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There are both physical and formal reasons to introduce the spin connection. Physically, we know that there are spin 1/2 particles. A spin 1/2 field cannot be described by anything built from 4-vector fields. You can realize this for example by that 4-vector fields (and so anything built from them) returns to their original value after a $2\pi$ rotation ...


0

The form of the viscosity term can be derived on physical grounds. You can most likely do no better than to consult Landau and Lifshitz, volume 6, Chapter II. The Euler equation is the conservation law for momentum, it can be written in the form $$\frac{\partial}{\partial t} (\rho v_i) = -\partial_i \Pi_{ij}$$ that is the change of momentum density is the ...


1

There are three distinct things: 1) The rate-of-strain tensor is $ 2\underline{\underline{D}} = \nabla u + \nabla u^T $. Thus viscous stresses of homogeneous viscosity will be $\eta\nabla\cdot D$, which is equal to $\eta \nabla^2 u$ only if $\nabla\cdot u =0$ (incompressible and constant density). 2) The viscosity may be heterogeneous in space, in case ...


2

I found the mistake. (1) The state $(L_{-2}^{(1)} + L_{-2}^{(2)})|0\rangle$ does correspond to the total stress tensor for the product CFT. Acting on this state with the lowering operator $L_{2}^{(1)} + L_{2}^{(2)}$ gives a state proportional to the vacuum (except in the case where the central charge $c=0-$then the stress tensor is primary. (2) The state ...


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The question is somewhat ill-defined. This answer is based on what I gather the OP is asking. No. Generically, the covariant derivative of a tensor is zero only if the entire tensor is zero. Note how the covariant derivative is defined (taking example of a 1-form) $$ (\nabla_\mu A)_\nu = \partial_\mu A_\nu - \Gamma^\lambda_{\mu\nu} A_\lambda $$ Now, ...



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