892 reputation
115
bio website joshphysics.com
location Los Angeles
age 28
visits member for 1 year, 11 months
seen Dec 19 at 23:04

New project: phermi.com

Let me know if you know of any hard physics problems with clever solutions. (email listed to the left)

Personal website: joshphysics.com

Currently a lecturer at the UCLA Department of Physics and Astronomy.

Ph.D. theoretical high energy physics, UCLA.

BA/BS in physics/math, UC Berkeley.


Mar
28
comment What extra assumption makes this transformation affine?
@BISHD Just a map.
Mar
28
asked What extra assumption makes this transformation affine?
Mar
8
comment Inverse function with Dirac Delta
@TZakrevskiy Could you clarify what "completely describes" means? What if you were to take $\{\delta_x\}_{x\in\mathbb R}$?
Mar
7
revised Inverse function with Dirac Delta
added 77 characters in body
Feb
24
comment $g''(t_0)$ where $g(t) = f(t, 1-t) , t_0 = 0$
@user129120 Your guess is as good as mine.
Feb
24
answered $g''(t_0)$ where $g(t) = f(t, 1-t) , t_0 = 0$
Feb
17
comment Inverse function with Dirac Delta
@Garrett See math.ucsd.edu/~wgarner/math4c/textbook/chapter2/…, for example. Notice also that saying "the" function $f(x) = x^2$ is misleading. One needs to both specify the domain of the function, and what it does to each element of the domain. If we define $f$ as the function whose domain is $[0,\infty)$ and for which $f(x) = x^2$ for all $x\in [0,\infty)$, then $f$ is invertible. But if we extend its domain to $(-\infty, \infty)$, then it is not invertible.
Feb
8
accepted Cochains: terminology
Feb
8
comment Cochains: terminology
@TedShifrin I most certainly will, especially since you're the second person who's suggested that to me, the first being a math grad student. I as a theoretical physics grad just learning this stuff, I really appreciate your help.
Feb
8
comment Cochains: terminology
@OlivierBégassat Thanks again.
Feb
8
comment Cochains: terminology
Thanks very much for this Olivier. Would you mind pointing me to a reference that discusses this stuff; I would very much appreciate it.
Feb
8
comment Cochains: terminology
@TedShifrin As a matter of terminology, would you happen to know if elements of $\mathrm{Hom}(M,\mathbb R)$ are ever referred to as cochains when $M$ is the $\mathbb Z$-module of chains? The $k$-form example you gave is precisely what I had in mind when I wrote this question; such objects arise naturally in physics (especially thermodynamics when modeling heat). By the way, I'm a fan of "multivariable mathematics," so I was, I must admit, somewhat star-struck when I saw your answer.
Feb
8
revised Cochains: terminology
edited body
Feb
8
comment Disprove a linear mapping
@SergioParreiras See the earlier version. For some reason, the OP removed the definition of $g$.
Feb
8
comment Disprove a linear mapping
@Slavica Yep. Sure thing.
Feb
8
comment Disprove a linear mapping
@Slavica What do you mean by a "general" disproof? This demonstrates that the statement "$g$ is linear" is false, because it is not the case that $g(\alpha x) = \alpha g(x)$ for all $\alpha$ and $x$. It's not as though "falsness" has multiple degrees.
Feb
8
answered Disprove a linear mapping
Feb
8
asked Cochains: terminology
Feb
7
comment Translations in two dimensions - Group theory
@user35952 Unfortunately I think not.
Feb
7
comment Translations in two dimensions - Group theory
@user35952 Yes it's a circle too. It's a circle with the same radius but whose center is at $(-a,-b)$. We could only have said that the equation for the circle was preserved in form (at least if we're using standard terminology) if there were some $R$, say, for which the transformed equation were $x^2+y^2 = R^2$; that equation has the same form. So, for example, a scaling $x\mapsto \alpha x, y\mapsto\alpha y$ would preserve the form of the equation for a circle at the origin because the new equation would be $x^2 + y^2 = (r/\alpha)^2$. Are thinking of a different notion of "preserve."