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There is no a-priori sense in having an infinitely large potential, you can't calculate with infinite numbers (except for some very special mathematical considerations). So, the physical meaning that is given to such a potential must be implicitely contained in the accompanying text that explains the infinite potential well. An this meaning is commonly: &...


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All the literature says that the physically meaningful solutions to the Schrödinger equation in an infinite potential well must fulfill the boundary condition that the wave function is 0 at the walls of the well, otherwise the wave function wouldn't be continuous. That's not the reason at all: there are plenty potentials $V(\mathbf{r})$ for particles with '...


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He mentions that the shape of the cross section is irrelevant but it should be the same across the resistor. So why [is this assumption] necessary? Also is it necessary that cross sections be perpendicular to the z-axis? These assumptions are necessary to simplify the problem so that the field lines are straight (so that a simple relationship between ...


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Yes, Zwiebach is in section 6.9 talking about open strings in the static gauge (6.62) with free endpoint conditions (6.56), aka. Neumann BC, not Dirichlet BC (6.55). In other words, if there is a D-brane, it should be space-filling. Such endpoints move with the speed of light transversely to the string. On the other hand, Dirichlet BC is discussed in problem ...


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Is this integral equation valid if applied across a boundary of a magnetized material,(that is one end of the integral is inside the magnetised body and the other is out of it)? Or does it inherit any problems from its differential form? There is an assumption regarding this boundary condition that is often overlooked for various reasons, but can sometimes ...


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The contour integral across the boundary should not be considered as the direct application of Stoke's theorem in this area, but in a spirit similar to the principle of analytical continuation. The Stoke's theorem $$ \oint_C \bf{H}\times d\bf{l} = \int\int_A \nabla \times \bf{H}\cdot d\bf{a} = \bf{J}_{f,enclosed}. $$ This relation is applicable when $\nabla \...


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If I understand correctly the question, the answer lies in the 6.3.3 paragraph of Griffiths' book. Yes, it does inherit the discontinuity. In this case, you could still use the differential equation just above and just below the discontinuity, and taking an integration path that crosses the boundary will give you boundary conditions for the magnetic field ...


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