Reputation
940
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
4 13
Newest
 Yearling
Impact
~15k people reached

Mar
8
comment Topological Quantum Field theories
In, e.g., 1+1 dimensional TQFT, dynamics is done by specifying the number of loops you begin with at time $t=0$, and how many you have at $t=1$, as well as the topology of the world-sheet for $0<t<1$. BUT the partition function (controlling dynamics) then becomes a function of the topological invariants (which invariants depends on the TQFT). This is good for, e.g., BF-theory since computing topological invariants is simpler than, say, solving the Wheeler-DeWitt equation :)
Mar
8
answered Topological Quantum Field theories
Mar
8
comment Topological Quantum Field theories
@SanathDevalapurkar, time reparametrization invariance forces the Hamiltonian to be a constraint; a great review of this can be found in Henneaux and Teitelboim's Quantization of Gauge Systems, viz. chapter 4.
Mar
3
awarded  Revival
Mar
3
answered Superspace as the Hilbert Space for Quantum Gravity
Mar
3
comment Superspace as the Hilbert Space for Quantum Gravity
Well, no, you never do that in -- say -- using geometric quantization for the canonical formalism for electromagnetism (or, a simpler example, a scalar field). You don't do it when geometrically quantizing a mechanical system either. The symplectic manifold $\mathcal{M}$ is a necessary ingredient for constructing the Hilbert space (or Fock space for field theories), but not sufficient to qualify as the Hilbert space (resp. Fock space) for the full quantum theory. NB: you can use phase space path integrals, but this is an irrelevant fun fact rather than pertinent information.
Mar
3
comment Superspace as the Hilbert Space for Quantum Gravity
You might want to read Woodhouse's Geometric Quantization. You don't turn $\mathcal{Riem}$ into a Hilbert space: you use it (well, technically $\mathcal{Riem}/\mathcal{Diff}$) as the configuration space, then construct an infinite-dimensional symplectic manifold $\mathcal{M}$ which you use as the underlying manifold for a complex line bundle. The space of sections on this line bundle then in (insert magical full quantization step here) and you've got a Fock space.
Feb
20
revised Identity $\int_{-\infty}^{\infty}\frac{e^{uz}}{1+e^u} \mathrm{d}u=\frac{\pi}{\sin(\pi z)}$
Modified title, updated some TeX
Feb
20
suggested approved edit on Identity $\int_{-\infty}^{\infty}\frac{e^{uz}}{1+e^u} \mathrm{d}u=\frac{\pi}{\sin(\pi z)}$
Feb
20
comment How to find a closed form for the derivatives of $F(x)=\frac1x\int_0^x\frac{1-\cos t}{t^2}\,dt,$ $F(0)=\frac12$?
Hint: consider $g(x)=xF(x)$ and Differentiate under the Integral Sign
Feb
20
comment Simple proof exercise recommendation, with full answers
And play with the first equation. The $\sqrt{ab}$ strongly hints at squaring both sides of the equation, just to see what happens...
Feb
20
comment Simple proof exercise recommendation, with full answers
"It's 2am, and I've spent more time than I really want to think about on an exercise which was clearly not meant to cause someone this much distress." You might be pushing too hard. Instead of forcing yourself to think about the problem, try taking a walk. I've noticed it help considerably, and many mathematicians recommend it (e.g., Alain Connes).
Feb
14
comment Symbolic math engines barf on this ostensibly tractable integral.
The trick is to expand this out using $2\cos(x)=\exp(ix)+\exp(-ix)$, consolidate the terms as $\int\exp(Ct)\,\mathrm{d}t$ then execute it.
Feb
13
comment integral of 1/(sqrt(e^x)) from 0 to infinity(Improper integral)
For 3, consider $\int (1+e^{ix})e^{-x}dx$ since $\exp(ix)=\cos(x)+i\sin(x)$ we take the imaginary part of the resulting integral...
Feb
12
comment Symbolic math engines barf on this ostensibly tractable integral.
What is the variable you're integrating? (There's no $dx$ factor in the integrand)
Feb
12
comment Vector Delta Function Identity
These are all simple roots, too, so none of them would have a multiplicity greater than 1...
Feb
12
comment Applications of Operator Algebras to modern physics
@Jeff Since you're at Caltech (I gather), go find Matilde Marcolli and speak with her. She's an expert on Noncommutative geometry, which basically amounts to applying operator algebras to...everything...
Feb
4
comment Is computer science a branch of mathematics?
@CarlMummert: true, but I figured everyone else would give rigorous books related to the formal field "the theory of computation", like the ones you noted. So, I wanted to just note there's a programming book that uses the axiomatic method :)
Feb
2
comment How must I understand concepts equations of physics?
It might be wise to study (a) differential geometry, and (b) partial differential equations. Differential geometry for the obvious reasons. But (b) because classical field theory, for the most part, boils down to PDEs [or, depending on your outlook, some "abstract nonsense" involving sections of fiber bundles ;)].
Jan
29
comment If 1 $\leq x$, then $\sqrt{x} \leq x $
Just a style thing, I would begin the proof writing "Proof. Assume for contradiction $x<\sqrt{x}$. Square both sides of $x\geq1$ gives us $x^{2}\geq x$. But squaring both sides of our assumption gives us $x^{2}<x$ which is a contradiction. We reject our assumption, and conclude $\sqrt{x}\leq x$." Just to make crystal clear which assumption we are trying to contradict. But I'm a windbag, so...