2,496 reputation
11334
bio website berkeley.academia.edu/…
location Berkeley, CA
age 25
visits member for 3 years, 11 months
seen Mar 28 at 17:33

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).


Sep
3
accepted Divergence of One and Two Graviton Exchanges
Sep
3
comment Divergence of One and Two Graviton Exchanges
One last question, and then I think I got it. How do we determine the kinetic term for $h_{\mu \nu}$? (I ask because this will allow me to determine the appropriate dimensions of a spin $2$ field, and hence the appropriate dimensions of the coupling constant.)
Sep
3
answered A rock connected to one end of string in circular motion gets released.. and what happens?
Sep
2
comment Divergence of One and Two Graviton Exchanges
Are you assuming that the particle is a scalar, so that the interaction is of the form $\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}$? This is the only way I could see you getting momenta coming into what appears to be your vertex factors. But if my dimensional analysis is correct, $[\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}]=L^{-4}$, so that the corresponding coupling constant would be dimension-less, and in particular, could not be proportional to $1/M_P$. What's going on?
Sep
2
revised Divergence of One and Two Graviton Exchanges
deleted 1 characters in body
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
. . . In practice, you do this by writing the Lagrangian in terms of $\phi :=\phi _0-v$, where $v$ is some minimum of the potential and $\phi _0$ is the original field. My question could then be equivalently phrased as "Why does this guarantee that $\langle 0|\phi (x)|0\rangle =0$?".
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
@MichaelBrown Indeed, I was under the impression that you had to have this as a re-normalization condition in order to apply LSZ (that is, a hypothesis require for the LSZ Reduction Formula to hold was that $\langle 0|\phi (x)|0\rangle =0$). In fact, I thought this was the entire idea behind the symmetry breaking: you must re-write your Lagrangian in terms of the re-normalized field (with vanishing VEV), and if the bare field had a non-vanishing VEV, this will 'break' the symmetry . . .
Sep
1
comment Vacuum Expectation Value and the Minima of the Potential
So then what is the proof of this leading-order approximation?
Sep
1
asked Divergence of One and Two Graviton Exchanges
Aug
31
asked Vacuum Expectation Value and the Minima of the Potential
Aug
31
revised The vacuum in quantum field theories: what is it?
Expanded the question
Aug
31
revised The vacuum in quantum field theories: what is it?
Expanded the question
Aug
31
asked The vacuum in quantum field theories: what is it?
Aug
29
comment Fine-Tuning, the Hiearchy Problem, and Mass in the Standard Model
@DavidZ You mean the next paragraph in my question? No, that was written by me, included for the purposes of clarifying just what I mean by mass (and what I mean by "my understanding of mass") in this context (as opposed to, for example, just some coefficient in the Lagrangian).
Aug
29
asked Fine-Tuning, the Hiearchy Problem, and Mass in the Standard Model
Aug
21
awarded  Popular Question
Aug
6
revised How do I find constraints on the Nambu-Goto Action?
added 85 characters in body
Jun
8
awarded  Popular Question
May
27
accepted Noether charge of local symmetries
May
27
revised Noether charge of local symmetries
added 14 characters in body