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Can I prove conservation of mechanical energy?

  1. Jun 9, 2012 #1
    I am trying to consolidate my understanding of kinematics and mechanics.

    If there is no such thing as friction, and the only force in the universe is gravity, then can we prove using the laws of kinematics and the definition of potential energy/kinetic energy that the mechanical energy of every object is conserved?

  2. jcsd
  3. Jun 10, 2012 #2


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    With Newtonian gravity, it is possible. You can use that every force on an object has a corresponding force on another object, and after some lengthy calculation (at least I think it was lengthy), you can show that the time-derivative of the total energy is 0, which means conservation of energy. This is not restricted to gravity, you can include the electromagnetic interaction as well.
    With General Relativity, it is problematic - as you have no universal time which is the same everywhere, what is meant with "the energy at a specific time"?
  4. Jun 10, 2012 #3
    I see. For the sake of curiosity, I am interested in seeing the mathematical proof in the case of Newtonian gravity, obeying the laws of Newtonian mechanics. Do you know where this proof can be found?

  5. Jun 10, 2012 #4
    Perhaps starting with a simple long (several meters) pendulum and a stop watch can show that T (kinetic energy)+ V (potential energy) is a constant to a small fraction of a percent.
    [added] You also have to monitor the amplitude of the pendulum as a function of time.
    Last edited: Jun 10, 2012
  6. Jun 11, 2012 #5


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    I saw one on a blackboard and I think that some introductory physics books should have that.

    Anyway, I can derive it here, too:
    With Newtonian gravity, each pair of objects i,j has a potential [itex]V_{ij}=m_i m_j \frac{G}{|r_{ij}|}[/itex] where rij is the distance between the objects.

    The total energy of the system is then given by
    [tex]E=\frac{1}{2} \sum_k m_k\, v_k^2 + \frac{1}{2} \sum_{i,j,i \ne j} V_{ij}[/tex]
    where the second 1/2 is required as the sum counts each pair twice. The time-derivative is then given by
    [tex]\frac{dE}{dt}=\sum_i m_i\, v_i \cdot a_i + \frac{1}{2} \sum_{i,j,i \ne j} m_i\, m_j\, G \frac{d}{dt}\frac{1}{|r_{ij}|}[/tex]
    The dot indicates a scalar product. The chain rules now gives
    = \frac{1}{|r_{ij}|^2} \frac{d}{dt} |r_{ij}|
    = \frac{1}{|r_{ij}|^3} (x_i-x_j) \cdot (v_i-v_j)[/tex]

    The force on an object is now given by the gravitational forces of all other objects, which is the negative gradient of the potential.
    [tex]m_i a_i = - \sum_{j \ne i} m_i\, m_j\, G\; \nabla \left(\frac{1}{|r_{ij}|}\right) = - \sum_{j \ne i} m_i\, m_j\, G \frac{x_i-x_j}{|r_{ij}|^3}[/tex]

    If you insert both expressions in the expression for dE/dt, you will see that both parts just cancel and you get 0.

    This is not restricted to potentials which are proportional to 1/r. In the general case of V(|r|), you get [tex]\frac{d}{dt}V(r)=\left(\frac{d}{d|r|}V(|r|)\right) \frac{1}{|r|} (x_i-x_j)\cdot(v_i-v_j)[/tex] and [tex]m_i a_i = - \sum_{j \ne i} m_i\, m_j\, G \left(\frac{d}{d|r|}V(|r|)\right) \frac{x_i-x_j}{|r_|}[/tex]
  7. Jun 11, 2012 #6
    Truly amazing! To think that what you just wrote forms the basis of all of classical mechanics!

  8. Jun 11, 2012 #7
    It should be noted that in fact potential energy is in a sense defined so that overall energy of the system is conserved. On the level of newtonian formulation of mechanics, there is no explicit need for concept of potential energy. The proof given above is precisely speaking proving only the thing that we have chosen right form for the potential energy so that it indeed does reproduce the Newton law of the gravitation.
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