Consider two point masses colliding with each other, then, by conservation of momentum, $$m_1v_{11}+m_2v_{21}=m_1v_{12}+m_2v_{22}$$ But this doesn't characterize the complete movement of the two masses and another equation must be used, which is the conservation of kinetic energy. However, it is not obvious why should kinetic energy be conserved in a collision, so I thought about proving it first using the fact that the collision is the same backwards, that is, if you flip/negate the velocities after the collision and colide the masses again, the final velocities should be the original starting ones. Unfortunately I was not able to prove this yet, as using this fact in the conservation of momentum equation doesn't bring any new information. Also, it is true that the conservation of energy is compatible with this symmetry, but this is not sufficient to prove it must be derived from the symmetry.

  • $\begingroup$ It's not clear why you state 'this doesn't characterize the complete movement of the two masses'. I submit that conservation of momentum is sufficient for complete characterization. There's nothing to add. As we know: conservation of momentum is galilean invariant. For every coordinate sytem of the equivalence clsss of inertial coordinate systems: the attributed momentum is different for each, but always conserved. Kinetic energy has that galilean invariance property too. Depending on the (inertial) coordinate system the attributed kinetic energy is different, but always conserved. $\endgroup$
    – Cleonis
    Jun 12, 2023 at 15:44
  • $\begingroup$ The kinetic energy is not conserved in collisions, in general. So there is nothing to prove. If it happens to be conserved, for very special types of colliding bodies, then it is not a result of a general principle. You either assume it or you observe that it happens for some collisions. $\endgroup$
    – nasu
    Jun 12, 2023 at 16:05
  • $\begingroup$ @nasu The question is not about properties of collision, the question is about properties of kinetic energy. In thought experiments it is common to grant perfectly elastic collision. In statistical mechanics: it is granted that collision between atoms can be treated as perfectly elastic. So it is relevant to assess the properties of kinetic energy for the case of perfectly elastic collision $\endgroup$
    – Cleonis
    Jun 12, 2023 at 16:12
  • $\begingroup$ Cleonis, when I say it doesn't characterize the complete movement, I mean that you can't predict the velocities after collision with only momentum conservation, you need to have more information, such as energy conservation. @nasu, I know it is not conserved in general, but my idea is to prove it in an ideal situation. Of course that one has to define what "ideal" is, which in this case is having the symmetry I mentioned. I didn't want to state that the collisions were elastic because an elastic collision is defined as the one that conserves energy, so there would be nothing to prove. $\endgroup$
    – WordP
    Jun 12, 2023 at 16:13
  • $\begingroup$ It has nothing to do with "ideal". And elastic collisions conserve Kinetic energy, not energy. The symmetry is related to conservation of energy, not of KE. There is no law of conservation of KE. $\endgroup$
    – nasu
    Jun 12, 2023 at 17:05

1 Answer 1


Consider the most general Lagrangian for this system:

$$\mathcal{L} = \frac{1}{2}m_1\dot{x_1}^2 + \frac{1}{2}m_2\dot{x_2}^2 - V(x_1,x_2)\tag{1}$$

where the first two terms are the kinetic energy of each particle and $V$ is the inter-molecular forces. Now if we are considering a simple elastic collision, these balls (which I assume to be point-like) are only interacting when they are at the same position (i.e. during a collision). Therefore, the potential energy must be of the form

$$ V(x_1,x_2) = V_0\delta(x_1-x_2)$$

Therefore, we can write the full Lagrangian as

$$L = \frac{1}{2}m_1\dot{x_1}^2 + \frac{1}{2}m_2\dot{x_2}^2 - V_0\delta(x_1-x_2)\tag{1}$$

We can now easily find the Hamiltonian of the system (which corresponds to the total energy) by performing a suitable legendre transformation:

$$ H = \sum_{i=1}^2 \dot{x}_i \frac{\partial L}{\partial \dot{x}_i} - L = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + V_0\delta(x_1-x_2)\tag{1}$$

Note since $\frac{\partial L}{\partial t} = 0$, The energy is conserved and therefore the total energy will not change after a collision. Since $\delta(x_1 - x_2)$ is strictly zero iff $x_1 \neq x_2$, the sum of kinetic energies should not change before and after a collision.

The proof lies on the fact that a simple elastic collision can be modelled by a dirac delta and since it is time independent and activated only during the collision; it follows from energy conservation that the initial kinetic and final kinetic energies must be the same.

  • $\begingroup$ I was hoping for a more mathematically simple proof, but if this is the only way then I guess it is okay $\endgroup$
    – WordP
    Jun 12, 2023 at 14:54
  • $\begingroup$ If you want a proof using Newtonian Mechanics or a more intuitive explanation, I can definitely make an edit. Just let me know. $\endgroup$
    – emir sezik
    Jun 12, 2023 at 14:55
  • $\begingroup$ Such a proof would be very appreciated! And it would also be very interesting to see a proof that doesn't assume the existence of kinetic energy from the start, because then it would present a way someone could have "discovered" kinetic energy by themself. $\endgroup$
    – WordP
    Jun 12, 2023 at 15:03
  • $\begingroup$ You prove that an ellastic collision is ellastic. By definition, ellastic collision implies that the kinetic energy is conserved. In general, the kinetic energy is not conserved in a collision. You cannot prove the conservation of kinetic energy as it is not a general property of collsions. $\endgroup$
    – nasu
    Jun 12, 2023 at 16:03
  • $\begingroup$ I agree that elastic collision is by definition a collision that preserves Kinetic Energy though, the calculation shows in which scenarios this can be the case which is something non-trivial. I also agree you cannot prove it but in certain cases and assumptions, this can be shown to be true. $\endgroup$
    – emir sezik
    Jun 12, 2023 at 19:12

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