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fqq
fqq
12:39 AM
@NikeDattani I know the formalism but have not heard of specific packages for it.
 
10 hours later…
raf
raf
10:38 AM
If the constraint equation is expressed with inequality ( like < or <=), is it considered non-holonomic?
like for particles in closed spherical shell?
 
2 hours later…
12:58 PM
Yes, that's a classic example of a non-holonomic constraint iirc
 
2 hours later…
2:35 PM
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Q: Please reopen my question about a beam ODE

euler132 I have a question about my Physics Stack Exchange post: ODE of a beam on an elastic foundation Please reopen my edited question. This was not a homework-type question! I have presented a solution from a book, which I do not understand. Thank you!

3:32 PM
I don't really get the point of the "candidate score" thing for elections. Seems like it's trying to be a tl;dr
 
1 hour later…
4:42 PM
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Q: Please reopen my question about a physical limit

InductiveloadI asked this question about what it is (if anything) that provides and upper limit to the speed of a turbine in a fluid stream: Maximum rotation rate of a turbine in a fluid with a given speed It was closed as "engineering". The linked meta post says: Are any engineering questions allowed? Quest...

I think rigid bodies allow for time asymmetric potentials :/ Is this a known fact?
5:05 PM
No reason why you can't just throw a delta potential in to a non-relativistic Lagrangian I don't think
@MoreAnonymous Skimming your post it looks like you set a constant $\alpha$ equal to a function, but ignored differentiating $\alpha$ in the EL equations
$\alpha$ is a constant
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Q: Reopen my question on physics

Carlos WerbockI'm sorry for being asking non clarified questions. That won't be repeated. Is it ok to remove the ban?

The last equality in your post doesn't look like a constant, and if it is just the value of that function at a point then it's not the equations you were trying to get in the first place
5:19 PM
@bolbteppa The classic slap
Interaction that lasts only a moment in time
5:38 PM
Right that's a delta function centered at $t_{unfortunate}$
5:56 PM
@bolbteppa If you solve it you get the collision you wanted. I agree this is not a time symmetric potential though
Note if you go backward in time $ \alpha \neq \frac{m_1 m_2}{m_1 + m_2} \frac{(\vec r_1 - \vec r_2) \cdot (\vec v_1 ' - \vec v_2 ') }{|\vec r_1 - \vec r_2|^2}$
the post collision velocities are $\vec v_1'$ and $\vec v_2'$
6:19 PM
You derived $m \ddot{x}_1 = - 2 \alpha(x_1 - x_2)\delta(t - t_c)$. This is not $\frac{d \vec p_1}{dt} = - \frac{2 m_1 m_2}{m_1 + m_2}\frac{(\vec r_1 - \vec r_2) \cdot(\vec v_1 - \vec v_2)}{|\vec r_1 - \vec r_2|^2} (\vec r_1 - \vec r_2)\delta(t-t_c)$ and you can't just set $\alpha$ to be equal to the missing part $\alpha = \frac{m_1 m_2}{m_1 + m_2} \frac{(\vec r_1 - \vec r_2) \cdot (\vec v_1 - \vec v_2) }{|\vec r_1 - \vec r_2|^2}$ because its a function but you treated it like a constant in $L$
6:42 PM
@bolbteppa So I think of things physically .. .Is there any experiment you can do distinguish between both cases? Let $\tilde \alpha = \frac{2 m_1 m_2}{m_1 + m_2}\frac{(\vec r_1 - \vec r_2) \cdot(\vec v_1 - \vec v_2)}{|\vec r_1 - \vec r_2|^2}(t) $ and $ \alpha = \tilde \alpha(t_c) $ ?
Technically $\delta(t - t_c)$ is zero everywhere except at $t_c$ where it's infinite so not sure what it means physically, usually there's an integral somewhere letting us avoid these issues but here I don't see how you can invoke one. Further I'm not sure what you're saying but it sounds like what you're saying could mean we can replace $\alpha$ by anything not just one specific form
7:08 PM
@bolbteppa I think the force experienced in an ideal collision is infinite (this is well known) ...$\Delta p / \Delta t$ ... Change in momentum is finite while delta t goes to $0$ ... And yes you could write any $\alpha$ but then you wouldn't be modelling a collision
 
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9:47 PM
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Q: What would determine applications of established science to new systems, to which they clearly apply?

Anders GustafsonI read in an answer to a question I read that a defense for if a question is accused of being non mainstream is applications of established science to new systems where they clearly apply. One naive interpretation I can think of is that if it's possible to use the mathematics of a model in mainst...

fqq
fqq
10:37 PM
@MoreAnonymous collisions are when the particles touch, not a fixed time regardless of particle position
as someone said in the comments, you probably want to look at this en.wikipedia.org/wiki/Hard_spheres

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