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bio website berkeley.academia.edu/…
location Berkeley, CA
age 25
visits member for 4 years, 4 months
seen 9 hours 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).


Nov
26
comment How is it that Quantum entanglement does not let you transmit infomation?
Also, this: wikiwand.com/en/Quantum_decoherence
Nov
26
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!
Nov
26
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.
Nov
26
answered How is it that Quantum entanglement does not let you transmit infomation?
Nov
25
reviewed Approve Total and partial derivatives in thermodynamics and Maxwell relations
Nov
25
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.
Nov
25
answered All geodesics are inextendable?
Nov
25
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.
Nov
25
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.
Nov
24
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.
Nov
24
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.
Nov
24
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
24
answered Do bad clocks measure proper time?
Nov
24
revised Normalization constant of the Vacuum polarization
added 2 characters in body
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
awarded  Notable Question
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
21
answered Work-Energy theorem vs conservation of mechanical energy?
Nov
21
reviewed Approve Is it possible to flow current in open circuit?
Nov
20
awarded  Custodian