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General Relativity in the absence of observers
At a very very fundamental level, there is no entropy!!!!
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General Relativity in the absence of observers
But if we want to keep ourselves "sane" we better keep track with the mainstream version of the events too. This is why I am here. To stand on the boundary of magic and hardcore science, hoping to find something fantastic.
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General Relativity in the absence of observers
We enter here the antropic principle or the fine tuning for life of the Universe. Both very interesting ideas. I think, that the space-time is made of an infinity of quantum observers and that Relativity is a very much inflated and washed out version of the quantum uncertainty, because they both need someone to make an experiment. It doesn't matter that in relativity the observers are some rulers and cameras and clocks, as long as all the gathered data from these rulers and clocks will be ultimately used and interpreted by human observers.
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General Relativity in the absence of observers
I will take into consideration your advice. Perhaps spontaneous or self-induced decoherence makes the presence of observers futile. I am familiar with the apparition of mass terms due to symmetry breaking, I can relate to what you said. Just that, as it happens with anybody, it is hard to keep the classical part and the quantum part separated in the same brain. The mind wants to know things between which it can make connections, and when it fails there is a problem. I just tried to push the boundaries a little.
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General Relativity in the absence of observers
Nice answer. This phrase is useful : "expectation values for observables for free particles still follow geodesics." Tidal forces are felt because of electromagnetic forces trying to keep the test body in one piece. If we could shut those down, no gravity would be felt. I know the space-time metric is not uniquely defined and for every Feynamn path there must be a given metric with a certain action. So, observation is making the metric to collapse too...so there is a slight probability also for escaping the geodesic. Thank you anyway, I'll conceive more coherent questions from now on.
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General Relativity in the absence of observers
In special relativity, an observer is a frame of reference from which a set of objects or events are being measured. Usually this is an inertial reference frame or "inertial observer". (Wikipedia) So measurement IS important for deducing curvature. And measurement means particles interacting and wave functions collapsing. What I am asking is, again, does measurement determine to some extent the space-time curvature ?
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Weyl Spinors and Lorentz Invariance
Look, could this help you? phas.ubc.ca/~mav/p526/lec11b.pdf
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Can someone explain to me what the phrase "Everything moves through space-time at the same speed" means?
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Can someone explain to me what the phrase "Everything moves through space-time at the same speed" means?
This issue becomes important in cosmology when it seems that time can not be used anymore, becoming seemingly inexistent for an external observer, so instead of the time coordinate, they use another type of axis to define time, and that axis is in fact the history of the evolution of the Universe. Also in rotating black-holes, time and space exchange places.
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Can someone explain to me what the phrase "Everything moves through space-time at the same speed" means?
Aaaa...yeah, time is viewed as a component of the 4D space. I can understand your problem, I had it too once. The notion of speed is seen in mathematics usually as the rate of change of any type of variable and not necessarily as a movement through space. In general relativity we must however use clocks to make observations. You can not determine your own speed along the time axis without comparing it to an external clock or reference. My example says that change is what determines motion along time, hence it determines what we call time. I just gave you the way time emerges.
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Weyl Spinors and Lorentz Invariance
The Higgs mechanism occurs whenever a charged field has a vacuum expectation value. In the nonrelativistic context, this is the Landau model of a charged Bose–Einstein condensate, also known as a superconductor. In the relativistic condensate, the condensate is a scalar field, and is relativistically invariant. (Wikipedia)
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Weyl Spinors and Lorentz Invariance
Also, please see this interesting article about two-fermions couplings. It goes about the same way I've explained to you. arxiv.org/pdf/0801.1615.pdf
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Weyl Spinors and Lorentz Invariance
The mixed product should do the trick, because when one flips in some direction the other one will flip in the opposite and viceversa, keeping the over all spin zero. You do not have a mixed product there, you just have product of a spinor with itself, this is why they do their own thing and not being invariant. They must become a single entity and the only way to relate two things in this universe is to have some form of product between them, sometimes mediated by a constant showing how strong that action of one over another, really is.
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