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This turns out to be more involved than I though. The problem is that the boundary conditions I gave can't be matched by a stationary solution. You need a source term to force the ends together. The solution with a constant source is given in this paper. In brief, if we define a curvature $\kappa$ so that $\kappa^2=(\mathbf{R}'')^2$, then $$-\kappa'' - ... 1 Oh, I like this. Hopefully I can avoid stupid math mistakes. (No warranty is made, express or implied, that I will not make stupid math mistakes. Read at your own risk) We can attack this problem with calculus of variations. Specifically, we can map this to a problem in Lagrangian mechanics using a Lagrangian proportional to \mathcal{L}=\ddot{\mathbf{R}}. ... 0 Clearly, if the tension in one part of a rope is different than in another part, there will be a gradient in tension - which in turn means that if you look at a particular part of the rope at location x where there is a gradient in tension \frac{dT}{dx}, then there is a net force on an element of length \Delta x:$$F = \frac{dT}{dx}\Delta x$$For a ... 0 It also depends on whether or not the rope is inextensible. Assume it isn't. You tie one end to a rigid wall and pull on the other end with a constant tension T. Then at some time t_0, you start to pull with a slightly higher tension T_1 > T. It will take some time \tau of order \frac{L}{c}, c being the typical celerity of mechanical waves ... -1 Since the LINE, a section of ROPE, has no attached mass or restraint and it seems it is accelerated as a whole, there is no tension. If one end is pulled to accelerate it then F=ma=0.0 N. 0 The usual definition of normal ordered product is:$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) - \langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle $$As you said, this is the regular part of the OPE, since only the divergent part of two operators gives non vanishing contribution to the correlator. Of course$$\langle ...