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Natural Science Forum / Physics / Particle Physics / August 2006Cosmological Constant problem
I hear that the calculated value of the vacuum energy using QFT is 120 orders of magnitude more than what is observed. But I wonder if this calculation was done in a very slow locally expanding spacetime. Or was it done with a strictly non-expanding metric? Perhaps when the QFT calculation is done in a very slowly expanding spacetime metric that the calculation might come out more equal to observation. Perhaps the small differential ends up multiplying the result and lower it by 120 orders of magnitude. Anyone have any insight into these things? Thanks. > I hear that the calculated value of the vacuum energy using QFT is 120 > orders of magnitude more than what is observed. But I wonder if this [quoted text clipped - 4 lines] > ends up multiplying the result and lower it by 120 orders of magnitude. > Anyone have any insight into these things? Thanks. A very slow local expansion should (with my understanding) at most give a very small correction and you would still get about 120 orders of magnitude wrong. You could always transform your calculation to a *local* flat (i.e non-expanding) space time and do your vacuum energy calculation as usual. If you want to do it in a curved space-time one would only get extremely small corrections due to the higher derivatives of the metric. Do supersymmetry and you instead get 60 orders of magnitude wrong. Normal order your operators and you get the vacuum energy to be 0 (infinitely times to low). best >> I hear that the calculated value of the vacuum energy using QFT is 120 >> orders of magnitude more than what is observed. But I wonder if this [quoted text clipped - 8 lines] > very small correction and you would still get about 120 orders of > magnitude wrong. It would certainly have to be only a small correction to interactions. I wonder, though, if the same can be said of the vacuum energy? |
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