# Help with anti-derivative

1. Nov 8, 2013

### jjustinn

1. The problem statement, all variables and given/known data

I'm trying to find the Hamiltonian for a system consisting of a single particle moving in 1D elastically colliding with an infinite potential barrier.

By conservation of energy, we know the magnitude of the momentum must be the same before and after the collision; for simplicity, assume the collision takes place at x = t = 0. So, p(t < 0) is p0, and p(t > 0) is -p0.

So, the total change in momentum is -2p(t) at t=0.

Since the entire change of momentum takes place at x=0, the force (dp/dt) is a delta function:

dp/dt = -p(t) δ(x(t))

Hamilton's equations of motion are dH/dx = -dp/dt, and H = T(p) + V(x); so because T (p2/2m) is independent of x, we have

dH/dx = dV/dx = -dp/dt = 2p(t) δ(x(t))

We also have dx/dt = dH/dp = p/m.

So -- what is V?

2. Relevant equations

H(p, x) = T(p) + V(x)

T = p2/2m

∂H/∂x = -dp/dt = 2p(t)δ(x(t))
∂H/∂p = dx/dt = p(t)/m

3. The attempt at a solution

Last edited: Nov 8, 2013
2. Nov 8, 2013

### UltrafastPED

V is an infinite potential barrier ... so if it is approachable from only one side (from the right, x>0) then it is a step function: V(x>0) = 0, V(x<=0) = infinity.

If you are familiar with the Dirac delta function - it works very well for this situation.

3. Nov 8, 2013

### jjustinn

Now that's the answer I keep finding everywhere, but I cannot get it to come out right; if V is a *finite* step function (say v0 * step(x)), then dV/dx is v0 * delta(x)...and so dp/dt is -v0delta(x), so the final momentum will be p0 - v0...which is only elastic/energy-conserving if v0 isn't a constant, but is equal to -2p(t).

It gets even worse if v0 is infinite, because then you've got infinity times the delta function ( see the first link in my attempted solutions for an extended discussion on this).

Now , I've seen this answer so many places it *has* to be right, but I can't see how to make it work. What am I doing wrong?

4. Nov 8, 2013

5. Nov 8, 2013

### jjustinn

Yeah, that's what irks me : there are thousands of derivations for a quantum wave function in an infinite well -- which is by definition unobservable and useful only in very specific, highly technical situations...but there's not a single derivation for a cue ball bouncing against the side of the table (in Hamiltonian/Lagrangian form, that is; even when they give the potential function, the actual equations of motion are invariably found via energy / momentum conservation.

6. Nov 8, 2013

### UltrafastPED

Thanks ... maybe I will include that in the problem set when I teach analytical mechanics again!

Even this tome on the physics of pool doesn't due it:
ftp://ftp.tcg.anl.gov/pub/shepard/pool/old_versions/physics.PDF

7. Nov 8, 2013

### jjustinn

No problem -- but you need to PM me the solution when you work it out, because this is driving me nuts ;)

8. Nov 8, 2013

Will do!

9. Nov 9, 2013

### MisterX

Perhaps the problem here is what does it mean that the force is a Dirac delta of x?

I think one should consider the limit as something approaches this.

Let the walls of the well be ramps and let the slope go to infinity. Then perhaps it is more clear - with E< V the object only gets so far and then comes again. Increasing the slope doesn't change the maximum potential energy achieved by the object before it comes back. Rather, increasing the slope merely expedites the process of turning around.

Similarly you may think of force as functions of x for some sequence of functions which approaches the appropriate Dirac deltas. The object will only pass through a certain percentage of the area under the delta. As the function is made skinnier this will take less and less time.

10. Nov 9, 2013

### jjustinn

Sure, that's how it is defined; but in the actual equations of motion for which we're trying to find the Hamiltonian, the force (dp/dt) is *exactly* a delta function (since the entire interaction takes place at a single point/instant)...and as far as I can tell, using a function (e.g. Gaussian) that becomes a delta function in a limit gives the same result and has the same problems, but just has messier integrals...but, it did seem like someone in another thread was coming around to a similar conclusion -- so am I missing something?

11. Nov 9, 2013

### jjustinn

Interesting: this reference seems to have found something very similar to what I've been saying: http://books.google.com/books?id=xV...=lagrangian "step function potential"&f=false

It's using a continuous step function rather than a true step, but for all practical purposes it's the same (the reasons for choosing a continuous one were for numerical simulation) -- they find that to get a proper "reflection", the step has to be exactly -2 times the initial momentum...in other words, contrary to virtually everything else I've seen (including e.g. Huang's Statistical Mechanics), they say an infinite potential wall will *not* give an elastic (energy and momentum conserving) collision (even though the potential does not depend explicitly on time)...

12. Nov 9, 2013

### UltrafastPED

Thinking about this problem ... at the heart of the weak variational principle of the calculus of variations is the requirement that virtual displacements be reversible, and that they be subject to the constraints of the system.

For example, the bead on a wire is constrained to the wire, but can move back and forth along it. The displacement of the bead along the wire is a perfectly good generalize coordinate ... and the velocity along the wire is well-defined everywhere.

But if we have a bead on a table the motion is not reversible when it comes to the edge ... one step in the wrong direction and the bead falls off. This type of problem requires machinery which is beyond the scope of "the weak variational principle", and so you will not see it in any of the usual treatments of Lagrangian mechanics. And if you cannot write the Lagrangian, you will not be able to write the Hamiltonian.

The same issues come up with "infinite barriers" or "hard walls" and mechanical contact collisions; they run aground in the shoal waters of "the derivative of position wrt time is undefined at the point of contact".

