We know that $∂x^ρ /∂x^μ = δ^ρ_μ$
Τhen, $∂x_ρ /∂x^μ = η_{ρμ}$ Should be correct, right?
Similarly, $\frac{∂x_ρ} {∂x_μ} = δ^μ_ρ$
Also, if
$x'^μ = e^α x^μ $, then
$∂'_μ$ should be $e^α ∂_μ$
I am new to this topic and I don't know if I am in the right direction. Can someone please help clarify?
I am also new in this topic and a high school student, my go on this is as:
$1.$ $\displaystyle \frac {\partial x^{\rho}}{\partial x^{\mu}} = \delta_{\mu}^{\rho} $ when we say that $x^\nu$ is component of vector.
$2.$ $\displaystyle \frac {\partial \vec x_{\rho}}{\partial x^{\mu}} = \Gamma^{\nu}_{\rho \mu} \ \ \vec x_{\nu} \ $ where $ \ \displaystyle \vec x_{\nu} \ $ is Basis vectors as lower index show covariance.
Moreover, $\displaystyle \Gamma^{\nu}_{\rho \mu} = \frac {\partial^{2}}{\partial x^{\mu} \partial x^{\rho}} \cdot \partial_{j} \ g^{j\nu} $
(I may be wrong in this part cause we generally not use lowered index for components, but if it is the case : $\displaystyle \frac {\partial x_{\rho}}{\partial x^{\mu}} = g_{\rho \nu} \frac {\partial x^{\nu}}{\partial x^{\mu}} = g_{\rho \nu} \ \delta^{\nu}_{\mu} = g_{\rho \mu} $)
$3.$ $\displaystyle \widetilde {x^{\mu}} = e^{\alpha} x^{\mu} $ This shows us the contravariant old to new transform, using inverse Jacobian. Thus $\displaystyle e^{\alpha} \widetilde {\partial_{\mu}} = \partial_{\mu} $
