2,520 reputation
11535
bio website berkeley.academia.edu/…
location Berkeley, CA
age 25
visits member for 4 years
seen May 21 at 16:58

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).


Oct
5
comment Is the Lorentz group compact (and if not, is U(1)?)
In some sense periodic identification is absolutely crucial. There are precisely two connected $1$-dimensional Lie groups: $S^1$ and $\mathbb{R}$. These two are distinguished precisely by 'periodic identification', and furthermore, $S^1$ is compact while $\mathbb{R}$ is not. Even though $[0,c]$ or $[-c,c]$ might be compact, they are not Lie groups.
Sep
27
comment Renormalizability of the Polyakov Action
@user10001 Thanks much. Do you want to write up what you said in a comment as an answer? At this point, I could write it up myself, but I feel as if it would make more sense to give you the credit for it.
Sep
23
comment Renormalizability of the Polyakov Action
@user10001 This works (in fact, with just a bit more detail, I would accept this as an answer), but there is at least one thing that worries me about this. Presumably by $1$ in $G=1+aX+(aX)^2+\cdots$, you mean $\eta _{\mu \nu}$. If this is the case, you find that the kinetic term is given by $-\frac{1}{2}\partial _\alpha Y^\mu \partial ^\alpha Y^\nu \eta _{\mu \nu}$, which means that the field $Y^0$ has the wrong sign for its kinetic term.
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
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
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
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
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?
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).
Mar
29
comment Confusion between Electric field and Magnetic field of a charged particle.
@A4KASH Just think of a particle moving throughout space. In my reference frame, the particle might be at rest, but in your reference frame, you might observe the particle to be moving with velocity $\mathbf{v}$, because you are moving with respect to me with velocity $-\mathbf{v}$. Thus, even though we measure different numbers, we are still observing the same physical phenomenon. A similar thing happens with the electromagnetic field: you might measure different numbers than I do, but only because our perspectives are different: we are still observing the same physical phenomenon.
Mar
15
comment The interpretation of mass in quantum field theories
For what it's worth, I mixed up sign conventions in my previous comment. It should be that $p^2=-m^2$, with the convention used in the Lagrangian.
Mar
15
comment The interpretation of mass in quantum field theories
If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one. Does that make sense?
Mar
15
comment The interpretation of mass in quantum field theories
Ironically, the word "precise" here is not meant to be precise; it is open to interpretation. The word that really matters here is "operational": to define mass via some sort of (thought) experiment. In section 2.2 of academia.edu/829613/… , I give a classical definition in the spirit of which I am looking, except now, I want to do this in a relativistic setting. I never thought of this until just now, but maybe the idea behind the classical definition could just be modified?
Mar
15
comment The interpretation of mass in quantum field theories
We can define mass this way, and I already know how to relate this definition to the term appearing in the Lagrangian. The question is, how do we relate this mathematical definition of mass to a precise, physical, operational definition of mass. Does that make sense?
Mar
15
comment The interpretation of mass in quantum field theories
If you go this route, then it's really a question about special relativity, not quantum field theory, but I was thinking there were other, deeper reasons that actually require the framework of QFT to understand.
Mar
15
comment The interpretation of mass in quantum field theories
Perhaps you're referring to the fact that $P|p\rangle =p|p\rangle$ and $p^2=m^2$? ($P$ is the element of the Lorentz algebra in the particle's representation and $p$ is a $4$-vector). That's fine, but this just reduces the question to: if a particle has 4-momentum $p$ such that $p^2=m^2$, why do we interpret $m$ as the mass of the particle?