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11232
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 7 months
seen 20 mins ago

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).


21h
comment How is it that Quantum entanglement does not let you transmit infomation?
Also, this: wikiwand.com/en/Quantum_decoherence
21h
comment How is it that Quantum entanglement does not let you transmit infomation?
Also, I don't think collapse is truly instantaneous. As a matter of fact, I don't think the notion that it be instantaneous even really makes sense. How would one show, experimentally, that this is instantaneous? You can measure again as fast as you can immediately after your first measurement, but there will always be a non-zero amount of time between these measurements. One would need to show that the correlation between the two measurements increased as we decreased the time between the measurements . . . but Heisenberg says that this cannot happen for the uncertainty will diverge!
21h
comment How is it that Quantum entanglement does not let you transmit infomation?
I wouldn't really say this is in agreement with relativity. More like it doesn't disagree with relativity. In particular, there is no hard speed limit here. All this says is that you can't transmit information instantaneously, but this argument doesn't show that you can't send it faster than the speed of light.
2d
comment All geodesics are inextendable?
Well, then I would ask what they did in the comments: what do you mean by in-extendable? My definition would imply that a geodesic defined on all of $\mathbb{R}$ is automatically in-extendable.
2d
comment Do bad clocks measure proper time?
I'm also not sure if I understand your question regarding how to measure the metric. I am not an experimentalist, so I could only be able to give a "in principle" answer to this. I am also un-familiar with this five-point curvature detector.
2d
comment Do bad clocks measure proper time?
Well, quite honestly, I had never even heard the term "bad clock" until I read this question. I think it's safe to say that, unless otherwise stated, if someone says "clock" they mean "good clock". With this assumption, then all three quotes are correct and they would not retract their claims.
2d
comment Do bad clocks measure proper time?
In particular, in general, no single observer can measure the entire space-time metric everywhere. You can only measure the space-time metric locally.
2d
comment Do bad clocks measure proper time?
I'm afraid I will not be able to explain how one can measure the metric in an understandable way without being able to draw a diagram, but the basic idea is one that comes up in a derivation of length-contraction/time-dilation. Briefly, you send a light beam to a point you want to determine the distance to (after you've placed a perfect mirror there), and by measuring the time (proper time, according to you) it takes to get back to you, you can determine the distance to that point. It's not quite this simple in curved space-time, but that's the basic idea.
2d
comment Do bad clocks measure proper time?
By definition, bad clocks do not measure proper time. The argument I gave should show that this is equivalent to 'making the world-lines of free particles locally look curved'.
Nov
22
comment Work-Energy theorem vs conservation of mechanical energy?
Sure, but this is not the integral of $-kx$, it is the integral of $-kx(s)$ with respect to s.
Nov
21
comment Work-Energy theorem vs conservation of mechanical energy?
How do you get that result for the integral? It should not be the case that $x(s)=s$. Remember, the spring itself is moving as well, so it should be the case that $s\geq x(s)$.
Nov
14
comment A question from Schwinger's particles, sources and fields monograph
I am confused. In some places you write $\phi ^\mu$ and in others you write simply $\phi$. Is there a scalar field $\phi$ and a vector field $\phi ^\mu$?
Nov
14
comment Calculating the Sun's emitted power in a wavelength range?
See wikiwand.com/en/Stefan%E2%80%93Boltzmann_law#/… . I believe this calculation is very similar to what you would like to do.
Nov
14
comment Which renormalisation techniques are available for 3+1 QED?
You should probably distinguish between re-normalization and regularization. Pauli-Villars is a method of regularization. Two other ones that come to mind are dimensional regularization and the usual cut-off regularization.
Nov
12
comment Momentum conservation in an electromagnetic system?
Well, I honestly don't know why you would need to do this. Perhaps I misunderstand your question. I thought that you took note of the fact that momentum was not conserved, and was confused as to why that was not the case. And the reason of course is because we are applying a non-zero net force to the system. As a 'sanity check', I suppose you could actually calculate the change in momentum of those three things to verify that it does indeed come out to $f\Delta t$, but given that it's so tedious to do so, I don't know why you would want to do that.
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 $r$.