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Conside a contact interaction given by a delta function on their worldlines. Use a gauge fixed Lagrangian for two point particles in terms of their proper times $t$ and $t^{\prime}$. Is it possible to find proper equations of motion for this system? For e.g. $$ S=\int dt \; \dot {x}^{2}+\iint\! dt \;dt^{\prime}\; \dot{x}(t)\cdot\dot{y}(t^{\prime})~ \delta^{D}(x(t)−y(t^{\prime}))+\int dt^{\prime} \; \dot{y}^{2} $$ working in, say, $D$ target space dimensions. This is written as a simplification of the relativistic / curved space system to illustrate the point. I think it will be possible to integrate the delta function out in $D=1$, but not in higher target space dim. The problem I have is in how to do the variation of the delta function term. Physically it is producing an interaction every time the worldlines of the particles intersect and I've tried writing this as a sum over such points - where $x(t_{0})=y(t^{\prime})$ - of $\frac{\delta(t−t_{0})}{\dot{x}(t_{0})}$ but this is valid only in $D = 1$ and I'm still not sure I can get the variation correct.

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  • $\begingroup$ Nice singular action you have there :D. You could always try regulating the delta function, something like $\exp(-(x(t)-y(t'))^2/2\sigma)$. I tried introducing a target space auxiliary field $\sigma_i(x)$ to split the interaction into local terms $\sim \int \mathrm{d}t \dot{x}_i(t) \int \mathrm{d}^D \chi \sigma_i(\chi) \delta^D (\chi-x(t)) + \cdots - 1/2 \int \mathrm{d}^D \sigma_i(\chi)\sigma_i(\chi)$. $\endgroup$ – Michael Brown Jun 11 '13 at 14:36
  • $\begingroup$ Integrating out $\sigma_i(\chi)$ I get your action plus nonlocal self-interactions $\int\mathrm{d}t\mathrm{d}t' \dot{x}_i(t) \delta(x(t)-x(t')) \dot{x}_i(t')$. But I'm not sure if these can be removed or even that to do so would be physical... $\endgroup$ – Michael Brown Jun 11 '13 at 14:36
  • $\begingroup$ Similarly here - the extra terms would represent point on the worldline interacting with itself so would need removing, perhaps in an ad-hoc fashion. $\endgroup$ – lux Jun 12 '13 at 12:22
  • $\begingroup$ Is it absolutely necessary to use an auxilliary field though? $\endgroup$ – lux Jun 12 '13 at 12:23
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Mathematically speaking, the non-local two-particle action in a $D$-dimensional flat target space

$$\tag{1} S[x,y]~:=~\int \! dt \frac{m}{2} \left( \dot{x}(t)^2 + \dot{y}(t)^2 \right) + \iint \! dt~dt^{\prime} ~\dot{x}(t) \cdot \dot{y}(t^{\prime})~V(x(t),y(t^{\prime})) $$

has equations of motion

$$\tag{2x} m\ddot{x}^{\mu}(t) ~=~ \int \! dt^{\prime} \left(\dot{y}^{\mu}(t^{\prime}) \frac{d}{dt}V(x(t),y(t^{\prime}))- \dot{x}(t) \cdot \dot{y}(t^{\prime})\frac{\partial}{\partial x^{\mu}}V(x(t),y(t^{\prime})) \right), $$ and $$\tag{2y} m\ddot{y}^{\mu}(t) ~=~ \int \! dt^{\prime} \left(\dot{x}^{\mu}(t^{\prime}) \frac{d}{dt}V(x(t^{\prime}),y(t))- \dot{x}(t^{\prime}) \cdot \dot{y}(t)\frac{\partial}{\partial y^{\mu}}V(x(t^{\prime}),y(t)) \right). $$ As a consequence, the speed of each particle is conserved in time

$$\tag{3} |\dot{x}|~=~{\rm const}, \qquad |\dot{y}|~=~{\rm const}. $$

Note that for $D=1$ the interaction term in the action (1) becomes a boundary term, which doesn't contribute to the equations of motion, i.e. the two 1D particles are free, and therefore each velocity is conserved in time.

A contact interaction of the form $V(x,y) \propto \delta^D(x-y)$ of multiple delta function is physically ill-defined since intersecting paths $x,y:\mathbb{R}\to \mathbb{R}^D$ render the action (1) singular. A contact interaction of the form $V(x,y) \propto \delta(|x-y|)$ of just a single delta function is expected to be better behaved.

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If you had two particles colliding head on what would they do? Well by symmetry the motion must remain 1D. By symmetry they must end with the same velocities. Since energy is conserved (correct me if I'm wrong), they must have the same final speed as initial speed. Thus they will bounce off each other and go back the way they came, having the opposite velocity.

Any other collision can be boosted to just be this collision, so this is really the only thing that can happen. Now consider that it just looks like the two particles went through each other. So this interaction gives the particles the same behavior as non-interacting particles, so you can really treat you particles as non-interacting unless you care about which is which. This is what is done when studying the thermodynamics of ideal gases all the time.

So the equation of motion basically says that the objects move in straight lines until they collide, and when the collide the swap velocities.

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    $\begingroup$ Your first statement is nonsense. Go to the CM frame, then you see that the same pair of particles can leave at any angle $\theta$. Fortunately, otherwise the LHC wouldn't exist as the two protons would just bounce off one another and go back into the beampipes! $\endgroup$ – Vibert Jun 11 '13 at 13:48
  • $\begingroup$ I'm also interested in general in the form of the eq. of motion and will want to consider the Hamiltonian formulation and consider setting up the Schroedinger eq. of this system when coming to quantise it... $\endgroup$ – lux Jun 11 '13 at 13:57
  • $\begingroup$ Varying the delta function is causing me the most trouble - i.e. variation of fields x -> x + dx etc as in the derivation of the eq. of motion from a Lagrangian $\endgroup$ – lux Jun 11 '13 at 13:58
  • $\begingroup$ @Vibert, now I think your statement about LHC is nonsense. LHC is a real-life device, where not only the interactions are not contact, but also everything is governed by QM, so nonzero $\theta$ is indeed possible, but with symmetry-respectful probability distribution (i.e. the distribution does not depend on $\phi$). In the OP's question the system is classical and with contact iteractions. $\endgroup$ – Peter Kravchuk Jul 11 '13 at 14:38

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