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!

Energy conservation in Lagrangian Mechanics

  1. Oct 4, 2015 #1
    In Lagrangian mechanics the energy E is given as :

    [tex] E = \frac{dL}{d\dot{q}}\dot{q} - L[/tex]

    Now in the cases where L have explicit time dependence, E will not be conserved.

    The notes I am referring to provide these two examples to distinguish between the cases where E is energy and it is not:

    1. Consider a particle of mass m sitting on a frictionless rod lying in x-y plane pointing in x direction. Now the rod starts to move towards -y direction with an acceleration a.
    [tex]\dot{y} = -at[/tex]
    and [tex]y = -at^2/2[/tex]

    so, [tex] L = 1/2m(\dot{x}^2 + a^2t^2) - mg(-at^2/2) ..........(1)[/tex]

    The Lagrangian has time dependence, so E is not energy and it is not conserved.

    2. Now, consider a particle undergoing projectile motion.

    hence, [tex] L = 1/2m(\dot{x}^2 + \dot{y}^2) - mgy ..........(2)[/tex]

    here there is no explicit time dependence, so E will be energy and it will be conserved.

    My question is, aren't the two cases similar. I can always replace a in case 1 with g and my equation (2) will then become equation (1). What is the difference here that I am missing?

    Also, in the cases where E is not energy, does it still mean that the energy of the system is conserved?
  2. jcsd
  3. Oct 4, 2015 #2
    I don't think you can say a=g in your first example, since the acceleration of the rod is not specified. The rod lies in the xy-plane (like on a table), and has a force acting upon it making it slide, and this force causing it to accelerate is not specified. So therefore your Lagrangian in example 1 has explicit time dependence.

    Also, I would be careful giving your first variable the name E. This is in reality the expression for the Hamiltonian of your system. The Hamiltonian is your total energy if your Lagrangian is on the form L = T - V (kinetic - potential energy), and all your constraints are time independent.
    If you Lagrangian is time independent, your Hamiltonian is always a constant of motion, but it does not always mean this constant is the total energy of your system.
  4. Oct 4, 2015 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The difference is that in the first case you have an open system, because you force the rod to be accelerated, and you don't include the dynamics of this acceleration in your Lagrangian but describe it as an external motion. Now your Lagrangian is explicitly time dependent, and the time dependence is not through a term of the form
    $$L=L'(q,\dot{q})+\frac{\mathrm{d}}{\mathrm{d} t} \Omega(q,t),$$
    because in the latter case, the Lagrangian ##L## is equivalent to the Lagrangian ##L'##, which is not time dependent.

    In the 2nd case all dynamics are described by the Lagrangian, and it is not explicitly time dependent.
  5. Oct 4, 2015 #4
    So the external force is doing some work on the article which is stored as potential, which is not accounted in the Lagrangian?

    And can you explain me this part. The equation is bit puzzling.
  6. Oct 4, 2015 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    If your time dependence is only through such a term, which is a total time derivative of a function that depends only on ##q## and explicitly on time, the system is equivalently described by the Lagrangian ##L'## which is not explicitly time dependent. Then you can define a conserved quantity
    $$H'=p \cdot \dot{q}-L', \quad p=\frac{\partial L'}{\partial \dot{q}}.$$
  7. Oct 5, 2015 #6
    Last edited: Oct 5, 2015
  8. Oct 6, 2015 #7
    The two lagrangians are equivalent because they lead to the same equations of motions, starting from the Hamilton's principle of stationary action: the variation of S', that is of the integral of L' dt between t1 and t2 it's equal to the variation of S minus the variation of ##( \Omega(q_2,t_2) - \Omega(q_1,t_1) )## where q1 = q(t1) and q2 = q(t2), but the last variation is zero, since the q are fixed at the limits of integration, by definition of that variation:

    ##\delta S' = \delta\int _{\mathbf{t_1}}^{\mathbf{t_2}} L' dt = \delta\int _{\mathbf{t_1}}^{\mathbf{t_2}} L dt - \delta (\Omega(q_2,t_2) - \Omega(q_1,t_1) ) = \delta\int _{\mathbf{t_1}}^{\mathbf{t_2}} L dt = \delta S##.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook