1,341 reputation
621
bio website
location
age
visits member for 3 years, 1 month
seen Jul 28 at 3:14

Jul
16
awarded  Nice Answer
Jul
2
awarded  Curious
Jun
15
revised Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
added 886 characters in body
Jun
14
awarded  Yearling
Jun
11
comment How close does light have to be, to orbit a perfect sphere the size and mass of Earth?
I agree with Chris White --- it would be nice to mention why light is being out of scope Newton's gravity theory, that is to specify the limitations of classical theory and when in general we do need to resort to GRT.
Jun
9
revised Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
minor formatting
Jun
9
accepted Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Jun
9
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
If you require only $U_\text{max}$ being greater or equal than the initial kinetic energy, then $0=\frac{m}{2} \nu^2(0_+) + [U_\text{max} − \frac{m}{2} \nu^2(0_+)]$ may not have solutions at all with $\nu(0_+) \in \mathbb R$. It will have a solution $\nu(0_+) = 0$ only if $U_\text{max}$ equals to a kinetic energy of an incident particle, but that makes it dependent on the particle momentum.
Jun
9
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
@auxsvr the problem is I do not see how they really solve the problem --- whenever I try to rigorously conduct the derivation I got stuck. And it seems to me that others hold the same opinion. I have commented on your answer on certain things I do not understand. Once again what is being sought --- Hamiltonian equations which can be solved as is, without resorting to any supplementary physical arguments. You see, Hamiltonian equations should be self contained, fully describing the system evolution, right?
Jun
9
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Oh, I got it, but still it would be nicer to be written. Moving along the derivation, as far as I understand you show that momentum when $x > 0$ is $0$ but it does not justify that is should reverse. Andy why is the energy is $1$ as you said? It should be $p^2/(2m)$ in any case. Moreover, with the potential fixed being equal $1$ does not prohibit the particle to enter the region if its kinetic energy is greater then $1$.
Jun
9
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Could you please elaborate on how you have integrated $\int_{0_-}^{0^+} \dot \nu \, dx$ ? This is not so obvious.
Jun
8
answered Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Apr
5
awarded  Benefactor
Apr
2
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
oops, I have awfully miscalculated the time. If somebody cares, the correct reasoning should be following: The distance traveled is $\varepsilon E_0 = v_0 (\tau / 2) - 1/2 \, \varepsilon (\tau/2)^2 $. If we divide this expression by $\varepsilon$, will get the answer in the form $\tau = \varepsilon f(E,v_0)$ which is linear in $\varepsilon$.
Apr
1
comment Hamiltonian function for classical hard-sphere elastic collision
Milton, have a look at my related question physics.stackexchange.com/questions/105318 As for you derivation, I bet your calculation of $\Delta P_1$ is wrong, you should conduct the integration more accurately with greater level of details.
Mar
31
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
If noone comes with an elegant framework which would avoid potential regularisation in the remaining bounty time, I will accept Qmechanic's answer.
Mar
31
comment Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Yes, this regularized potential works as a charm. It reverses the initial velocity and the penetration depth in the realm $x > 0$ is $\varepsilon E$, while the time spent here is $1/2 \, \pi \sqrt{\varepsilon m}$. Both these expressions go to zero with $\varepsilon \to 0^+$. I have one question though, how did you come up with this potential? Of course it is possible and probable to deduce it, but I bet you just had the right analogy from you experience. Is it so?
Mar
29
awarded  Promoter
Mar
27
asked Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Mar
13
awarded  Popular Question