Thomas
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 Jan 24 answered Why do superconductors have a maximum current density? Jan 21 revised Momentum of slowly spinning (viscous) fluid added 61 characters in body Jan 20 revised Momentum of slowly spinning (viscous) fluid edited body Jan 20 comment Momentum of slowly spinning (viscous) fluid I added the derivation to the answer. Jan 20 revised Momentum of slowly spinning (viscous) fluid added 999 characters in body Jan 20 comment Momentum of slowly spinning (viscous) fluid I think I finally figured out what is wrong with your equation, see my comment below. Jan 20 comment Momentum of slowly spinning (viscous) fluid Wow, this had me stumped for a while (could not find the mistake in what you did). We consider force balance (in $\hat{e}_\theta$ direction) on a cylindrical fluid element. The force/area on the angular sections is $\tau_{\theta r}\hat{e}_\theta$, and force balance gives the result you discuss. However, there is also a force/area on the radial sections $\tau_{\theta r}\hat{e}_r$, and this contributes an extra term because of $\partial \hat{e}_r/(\partial \theta)=\hat{e}_\theta$. Jan 20 comment Momentum of slowly spinning (viscous) fluid I can't quite figure out what you did there, but it clearly not right (you can just look up the Laplacian on a vector function). First of all, the equation of motion does not come from radial force balance (radial force balance determines the pressure), it comes from angular force balance. Jan 20 comment Momentum of slowly spinning (viscous) fluid The viscous drag is $\nu\nabla^2 v$. In cylindrical coordinates this must give Bessel functions. Jan 20 revised Momentum of slowly spinning (viscous) fluid added 227 characters in body Jan 20 revised Momentum of slowly spinning (viscous) fluid added 223 characters in body Jan 20 comment Momentum of slowly spinning (viscous) fluid Unless I made a mistake the sum should be the Bessel expansion of $r/R$. Jan 20 answered Momentum of slowly spinning (viscous) fluid Jan 19 comment “Irreversibility” of the RG flow What the paper tries to formalize is the notion that there is a direction to RG flow. In particular, by computing certain observables we can determine whether we are flowing to the IR or the UV. Intuitively, this observable counts the number of degrees of freedom, making precise the idea that at low energy there should always be fewer (or at most equal number) degrees of freedom than at high energy. Jan 19 comment “Irreversibility” of the RG flow This use of the word "irreversibility" has nothing to do with the irreversibility of time evolution and the second law. The RG flow is in principle fully reversible, although in practice there is some loss'' of information. For example, if your low energy effective action is a theory of Goldstone bosons then you may not be able to reconstruct the underlying microscopic theory that led to spontaneous symmetry breaking (some of its operators flowed to zero). Jan 19 answered Rigorous derivation of Fick's first law Jan 17 answered Why is the sound channel in the ocean especially good for low frequency sound? Jan 11 comment Renormalization group resummation Indeed, the (perturbative) RG does not sum all higher order terms, it sums classes of higher order terms such as large logs $[g^2\log(Q^2)]^n$. Jan 10 answered Baryon - Anti Baryon scattering Jan 6 comment Yang-Mills theories, confinement and chiral symmetry breaking No. The size of the hydrogen atom is $1/m_e$ not $1/m_p$, or $1/m_H$. Similarly, the size of a B meson is $1/\Lambda$. This is formalized in heavy quark effective theory (HQET).