Greg P
Reputation
793
Top tag
Next privilege 1,000 Rep.
Create new tags
 Feb 19 comment If I Blow smoke out my window, will any come inside my room? @lemon, careful when you say that air molecules bounce back and forth between the walls ~100 times per second. While it is true that the typical speed of a molecule is ~1km/s, the mean free path is very tiny, of the order of tens of nanometers! In other words, a molecule doesn't keep going in the same direction for very long! The (self) diffusion coefficient is of order cm^2/s. In one second, you can expect a molecule to diffuse something like a centimeter. I agree with you that some smoke will always get in the room, but mostly this is because of flows of air, not just diffusion. Jan 11 awarded Nice Answer Feb 3 revised Hilbert's sixth problem (current answers neglect the fact that $C_{U} \subseteq U$ ) deleted 3 characters in body Feb 3 answered Hilbert's sixth problem (current answers neglect the fact that $C_{U} \subseteq U$ ) Sep 24 awarded Autobiographer Sep 14 awarded Nice Answer Nov 10 awarded Yearling Nov 10 awarded Yearling Jun 7 comment why does perpendicular motion to the direction of someone' s approach does not affect the distance between them Suppose you are a mile directly north of me. At exactly the same time, I take one step towards you (north) and you take one step west. By how much did the distance between us decrease? Basically by one step. Not quite one step, actually (work it out, Pythagorean theorem). But in the limit of taking very small steps (like the bugs that are continually tracking each other) we would be getting closer by one step's distance. Similarly, the rate of decrease of the inter-bug distance is equal to the bugs' speed. Apr 10 comment Is fire plasma? A similar question is the basis for a recent project in science communication. Check it out: flamechallenge.org Apr 10 comment Why is $L^2$ norm of the gradient called kinetic energy? In quantum mechanics, the gradient operator represents momentum (to within a constant factor). That is why they would call the square of the gradient the kinetic energy (momentum squared, to within a constant factor). That is quite general and not confined to any particular system. But yes, the vector potential is added to deal with the electromagnetic field. Mar 20 comment Hamilton's equations in terms of initial conditions Yes, I see what you mean now. The give-away of course is that the equation is not correct as printed in the paper (see pcr's example). Mar 19 comment Hamilton's equations in terms of initial conditions The paper explicitly defines $z = {\bf Z(0)}$, and both $z$ and ${\bf Z}$ are used elsewhere in the paper, so I'm still not sure. Mar 19 comment Hamilton's equations in terms of initial conditions No, you didn't miss it ;-) I added it after your answer to clarify the question. Mar 19 revised Hamilton's equations in terms of initial conditions added 71 characters in body Mar 19 comment Hamilton's equations in terms of initial conditions In the paper, $z$ represents the initial values, that is, $z = {\bf Z(0)}$. I still do not see how to get these equations involving $z = {\bf Z(0)}$ from the usual Hamilton's equations which involve positions and momenta evaluated at the same time $t$ as on the left hand side. Mar 19 revised Hamilton's equations in terms of initial conditions added 261 characters in body Mar 19 comment Hamilton's equations in terms of initial conditions This is the standard vector version of Hamilton's equations. How are these related to the ones in the paper (and the question) which include derivatives with respect to the initial values of p and q? That was my question. Mar 19 comment Hamilton's equations in terms of initial conditions Thanks, but my question is specifically about the issue of evaluating the derivative with respect to the initial conditions. In other words, how do I get the "usual" Hamilton's equations starting with the ones in the question. Mar 19 revised Hamilton's equations in terms of initial conditions added 283 characters in body