Confuse-ray30
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Member for 11 months
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Is the uncertainty principle a consequence of classical reference frame?
@Ruffolo If I replace the box by a new particle species instead that does not interact, then we have the same argument. Now Im wondering if we can understand this better under this light... You commented thaz while I was typing away. I think you should, yeah
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Is the uncertainty principle a consequence of classical reference frame?
Yeah, I noticed you dont do much with the box. It doesnt even need to be around the particle. I guess my answer is wrong then. @Ruffolo
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Physics journals that make their peer review files public
Maybe post it into Academia
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Why Lagrangian mechanics cannot find the state of a system at any instant during the entire course of time? But Hamiltonian mechanics does
The hamiltonian approach does not start from EOM, in so far you accept (just as in the lagrangian formalism the least action) that H generates time translation, i.e. it is the hamiltonian function of the flow of phase space. From this, the EOM follow IN PARTICULAR for p and q.
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Why Lagrangian mechanics cannot find the state of a system at any instant during the entire course of time? But Hamiltonian mechanics does
In classical mechanics, how is the principle of least action any sort of problem? (IC, BC) What I mean is, the principle of least action does not constrain the paths considered at all. It only constrains the variations at the end points. But this does not imply we can only have boundary value problems as a result. We can still ask the for two IC and solve the EOM
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Why Lagrangian mechanics cannot find the state of a system at any instant during the entire course of time? But Hamiltonian mechanics does
Id be interested as well. Afaik, they should be equivalent. Even if there are some superfluous DOF it should still be equivalent
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Wave function collapse at particle detection
@wawa when reading your question, I assumed the detectors would be on opposite ends. I think the person answering did the same. If you modify the detectors to be so close that the localized wavepacket has enough support for both of them, then yes, removing another interaction (the detector) will affect the results.
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Wormhole metric by identification
Btw, I cannot access the paper, so... I dont think I can provide more insight. Unless you explain what "identification" means here. But I guess youd be able to answer the question at that point.
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Wormhole metric by identification
I don't see where a new coordinate is defined in the question. Let's say you want spacetime to be a cylinder. Then you identify two sides of a square. This gives you the condition that on the boundary of the square, the metric must be the same (since they are the same spacetime point) AND you identify the boundary with each other.
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Do two electrons with the same $(n, \ell, m_\ell)$ but different $m_s$ have different wavefunction?
But not taking the whole thing to be your wave function makes the exclusion principle hard to formulate, no?
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Wormhole metric by identification
Just because the metric is the same at two space-time points does not make these two points related. However, an identifaction essentially glues space time at one point, i.e. creates a bridge.
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Why Wick rotate at all? Why not just use $\mathbb{C}^4$?
Similar ideas are employed in Penroses twistors.
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A 12th-grade physics question about inductors
Deleted unphysical constants
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Cosmological constant from the view of renormalization
If we have to scales, say electron and neutrino mass, we integrate out to H_0 again. Is the Cosmological constant then the "bare" value plus m^4 of the lowest mass or the sum of m^4 of all masses and things like m^2(electron)m^2(neutrino)?
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Is entropy scale-invariant?
@NotAGroupTheorist 1st q: Beyond scopr of og question, but in short: from the microscopic Hamiltonian. 2nd q: Yes. 3rd q: No, thermodynamical equilibrium makes sense and is defined as when the system does not change its expectation value. (In this case ensemble expectation, i.e. the one you use in probability theory given a distribution). Well, I have no idea what you mean in the last one: Given a distribution p(x), the expectation value can be determined by N->infty, 1/N sum x, where each x is drawn from p. So no.