2,426 reputation
11332
bio website berkeley.academia.edu/…
location Berkeley, CA
age 25
visits member for 3 years, 8 months
seen Dec 23 at 23:12

Currently a graduate student in mathematics at the University of California - Berkeley.

Previously obtained a MASt. in Applied Mathematics from the University of Cambridge (2013), and a B.S. in Mathematics and a B.A. in Physics from the University of Chicago (2012).


May
3
comment Could a people do all sort of gymnastics movement in vacuum space?
I don't understand your question. Could you try clarifying?
May
3
comment QED BRST Symmetry
With respect to (2), does this not imply that the Lagrangian is not invariant under the BRST transformation? Why would he ask me to show this if this is not in fact the case?
May
2
comment The discontinuity of Electric Field
"Why there exist the tangencial component?" -- Well, why not? Unless I know that the tangential component is 0, I can't just assume it is, even though that wouldn't affect this result. I am pretty sure Griffiths does show that the tangential component has to be continuous, i.e., in the notation above, $\mathbf{E}_{\text{tangential,above}}=\mathbf{E}_{\text{\tangential,below}}$. Thus, if the surface is a conductor (which by definition means the electric field inside it must be $0$), we do require that the tangential component is $0$. This is not necessarily true in general however.
May
2
comment What areas of physics should a mathematician study to understand TQFT?
As a mathematics student currently studying QFT, I can say that, if indeed any understanding of QFT at all is required for TQFT, you will need to know both special relativity and quantum mechanics pretty well.
Nov
22
comment Angular Momentum Operator
@atomicpedals ". . . why would I want to use $L^3$ instead of $L$;" First of all, be careful when you write $L^3$. $\mathbf{L}$ is a vector, so if you write $L^3$, I'm going to think you mean $(L_x^2+L_y^2+L_z^2)^{3/2}$ (because it doesn't make sense to cube a vector, although it does make sense to cube the magnitude of a vector), which is not your meaning. As for why would you work with $L_i^3$ over $L_i$: you wouldn't. I'm pretty sure this exercise is just for the sake of practice.
Oct
28
comment Is the Lorentz group compact (and if not, is U(1)?)
@RonMaimon There are only two connected one-dimensional manifolds: $S^1$ and $\mathbb{R}$. Both of them happen to be Lie Groups. $S^1$ has many names: $S^1$, $U(1)$, $\mathbb{R}P^1$, $\mathbb{T}^1$, $SO(2)$, $Spin(2)$, and the list probably goes on. All of these have different definitions, which happen to coincide for these small cases. To the best of my knowledge, there is really only the one name for $\mathbb{R}$. Also, there aren't "two versions" of $U(1)$. $U(1)=S^1$, period. The "other version" you're thinking of is just $\mathbb{R}$. It is not a "version" of $U(1)$.
Oct
26
comment Does $E$ really equal $mc^2$?
This should be closed. It's a double-post, and not a particularly high-quality question anyways.
Oct
26
comment Is the Lorentz group compact (and if not, is U(1)?)
I completely agree with BebopButUnsteady. Every physicist should know enough point-set topology to understand at least basic manifold theory.
Oct
25
comment How must you spin the ball to make it alternate between 2 positions?
SHM=Simple Harmonic Motion
Oct
24
comment What is the definition of a 2-cocycle in Quantum Mechanics
en.wikipedia.org/wiki/… In particular, "The set of isomorphism classes of central extensions of G by A is in one-to-one correspondence with the cohomology group H2(G,A), where the action of G on A is trivial."
Oct
23
comment Defining a Riemannian manifold - made easy?
@RonMaimon Still though, even if you could avoid this chicken-and-egg problem, in the end, ZFC is not powerful enough to talk about something like category theory, e.g., the notion of "the category of groups" (to the best of my knowledge) is not a set in the sense of ZFC, and is hence just an intuitive idea. Thus, to the best of my knowledge, if I want to work with categories, I have to make use of the "naive" notion of a set anyways.
Oct
23
comment Defining a Riemannian manifold - made easy?
"The mathematical stuff is very simple--- do not be intimidated! The picture is there, the symbols are just a baroque but necessary language to communicate the picture."+"The learning curve is only superficially steep--- after you solve even the simplest problems in GR, it melts away."--I couldn't agree more.
Oct
23
comment Defining a Riemannian manifold - made easy?
@RonMaimon I eventually decided you can't get something from nothing. You have to start somewhere. As a result, I wound up, more or less, accepting "naive ZFC" as given. If you're interested in how I came up with this "philosophy of mathematics" in more detail, you can read chicago.academia.edu/JonathanGleason/Papers/759541/…. I initially wrote it just for myself, so I apologize if it's not quite as "polished" as it should be. In any case, I certainly do not have the belief that this is the only correct way to think about things.
Oct
23
comment Defining a Riemannian manifold - made easy?
@RonMaimon Perhaps you misunderstood what I meant. I don't mean to question what you said regarding the mathematics, just your interpretation of it. For example, choice implies Banach-Tarski. This is fact. That this implication is philosophically dubious is an opinion. In any case, I have decided to "reject" any form of axiomatic set theory because of the philosophical "chicken and egg problem". That is to say, an axiomatic formulation of set theory requires, at least, first-order logic. However, you require at least the naive notion of a set in first-order logic.
Oct
23
comment Defining a Riemannian manifold - made easy?
@Peter4075 "Probably my ideal textbook would contain rigorous maths and pretty cartoons!"--I definitely don't disagree with this. IMHO, to obtain the best understanding, you need to understand both the rigorous formulation and what it all means intuitively. You use the intuition to solve problems and the rigor to make it precise. Both go hand in hand :).
Oct
23
comment Defining a Riemannian manifold - made easy?
@RonMaimon Fine. As long as we can agree that this (the philosophical dubiousness (is this a word?)) is just your opinion, and not mathematical fact.
Oct
22
comment Defining a Riemannian manifold - made easy?
@RonMainmon Abraham and Marsden, in their Foundations of Mechanics require that the atlas be maximal. This is, by the way, a book on physics. You need maximality for, among others things, the charts in the maximal atlas to form a basis for the topology. "A maximal atlas is the set . . . transition maps."--This is just not true. What you describe is not even an atlas. "It is philosophically dubious . . . for any result."--The real numbers are of size continuum. Does that make them philosophically dubious?
Oct
21
comment Defining a Riemannian manifold - made easy?
I should mention that I don't think Ron's answer was a bad one. Just not my cup of tea.
Oct
21
comment Defining a Riemannian manifold - made easy?
Just my two cents, but I think the comparison of the two answers below, my own and Ron's, is the perfect example of how precise mathematical language can aid in understanding. In what took me 9 lines to define, using "naive" language, it took Ron 6 steps to define, one of which was just as long as my entire definition! Learning the math is not an easy task, but once learned, it will make things so much easier. By the way, I wouldn't be so intimidated by thigns like "Hausdorff", "second-countable", and "maximal". It sounds intimidating, but once you learn what they mean, it's not so bad:)
Oct
21
comment Defining a Riemannian manifold - made easy?
@RonMaimon Well, a manifold requires an atlas, and some basic results of manifold theory require the atlas to be maximal. It doesn't matter in the end whether in the definition you require your atlas to be maximal or not, because every atlas uniquely determines a maximal atlas. That being said, if this weren't the case, you would need to require the atlas to be maximal. And it certainly is not "philosophically dubious".