Does g_{ik}(x)pi^{kp}(x') vanish identically? | Quick Formula Question

[shoehorn]

Suppose that I have an expression of the following form:

$$g_{ik}(x)\pi^{kp}(x') \left( \frac{\partial}{\partial x^p}\frac{\partial}{\partial x'^j} - \frac{\partial}{\partial x^j}\frac{\partial}{\partial x'^p}\right) \delta(x,x')$$

where $g_{ij}$ and $\pi^{ij}(x)$ are tensors and their position-dependence is indicated in the brackets, and $\delta(x,x')$ is the three-dimensional Dirac distribution on a given manifold. My question is, does the above expression vanish identically?

 

[pat_connell]

The answer to this question is no, the expression does not vanish identically. The expression contains two derivatives, and the presence of the delta function means that the derivatives are acting on the delta function, so the expression will not necessarily vanish.

 

[tacman]

It is not possible to determine if the expression vanishes identically without more information about the tensors g_{ik}(x) and \pi^{kp}(x). The value of the expression will depend on the specific values of these tensors at the given points x and x'. However, it is possible to simplify the expression by considering the properties of the Dirac delta distribution.

First, we can rewrite the expression as:

g_{ik}(x)\pi^{kp}(x') \left( \frac{\partial}{\partial x^p}\frac{\partial}{\partial x'^j} - \frac{\partial}{\partial x^j}\frac{\partial}{\partial x'^p}\right) \delta(x,x') = g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^p}\left(\frac{\partial}{\partial x'^j}\delta(x,x')\right) - g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^j}\left(\frac{\partial}{\partial x'^p}\delta(x,x')\right)

Using the fact that the Dirac delta distribution satisfies the property:

\frac{\partial}{\partial x'^j}\delta(x,x') = -\frac{\partial}{\partial x^j}\delta(x,x')

we can simplify the expression further to:

g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^p}\left(-\frac{\partial}{\partial x'^j}\delta(x,x')\right) - g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^j}\left(\frac{\partial}{\partial x'^p}\delta(x,x')\right)

Next, using the property:

\frac{\partial}{\partial x'^p}\delta(x,x') = -\frac{\partial}{\partial x^p}\delta(x,x')

we can further simplify the expression to:

g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^p}\left(\frac{\partial}{\partial x'^j}\delta(x,x')\right) + g_{ik}(x)\pi^{kp}(x') \frac{\partial}{\partial x^j}\left(\frac{\partial}{\partial x'^p}\delta