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bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 2 years, 11 months
seen Apr 11 at 20:14

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
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
Nov
9
revised Why Negative Energy States are Bad
edited title
Nov
9
asked Why Negative Energy States are Bad
Nov
1
asked The Energy-Momentum Tensor and the Ward Identity
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
26
answered Is the Lorentz group compact (and if not, is U(1)?)