First, you are right in that non-Minkowski solutions to string theory, in which the gravitational field is macroscopic, it should be thought of as a condensate of a huge number of gravitons (which are one of the spacetime particles associated to a degree of freedom of the string). (Aside: a poi...
"First, you are right in that non-Minkowski solutions to string theory, in which the gravitational field is macroscopic, it should be thought of as a condensate of a huge number of gravitons"
can that be made at all precise?
I don't remember seeing anything like that in string theory
@0celo7 Yes. A modification of the spacetime metric has exactly the same form as the insertion of the vertex operators of gravitons when computing scattering amplitudes. So whether you consider a non-Minkowski metric to be a "background" or an expectation value of a bunch of gravitons is kind-of a matter of taste
@0celo7 This doesn't depend on what you start with - you can fix a non-flat metric and view modifications of it as a graviton condensate. What you declare a background and what a graviton effect is arbitrary
It's "natural" to consider a flat metric as the "graviton-less" state, but I don't think there's anything forcing that view upon us
Physica electromagnetica de campo electromagnetico et effectu suo in particulas onus electricum habentes tractat. Theoria Maxwelliana vim electricam et vim magneticam in formam theoriae relativitatis speciali congruentem unit. Aequationes Maxwellianae per se describunt campum electricum et campum magneticum et eorum causam imperfectam in particulis onus habentibus. Aequatio Lorentziana vim in particulas onus habentes a campis electricis magneticisque effectam describit. Aequationes Maxwellianae in vacuo sunt basis doctrinae lucis electromagneticae in quo celeritas lucis
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maybe?
gets the cake for weirdest corner of the internet I've found in a while
@BalarkaSen This $U$ is real weird. Its an annulus around the equator of a sphere, but antipodal points are identified. So for example rather than rotating by $2 \pi$ to get back where you starter, you can also rotate by $\pi$ and then go down through the equator. Which means if I cut along the equator, and rotate one of the annuli by $\pi$ then glue it back along the cut equator..... that makes it seem like its just $S^1 \times I$?
I'm working through the hardest (and most controversial) chapter now
Anonymous
@CaptainBohemian It's not you. Food with high water content makes everyone feel more full (compared to a an equivalent amount of dry food). I think this has some interesting science behind it. You could ask on Bio SE ;).
@BalarkaSen retarded question (unrelated to exam): I have $\lambda:[0,1]\to\Bbb R$ continuous in a neighborhood of each point of the domain. That means it's continuous on the whole thing and the image is connected, right?
actually i found using yahoo email to publish papers is better than using academic email if you are just a student or research assistant because academic email will be cancelled some short time after you graduate or leave your position