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68

Entanglement is being presented as an "active link" only because most people - including authors of popular (and sometimes even unpopular, using the very words of Sidney Coleman) books and articles - don't understand quantum mechanics. And they don't understand quantum mechanics because they don't want to believe that it is fundamentally correct: they always ...


44

I wish to complete @Luboš Motl's answer, to which I agree. My point is on why people continue to make this mistake of an active link. This mistake is connected with one of the most interesting properties of quantum mechanics, Bell's theorem. One can argue that any physical theory is an hidden variable theory, the hidden variable being the description of the ...


38

Since general relativity is a local theory just like any good classical field theory, the Earth will respond to the local curvature which can change only once the information about the disappearance of the Sun has been communicated to the Earth's position (through the propagation of gravitational waves). So yes, the Earth would continue to orbit what ...


24

Gravitational influences do propagate at the speed of light, not instantaneously. The question of what would happen if the Sun instantly disappeared is actually a funny one in general relativity. The equations of general relativity imply as a mathematical consequence that energy must be locally conserved. Therefore, there is no valid solution to the ...


18

The higher the number of derivatives the more initial data you have to provide. If you have some Lagrangian that contains an infinite number of derivatives (or derivatives appearing non-polynomially, such as one over derivative) then you have to provide an infinite amount of initial data which amounts to non-local info, in the sense explained below. If you ...


16

All observations are consistent with standard GR so far, but I don't think the speed of gravity, in particular, has ever been measured. Experimental measurements of the speed of gravity was quite a controversy a few years ago when a paper came out claiming that the speed of gravity was very close to $c$ as measured by the Shapiro delay. To see papers on the ...


11

In classical mechanics, the lagrangians of two particles may be added only if the particules do not interract. I wouldn't say that. You can always write a Lagrangian $L$ for a system of two particles. In general, it takes the form $$L = L_1 + L_2 + L_i$$ where $L_i$ is an interaction term that depends on the coordinates and/or velocities of both ...


9

In fact your view is quite close to the 'official' one; entanglement occurs just because both particles are described with one wave-function; the magic is in our classical habit of thinking that separate objects are described with separate "coordinates".


9

Just a nice analogue Prof. Jürgen Audretsch told me once: Imagine at home you put one glove in your coat without looking (and noticing it's only one of the two). After exiting the train you notice it's cold and you pull out that single glove. At this very instant you know it's either the left or the right glove, and you therefore know which one is left ...


9

Everything doesn't expand equally because of cosmological expansion. If everything expanded by the same percentage per year, then all our rulers and other distance-measuring devices would expand, and we wouldn't be able to detect any expansion at all. Actually, general relativity predicts that cosmological expansion has very little effect on objects that are ...


9

Your question was first asked by Laplace. The following is from the Wikipedia article on "The speed of gravity" Laplace The first attempt to combine a finite gravitational speed with Newton's theory was made by Laplace in 1805. Based on Newton's force law he considered a model in which the gravitational field is defined as a radiation field or ...


9

You assume that you can instantly measure the momentum to arbitrary precision, and this isn't the case. Let's consider a plane light wave to keep things simple, and suppose you want to measure the momentum so precisely that the position uncertainty becomes exceedingly large. How precisely do we have to measure the momentum? Well the uncertainty principle ...


8

Clearly, an interaction involving $\phi(x+h)$ deserved to be called nonlocal. But since $\phi(x+h)=\sum_{k=0}^\infty \phi^{(k)}(x) h^k/k!$, any nonlocal interaction can be expressed as a power series involving arbitrarily many derivatives. Therefore an action (or Lagrangian) is called nonlocal if it involves infinitely many derivatives. If there are only ...


7

You are confusing the concepts of "interactions" and "nonlocality". In realistic field theories, including all theories we ever used to study phenomena in the world around us, the interactions exist but they keep the physics local. As David mentioned, the Lagrangian density takes the form $${\mathcal L} = \sum_i \left[ (\partial_\mu \phi_i)^2 + m^2 \phi_i^2 ...


7

Disillusionment with systems described by higher order Lagrangians harks back to a 1950 paper by Pais and Uhlenbeck, in which they showed that such systems were prone to pathologies, including states with negative energy and states with negative norm. There's a more recent discussion of this in arXiv:hep-th/0408104.


