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Mar
6
comment What is the justification for continuing to model space-time as a manifold in string theory?
@RobinEkman Yup. Agreed.
Mar
6
comment What is the justification for continuing to model space-time as a manifold in string theory?
@RobinEkman Well, I suppose I am being sloppy here. If by "observer" we mean a world-line in space-time (the idea being that our observer is traveling along this world-line), then what I really meant is "observer together with a choice of smooth orthonormal frame along the world-line, with the single time-like vector in each basis being the tangent to the world-line".
Mar
6
comment What is the justification for continuing to model space-time as a manifold in string theory?
@ACuriousMind Well I suppose it doesn't need a priori justification, but in this case the "only [reason] that matters" (a posteriori justification via experiment) doesn't exist, and so I think a priori justification is more important here than it might otherwise be.
Mar
6
comment What is the justification for continuing to model space-time as a manifold in string theory?
@RobinEkman I don't disagree with what what you say (regarding coordinate-independent theories), but I also don't understand how this is in contradiction with that I previously stated. The laws of physics must be coordinate-independent, but in order for an experimentalist to verify the laws, they must first pick coordinates. The statement that the laws are coordinate-independent then means that it doesn't matter what choice of coordinates the experimentalist happens to make: his physical conclusions should be the same regardless.
Feb
21
comment Density matrix: error with diagonalization claim and fixing it
@NorbertSchuch Thanks again. I've updated the answer with the slicker proof.
Feb
21
comment Density matrix: error with diagonalization claim and fixing it
Actually, this conversation you guys have had made me realize I made quite a serious error. When I wrote down the example I alluded to in which $\sum _k\lambda _k\neq 1$, I forgot to normalize the original $\left| \psi ^{(k)}\right>$. If you do this, it seems that indeed you will have $\sum _k\lambda _k=\operatorname{tr}(\rho )=1$. I will update the answer accordingly. Thanks!
May
15
comment How do I find constraints on the Nambu-Goto Action?
Honestly, there is probably a much more clever way to do this, but I personally don't think it is a wise use of time to try to find a better solution to something like this.
May
15
comment How do I find constraints on the Nambu-Goto Action?
Then you can just have Mathematica row-reduce this matrix so that you can just read-off the multiplicity of $0$ as an eigen-value. If you do this for several different values of $p$, you should see a pattern. This of course gives you the answer. To prove it, you would have to turn this calculation into a proof; I am quite confident that I never actually bothered to do that.
May
15
comment How do I find constraints on the Nambu-Goto Action?
Write out the formula for $\det (g)$ using the co-factor expansion along the first column (or row) to obtain an explicit formula for $L$. For concrete-ness, take the space-time dimension to be, say, $D=4$. This way, you will be able to write down an explicit formula for the matrix $\frac{\partial ^2L}{\partial (\partial _tX^\mu )\partial (\partial _tX^\nu )}$ (of course if you want a general proof you will need to do it for arbitrary $D$). The point of doing this is that now we will have a very explicit $4\times 4$ matrix that you can just plug-in to Mathematica (or some other CAS) . . .
May
15
comment How do I find constraints on the Nambu-Goto Action?
Actually, since it's been a little over two years now since I actually did the computation, I admittedly don't remember exactly what I did, and I can't say for sure unless I just re-do everything. I can, however, tell you the first thing I would try . . .
Nov
30
comment Fast and slow modes, and the vanishing of certain diagrams during re-normalization
"Within a perturbative RG scheme, by which I mean the expansion parameter remaining small all the way to the fixed point and at the fixed point, engineering scaling dimension is enough to prove the irrelevance of the $\phi ^6$ vertex." ---- How do I actually go about proving this statement? With a proof of this, I believe my question will be completely answered. (Also, am I correct in presuming that this is not specific to $\phi ^6$, but is also true for general $\phi ^{2n}$ with $n\geq 3$?)
Nov
30
comment Fast and slow modes, and the vanishing of certain diagrams during re-normalization
In particular, am I correct in saying that we can ignore the diagram I am worried about, not because it vanishes, but because its contribution is irrelevant? On another note, how does one relate this to "momentum conservation"? Is the meaning of this just that one cannot produce vertices with an odd number of slow modes?
Nov
30
comment Fast and slow modes, and the vanishing of certain diagrams during re-normalization
How do we know a priori that vertices of the form $\phi ^{2n}$ with $n\geq 3$ will be irrelevant. Once again, don't we actually need to complete the re-normalization procedure to verify this precisely? Sure, one can do dimensional analysis and obtain that the 'engineering degree' suggests that they will be irrelevant, but to actually prove this expectation is correct, do we not have to do the full re-normalization procedure?
Nov
30
comment Fast and slow modes, and the vanishing of certain diagrams during re-normalization
"Clearly, $E$ cannot be odd since $V$ and $I$ are positive integers." ---- Sure, but what about, e.g., $E=6$. For example, consider a diagram with two $\phi _{\text{s}}^3\phi _{\text{f}}$ vertices with the two $\phi _{\text{f}}$ legs contracted. This yields a diagram with $6$ external slow legs, no? In fact, it is this specific diagram which I have been struggling to show vanishes for the past couple of days now.
Nov
30
comment Fast and slow modes, and the vanishing of certain diagrams during re-normalization
"However, $\phi _s^3$ vertex is not generated because there is no process in a $\phi ^4$ theory that can produce a $\phi ^3$ vertex." ---- I feel as if this argument is circular . . . Part of what we would like to show is that, after a RG transformation, we do not generate any new terms in the action (at least to the order we are working). It is conceivable that we could generate a $\phi ^3$ term after re-normalization, in which case the theory obviously could produce a $\phi ^3$ vertex. Am I seriously mis-understanding something?
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
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
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.