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If we take the simple approach of determining the state of the "jerk" today by assuming an exponential expansion (e.g., $a(t)\sim\exp(H_0 t)$), then $$ \dot a=H_0a\tag{1} $$ The derivative of this is then, $$ \frac{d^2a}{dt^2}=H_0\dot{a}=H^2_0a $$ And now for the "jerk," $$ \frac{d^3a}{dt^3}=H^2_0\dot{a}=H^3_0a\tag{2} $$ The Hubble constant is already pretty ...


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Clumps of just anti-matter will have the same gravity field around them as clumps of matter. There was an experiment at either Fermilab or SLAC in the 1970s or early 1980s where the falling of a beam of anti-protons was measured. I was trying to look up details on this a couple years ago, and didn't find it. But I know I read about it long ago. Bottom ...


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So what about antimatter - since charges are opposite, perhaps it also clumps together to form anti-gravity superpositions. As Red Act says in a comment, gravity is too weak to be important on the scale of individual particles. However charge does group antiparticles together. For example an anti-proton and a positron will form an antihydrogen atom. In ...


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One. The inflationary period is thought to have lasted from around $t = 10^{-36}$ seconds to $t = 10^{-33}$ seconds after the Big Bang. So while you're technically correct to say it lasted less than a second that's a bit of an understatement. Two. See my answer to What was the density of the universe when it was only the size of our solar system? for the ...



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