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Student.


Jan
25
reviewed Reviewed Multivariable Limits
Jan
25
comment Multivariable Limits
Would you mind writing in TeX format? please?
Jan
23
revised Is length adimensional when space is not flat?
added 209 characters in body
Jan
23
revised Describing the algebra of functions on $S^2$
added 39 characters in body
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
23
revised Describing the algebra of functions on $S^2$
added 39 characters in body
Jan
23
answered Describing the algebra of functions on $S^2$
Jan
23
revised Is length adimensional when space is not flat?
added 1 characters in body
Jan
23
awarded  Fanatic
Jan
22
revised Is length adimensional when space is not flat?
added 15 characters in body
Jan
22
revised Is length adimensional when space is not flat?
added 234 characters in body
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
22
revised Is length adimensional when space is not flat?
added 234 characters in body
Jan
22
revised Is length adimensional when space is not flat?
added 140 characters in body
Jan
22
revised Is length adimensional when space is not flat?
added 277 characters in body
Jan
22
answered Is length adimensional when space is not flat?
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
revised two identical point charges can't collide
added $n=2$ case.
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.