Classical Square Well: Hamiltonian Form & Elastic Collision

In summary, a classical idealized particle, moving in one dimension, with momentum p and kinetic energy T comes into contact with an infinite step-function potential V, there will be an (instantaneous) elastic collision - the particle's momentum becomes -p, so its energy remains constant ( momentum is conserved by transferring 2p additional momentum to the barrier, but if its mass is taken to be infinite, this will not change its velocity/energy).
  • #1
jjustinn
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My understanding is that a classical idealized particle, moving in one dimension, with momentum p and kinetic energy T comes into contact with an infinite step-function potential V, there will be an (instantaneous) elastic collision - the particle's momentum becomes -p, so its energy remains constant ( momentum is conserved by transferring 2p additional momentum to the barrier, but if its mass is taken to be infinite, this will not change its velocity/energy).
Further, if I'm not mistaken, the same thing will happen with any finite V, where V > T.

First: are both of those statements accurate?

If so , I'm having trouble showing it in Hamiltonian form.

As I understand it, the whole system, where the barrier is at x=0, should be describable by

H = p2/2m - Vθ(x)

Where θ(x) is the unit step function (x<0 θ(x)=0; x>=0 θ(x)=1), and V is a constant > p2/2m.

So, Hamilton's equations give

∂H/∂p = dx/dt = p/m
-∂H/∂x = dp/dt = -Vδ(x)

Differentiating the first and setting it equal to the second gives us a Newtonian-style F=ma equation (F = dp/dt; a = d2x/dt2):

m a= -Vδ(x)

However, this can't be right, because as far as I can tell, the force at x=0 is proportional to V -- specifically, the total change in momentum between x-dx and x+dx is exactly V...while for an elastic collision, it would be -2p -- so I'm apparently missing something.

Any ideas?
 
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  • #2
you've got a Dirac-Delta function there. So the force at x=0 is not proportional to V, the force is infinite there.

edit: well, I guess it is proportional to V. But I don't fully understand where the problem is. The particle will approach the Dirac-Delta function, but at some point will lose all its kinetic energy (before it gets to x=0), and it will turn around. And when it gets back to the same position it started at, it will have the same kinetic energy, but opposite momentum, right? So I don't see the problem.
 
  • #3
BruceW said:
you've got a Dirac-Delta function there. So the force at x=0 is not proportional to V, the force is infinite there.

edit: well, I guess it is proportional to V. But I don't fully understand where the problem is. The particle will approach the Dirac-Delta function, but at some point will lose all its kinetic energy (before it gets to x=0), and it will turn around. And when it gets back to the same position it started at, it will have the same kinetic energy, but opposite momentum, right? So I don't see the problem.

Right -- it's a really trivial problem to solve with conservation of energy/momentum or Newton's laws...but I can't get it to come out right in Hamiltonian form.
 
  • #4
what are you having problems with?
 
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  • #5
BruceW said:
what are you having problems with?

If I use the Hamiltonian given above, the equations of motion don't come out right, because the force is proportional to the height of the well...so as far as I can tell, energy/momentum isn't even conserved.

Edit: finish answering question.

So, I'm trying to figure out if my assumptions are wrong, if the Hamiltonian is wrong, or if my subsequent math was wrong.
 
  • #6
If you have a potential that depends explicitly on the spatial variables of the particle, then momentum is not conserved. Energy is conserved because the potential is not dependent on time. This is well-known physics, momentum is not supposed to be conserved in this case.

As another example, if we have a potential energy like mgh (object falling through gravitational potential), then momentum in the vertical direction is not conserved.

edit: so what I'm saying is that it's OK that momentum is not conserved.
 
  • #7
BruceW said:
If you have a potential that depends explicitly on the spatial variables of the particle, then momentum is not conserved. Energy is conserved because the potential is not dependent on time. This is well-known physics, momentum is not supposed to be conserved in this case.

As another example, if we have a potential energy like mgh (object falling through gravitational potential), then momentum in the vertical direction is not conserved.

edit: so what I'm saying is that it's OK that momentum is not conserved.

Ah, right -- hadn't thought of that...but that's really the problem then, because in the equations of motion that we know this system produces, momentum *is* conserved (at least if the barrier is taken to have infinite mass -- and in any case, energy is conserved).

Also, it isn't just momentum that's not conserved, but energy, too (since dp/dt is proportional to V, the *only* way for energy to be conserved is if V exactly reverses the momentum, which is does not in general)-- and the Hamiltonian does NOT explicitly depend on the time.

And in any case, the lack of conservation isn't a "reasonable" amount here, either: for an infinite well, the change in momentum (and hence energy) is infinite...

Edit: by "the change", I mean total change (integral of dp/dt over t-dt ... t+dt), NOT dp/dt, which of course would be infinite (a delta function at t) in the correct case as well.
 
  • #8
well (as example), if we want to model a ball falling (but we don't care about how much the Earth moves up towards the ball), then we can use V=mgh and momentum is not conserved. This is fine. This makes a reasonable model to test experiments against.

But yeah, if you want a system that does conserve momentum, then you would need some kind of object with mass that causes a force on the particle. And so an object at x=0 with extremely large mass would give you this kind of system. (And the potential is now a function of the relative position of the two objects, which explains why momentum is conserved in this case).

So now going back to the system with a potential and just one particle. yes, I think it does conserve energy. The potential does exactly reverse the momentum.

well, if we allow the particle to end up at exactly x=0 then yes we will have a problem, because the force here is a Dirac-Delta function. The problem is caused by the fact that the potential is not continuous. If the potential was continuous, yet very steep, then we would get a very large force, but things would still be OK. So physically, a very steep potential might be more realistic than a completely vertical potential.
 
  • #9
BruceW said:
So now going back to the system with a potential and just one particle. yes, I think it does conserve energy. The potential does exactly reverse the momentum.

Any help showing this? I get dp/dt=-∂H/∂x, which is depends on V, not the momentum...so how does it reverse it (eg total change = -2p)?

well, if we allow the particle to end up at exactly x=0 then yes we will have a problem, because the force here is a Dirac-Delta function. The problem is caused by the fact that the potential is not continuous. If the potential was continuous, yet very steep, then we would get a very large force, but things would still be OK. So physically, a very steep potential might be more realistic than a completely vertical potential.

But in the Newtonian case, this is exactly what happens, without any issue: the momentum at x=-dx is p, at x=+dx it's -p, so the force (dp/dt) is -2pδ(x)...
 
  • #11
jjustinn said:
Any help showing this? I get dp/dt=-∂H/∂x, which is depends on V, not the momentum...so how does it reverse it (eg total change = -2p)?
your equation is m a= -Vδ(x) So... yeah there's not much you can do with it as it is here because the Dirac-Delta function does not help us at all. But, if you use the limiting approximation of a Dirac-Delta function, then you can see what happens for a very sharp change in potential (not a vertical potential). We can use a Gaussian function, centered on x=0 as an approximate Dirac-Delta function, in the limit of the width of the Gaussian function being very small and the integral of the Gaussian over all space being equal to 1. So, now we have m a = -Vf(x) where f(x) is an actual function (the Gaussian function). So now, the particle will initially be moving towards x=0 and then at some point close to x=0, the particle will lose all its velocity, and turn around, and when it gets back to the same initial point, it will have regained its initial momentum, but in the opposite direction.

So to show this, we have ##mdv/dt = -Vf(x)## and since ##d/dt=vd/dx## we get ##mvdv/dx=-Vf(x)## and now integrating both sides with respect to x we have ##0.5mv_f^2-0.5mv_i^2= -V \int f(x) dx## So if we say the particle returns to the same place, then the right hand side of the equation will equal zero. Therefore ##v_i^2=v_f^2##, i.e. the magnitude of velocity is the same. And since the particle is now returning from being near the potential, we know the particle now has equal and opposite momentum to what it had before.
 
