Monday, 11 May 2015

homework and exercises - Hamiltonian function for classical hard-sphere elastic collision




I'm trying to find the Hamiltonian function for a system consisting of a single particle in one dimension colliding elastically with a wall at x=0.


Everything I've read on the topic (e.g. this question Why can't collisions be elastic?) says that the wall can be represented by a step function potential barrier V=Kθ(x) with K any number higher than the maximum momentum of the particle -- often taken to be infinite for simplicity (e.g. V=0 if x<0, V= if x>0, so the Hamiltonian would be H=p2/2m+Kθ(x), where θ(x) is the step function, θ(x)=0 for x < 0, θ(x)=1 if x>0, and K is a constant that we eventually take to infinity)...but I cannot get that to work.


When I apply the Hamiltonian equations of motion to that Hamiltonian, I end up finding that the collision is only elastic and energy-conserving if K is not only finite, but also dependent on the particle's momentum.


I'm fairly sure that there's just something wrong with my derivation, because using an infinite potential barriers to exclude particles from certain positions is used everywhere in physics (e.g. hard-sphere dynamics in molecular dynamics or billiard dynamics, the Van Der Walls forces, etc)...but I just can't see where I'm going wrong. So, below is my full derivation (I start with two particles of diameter D, and then take the mass of the second particle to be infinity at the end; however, I get the same results if I start with just one particle of diameter D2).


Note: the problem is specifically getting the Hamiltonian; the equations of motion themselves are trivial to get via e.g. conservation of energy/momentum arguments; x is proportional to |t|, and p to sign(t)).


Variables


D := diameter of particle


K := height of "wall" (nonzero; assumed by most references to be a constant, often infinite, but always greater than the maximum kinetic energy; this derivation only assumes that it is position-independent)


mi := mass of particle i (for a wall, m2)



x1,x2,p1,p2 := coordinates and momenta of particles


r=|x1x2| := absolute distance between centers of particles (so (Dr)>=0 iff they are touching)


H=T+U


U(r)=Kθ(Dr)


T=p21/2m+p22/2m


Definitions / abbreviations


ϵ(x):=sign(x)


θ(x):=step(x)


r/x1=ϵ(x1x2)


r/x2=ϵ(x1x2)



θ/x=δ(x)


Derivation


(0) Hamilton's equations:


H/x1=dp1/dt=U/x1


H/x2=dp2/dt=U/x2


(1) Chain rule (prime means differentiation wrt appropriate x):


U/x1=Kθ(Dr)(r)=Kδ(D|x1x2|)ϵ(x1x2)


U/x2=Kθ(Dr)(r)=Kδ(D|x1x2|)ϵ(x1x2)


(2) note that the derivatives are equal/opposite


U/x1=dp1/dt=dp2/dt=U/x2



(3) find total change in momenta ΔPi (and final momenta pi(t2))


Let t1 := the time where x1x2=+D; let t0:=t1dt,t2:=t1+dt


Let p1(t1)=P1,p2(t1)=P2


Let ΔP1:=t2t0(dp1/dt) dt=K=ΔP2


Then p1(t2)=P1+K;p2(t2)=P2K


(4) But, energy must be conserved: T(pi)=T(pi+ΔPi)


(P1+K)2/2m1+(P2K)2/2m2=P21/2m1+P22/2m2


(P21+2KP1+K2)/2m1+(P222KP2+K2)/2m2=P21/2m1+P22/2m2


(2KP1+K2)/2m1+(2KP2+K2)/2m2=0


Multiply through by 2m1m2



(2KP1+K2)m2+(2KP2+K2)m1=0


2K(m2P1+1/2Km2m1P2+1/2Km1)=0


(5) By definition, |K| > 0. So,


m2P1m1P2+1/2K(m1+m2)=0


Conclusion:


K=2(m1P2m2P1)/(m1+m2)


So to conserve kinetic energy, K (and hence U) must depend on the momentum.


(6)


In the special case where m2:


1/2K=(m1P2)/(m1+m2)(m2P1)/(m1+m2)



1/2K=(m1P2)/(m2)(m2P1)/(m2)


1/2K=0P1,K=2P1, as expected.


(7) However, since the potential now depends on the momenta, dxi/dt is no longer simply pi/2mi; we have


H=p21/2m1+p22/2m2+2(m1P2m2P1)θ(Dr)/(m1+m2)


H/p1=dx1/dt=p1/m12m2θ(Dr)/(m1+m2)


H/p2=dx2/dt=p2/m2+2m1θ(Dr)/(m1+m2)


(8)


Or in the special case of the wall,


H=p21/2m1+p22/2m22P1θ(Dr)


H/p1=dx1/dt=p1/m12P1θ(Dr)



H/p2=dx2/dt=p2/m2+2P1θ(Dr)


...which is zero outside of the barrier, but there is ambiguity in the definition of the step function at zero; we can retain the normal equations of motion only if θ(0)=0.


Any thoughts would be greatly appreciated.




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