 25, 2013 #1
    So I was going through an ODEs textbook and in a section discussing physical problems, decided that it would be interesting to come up with the equations of motion using Lagrangian mechanics for the examples they posted. For the first example, a falling rock, this easily worked. The second example was a falling object with air resistance. The equation they came up with was ma + kv = mg (I'd use dot notation but I'm unsure how to do it on here) which makes perfect sense; my issue was that I couldn't think of how the air resistance would affect either the kinetic or potential energies. So I pose the question: how would air resistance affect the kinetic or potential energy and what would their equations look like, such that the Lagrangian method would yield the above equation of motion?
     
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  3. May 25, 2013 #2

    Vanadium 50

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    Air resistance is a non-conservative force, so it cannot be expressed by a potential.
     
  4. May 25, 2013 #3
    The Euler-Lagrangian equation itself is changed in this case, the lagrangian is still the same one in vacuum.
     
  5. May 25, 2013 #4

    dextercioby

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    For non-conservative systems, the Lagrangian needs to be <guessed>. What term would you add to the kinetic one, so that after applying the E_L equations you'd get the same eq. of motion as in Newtonian dynamics ? My guess for linear air resistence would be

    L = mv^2 /2 + mgx + kxv
     
  6. May 27, 2013 #5

    Jano L.

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    Maurice, air friction is a complicated phenomenon and it has no single and simple mathematical expression in terms of velocity. Usually, rapidly falling objects will create turbulent flow of the air and then the resistance force can be argue to be proportional to square of the velocity (Newton's formula for resistance). Resistance that increases proportionally to speed occurs for either very slowly moving bodies (falling dust grains), or very fast ones in diluted gas (higher atmosphere...)

    In general, dissipation is a phenomenon that is hard to describe by the standard Euler-Lagrange equations.

    Adding new terms to the Lagrangian is an interesting idea, but I have never seen such a new term which would give the simplest equation of motion with friction

    $$
    m\ddot x = -k\dot x.
    $$

    The term ##kxv## is a total time derivative of ##kx^2/2## and thus has no influence on the equations of motion.

    On the other hand, if you do not need to stick with the standard Euler-Lagrange equations (in other words, with the Hamilton principle), there is a modification of the Euler-Lagrange equations due to Rayleigh, which adds new terms to the E-L equations and describe linear friction. Search "Rayleigh dissipative function" in textbooks, for example Goldstein's Classical Mechanics.
     
  7. May 28, 2013 #6
    I agree with Jano L.'s post. I thought to add you will have difficulty trying to come with a Lagrangian that results in
    $$
    m\ddot x = -k\dot x.
    $$
    via the standard Euler-Lagrange equations. Consider how the the total energy of the system (the Hamiltonian) changes over time.
    $$
    \frac{d}{dt} \left(\sum_i \frac{\partial \mathcal{L}}{\partial \dot{x}_i}\dot{x}_i - \mathcal{L} \right) = -\frac{\partial \mathcal{L}}{\partial t}
    $$
    We can violate conservation of energy by, for example, having a time varying potential so [itex] \frac{\partial \mathcal{L}}{\partial t} \neq 0[/itex] but that doesn't seem to allow implementation of friction.

    I did have an idea once for attempting to implement friction without modifying the Euler Lagrange equations. The idea was to double the number of generalized coordinates, and try to get the new coordinates to account for the work done by friction. I was not successful.
     
    Last edited: May 28, 2013
  8. May 28, 2013 #7

    Jano L.

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    The friction is often described in a very similar way. One such simple model is one oscillator in interaction with many other oscillators. Although the system as a whole conserves energy, the energy of the first oscillator can change. Under certain conditions the systematic result of the action of the other oscillators is that the first one loses energy.

    You can find details in this paper:

    http://jmp.aip.org/resource/1/jmapaq/v6/i4/p504_s1?isAuthorized=no [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. May 28, 2013 #8
    I'm not sure how much time you want to throw at this problem, but if it continues to interest you then here is a paper which reformulates Hamilton's principle to allow for non-conservative forces. There might be a more updated version in the Physics Review somewhere, but I had this version easy at hand.

    http://arxiv.org/pdf/1210.2745.pdf
     
  10. May 29, 2013 #9

    stevendaryl

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    I'm not sure I understand the significance of this, but it's possible to choose a Lagrangian whose equations of motion include air resistance this way:

    [itex]L = e^{b t}(\dfrac{1}{2} m v^2)[/itex]

    Then the Lagrangian equations of motion are:

    [itex]\dfrac{d}{dt} (\dfrac{\partial L}{\partial v}) = \dfrac{\partial L}{\partial x}[/itex]
    [itex]\dfrac{d}{dt} (e^{b t} m v) = 0[/itex]
    [itex]b e^{b t} m v + e^{b t} m \dfrac{d v}{dt} = 0[/itex]
    [itex]m \dfrac{d v}{dt} + b m v = 0[/itex]
     
    Last edited: May 29, 2013
  11. May 29, 2013 #10
    That's quite interesting, what strikes me as odd is that b is positive instead of being negative, you would think the new "kinetic energy" term should die off exponentially.
     
  12. May 29, 2013 #11

    Jano L.

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    Interesting. This Lagrangian can be used for particle whose mass increases exponentially in time while the momentum remains unaltered. This can happen if the mass comes from stationary reservoir. Since L is independent of x, we still have momentum conservation, and thus velocity is decreasing.
     
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