  • #12
After looking at your other threads, I see that what you are looking for is an equation for completely elastic collision. hmm. I know that for 'hard spheres', physicists use a potential that is infinite inside the sphere and zero outside. So yeah, this would be similar to your equation, except V should be infinite. (I guess this is to make sure that no particle has enough momentum to break through into the hard sphere).

http://en.wikipedia.org/wiki/Hard_spheres

Or maybe google hard spheres.

edit: I guess maybe you are still worried about whether the collision is elastic or not. Well, if we say elastic collision is simply conservation of energy, then that's easy since as long as the Hamiltonian is not explicitly time-dependent, then we have conservation of energy. But, if we say elastic collision is conservation of kinetic energy (and this is the most common definition I think), then it is a bit more complicated. Well, maybe the most common definition is that elastic collision means that the kinetic energy once the particles get far enough away from each other that they become effectively free again, that the kinetic energy is again the same as it initially was. So in this definition, you'd need to make sure that the potential energy isn't being stored in some other weird way. But luckily, since you are creating the exact form of the potential yourself, it is easy to check where the potential energy is being stored, since you can just look at the terms of the equation that you've written.
 
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  • #13
BruceW said:
After looking at your other threads, I see that what you are looking for is an equation for completely elastic collision.
Correct -- well, specifically, the Hamiltonian for it; the equations of motion, and the force, are trivial. I just cannot find a Hamiltonian that gives them.
I started by looking for the hard-sphere interaction for N particles in 3D, but as I ran into frustration after frustration, I started looking for simpler problems : first point particles, then just two particles , then restricting to 1D, and finally settling for the simplest non-trivial physics problem there is: a single point particle bouncing off of a perfect wall, once, in one dimension, with no external forces.

hmm. I know that for 'hard spheres', physicists use a potential that is infinite inside the sphere and zero outside. So yeah, this would be similar to your equation, except V should be infinite. (I guess this is to make sure that no particle has enough momentum to break through into the hard sphere).

http://en.wikipedia.org/wiki/Hard_spheres

Or maybe google hard spheres.
Using the hard-sphere potential is almost identical to what we've been discussing -- in particular, it has the same problem whereby the infinite potential leads not only to an infinite force ( which any discontinuous potential would), but an infinite change in momentum -- the difference being that the interaction is a step function (theta((x - y)^2 - r^2), rather than a delta function (delta(x - y))..

edit: I guess maybe you are still worried about whether the collision is elastic or not.

I'm worried that when I take the derivative of the infinite step function, I get an infinite change in momentum...if V = Kstep(x), then dV/dx = Kdelta(x), so the final momentum will be the initial momentum minus K...and if K is infinite, then the momentum is negative infinity...if K = 2p, then you do get an elastic collision, but it doesn't make sense to me that the potential would depend on p (not to mention every site I've seen agrees with you).

Well, if we say elastic collision is simply conservation of energy, then that's easy since as long as the Hamiltonian is not explicitly time-dependent, then we have conservation of energy.

That's the definition I'm using; but again, unless I'm making a mistake, energy won't be conserved, even though the Hamiltonian isn't explicitly time-dependent...as mentioned above, the final momentum will be minus infinity, and the kinetic energy is p^2/2m. I know that can't be right, but I can't get it to work -- do you get a different result?
 
  • #14
BruceW said:
your equation is m a= -Vδ(x) So... yeah there's not much you can do with it as it is here because the Dirac-Delta function does not help us at all. But, if you use the limiting approximation of a Dirac-Delta function, then you can see what happens for a very sharp change in potential (not a vertical potential). We can use a Gaussian function, centered on x=0 as an approximate Dirac-Delta function, in the limit of the width of the Gaussian function being very small and the integral of the Gaussian over all space being equal to 1. So, now we have m a = -Vf(x) where f(x) is an actual function (the Gaussian function). So now, the particle will initially be moving towards x=0 and then at some point close to x=0, the particle will lose all its velocity, and turn around, and when it gets back to the same initial point, it will have regained its initial momentum, but in the opposite direction.

So to show this, we have ##mdv/dt = -Vf(x)## and since ##d/dt=vd/dx## we get ##mvdv/dx=-Vf(x)## and now integrating both sides with respect to x we have ##0.5mv_f^2-0.5mv_i^2= -V \int f(x) dx## So if we say the particle returns to the same place, then the right hand side of the equation will equal zero. Therefore ##v_i^2=v_f^2##, i.e. the magnitude of velocity is the same. And since the particle is now returning from being near the potential, we know the particle now has equal and opposite momentum to what it had before.

