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1

Well, you do but the details are generally irrelevant for very low energies, like the LHC. The rational is as this: SUSY partners have to be heavier than regular matter, otherwise they would have been observed $\Rightarrow$ if realised in Nature, SUSY is broken. Like any other symmetry, if spontaneously broken leaves a massless (Nambu-)Goldstone mode. In ...


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Well, in his book "Symmetries in Science" (see attached picture), B. Gruber describes SSB without mentioning the ground state at all. His words describe exactly what I found in my system, so I conclude this is indeed a SSB, even if I don't have an energy function. It is maybe a slightly more general definition of SSB compared to that of Hamiltonian ...


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Spontaneous symmetry breaking requires more than two distinct solutions to a system of equations, because there are often physical considerations not captured in the equations that determine the solution or combination of solutions chosen. For example, take the 1D elastic collision between two objects of equal mass, one stationary and one initially moving ...


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This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization. A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the ...


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The state of the universe is not homogeneous and isotropic, but the laws of physics are. For example, the speed of light propagation is the same in all directions, and the mass of the electron is not a function of position. Writing down a Lagrangian requires an assumption about the laws of physics (or more precisely, an assumption about the dynamics). There ...


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Suppose that we align a perfectly cylindrical pencil with the z-axis. If the initial conditions are rotationally symmetric about that axis, then because the laws of physics are rotationally symmetric, the final state must also be symmetric under rotations about the z-axis. This means that the wavefunction of the universe will have to evolve into a ...


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I'll look at the question from multiple aspects. Classical mechanics, exact measurements, no thermodynamics, no perturbing forces In this fictitious universe it is possible to stand our perfectly balanced pencil exactly vertically and perfectly stationary. This is an unstable equilibrium position. With no perturbing forces, no thermodynamics, no quantum ...


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In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity. But there are forces afoot inside the pencil. Unless you also chill the pencil to absolute zero and thus stop all molecular activity, the trillions of atoms in the pencil are vibrating in random directions. It won't take long for the ...


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What you've fabricated is of course unrealistic in the physical world as CuriousOne stated, but not so in the virtual world of simulation. All the conditions you ask for can be arranged in a simulated universe. If perfectly balanced as its initial condition, the virtual pencil will not fall in this virtual world. It is unstable, but it will not fall until a ...


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Which way will the pencil fall, after you let go? Facetious answer: it would fall in the direction to which it was leaning when you let go. Okay, now to justify that: you cannot balance the pencil before letting go. For an object resting on a surface to be balanced, its centre of mass (a single point) must be directly above a point within the area of ...



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