452 reputation
1415
bio website none
location
age
visits member for 2 years, 1 month
seen Jun 24 at 21:30

Student.


Apr
16
comment Is this already an equation/law that has been found?
I've only seen that attribution in "Die Vermessung der Welt".
Apr
1
comment Formal Variational Calculus Reference Request
Arnold's book "Mathematical Methods of Classical Mechanics" has something to say about that. You might be also interested in Bleecker's "Gauge theory and variational principles"
Jan
27
comment Multivariable Limits
Well, it's replacing, say, log(a^z) by $\log(a^z)$, etc. You can edit your own answer correcting that. Take a look at tobi.oetiker.ch/lshort/lshort.pdf
Jan
26
comment Multivariable Limits
It'S not that hard :). What you wrote here is mainly only missing a pair of $ enclosing the math.
Jan
25
comment Multivariable Limits
Would you mind writing in TeX format? please?
Jan
23
comment Describing the algebra of functions on $S^2$
indeed, sorry :), well, it's negative. Have a look at M. Paschkes' Ph.D. thesis., page 43. Browse "paschke" here wwwthep.physik.uni-mainz.de/Publications/theses
Jan
22
comment Is length adimensional when space is not flat?
Thanks 5PM, my apolgies, I should have written the "(squared) length element $dr^2+r^2dθ^2$ has the same units as $dx^2+dy^2$".
Jan
14
comment two identical point charges can't collide
I agree with Haskell's claim. To clarify, I've been talking only about two inertial frames: $S$, w.r.t which the particles have position $\mathbf q_i$, and $S'$, defined by having the origin at $Q:=(\sum_i m_i \mathbf q_i)/(\sum_i m_i)$. If you write down Newton's second law for $Q$, it's equation of motion is $\ddot{Q}=F_{12}+F_{21}\equiv 0$ (by third law), whence $Q(t)=Vt$ for some constant vector $V$. If one of them is inertial, both are.
Jan
14
comment two identical point charges can't collide
@uncookedfalcon Yes, but if you don't like that frame, you can replace $\mathbf{r}$ everywhere by $\mathbf q_1-\mathbf q_2$. The same analysis caries over to this "more general" frame. Regarding your question, both frames are inertial (because setting the force to zero, you get a particle that follows a stright line with constant velocity as the solution to the differential equation). Does that answer your question about frames?
Jan
13
comment two identical point charges can't collide
@uncookedfalcon if you wish to analyze the problem in arbitrary dimension, the field is no longer inverse-square dependent.
Jan
13
comment two identical point charges can't collide
If your law is to have physical meaning, it must be $n=3$ and in Newton+Coulomb law it's not the norm squared, but to the third power.
Jan
12
comment Metric tensor of complex numbers & Hamiltonian Mechanics
I've found it. It's "Introduction to Mechanics and Symmetry" by Marsden and Ratiu, Exercise 2.1-1.
Jan
11
comment Metric tensor of complex numbers & Hamiltonian Mechanics
the $2$ factor comes from the definition of the partial with respect to $\bar{z}$. // These equations are $N$ complex equations, so $2N$ real eqs. I think I saw that once on one of Marsden's books. Let me search...
Jan
11
comment Metric tensor of complex numbers & Hamiltonian Mechanics
Ah, I see. But curvature has double derivatives of the metric tensor. In this case $\eta$ is the metric tensor on the nose :) en.wikipedia.org/wiki/Riemann_curvature_tensor
Jan
11
comment Metric tensor of complex numbers & Hamiltonian Mechanics
@AimForClarity I'm not sure I understand why you write "curvature" for $\eta$. I do not understand that part of the question.
Jan
5
comment Earth-Sun distance equation
@kEoz meta.stackexchange.com/questions/5234/…
Jan
1
comment 8 periodicity: Clifford clock- Bott periodicity - KO-dimension in noncommutative geometries
Very rich answer indeed! I appreciate specially the reference to Landsman's notes, which I didn't knew.
Jan
1
comment Trouble simplifying a tough equation
$1=HappY(Ne^\omega)^{-1}Ye^{ar}$ to be more accurate.
Dec
24
comment Identifying a surface $\rho^2\cos(2\phi)-1=0$
@Rakisbro I updated...
Dec
24
comment Identifying a surface $\rho^2\cos(2\phi)-1=0$
you mean, what you wrote multiplied by $\rho$ on the right-hand side, right? or is $\rho=\sqrt{x^2+y^2}$ ? (my confusion arises 'cause one usually reserves $\rho$ for cylindric coordinates.)