1,991 reputation
1929
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 3 months
seen Jul 22 at 22:45

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
23
comment Quantization of Nambu–Goto action in multiples of Planck's constant?
@LubošMotl When you say ". . . the (theory defined by) the Nambu-Goto action may be quantized and and what one gets is known as string theory . . .", do you mean, in particular, that not only is the is the Nambu-Goto action equivalent to the Polyakov action at the classical level, but it is in fact equivalent to the Polyakov action even at the quantum level? Is this unique to the string ($p=1$), or is this result true for the corresponding actions for $p$-branes as well? In particular, is it true for point-particles ($p=0$)?
Sep
23
awarded  Popular Question
Sep
14
asked Renormalizability of the Polyakov Action
Sep
12
accepted Derivation of the Polyakov Action
Sep
12
revised Derivation of the Polyakov Action
deleted 56 characters in body
Sep
12
comment Derivation of the Polyakov Action
. . . Of course, I guess you might just say that the simplest one that works is the way to go (in this case, an action with just one extra field). Still, I have to admit, I'm not completely satisfied with this answer.
Sep
12
comment Derivation of the Polyakov Action
. . . And hell, if we're going to add in yet another field, we could probably find get another symmetry along with it.
Sep
12
comment Derivation of the Polyakov Action
I like this, but this method doesn't really convince me that putting in another auxiliary field is the way to go. Let's say in QFT we want a theory with a complex scalar field with a $U(1)$ symmetry. Of course, we could always introduce other fields into the theory, but that's not what one usually does unless you wanted the extra fields to begin with. And even if you decide that introducing a new field is the way to go, why stop at one? Surely we could add two new fields that respected all the symmetries we wanted . . .
Sep
12
revised Derivation of the Polyakov Action
added 78 characters in body
Sep
12
asked Derivation of the Polyakov Action
Sep
6
revised Fine-Tuning, the Hiearchy Problem, and Mass in the Standard Model
edited body
Sep
4
asked Symmetry Breaking and Vacuum Expectation Values
Sep
3
revised A rock connected to one end of string in circular motion gets released.. and what happens?
added 156 characters in body
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
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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 . . .