2,031 reputation
1929
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 4 months
seen 19 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).


May
20
revised The Faddeev-Popov Lagrangian
Typo in an index.
May
20
comment The Faddeev-Popov Lagrangian
In the Scholarpedia article, I am looking at the part that begins with "One more improvement was introduced by 't Hooft . . .". It almost seems as if the term I'm wondering about was inserted by hand by allowing a more general gauge fixing condition. With this more general condition, the relevant delta function contributes a nonzero term to the Lagrangian. If I understand this correctly, that's all well and good, but why the need for the more general condition? Is $\partial _\mu A^{\mu k}$ not sufficient? Does this not just complicate things further by introducing an extra term?
May
20
asked The Faddeev-Popov Lagrangian
May
9
answered The meaning of scale invariance in power law distribution
May
8
revised The meaning of scale invariance in power law distribution
Fixed spelling, grammar, and wording to increase clarity of the post.
May
8
suggested suggested edit on The meaning of scale invariance in power law distribution
May
8
awarded  Citizen Patrol
May
4
awarded  Popular Question
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
awarded  Yearling
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
3
revised QED BRST Symmetry
added 2 characters in body
May
3
revised QED BRST Symmetry
added 1 characters in body
May
3
asked QED BRST Symmetry
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.
May
2
answered The discontinuity of Electric Field
Jan
29
accepted How does Newtonian gravitation conflict with special relativity?
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.
Nov
22
answered Angular Momentum Operator