Tag Info

New answers tagged

11

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1. II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical ...


1

Starting from the boundary condition $$ \partial^{\sigma}X^{\mu}(\tau,0)=0 $$ and lowering the index on the derivative using the metric gives $$ \gamma^{\sigma\tau}\partial_{\tau}X^{\mu}(\tau,0)+\gamma^{\sigma\sigma}\partial_{\sigma}X^{\mu}(\tau,0) =0. $$ Apparently Polchinski wants to express this in terms of the metric $\gamma_{ab}$ with its indices ...


0

In a viscous fluid the shear stress is proportional to the velocity gradient. $\sigma=\eta \frac{dv}{dy}$ where $\eta$ is the viscosity, and $v$ is the fluid speed at right angles to the $y$ axis. Therefore as the small distance $dy$ tends to zero, the change of fluid speed $dv$ also tends to zero, for any non-zero viscosity. Let us now follow Navier ...


0

Usually we think of friction as something like this (not a formal definition, but I think it's close enough to be understood as "friction"): when two objects move past one another and are in contact, the differential velocity between them leads to a force we call "friction" At the boundary between a liquid and a solid, if we permit a different velocity ...


0

The little group is the subgroup of the Lorentz group that leaves an arbitrary four-momentum vector invariant, i.e. for an element of the group $g$ and momentum $V$ we have $gV=V$. This group is in general different for massive and massless particles. If you now find that the little group of your holomorphic primaries corresponds to that of massive states, ...



Top 50 recent answers are included