Covariant derivative acting on Dirac delta function in curved space

haj
Messages
1
Reaction score
0
Homework Statement
$$\nabla_\beta \nabla_\nu \delta(x-y)=?$$
Relevant Equations
$$\nabla_\mu \phi=\partial_\mu \phi- w \Gamma^\nu_{\nu \mu} \phi$$
Pardon my naive computational question. In my calculations, I encounter the following expression:
$$
\frac{\delta}{\delta g^{\gamma \epsilon}(z)} \left( g_{\mu \alpha}(x) \nabla^x_\beta \nabla^x_\nu \delta(x-y)\right)
$$
To take the functional derivatives, we first need to determine what $$\nabla_\beta \nabla_\nu \delta(x-y)$$ is. Specifically, what is $$\nabla_\mu \delta(x-y)$$ ?

In the textbook it is written that the covariant derivative of a scalar density is given by:
$$
\nabla_\mu \phi=\partial_\mu \phi- w \Gamma^\nu_{\nu \mu} \phi
$$
Does this relation also hold for the delta function, given that the delta function is a scalar density of weight +1?
 
Last edited:
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top