- #1
haj
- 1
- 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?
$$
\frac{\delta}{\delta g^{\gamma \epsilon}(z)} \left( g_{\mu \alpha}(x) \nabla^x_\beta \nabla^x_\nu \delta(x-y)\right)
$$
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: