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Lagrangian in Newtonian mechancs

  1. Oct 16, 2015 #1
    Lagrangian in classical mechanics equals L=T-V, where T is kinetic energy and V is potencial energy.
    But, how to compose such a Lagrangian? Everywhere, where I found, it is only assumed and then equation
    ##d/dt (\partial L/\partial \dot{x})-(\partial L/\partial x)=0## is used.
    But, why L=T-V, is this only guessing, or is there any logic, because H=T+V is calculated afterwards. Why T and V are already specified at L? Is it possible to equate L with something, when we do not know that kinetic and potencial energy exist?
     
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  3. Oct 16, 2015 #2

    Orodruin

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    This is a bit like asking why F = ma in Newtonian mechanics. It simply turns out to be a good description of how Nature works. Ot also turns out to be equivalent to F = ma for a large class of cases. At a more fundamental level, the Lagrangian is what actually describes your physical model in Lagrange mechanics. You can have different Lagrangians but a very particular class of them have turned out to be a fairly accurate description of observations.

    When it comes to the transition to Hamiltonian mechanics, it is again the same thing. The world might have been such that the equations of motion were not given (or well approximated) by the Poisson bracket with the Hamiltonian, but experiments tell us that it is. Physics is an empirical science.
     
  4. Oct 16, 2015 #3
    My question is more about cyclic logic: We first define kinetic energy and potential energy, and the we define conservation of energy with help of Hamiltonian and with Noether theorem. Is possible to define H without explicit input of T and V?
     
  5. Oct 16, 2015 #4
    Yes. H = ( dq/dt)( dL/dv) - L. The dL/dv is partial derivative. H is defined, no mention of T or V. It's not really a guess that L = T = V because the consequence ( Newton's Law ) is anticipated. Newton's Law is correct which validates the Lagrangian.
     
  6. Oct 16, 2015 #5

    Orodruin

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    No, this is not the case. Noether's theorem implies that the Hamiltonian is conserved for systems which are invariant under time translations. In general, the Hamiltonian is not conserved if there are time-dependent external forces acting on the system (as described in Lagrangian mechanics by an external potential which depends on time explicitly). In other words, if you have a particular form of the Lagrangian, generally taken to be of the form ##\mathcal L = g_{ij}(x) \dot x^i \dot x^j - V(x)##, then ##\mathcal H = \dot x^i (\partial \mathcal L/\partial \dot x^i) - \mathcal L## is a conserved quantity. In this case, the conserved quantity ##\mathcal H## is computed from the results of Noether's theorem, i.e., if the Lagrangian is on the given form, then there is a quantity ##\mathcal H## which is going to be equal to a constant, which we can call ##E##. The same goes for, e.g., spatial translations and momentum - if the Lagrangian is invariant under translations in the ##x^i## coordinate, then there is a quantity ##\partial \mathcal L/\partial \dot x^i## is conserved and we may call its value ##p_ i##.


    In Hamiltonian mechanics, you can essentially define any Hamiltonian function you wish of your phase space variables. The question is whether or not they describe a physical system. The thing to remember is that Newtonian, Lagrangian, and Hamiltonian mechanics are all equivalent for a large class of physical systems - as shown in most of the literature on the subject of analytical mechanics.

    This is true for systems which in Newtonian mechanics may be described solely by a potential (and hence a conservative force field), which may be enough for the OP's current purposes. There are other systems where the Lagrangian will not be of this form.
     
  7. Oct 17, 2015 #6
    Is it possible to calculate with L and H, that kinetic energy is proportional with ##v^2##? Because, if we insert L=T-V, and then we calculate H, this is not determination of ##E\propto v^2##, but it is more assumption?

    I think that calculation of ##Fdx = m(dv/dt) dx = mv dv## is only cyclic calculation. ##mv^2## should be calculated without use of potential energy. We can find on other ways that, for instance, ##mv^3## or ##mv^4## are not conserved quantities, but ##mv^2/2## IS conserved quantity.
    Namely, if we have a collision of two bodies ##\Delta E## is conserved only at ##E\propto mv^2##, but not at ##E\propto mv^3##. ##\Delta E## means difference between energyies before and after collision.
    More in my question:
    https://www.physicsforums.com/threads/quadratic-forms-and-kinetic-energy.826884/
     
    Last edited: Oct 17, 2015
  8. Oct 17, 2015 #7

    Orodruin

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    You seem to want to derive everything from first principles. This is not what physics is about. Physics is about finding a description of Nature which works as well as possible.

    "Kinetic energy" is a concept we introduce to help us in that description and it so happens that it is conserved in some situations.
     
  9. Oct 17, 2015 #8
    Maybe you want to say:
    We define kinetic energy as ##T=mv^2/2## and then use of ##L=T-V## and of differential equation for ##L## (above written by myself) gives that ##H=T+V## and is conserved quantity, thus at ##V=0##, ##T## is a conserved quantity?
     
  10. Oct 17, 2015 #9

    vanhees71

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    The Lagrangian or Hamiltonian is indeed defining a model for the motion of mechanical systems (but the Hamilton principle of least action plays a role in all of physics, not only point mechanics). You cannot derive it from first principles. This is also the case for Newtonian mechanics which is based on observations, as any natural science. The action principle, however has a great advantage compared to naive Newtonian mechanics, because it can help us to guess appropriate equations of motion based on symmetry principles. E.g., in relativistic field theory you can systematically study Lagrangians that are covariant under Poincare transformations, defining all kinds of fields based on the representation theory of this group.

    Another great help is that you can also guess how to build quantum theories from classical models. Again, it's of course impossible to derive a specific quantum model from a classical model, but it provides hints to heuristically guess useful Hamiltonians for a given problem. So the action principle is just an alternative formulation of the basic laws, found from observations and quantitative measurements, which provides sharp mathematical tools to apply these basic laws to concrete problems.
     
  11. Oct 17, 2015 #10
    Not always, that result is true in some specific cases only: holonomic, conservative systems (if I remember well).
    From what I heard, Lagrangian, in general, is defined exactly as the function which gives the correct equations of motion through the equations you write here (lagrange equations), or through the stationary action principle, so, *in general*, it can be quite difficult to find the exact lagrangian function (sometimes it can be found with attempts).

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