Sorry, missed this earlier. This looks just like the normal Newtonian derivation, except with the delta function replaced with a gaussian...I don't quite follow what you mean by the delta function not helping -- since it's both exactly what we want (instantaneous localized force) and easier to integrate (Δp = ∫dp/dt dx = ∫-Vδ(x)dx = -V)...but for argument sake, how would you put your equations into Hamiltonian form? Maybe we can then take the limit and get the delta version?
 
  • #15
jjustinn said:
m a= -Vδ(x)

However, this can't be right, because as far as I can tell, the force at x=0 is proportional to V -- specifically, the total change in momentum between x-dx and x+dx is exactly V...while for an elastic collision, it would be -2p -- so I'm apparently missing something.

Any ideas?

It's important to make a distinction between force with t as the dependent variable, and force with x as the dependent variable. Both should be delta functions. However, the one with time as the dependent variable will not be multiplied by V, unlike the one with x as a dependent variable.

The delta function itself is a difficulty here. The delta function of x integrates to one, but the object in question does not fully go through it. So it seems easier to think of this as the limit of a sequence of functions. The rectangle definition for a Dirac delta may be easier to use than the Gaussian one.
 
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  • #16
MisterX said:
It's important to make a distinction between force with t as the dependent variable, and force with x as the dependent variable. Both should be delta functions. However, the one with time as the dependent variable will not be multiplied by V, unlike the one with x as a dependent variable.

The delta function itself is a difficulty here. The delta function of x integrates to one, but the object in question does not fully go through it. So it seems easier to think of this as the limit of a sequence of functions. The rectangle definition for a Dirac delta may be easier to use than the Gaussian one.

I think we might be getting somewhere; that's where a lot of my technical difficulty has come in -- dealing with x/t...particularly since in the elastic case, dp/dt = -2p(t)δ(x(t))...I couldn't figure out how to integrate that without having the explicit form of x(t).

I tried to skip around it by looking instead at dp/dt = -∂V/∂x, so V = ∫2pδ(x)dx = 2pθ(x), but that obviously doesn't match the requirements (e.g. V being an infinite, momentm-independent step function).

I still don't quite see how the various approximate dekta functions could solve/simplify this problem, though, since it seems that as long as dp/dt = -∂V/∂x integrates to a momentum-independent quantity -- no matter how fast/slow -- you cannot end up with an elastic collision.
 
  • #17
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)...
 
  • #18
jjustinn said:
Using the hard-sphere potential is almost identical to what we've been discussing -- in particular, it has the same problem whereby the infinite potential leads not only to an infinite force ( which any discontinuous potential would), but an infinite change in momentum -- the difference being that the interaction is a step function (theta((x - y)^2 - r^2), rather than a delta function (delta(x - y))..
eh? what do you mean by 'interaction' ? In your case, the potential energy is a step function, and in the case of hard spheres, the potential energy is also a step function. The only difference is that they use a step function with infinite height. This is a bit weird, I admit. So you could instead say that the step function has height at least as large as the kinetic energy of the approaching particle, so that the particle does not pass into the hard-sphere. This should give the same result.



jjustinn said:
I'm worried that when I take the derivative of the infinite step function, I get an infinite change in momentum...if V = Kstep(x), then dV/dx = Kdelta(x), so the final momentum will be the initial momentum minus K...and if K is infinite, then the momentum is negative infinity...if K = 2p, then you do get an elastic collision, but it doesn't make sense to me that the potential would depend on p (not to mention every site I've seen agrees with you).
Think about a very high and very narrow Gaussian function as an 'approximate' Dirac Delta function. The particle will travel along, and will encounter this Gaussian potential. The particle will only keep moving as long as its initial kinetic energy is greater than the potential at the point where it is. In other words, it will never get to the point of infinite potential, or infinite force. It will turn around as soon as it has converted its kinetic energy into potential energy, and when it comes back down the potential 'hill', it will turn this potential energy back into kinetic energy.
 
