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by cfgauss2718
670 days ago
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Here’s maybe a useful example. Consider a scalar potential function F on R^3 that describes some nonlinear spring law. At a point p=(x,y,z), the differential dF can be thought of as a (1,0) tensor measuring the spring force. It acts on a particle at p moving with velocity v to give the instantaneous work of the particle on the spring dF(p)(v). Now, suppose that we want to know how this quantity changes when we vary the x coordinate. The x coordinate is also a function of p, we can represent its differential as dx, which is a co-vector(field). The quantity that captures this change can be thought of as a (1,1) tensor field, which is related to the stiffness of the spring potential in the x direction at each point p. In the usual undergraduate setting, this tensor field is given as the hessian of F, call this H. The action of this tensor looks like the product u^T H(p) v, where in our case, u^T = dx(p) = [1 0 0]. A good giveaway for when a “co-vector” appears in a tensor calculation is whenever there is a “row vector” in a matrix operation (most people identify “column” vectors with proper vectors). It’s helpful in this case that “row” rhymes with “co-“. |
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