6

1) Well, the notion of locality depends on context. Usually in the context of QFT, a local function means a function of the form $$ f(\varphi(x), \partial\varphi(x), \partial^2\varphi(x), \ldots,\partial^N\varphi(x) ;x), $$ where $N\in\mathbb{N}_0$ is some finite order. See also this and this Phys.SE post. 2) Concretely, in the mentioned place almost at ...


6

It looks like this loophole is not explicitly discussed in the "axioms", but it is mentioned in the paragraph before equation (2) which I copy here: A symmetry transformation is said to be an internal symmetry transformation if it commutes with P. This implies that it acts only on particle-type indices, and has no matrix elements between particles ...


6

Let me give a tentative answer. I agree that these terms —especially 'locality'— are used for different concepts and this is annoying. I will list several notions of locality. Causality (or Einsteinian locality): Results of experiments carried out at a space-like distance are not correlated. This assumes that there are not previous correlations before ...


6

I think that the best picture to understand this correlation is given by many-worlds interpretation: A singlet decomposes in a coupled pair of particles superposition $|+⟩_A|-⟩_B + |-⟩_A|+⟩_B$, so observer A sees a simple superposition of $|+⟩ + |-⟩$ (which is a partial trace of the global density matrix) and so does B. In the many worlds interpretation, ...


6

The issue is locality of what? Which quantities are assumed to be local? If you say the reults of all experiments, hypothetical as well as actual, and say that these must be assignable definite values, then this is in conflict with quantum mechanics. But this is the assumption that is called "hidden variables", the reason is that this is the assumption ...


5

This is a truly excellent question in my opinion. It is still being worked on. Here are some professional references that will somewhat clarify the issue, or perhaps even confuse you further: http://arxiv.org/abs/1102.4467 http://arxiv.org/abs/1007.5518 http://arxiv.org/abs/1006.3680 Michael J.W. Hall http://arxiv.org/abs/0808.2178 Travis Norsen ...


5

From Hacker News https://news.ycombinator.com/item?id=6253263 This is a far more interesting question than it might seem at first glance, and it deserves some attention because it tells us something fundamental and wonderful and just bloody awesome about the universe. But I don't know how to tell the story succinctly. So I'm going to do that thing I do. I ...


5

In your comment you ask: That question only addresses gravitational force. Is it the same for all kind of forces? Generally speaking a force transmitted by massless particles like the photon and graviton obeys an inverse square law while a force transmitted by massive particles falls off exponentially with distance. This means that only the forces ...


5

There are a few papers in which topological field theories are constructed in terms of nets of algebras. The idea generally is that a net of algebras gives you a model for the higher category associated to a point by an extended TQFT. (Physicists would say that a 2d conformal net describes a 2d CFT which is related to a 3d TQFT.) The first one that comes ...


5

In Newtonian physics, there was no problem with action at a distance, and indeed Newton explicitly formulated his theory of gravitation in such terms. It may be that this was criticised from a philosophical standpoint (I don't know whether it was or not), but there were no fundamental mathematical difficulties with the idea. However, in relativity the ...


5

This is really the same answer as Jamal's, but I wanted to express it in simpler terms. You may know that spacetime points that are simultaneous in one inertial frame are not necessarily simultaneous in a second inertial frame moving with respect to the first. This causes problems if we allow non-local interactions. Suppose we have some system $A$ that ...


4

A quantity is local if it is a finite linear combination $\sum_k g_k P_k(x)~~$ of products $P_k(x)$ (or other pointwise functions, such as $\sin \Phi(x)~$ for sine-Gordon theory) of field operators or their derivatives at the same point $x$. A quantum field theory is local if its classical Lagrangian density is local. (By abuse of terminology, an action or ...


4

1) Lagrangian Formulation. The Lagrangian density for a massive free scalar in the $(+,-,-,-)$ convention is $$\tag{1} {\cal L}~=~\frac{1}{2}d_{\mu}\phi ~d^{\mu}\phi-\frac{1}{2}m^2\phi^2. $$ The corresponding Euler-Lagrange equation is the massive Klein-Gordon equation $$\tag{2} (d_{\mu}d^{\mu}+m^2)\phi~=~0. $$ The momentum is $$\pi ~:=~ \frac{\partial ...



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