So, lets say that our system consists of two particles. Without the gravitational force or any other long-range force, the only possible way for these two particles to interact is to collide with each other directly, at least if the initial conditions are such that they lead to the collision.

This problem is easily solved with Newtonian mechanics, but trying to find the Potential of this system (for the sake of simplicity let's say its 2D, and same mass particles) should be something like $$ L=\dfrac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2+\dot{y}_1^2 +\dot{y}_2^2) - V$$ where $$V=V(|\vec{r_1} - \vec{r_2}|,|\vec{u_1} - \vec{u_2}|)$$ and my thought was $$V=\Theta (|\vec{r_1} - \vec{r_2}|)\cdot \delta(|\vec{u_1} - \vec{u_2}|)$$ but still can't get the results needed.

Note, that for same mass particles, the potential just changed velocities on $$\vec{r_1}=\vec{r_2}$$

Any ideas? I can still solve this with Newtonian mechanics, but Lagrangian approximation should work the same somehow

EDIT: Newtonian solution

With any given initial conditions (u1,u2,r1(0),r2(0)):

The x coordinate is given for both particles

$$x_1 = u_1 cos(\theta_1)\cdot t + x_{10}$$ $$x_2 = u_2 cos(\theta_2)\cdot t + x_{20}$$

so solving $x_1=x_2$ gives $$t_{x_1=x_2}=\dfrac{x_{20}-x_{10}}{u_1 cos(\theta_1)-u_2 cos(\theta_2)}$$ and same thing for $y_1=y_2$

$$t_{y_1=y_2}=\dfrac{y_{20}-y_{10}}{u_1 sin(\theta_1)-u_2 sin(\theta_2)}$$

and in both cases $\theta$ is the angle of velocity

Collision happens only if$$t_{y_1=y_2}=t_{x_1=x_2}$$ and in that case there is velocity exchange (same mass)

  • 1
    $\begingroup$ What are the $\vec u_i$ in your potential? Should the particles be point-like or have finite radii? $\endgroup$
    – Photon
    Commented Nov 5, 2017 at 10:11
  • $\begingroup$ \vec{u_i} is the velocity of each particle. Particles are point like and the collision is elastic (no energy loss). That potential is just my guess. Really dont know what form the potentila should have @Photon $\endgroup$
    – user174411
    Commented Nov 5, 2017 at 10:18
  • 1
    $\begingroup$ Can you describe in more detail what is it actually that you are trying to achieve? Are the particles hard spheres with contact interaction? Why do you mention gravity? Perhaps you can describe how you would do this in Newtonian mechanics and then I am sure it will be very easy to tell you how this is done in a Lagrangian formalism. $\endgroup$
    – Void
    Commented Nov 5, 2017 at 10:19
  • $\begingroup$ I would say, something proportional to $\delta(\vec r_1-\vec r_2)$. The velocities should not matter at all and your $\Theta$ ansatz is identically one unless the particles collide, this is reversed to what you probably intend to. $\endgroup$
    – Photon
    Commented Nov 5, 2017 at 10:22
  • $\begingroup$ @Void Newtonian solution is given. Any assistance is appreciated, finding the Potential of Elastic Collision and thus, solving it with Lagrangian Approximation $\endgroup$
    – user174411
    Commented Nov 5, 2017 at 10:47

1 Answer 1


The potential you are looking for is in the form $\delta^{(2)}(\vec{r}_1 - \vec{r}_2) = \delta (x_1 - x_2) \delta(y_1 - y_2)$ (the delta function is symmetric in its arguments, so we do not need to add the absolute value). The dynamical equations then read $$m\ddot{x}_1 = \delta'(x_1 - x_2)\delta(y_1 - y_2)$$ $$m\ddot{x}_2 = -\delta'(x_1 - x_2)\delta(y_1 - y_2)$$ $$m\ddot{y}_1 = \delta(x_1 - x_2)\delta'(y_1 - y_2)$$ $$m\ddot{y}_2 = -\delta(x_1 - x_2)\delta'(y_1 - y_2)$$ where $\delta'$ is the delta-function derivative. These are of course singular expressions which are best understood in integral form (I will only mention one of the $x$ equations from now on, the $y$ set is analogous) around the collision time $t_{coll}$ $$\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} m \ddot{x}_1 dt = \int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$$ $$\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} m \ddot{x}_2 dt = -\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$$ Now we take the limit $\epsilon \to 0$, eliminate the right-hand sides and see that we get the usual collision relations $$m \dot{x}_1(t_{coll} - 0) - m \dot{x}_1(t_{coll} + 0) = - (m \dot{x}_2(t_{coll} - 0) - m \dot{x}_2(t_{coll} + 0))$$ where the $\pm 0$ mean just before/after the collision. The integral $\int_{t_{coll}-\epsilon}^{t_{coll} + \epsilon} \delta'(x_1 - x_2) \delta(y_1 - y_2) dt$ itself cannot be evaluated directly but from the properties of the delta function and its derivatives we know that it is generally non-zero.

However, with more considerations it is easy to realize that the output of this collision is ill-determined. For instance, it is mathematically admissible that the particles fly through each other without a change of velocity. In other words, there is no well-defined model of point particles in contact interactions.

To eliminate this ambiguity, one can either compute the differential cross-section of the process and input it into every collision by hand, or introduce a different collision model. One of the simplest "contact" collision models is the hard sphere $V \sim \delta^{(d)} (|\vec{r}_1 -\vec{r}_2| - 2R)$ where $R$ is the diameter of the sphere-particles.

  • $\begingroup$ Really Really Thank you for the time you spent and is Much appreciated! So, form what you say, I think I have understood most of them, and that I can't really find the extra equation needed because of the not well determined contact interactions. $$ -- $$ What I don't understand is how "contact" collision models - $V \propto \delta ^d (|\vec{r1}=\vec{r2}|−2R)$ would solve this. I think you would still miss an extra equation right?? $\endgroup$
    – user174411
    Commented Nov 5, 2017 at 12:49
  • $\begingroup$ This is already a well determined model, see: physics.stackexchange.com/questions/107648/… Also, the hard-sphere collision has a well-defined limit of $R \to 0^+$: physics.stackexchange.com/a/184955/52394 (The point is, however, that taking a similar limit with other potentials may lead to different scattering behaviour...) $\endgroup$
    – Void
    Commented Nov 5, 2017 at 14:34
  • $\begingroup$ What you sent is Newtonian approach. I don't think that the potential is so easily defined. I mean, how can I get the results needed. (I made the maths and I still can't find the velocity exchange) $\endgroup$
    – user174411
    Commented Nov 6, 2017 at 19:22

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