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4h
comment Detecting negative energy products in particle accelerators
just a fancy way to refer to the actual physical particles that trigger the detectors. They are on-shell because their energy and momentum is bound to be in a momentum manifold called 'shell' that characterises their rest mass, which in the case of photons, makes that shell to collapse into a cone
4h
revised Detecting negative energy products in particle accelerators
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4h
comment Detecting negative energy products in particle accelerators
I'm not referring to off-shell fairy stuff. I'm referring to Berkeleyian-hard on-shell negative energy particles in the final, outgoing states that hit the detectors. My question (again) is; how would we know there was one, if there happened to be one in the data? If one in every.. say $10^{10}$ collisions a real, on-shell negative energy particle, were created, could we spot it in the data, or could we not?
4h
comment Detecting negative energy products in particle accelerators
but you deserve a direct answer: I'm not sure what I think, but I already know what the theory says. Now, I want to contrast that with what measurements say on the matter, or don't say, or only partially say
4h
comment Detecting negative energy products in particle accelerators
physics is not about what we think there should be, but about what there is. So what you or me think on the subject is irrelevant. What the data say, on the other hand, is not :-)
4h
revised Detecting negative energy products in particle accelerators
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5h
asked Detecting negative energy products in particle accelerators
Sep
11
revised H-theorem and Boltzmann equation applied to Boltzmann distribution
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Sep
11
asked H-theorem and Boltzmann equation applied to Boltzmann distribution
Jul
27
comment Is the future already determined?
MWI is not meant to solve any particular problem: It is just the natural, observer-invariant description of quantum mechanics that preserves unitarity at all times. Nothing more, nothing less
Jul
27
answered Is the future already determined?
Jul
13
awarded  Yearling
Jul
2
awarded  Curious
Jun
17
comment Diffeomorphism Invariance of General Relativity
"needs to lift to a general covariant transformation of Y, which are not mere coordinate transformations." yes, an example of the actual difference between one and the other would certainly help here
Jun
17
revised Are diffeomorphisms a proper subgroup of conformal transformations?
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Jun
17
comment Are diffeomorphisms a proper subgroup of conformal transformations?
Wait.. this could not be right.. General Relativity is invariant under diffeomorphisms, but it certainly is not invariant under conformal transformations, if conformal transformations where a subgroup of diff, you would have a contradiction
Jun
13
awarded  Enthusiast
Jun
7
comment Increasing Earth's albedo
dmckee while this is true in average, enhanced reflectivity in a dense urban area would lower the average absorbed power, and probably the average temperature. But on the other hand, if you are willing to paint all the roofs, why bother? put some solar panels and some solar freakin' roadways: youtube.com/watch?v=qlTA3rnpgzU&feature=kp
Jun
4
comment Are diffeomorphisms a proper subgroup of conformal transformations?
I try to understand this intuitively. A diffeomorphism can arrange points(events) in spacetime arbitrarily. Take a diffeo close to the identity such that it can be written as a small tangent flow on each point. So what you are saying is that if the tangent flow is curly (is not a gradient of a scalar) it can't be written as a conformal close to the identity, which always have their tangent flow to be the gradient of some scalar. Is this correct?
Jun
4
asked Are diffeomorphisms a proper subgroup of conformal transformations?