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As i understand, according to standard model forces are just mediating particles used to form heavy particles, atoms, molecules and other material from fundamental particles like quarks and leptons. Force particles alone do not constitute any form of matter. To over knowledge of science these forces consistent everywhere (where there is matter) in the ...

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I'm just going to add something orthogonal to Floris' correct answer, because the question is posed in a very general manner, which allows for relatively diverse yet correct kinds of answers. Here's a crude-intuitive way of seeing it: A wavelength is just the spatial period of a wave, be it of mechanical or electromagnetic nature. Meaning the distance ...

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In principle, a wave of any size will interact with a system of any size. The question should therefore be posed differently: how is the interaction of the two affected by their relative size? Let's take the simple example of scatter. You are familiar (whether you know it or not) with Rayleigh scatter - it's an elastic light scattering phenomenon that makes ...

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What I know is that for polarizing an atom, i.e. creating a dipole, we consider the wave-length of the e.m. field greater by a couple of orders of magnitude than the dimensions of the atom. Does it mean that we test the atom? On the other hand, if we study the structure of a crystal, we send on it e.m. waves of $\lambda$ bigger than the distance between the ...

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Thanks to the procedure suggested by Qmechanic, I have clarified myself. I need just to invert the matrix $m$, since it has for the momenta $\sigma_{em}=\Sigma :\nabla u+P^T\nabla \phi$ $d_{em}=P:\nabla u-\epsilon\cdot\nabla\phi$ or $\left(\begin{array}{c}\sigma_{em}\\d\end{array}\right)= \left(\begin{array}{cc}\Sigma & P^T\\ P & ... 3 The heat radiates away from the source equally in every direction (until it hits something). Imagine the sun. It is roughly spherical and radiates energy (some as heat) in all directions. The energy is radiated in a sphere. As the energy travels away from the source (the sun) the sphere expands but the amount of energy remains the same. When the sphere is ... 2 The simple answer is the inverse square law which states that the intensity (power per unit area) of a heat source drops off with$1/r^2$where$r$is the distance (assuming a point source). Looking at your camp fire: imagine people standing side by side around the fire. Each gets a share of the heat. When they make the circle wider, more people can get ... 3 The heat simply disperses and the further away from the heat source you get the more space the heat has to disperse into 11 Now I am left wondering why does the heat become lost as if travels slightly. It is not lost. It is spread more out. If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100. The heat radiation sent from the source ... -3 Informally speaking heat is determined by the (average) momentum and density of moving particles. So if particles go slower, you get less heat. Also if there are less particles in a fixed volume its less hot. Now from simple geometry you can easily show that going away from the heat source means cooling down. Hope that helps. 0 My guess is that this expression is more physically presentable when we put this two terms in the same log. Now we have a log of a braket: $$\frac{ \langle \Psi_0 | U(\infty,0) }{\langle \Psi |\Psi_0\rangle}\frac{ U(0,-\infty)|\Psi_0\rangle}{\langle \Psi_0|\Psi\rangle}$$ This ket is more close to the ket of the Gell-Mann and Low theorem. 1 In order to perform the (possibly singular) Legendre transformation, it is necessary to have information about pertinent rank conditions of the structure constants$\rho$,$\omega$,$\Sigma_{ij,k\ell}$,$\epsilon_{ij}$and$P_{ijk}$. In this answer, we will sketch how the (possibly singular) Legendre transformation is performed in principle: We will use ... 1 The answer is no, we cannot rule out forces not directed along the line joining particles with current theory. As you point out, magnetic monopoles are a counterexample, and, as CuriousOne points out, magnetic monopoles both fit into a classical framework, are consistent with the Standard Model and are actively, experimentally sought. Some historical ... 5 I don't intend to repeat the points already made in the other answers, consider this as a small addition to those, with the intention to give a more practical description (reminding some of the basic ideas) without getting into observer-philosophies. Quite clearly the act of observing, i.e. measuring a quantum system can be done via many different ... 5 When does an interaction drop the system into an eigenstate? (i.e. when is a measurement=) This is an ill-posed question because, first of all, the system$S$doesn't drop into any state but each observer$O$has a state about it, as a state$\rho$is nothing but the coding of past measurements (so it should be named with reference to being dependent on ... 0 The Anthropic answer to this question is that if gravity were a lot stronger, then the evolution of the universe would have proceeded in a different way, it would have collapsed just after the Big Bang. One can speculate that all possiblities really exists, but we can obviously only find ourselves in those universes with laws of physics that are compatible ... 2 The static electrical force between two particles of electric charges$q_1$and$q_2$separated by a distance$r$is given by Coulomb’s law of electricity, which says that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force is repulsive for like ... 0 Quantum optics demonstrates the existence of interaction-free measurements: the detection of objects without light—or anything else—ever hitting them. Paul Kwiat, Harald Weinfurter and Anton Zeilinger SciAm November 1996 http://www.arturekert.org/sandvox/quantum-seeing-in-the-dark.pdf 10 This is not a settled question. Just as it is still debated whether or not there is wavefunction collapse, so is it debated what exactly we should understand by a measurement. In the following, we will go through the ideas behind the von Neumann measurement scheme, which is one way to try and talk about measurement in quantum mechanics. An interaction ... 7 All measurements are based on interactions , but not all interactions are measurable. The "set of measurements" is a subset of the "set of interactions". The simplest mathematical way to see this is by Feynman diagrams. Feynman diagrams have one to one correspondence with integrals, and when they describe a measurement, the crossection or lifetime can be ... 3 A "fifth force" can be given by a scalar field. This field arises in many theories, for example from the dilaton scalar field of string theory, that is non minimally coupled to$R\$ in the action. In the low energy limit we can phenomenologically treat string theory as a Scalar Tensor Theories of Gravitation (see the Wikipedia Link for details) in which you ...

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some theories propose that the infalton field is actually the same a dark energy,, if that were correct, then we would have a fifth field (force might not be correct in this context) that ineracts with the other four. And who knows what else, dark matter might have lots of forces that are invisible to us because they just do not inetract with the four we do. ...

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