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The strings in string theory are very different to guitar strings. They are not clamped at the ends like a guitar string. Instead they have free ends or they are formed into a loop. Also they have a fixed string tension that isn't dependent on their length. Classically if you have a free string with a constant tension that tension would immediately make it ...


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Here $x'$ denotes a point on the boundary, and $x$ denotes a point in the bulk. The problem he is looking at is given some boundary value $\phi_0(x')$ of the scalar field $\phi(x)$, how do you solve the wave equation. $$L \phi(x) = \phi_0(x')$$ In order to do that he is solving $$L_x K_{xx'} = \delta(x-x').$$ One way to solve this is to just consider ...


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An idea like yours was seriously proposed in the late 1990s as an explanation for the "hierarchy problem," which is the observation that the three fundamental forces which have working quantum field theories (electromagnetism and the strong and weak nuclear forces) have similar coupling constants at high energy, while gravity is much weaker. The suggestion ...


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Any transformation of the coordinates $x \mapsto x' = f(x)$ transforms the metric as follows $$ g_{\mu\nu}(x) \mapsto g'_{\mu\nu}(x') = \frac{\partial x^\rho}{\partial {x'}^\mu}\frac{\partial x^\lambda}{\partial {x'}^\nu} g_{\rho\lambda}(x)\,. $$ For the case at hand $$ x = (\sigma,\tau) \mapsto x' = (\tau+\sigma,\tau-\sigma) $$ So the inverse ...


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The ad-hoc quantization procedures for going from the classical theory of the superstring to the quantum superstring theory are just that - ad-hoc. We're not really looking at a single string with chosen boundary conditions and quantizing it in some clearly prescribed canonical fashion, we're playing around with the superstring and are trying to get to a ...


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1) Yes. 2) The second term comes from the variation of $\sqrt{\det g}$.


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