  • #19
jjustinn said:
Sorry, missed this earlier. This looks just like the normal Newtonian derivation, except with the delta function replaced with a gaussian...I don't quite follow what you mean by the delta function not helping -- since it's both exactly what we want (instantaneous localized force) and easier to integrate (Δp = ∫dp/dt dx = ∫-Vδ(x)dx = -V)...but for argument sake, how would you put your equations into Hamiltonian form? Maybe we can then take the limit and get the delta version?
You are integrating from one side of the step function, over to the other side. But this is not what you want. This is like saying how much momentum does the particle have when it has gone inside the hard sphere (or wall or whatever). But you don't want the particle to go inside the wall, you want the particle to be reflected by the wall. Therefore you should not be integrating from one side to the other. And I guess now you see the problem with the Dirac Delta function, is that since you are not integrating over it, it is not really useful to you. So if you use a gaussian instead, you can say that the particle gets close to x=0 but never actually reaches x=0, and the particle turns around, and goes in the opposite direction. So, in Hamiltonian form, the potential would be the error function. This looks like an s-shape. And we can take the limit of a very narrow Gaussian, then the error function will look almost like a step function, but it will be continuous. Or you can use some other function that has a roughly s-shape, and take the limit that the curve is very sharp. But the error function is the potential which would give a Gaussian as a force, i.e. an approximate Dirac-Delta function as the force, which is what you wanted.
 
  • #20
jjustinn said:
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)...
This reference is saying that you can use an 'approximately' step function which has height V0. And to get a proper reflection, V0 simply needs to be greater than the initial kinetic energy of the particle. They do not say that V0 has to be twice the initial kinetic energy for the collision to be elastic. The collision is elastic for any value of V0. What they were saying is that the particle will exactly reach the point x=0 and turn around if V0 is twice the initial kinetic energy. And so if V0 is slightly more than twice the kinetic energy, then the particle will will turn around before it gets to the point x=0.

It is the same for my 'error function' potential. If the kinetic energy is exactly half the height of the approximate step function, then the particle will turn around at exactly x=0. And If the kinetic energy is less than half of the height of the approximate step function, then the particle will turn around before x=0. But the collision will be still be elastic. And the fact that we take the limit that the approximate step function is very steep, i.e. almost vertical but not quite, this means that even though the particle turns around before x=0, it will turn around at some point very close to x=0. And as long as it turns around at approximately x=0, we don't really care that it actually turns around at some point which is very close to zero, but not actually zero.
 
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  • #21
BruceW said:
eh? what do you mean by 'interaction' ? In your case, the potential energy is a step function, and in the case of hard spheres, the potential energy is also a step function.

Sorry, I was mixing topics -- for "interaction" there I meant for two particles, rather than a particle and a wall (hence the delta function for point particles and a two-sided step function for spheres). It was off-topic, but I included it for context,

The only difference is that they use a step function with infinite height. This is a bit weird, I admit. So you could instead say that the step function has height at least as large as the kinetic energy of the approaching particle, so that the particle does not pass into the hard-sphere. This should give the same result.

Yes! That's what I thought, too (and what just about everyone says without actually showing how it works). But, when I do the math, I don't get conserved energy *or* momentum unless the "height" is *exactly* twice the momentum at the time of the collision.
I was convinced I had to be wrong, but then I found this: http://books.google.com/books?id=xV...=lagrangian "step function potential"&f=false -- and the author seems to have found the same thing...with the added bonus that he got the same result not with a delta/step function, but with a continuous one.
Think about a very high and very narrow Gaussian function as an 'approximate' Dirac Delta function. The particle will travel along, and will encounter this Gaussian potential. The particle will only keep moving as long as its initial kinetic energy is greater than the potential at the point where it is. In other words, it will never get to the point of infinite potential, or infinite force. It will turn around as soon as it has converted its kinetic energy into potential energy, and when it comes back down the potential 'hill', it will turn this potential energy back into kinetic energy.

Right -- and it works if you do a conservation of energy (or momentum) argument like that, but I just cannot find a hamiltonian that gives the correct results (even using the double momentum as the height doesn't work, because that then changes the ∂H/∂p = dx/dt, because it's no longer just the kinetic energy that contains the momentum, as was assumed to find that value.
 
  • #22
BruceW said:
You are integrating from one side of the step function, over to the other side. But this is not what you want. This is like saying how much momentum does the particle have when it has gone inside the hard sphere (or wall or whatever). But you don't want the particle to go inside the wall, you want the particle to be reflected by the wall.

