2,411 reputation
11332
bio website berkeley.academia.edu/…
location Berkeley, CA
age 25
visits member for 3 years, 7 months
seen Dec 19 at 4:29

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).


Nov
12
comment Momentum conservation in an electromagnetic system?
So in particular, it shouldn't be balanced. They should differ by $f\delta t$. Also, I admittedly don't understand why you think that the momentum of the E&M field won't change unless the top charge accelerates to a non-negligible degree. The Liénard–Wiechert depend on the velocity, which, for the bottom particle, is changing to a non-negligible degree (though I guess it's possible that you might find this dependence 'cancels' when you do the actual computation).
Nov
12
comment Momentum conservation in an electromagnetic system?
. . . Then the momentum change of the E&M field would just have to be so that, after you add these three together, you get $f\Delta t$.
Nov
12
comment Momentum conservation in an electromagnetic system?
So I am admittedly trying to avoid details because I believe the actual computation will be at least slightly tedious. Nevertheless, I think it's easy to see that this is possible. The system has three things contributing to its momentum: the mechanical momentum of the top particle, the mechanical momentum of the bottom particle, and the momentum of the E&M field. If we declare that the positive direction is to the right, then the momentum of the bottom particle is increasing over time and the momentum of the top particle is decreasing over time . . .
Nov
12
comment Momentum conservation in an electromagnetic system?
@JohnEastmond That's right. The momentum of your entire system should change (in magnitude) by $f\Delta t$ over a time period of $\Delta t$.
Nov
11
comment Why do we have to use an integral in this scenario to figure out $v_{max}$?
@user42141 The kinetic energy does vary . . . $v$ depends on $r$.
Nov
11
comment Why do we have to use an integral in this scenario to figure out $v_{max}$?
@NeuroFuzzy Thanks for catching that.
Nov
10
comment Derivation of the irreducible representations of SO(3)
The same principle applies to both $SO(3)$ and $SU(2)$. All this essentially comes down to the trivial fact that, if $x=y$, then we must have that $f(x)=f(y)$. In $SO(3)$, the identity element and the element that rotates by $\pi$ are the same element. Hence, they must do the same thing in a representation. Obviously, this principle is still true for $SU(2)$; however, in this case, we have distinct $g,h\in SU(2)$ such that $\rho (g)=\rho (h)=1\in SO(3)$ ($\rho :SU(2)\rightarrow SO(3)$ is the map I mentioned in my answer).
Nov
9
comment Derivation of the irreducible representations of SO(3)
Also, it cannot be related to $L_i$ at all. $L_i$ is an element of the Lie algebra, and we know that $\mathfrak{su}(2)=\mathfrak{so}(3)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
This is exactly what I am suggesting. This has nothing to do with spin vs. orbital. If somebody has told you that spin in $SO(3)$ can change a sign after a rotation by $2\pi$, they were wrong. This can only happen for spin with $SU(2)$. The point is is that, in quantum mechanics, we don't really care about representations, but rather projective representations (basically representations up to phase). It turns out that projective representations of $SO(3)$ are in canonical one-to-one correspondence with actual representations of $SU(2)$.
Nov
9
comment Derivation of the irreducible representations of SO(3)
Suppose that the representation has spin $n/2$ with $n\in \mathbb{Z}^+$. Then, in particular, this representation will contain an element $v$ such that $S_zv=\frac{n}{2}v$. This is at the level of the Lie algebra. At the level of the Lie group, $R_z(\theta )=\exp (-i\theta S_z)$ is a rotation about the $z$-axis by an angle $\theta$. Hence, $R_z(\theta )v=\exp (-i\theta \frac{n}{2})$. For $\theta =2\pi$, we had better have that $R_z(2\pi )=R_z(0)$; however, using the formula above, $R_z(2\pi )=\exp (-i\pi)=-1$. Thus, this does not give us a representation of $SO(3)$ for half-integer spin.
Nov
9
comment Derivation of the irreducible representations of SO(3)
The spherical harmonics give you all the irreducible representations of $SO(3)$, but it does not give you all those of $SU(2)$. You'll note this does not make any direct reference to the Lie algebra of $SO(3)$
Feb
26
comment Unitary representations of the diffeomorphism group in curved spacetime
If this is the case, then I feel we should apply the Faddeev-Popov trick to the Einstein Hilbert action so as to find the corresponding BRST charge associated with this diffeomorphism symmetry, and perform the usual cohomological construction to find the space of states. Has this not been done before? What is the result?
Nov
19
comment Amplitudes in renormalized perturbation theory
@Neuneck Yes, but I thought the point is that the total result should be proportional to $\lambda ^2$, not just $\lambda$.
Nov
19
comment Amplitudes in renormalized perturbation theory
@Neuneck I don't understand. The Feynman rule corresponding to the $4$-point counterterm vertex is just $-\mathrm{i}\, \delta _\lambda$, not $-\mathrm{i}\, \delta _\lambda \lambda$ or anything like this, so this entire counterterm diagram should just be of order $\lambda$. What am I missing?
Nov
16
comment Amplitudes in renormalized perturbation theory
This was more or less what I was expecting, but when I went to check this myself, I didn't see how $\delta _\lambda$ was of order $\lambda ^2$. Indeed, according to Peskin and Schroeder's Eq. (10.17) on pg. 324, $\delta _\lambda =\lambda _0Z^2-\lambda$, which is just of order $\lambda$ . . . or am I missing something?
Oct
5
comment Is the Lorentz group compact (and if not, is U(1)?)
I also think it is mis-leading to only say that Lorentz boosts are parametrized by $v\in (-c,c)$. They are, but you mustn't forget that the group law is not simply addition in $\mathbb{R}$; instead, it is given by the usual velocity addition formula in special relativity. It turns out that this group law makes $(-c,c)$ into a Lie group which is isomorphic (as Lie groups) to $(\mathbb{R},+)$ via $\mathrm{arctanh}$.
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 ($p=0$)?