2,501 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).


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
12
answered Momentum conservation in an electromagnetic system?
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
11
revised Why do we have to use an integral in this scenario to figure out $v_{max}$?
added 71 characters in body
Nov
11
answered Why do we have to use an integral in this scenario to figure out $v_{max}$?
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
10
awarded  Favorite Question
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
revised Derivation of the irreducible representations of SO(3)
added 19 characters in body
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 $SO(3)$
Nov
9
answered Integral over scalar product of eigenfunction of momentum operator and harmonic oscillator one
Nov
9
answered Calculation mistake some place in finding stress-energy tensor
Nov
9
revised Calculation mistake some place in finding stress-energy tensor
added 10 characters in body
Nov
9
answered Derivation of the irreducible representations of SO(3)