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11031
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 5 months
seen 13 hours ago

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).


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".
Oct
21
comment Defining a Riemannian manifold - made easy?
@Peter4075 Well, since you're doing this for fun, which is great, instead of making a living at this stuff, I won't stress the "learning it the right way" approach. I still do feel it would give you a deeper insight into the physics. I would say your description of manifold is not bad as far as intuition goes. As for Question 2, you can make up any metric you want, but, depending on what you do, it might not be a metric. For it to be called a metric, it has to be symmetric, nondegenerate, and smooth. This essentially means that your matrix has to be symmetric and invertible.
Oct
21
comment How does Newtonian gravitation conflict with special relativity?
@genneth But isn't this based on the assumption of the equivalence of gravitation mass and inertial mass? It should modify their inertial mass, but if these two concepts are in fact distinct, then there is no contradiction. How much evidence is there supporting their equivalence? Are there any compelling theoretical reasons for their equivalence? In my view, their equivalence, if it is true, is the single most compelling reason for gravitation and electromagnetism to not be formulated on the same footing.
Oct
21
comment Non-unitarity of wave function collapse
Here's the computation: $\langle \delta ^2|f\rangle =\langle \widehat{\check{\delta}\ast \check{\delta}}|f\rangle =\langle \check{\delta}\ast \check{\delta}|\hat{f}\rangle =\int \check{\delta}(x)\int \check{\delta}(y)\hat{f}(x+y)dydx=\int \check{\delta}(x)\int \delta(y)\hat{f}\left( x+\check{y}\right) dydx=\int \check{\delta}(x)\hat{g}(x,\check{0})=\hat{g}(\check{0},\check{0})=g(0,0)=f(0)$, where we have defined $g(x,y)=f(x+y)$ so that the notation was more understandable.
Oct
21
comment Non-unitarity of wave function collapse
By the way, I think this is the perfect example of how mathematical rigor can add to the techniques of theoretical physics instead of hidering them, as it seems some physicists believe. There is certainly a difference between being rigorous and being pedantic. Rigor helps. Pedantry doesn't.
Oct
21
comment Non-unitarity of wave function collapse
@LuboŇ°Motl You argument does not make mathematical sense because you are treating $\delta$ as if it were a function. As I said before, the correct way to define the product of two tempered distributions is as the Fourier Transform of the convolution of the Inverse Fourier Transforms. With this definition, for $f$ Schwartz, $\langle \delta ^2|f\rangle =f(0)$, so that $\delta ^2=\delta$.
Oct
21
comment How does Newtonian gravitation conflict with special relativity?
Let $\mathbf{G}$ be the gravitational field, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitational Constant, and let $\rho$ be the mass density. Then, the equivalent of Gauss's Law would read $\nabla \cdot \mathbf{G}=\frac{\rho}{\gamma _0}$. You say that this does not hold. Why? Is there experimental evidence that contradicts this?
Oct
20
comment Confusion between Electric field and Magnetic field of a charged particle.
@claws I believe that the discussion in Grffith's Electrodynamics text that begins on pg. 522 should be what you're looking for.
Oct
20
comment Confusion between Electric field and Magnetic field of a charged particle.
"If the arrangement of the charge inside the conductor makes the electric field zero everywhere inside, then it follows they contribute a zero electric field everywhere outside . . ." I don't think this is true. Consider an infinite cylindrical shell with nonzero surface charge density $\sigma$. By symmetry and Gauss's Law, the electric field is zero inside, but nonzero outside.
Oct
20
comment Non-unitarity of wave function collapse
@Lubos This is actually not correct. You can show that $\delta ^2=\delta$, so that "$\sqrt{\delta}=\delta$". I put this equation it quotes because I don't know anything about uniqueness of squaure-roots of nonnegative distributions. The point is, there is at least some square-root of $\delta$, which is itself $\delta$.
Oct
19
comment Confusion between Electric field and Magnetic field of a charged particle.
@claws This is a great question. You are hitting upon something that eventually led to the idea that there is no electric field, nor is there the magnetic field, but only the electromagnetic field. That is to say, whether you observe an electric field, a magnetic field, or a "mixture" of the two is dependent on your frame of reference, i.e., the electric and magnetic fields are really just two sides of the same coin. Regardless of whether you see the electric field or the magnetic doing the job, i.e. regardles of your frame of reference, you will still observe the same physical results.
Oct
19
comment Non-unitarity of wave function collapse
@lurscher This is kind of off topic, but for what it's worth, you can define the square of the delta function. In fact, you can define the product of any two tempered distributions as the Fourier transform of their convolution (which can be defined directly). I have a feeling that you can also make precise sense out of the square-root of a nonnegative tempered distribution, but I'd have to think about how to do that.