You're right. I should have integrated over t surrounding the collision time...if x=0 at t=t0, then the integral would be any interval containing t0 -- eg t0-dt...t0+dt. The end result us still the same -- energy is only conserved for a specific momentum-dependent height (whether that height is distributed over a gaussian, triangle, atanh, or step)...the delta function just let's us use t0+/-dt, rather than needing to figure out where the potential starts/stops (unless I'm missing something)
 
  • #23
jjustinn said:
Yes! That's what I thought, too (and what just about everyone says without actually showing how it works). But, when I do the math, I don't get conserved energy *or* momentum unless the "height" is *exactly* twice the momentum at the time of the collision.
I was convinced I had to be wrong, but then I found this: http://books.google.com/books?id=xV...=lagrangian "step function potential"&f=false -- and the author seems to have found the same thing...with the added bonus that he got the same result not with a delta/step function, but with a continuous one.
no, that is not what the author is saying. The author is saying that energy is conserved for a potential with any height. please read it again.

jjustinn said:
Right -- and it works if you do a conservation of energy (or momentum) argument like that, but I just cannot find a hamiltonian that gives the correct results (even using the double momentum as the height doesn't work, because that then changes the ∂H/∂p = dx/dt, because it's no longer just the kinetic energy that contains the momentum, as was assumed to find that value.
The hamiltonian given by the author of the book you were looking at is a hamiltonian that conserves energy. And it conserves energy for any height. It does not conserve momentum because it is not supposed to conserve momentum. If instead you introduce another particle with mass, and say that the potential depends on the relative distance between the two particles, then momentum would also be conserved.
 
  • #24
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.
 
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  • #25
jjustinn said:
(2) note that the derivatives are equal/opposite
##dp_1/dt = -dp_2/dt = ∂U/∂x_1##
I think that should be ##dp_1/dt = -dp_2/dt = - ∂U/∂x_1##. I think this is just a mis-type. It looks like you have it the correct way in the rest of the derivation.

jjustinn said:
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##
where does this step come from? why does ##∫(dp_1/dt)dt## from ## t_0...t_2## equal K ?

p.s. your derivation is really nicely written out. Thanks. It makes it a lot easier to read when people do that :)
 
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  • #26
BruceW said:
I think that should be ##dp_1/dt = -dp_2/dt = - ∂U/∂x_1##. I think this is just a mis-type. It looks like you have it the correct way in the rest of the derivation.
Ah hell, you're right. I'll have to go over it all again to be sure...but I guess the good news is that as long as they have opposite signs, it should still come out right.

where does this step come from? why does ##∫(dp_1/dt)dt## from ## t_0...t_2## equal K ?
So that's just a restatement of the definition of the delta function...the integral over any domain containing a zero of the delta funxtion's argument is equal to its coefficient evaluated at that zero (and since K is taken to be constant, it's K everywhere).

That is probably the most hand-wave-y part of the derivation...but I think it's sound because x(t) reaches zero at t1 (by definition), and then doubles back, and the delta function is symmetric about that zero.

I think it might be on firmer footing if I did a change of variables and integrated over x...but I figured that was just another place for an error to slip in. I'll take another look tonight.

p.s. your derivation is really nicely written out. Thanks. It makes it a lot easier to read when people do that :)
You've been kind enough to devote your time to my ridiculous problem, so the least I can do is try to make reading it less painful. Thanks again.
 
  • #27
BruceW said:
where does this step come from? why does ##∫(dp_1/dt)dt## from ## t_0...t_2## equal K ?

p.s. your derivation is really nicely written out. Thanks. It makes it a lot easier to read when people do that :)

If I actually do try to change the variables, that introduces a p into it as follows:

##ΔP_i = ∫±Kδ(D - |x_1 - x_2|)ε(x_1 - x_2)dt##
But, ##dx_i/dt = ∂V/∂p_i##, which by the initial assumption, was simply ##p_i/m##
So, ## dt = dx_i/(p_i/m_i) ##

Thus,
##ΔP_i = ∫±Kδ(D - |x_1 - x_2|)ε(x_1 - x_2)m_i/p_i dx_i##

But now, the integral isn't so easy...because p does vary with x...I was initially tempted to assume since the integral is over an infinitesimal domain to assume it was constant, but by definition it's not -- we're calculating Δp, but it depends on p.

But going a step further, it's not clear that integrating over x even makes sense, because both the upper and lower bounds are by definition on the same side of the barrier.

