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May
27
comment Cyclist's electrical tingling under power lines
Right, thanks, +1, exactly, those 30 kV per meter which is huge even if one only gets a small remnant of it is something I am thinking about. Surprising that not too many people get killed in various situations under the power lines...
May
27
comment Cyclist's electrical tingling under power lines
Oh, I see, so this could also depend on my having body hair? ;-)
May
27
comment Cyclist's electrical tingling under power lines
It's very interesting but from my basic school years, I became convinced that the effect of static electricity on hair is only relevant if the hair is dry etc. This sensation on the bike only occurs near the buttocks and groin area, I don't have so much hair to rely upon, and they're wet because I kind of sweat, anyway. So I don't believe the static electricity is really too relevant here.
May
27
comment Cyclist's electrical tingling under power lines
Dear Anna, another video on the antenna - stealing electricity: youtube.com/watch?v=BxzSZ78cM-4&feature=related - it's pretty cool, thinking about ways to do it as business. ;-)
May
27
comment Cyclist's electrical tingling under power lines
Wow, the light bulbs are amazing! And people are telling me about microamperes. I will probably have to downvote that answer. ;-) Is the light bulb stuff real? Isn't it a hoax?
May
27
comment Cyclist's electrical tingling under power lines
Yes, Anna, it appears when I am crossing but I suspect that if I were riding in parallel on the right place, it could be the same effect. And maybe not. Maybe there's some current running around the bike and the polarizations matter. ... I should make an experiment, like stopping at that point. But it has happened to me about 5 times in my life - although it's pretty safely guaranteed and regular with that bike - and it's unpleasant enough a feeling that I just don't want to repeat it again! But maybe i will do the sacrifice at some point haha.
May
27
comment Cyclist's electrical tingling under power lines
Dear anna, something like what you say has to be right. Can one estimate it? What is the voltage that may be in the antenna? What is the electric field in the electromagnetic wave? When one multiplies it by one meter, one has to get the voltage that may be attached to the body. I am pretty sure that Brian's estimate is smaller by many, many orders of magnitude.
May
27
comment Group Cohomology and Topological Field Theories
Yup, removing the ambiguity of $S$ is equivalent to finding $K$, more precisely finding $K$ mod $n$. But in the full quantum theory, $S$ only matters mod $1$ (in normal normalizations of physics, $2\pi$), because it appears in $\exp(2\pi i S)$ in the path integral only.
May
26
answered Particle mixing and indistinguishability
May
26
comment Introduction to string theory
See this list of textbooks on string theory: motls.blogspot.com/2006/11/string-theory-textbooks.html - A recent explosion of books on strings etc.: motls.blogspot.com/2012/04/…
May
26
revised Cyclist's electrical tingling under power lines
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May
26
revised Cyclist's electrical tingling under power lines
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May
26
revised Cyclist's electrical tingling under power lines
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May
26
revised Cyclist's electrical tingling under power lines
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May
26
asked Cyclist's electrical tingling under power lines
May
26
comment How to prove Wick's Theorem (Zee's eq. I.2 (16)) via Gaussian integration?
I agree with Raskolnikov. Your "proof" that it's zero seems to be based on taking an "example" with one $x$ insertion only. You got zero - which is right (and you get zero from the right, Zee's formula as well) - but you apparently extrapolated (by "math induction"?) that the result is zero for many $x$'s, too. But it's not. Why don't you try to insert at least 2 - and then 4 - $x$'s in the product? And don't forget to differentiate the factor $\sum A^{-1} J$ with respect to $J$!
May
25
comment Understanding units and the units of the derivative operator
For translations, you may have translations by 1 meter or 1 inch and there is no way to say that one of them is more universal or important than others. They're just translations by different lengths or different units of lengths. Which translations are viewed by someone as "more canonical" depends on his subjectively preferred units (for scaling, it's not the case: there are no units). In infinite space, the generator of translations is (just like the distance) dimensionful so there is no way how you could uniquely identify it with the generator of scalings.
May
25
comment Understanding units and the units of the derivative operator
Scalings are related to a multiplication by a dimensionless number such as $q=\exp(s)$ while translations are translations by a dimensionful quantity, the distance. So there is a huge difference between the units (and ability to use dimensional analysis) related to scaling and to translations. You seem to dislike this basic fact. But this basic fact is the very reason why we use units and dimensional analysis in the first place. In scaling, there are preferred objective scalings everyone agrees are universally important, like 1-times or $e$-times, but for translations, it's not the case.
May
25
comment Understanding units and the units of the derivative operator
If there is any natural reason to write a mathematical object in physics as $\exp(s)$ where $s$ is something, $s$ simply has to be dimensionless because of the dimensional analysis. That's what the dimensional analysis is all about. So if something gets scaled by $q=\exp(s)$, then both $q$ and $s$ are dimensionless numbers which means that there is nothing to adjust about their "units" and indeed, there can't exist any "system of units" that would be able to tell you anything about such $q$ or $s$. I still insist that I have answered all your questions; you just don't seem to like the truth.
May
25
comment Understanding units and the units of the derivative operator
Dear Jules, what is true for multiplication (or division) isn't true for subtraction (or addition). These are different operations. Your "argument" is equivalent to the argument that 2+2 cannot be 4 because 2+2 cannot be 5, either, and 4 and 5 are analogous numbers. The units are just some "universal factors". Take 2 meters plus 3 meters. It's $2m+3m=5m$. I used italics fonts because $m$ literally behaves as a mathematical variable. By the distributive law, $2m+3m=(2+3)m$ so the sum has the same units. But $2m/3m=2/3$ has no units - different than $m$ - because $m$ simply cancels.