13. Nov 9, 2013

### jjustinn

That does sound plausible, and It does match up with the intro paragraph of that last reference -- that walls and other immobile barriers are usually treated as constraints, not as potentials.

However, notwithstanding your persuasive argument about the reversibility of virtual displacements, virtually every source I've seen (Including the last one ) at least implies that such barriers *can* be treated as potentials; for instance , the entire field of molecular dynamics is based on a "Hard sphere" potential, as are Van Der Waals forces...maybe it's theoretically sloppy, but it seems to be common and well-accepted.

14. Nov 10, 2013

### UltrafastPED

I took a graduate seminar in molecular dynamics a few years ago; we did a lot of computer simulations.

IIRC we did everything with Newton's laws of motion, then calculated properties from the statistics. The potentials were only used to find the appropriate force. But there may be other methods.

I have attached some notes on analytical mechanics which go into detail on how the methods are constructed, and the nature of constraints ... but I did not investigate the question which you have brought up!

Last edited: Nov 10, 2013
15. Nov 10, 2013

### UltrafastPED

16. Nov 10, 2013

### jjustinn

Re-derivation

So I tried working this out from scratch...starting with two particles, then taking the mass of the second to be infinite to get an immobile wall, and confirmed that only if the condition on the height of the barrier is met is energy conserved -- and in that case, the correct Newtonian equations are reproduced... So I was hoping someone might see where I went wrong.

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)
$m_i$ := mass of particle i (for a wall, $m_2 → ∞$)
$x_1, x_2, p_1, p_2$ := coordinates and momenta of particles
$H = T + U$
$U(r) = Kθ(D - r)$
$T = p_1^2/2m + p_2^2/2m$

Definitions / abbreviations
$ε(x) := sign(x)$
$θ(x) := step(x)$
$r = |x_1 - x_2|$

$∂r/∂x_1 = ε(x_1 - x_2)$
$∂r/∂x_2 = -ε(x_1 - x_2)$
$∂θ/∂x = δ(x)$

Derivation

(0) Hamilton's equations:
$∂H/∂x_1 = -dp_1/dt = ∂U/∂x_1$
$∂H/∂x_2 = -dp_2/dt = ∂U/∂x_2$

(1) Chain rule (prime means differentiation wrt appropriate x):
$∂U/∂x_1 = Kθ'(D - r)(-r') = -Kδ(D - |x_1 - x_2|)ε(x_1 - x_2)$
$∂U/∂x_2 = Kθ'(D - r)(-r') = Kδ(D - |x1_ - x_2|)ε(x_1 - x_2)$

(2) note that the derivatives are equal/opposite
$-∂U/∂x_1 = dp_1/dt = -dp_2/dt = ∂U/∂x_2$

(3) find total change in momenta $ΔP_i$ (and final momenta $p_i(t_2)$)
Let $t_1$ := the time where $x_1 - x_2 = +d$; let $t_0 := t_1 - dt, t_2 := t_1 + dt$
Let $p_1(t_1) = P_1, p_2(t_1) = P_2$
Let $ΔP_1 := ∫(dp_1/dt)dt$ from $t_0...t_2 = K = -ΔP_2$
Then $p_1(t_2) = P_1 + K; p_2(t_2) = P_2 - K$

(4) But, energy must be conserved: $T(p_i) = T(p_i + ΔP_i)$
$(P_1 + K)^2/2m_1 + (P_2 - K)^2/2m_2 = P_1^2/2m_1 + P_2^2/2m_2$
$(P_1^2 + 2KP_1 + K^2)/2m_1 + (P_2^2 - 2KP_2 + K^2)/2m_2 = P_1^2/2m_1 + P_2^2/2m_2$
$(2KP_1 + K^2)/2m_1 + (-2KP_2 + K^2)/2m_2 = 0$

Multiply through by $2m_1m_2$

$(2KP_1 + K^2)m_2 + (-2KP_2 + K^2)m_1 = 0$
$2K(m_2 P_1 + 1/2 K m_2 - m_1 P_2 + 1/2K m_1) = 0$

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

$m_2 P_1 - m_1 P_2 + 1/2K(m_1 + m_2) = 0$

Conclusion:
$K = 2(m_1 P_2 - m_2 P_1)/(m_1 + m_2)$
So to conserve kinetic energy, K (and hence U) must depend on the momentum.

(6)
In the special case where $m_2 -> infinity$:
$1/2K = (m_1 P_2)/(m_1 + m_2) - (m_2 P_1)/(m_1 + m_2)$
$1/2K = (m_1 P_2)/(m_2) - (m_2 P_1)/(m_2)$
$1/2K = 0 - P_1, K = -2P_1$, as expected.

(7)
However, since the potential now depends on the momenta, $dx_i/dt$ is no longer simply $p_i/2m_i$; we have

$H = p_1^2/2m_1 + p_2^2/2m_2 + 2(m_1 P_2 - m_2 P_1)θ(D - r)/(m_1 + m_2)$

$∂H/∂p_1 = dx_1/dt = p_1/m_1 - 2m_2θ(D - r)/(m_1 + m_2)$
$∂H/∂p_2 = dx_2/dt = p_2/m_2 + 2m_1θ(D - r)/(m_1 + m_2)$

(8)
Or in the special case of the wall,
$H = p_1^2/2m_1 + p_2^2/2m_2 -2 P_1θ(D - r)$
$∂H/∂p_1 = dx_1/dt = p_1/m_1 - 2P_1 θ(D - r)$
$∂H/∂p_2 = dx_2/dt = p_2/m_2 + 2P_1θ(D - r)$

...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.

So...what's not right?

Last edited: Nov 10, 2013