I'm guessing I'm still doing something wrong even here...when you brought that item up, did you have any idea what you expected it to look like?
 
  • #28
yeah, since you must integrate over t (and not x), it is more complicated. And yes, even if you were integrating over x, if you want the particle to turn around at exactly the x=0 then you would still have problems, because you have a Dirac-Delta, so it is not really defined for when the integral neither goes over it, nor doesn't include it. (i.e. when the integral 'just' touches it, which is what happens in your case).

All your problems will go away if you use some kind of potential that is not infinitely steep. (like in the book you saw). If you have a potential that is not infinitely steep, then it is possible to calculate exactly what happens. And then you can take the limit that the potential is very steep but not infinitely steep, and you should get roughly the correct physics.
 
  • #29
BruceW said:
yeah, since you must integrate over t (and not x), it is more complicated. And yes, even if you were integrating over x, if you want the particle to turn around at exactly the x=0 then you would still have problems, because you have a Dirac-Delta, so it is not really defined for when the integral neither goes over it, nor doesn't include it. (i.e. when the integral 'just' touches it, which is what happens in your case).

All your problems will go away if you use some kind of potential that is not infinitely steep. (like in the book you saw). If you have a potential that is not infinitely steep, then it is possible to calculate exactly what happens. And then you can take the limit that the potential is very steep but not infinitely steep, and you should get roughly the correct physics.

OK -- you won me over ;)

HOWEVER, as I still suck at integrals, I'm having an identical problem integrating a finite (ramp) potential.

## V(x) = Kθ(-x)θ(x + 1)(x + 1) + Kθ(x)θ(1 - x)(1 - x) ##

The first pair of steps restricts the positive slope to (-1,0), while the second restricts the negative slope to (0, 1)...so you end up with a 45/45/90 triangle with total area 1 (multiplied by K). Then the derivative:

## ∂V/∂x = -dp/dt = Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x) ##

Or, K when -1 < x < 0, -K when 0 < x < 1 -- as expected (the delta functions cancel as they should).

So, our equations of motion are:

## ΔP = ∫(dp/dt)dt = ∫(Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x))dt ##

And, as usual, ## dx/dt = ∂H/∂p = p/m ##, so ## dt = m/p dx ##

Thus

## ΔP = ∫(dp/dt)dt = ∫(Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x))m/p dx ##

So, again, I'm stuck .. I cannot integrate over t to find p without an explicit form of x...and I cannot change variables to x without an explicit form of p(x).

Edit:
Or, if you go directly for x:

## dx/dt = p/m; d^2x/dt^2 = (dp/dt)/m ##
## d^2x/dt^2 = (Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x))/m ##
## dx/dt = 1/m∫ (Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x))dt ##

...which is exactly the same (except divided by m). Changing to x brings p back to the denominator (though the ms do nicely cancel). My only last hope is that somehow I can find p as a function of x...but that seems like exactly what Δp was supposed to be.
 
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  • #30
Hey, sorry it's been a while since I last replied. ah, yeah, so you're looking for an explicit equation for x as a function of time, right? You've got the equation ##1/2mv^2+V(x)=E## already. This can give you the magnitude of velocity at any position. But I'm guessing you want an explicit equation for x(t). OK, so you got to this equation:
jjustinn said:
## ∂V/∂x = -dp/dt = Kθ(-x)θ(x + 1) + Kθ(x)θ(1 - x) ##
There is a miss-type here, I think the second term should have a negative sign. But once that is fixed, you have your equation for x(t) right here! For a given range of x values, you have:
[tex]\frac{d^2x(t)}{dt^2} = \mathrm{constant}[/tex]
So for any motion within that range of x values, you simply have constant acceleration. And the motion of the particle will cause it to go into different regions, and have a different acceleration in each region. So yeah, it is a bit annoying to have to think about each region. But if you think about the different possible cases, for a given set of initial conditions, you can write down the entire function ##x(t)## for all ##t>0##. (The different possible initial conditions will fall into one of a few 'categories', which will each have a different equation for ##x(t)##). p.s. it will help to draw a few graphs of velocity against position. Also, it will help if you think about the kinetic energy of the particle.
 
