2,096 reputation
11031
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 5 months
seen Oct 21 at 19:10

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
22
accepted Radiative Corrections and Bremsstrahlung
May
22
revised Radiative Corrections and Bremsstrahlung
Typo
May
22
asked Radiative Corrections and Bremsstrahlung
May
21
accepted QED BRST Symmetry
May
20
answered Coulomb's Law: why is $k = \dfrac{1}{4\pi\epsilon_0}$
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.