  • #31
BruceW said:
you have your equation for x(t) right here! For a given range of x values, you have:
[tex]\frac{d^2x(t)}{dt^2} = \mathrm{constant}[/tex]
So for any motion within that range of x values, you simply have constant acceleration. And the motion of the particle will cause it to go into different regions, and have a different acceleration in each region. So yeah, it is a bit annoying to have to think about each region. But if you think about the different possible cases, for a given set of initial conditions, you can write down the entire function ##x(t)## for all ##t>0##. (The different possible initial conditions will fall into one of a few 'categories', which will each have a different equation for ##x(t)##). p.s. it will help to draw a few graphs of velocity against position. Also, it will help if you think about the kinetic energy of the particle.

So this seems like the same problem as before, except here the "constant" force is over a region instead of at a point...in both cases, you have an equation of the form ##d^2x/dt^2 = -F(x(t)) ##...and in this case, the "regions" are over the dependent variable.

I'll try a few graphs later on today and see if anything clicks, but it seems like it's going to be a ridiculously-complicated equation highly dependent on the initial conditions (as opposed to the "point" case, where the solution has the trivial closed form ##x(t) = |t|##)

Also, note that the velocity can only be given as a function of position up to a sign (in the case of a reflection it will pass previous points in the opposite direction) -- and even that is only because it's only one particle in 1D; if there was another particle or dimension, the total energy could be distributed among the extra degrees of freedom in an infinite number of ways...
 
  • #32
before, you had a force that was a Dirac-Delta in space... so yeah, I guess that force was not explicitly dependent on time either. It would be pretty unusual if you did have a force that was explicitly dependent on time. I think that would imply non-conservation of energy.

It's not that complicated to get an exact answer. But yes, it does depend on the initial conditions. And what do you mean the point case? you mean when the triangle is an isosceles triangle which is very tall and very thin? The equation would not be ##x(t) = |t|## (although, It is true for ##t \rightarrow \infty##)

well, if you specify the initial conditions, then you will know the sign of the velocity at any given time. Yes, the sign may change with time in some cases, but this is not a problem, since the velocity does change with time.

in 2D It is not that much more complicated. I'm imagining the potential as a right circular cone. (where the height corresponds to potential and the x and y coordinate correspond to position). Yes, it is a bit more annoying to work out the function ##(x(t),y(t))## but it is possible as long as you specify the initial conditions. There is no possible ambiguity about which direction the particle will move in. Also, in 3D, it is still OK. Just more difficult to visualise the potential, since you'd need to imagine a 4D cone.
 
  • #33
The "point case" was where the force was a delta function instead of a step function...the only point I was trying to make was that it seems the "simpler" finite triangle barrier (with step function force ) seems to have a much more complicated solution when compared to the step function barrier (with delta function force)...but before pre-judging any further, I'm going to take your advice and at least try to get a qualitative solution for the triangle case.
 

1. What is the Classical Square Well potential?

The Classical Square Well potential is a mathematical model used to describe the interaction between particles in a system. It is characterized by a deep, narrow potential well that resembles a square shape.

2. What is the Hamiltonian form of the Classical Square Well potential?

The Hamiltonian form of the Classical Square Well potential is a mathematical expression that describes the total energy of a system in terms of the positions and momenta of the particles within the system. It is commonly used in classical mechanics to study the dynamics of particles.

3. How is the Classical Square Well potential used to model elastic collisions?

The Classical Square Well potential can be used to model elastic collisions by assuming that the particles in the system have no internal structure and interact only through the potential well. This allows for the conservation of energy and momentum during collisions, resulting in elastic behavior.

4. What are the limitations of using the Classical Square Well potential?

One limitation of using the Classical Square Well potential is that it assumes particles have no internal structure and interact only through the potential well. This may not accurately reflect the behavior of real particles, which can have complex internal structures and interact through various forces.

Additionally, the Classical Square Well potential does not account for quantum effects, making it less accurate when studying systems at very small scales.

5. How does the depth and width of the potential well affect the behavior of particles in a system?

The depth and width of the potential well can greatly affect the behavior of particles in a system. A deeper well will result in stronger interactions between particles, while a wider well will allow for more freedom of movement. These factors can impact the stability and dynamics of the system, as well as the likelihood of particles undergoing elastic collisions.

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