# Why is the partial derivative a contravariant 4-vector?

The contravariant partial derivative is defined as following: $$\partial ^\mu = \frac{\partial}{\partial x_\mu}$$ where the index $$\mu$$ runs from 0 to 3. A contravariant vector under Lorentz transformation (at leas in Physics textbooks) is defined as: $$q'^\mu = \Lambda ^\mu _\rho q^\rho$$

Now what I don't get is why is the partial derivative above a contravariant 4-vector (the contravariant part, not the factor that it is a 4-vector).

• Note that the coordinates $x^{\mu}$ in SR are conventionally taken to have upper indices, so that the partial derivative $\partial_{\mu}$ has lower indices. – Qmechanic May 27 at 14:37
• More precisely, the "standard" partial derivative is covariant. The contravariant partial derivative that you have written above is the contraction of the partial derivative with the inverse metric, and is a less naturally fundamental operator. – tparker May 27 at 14:57

Under a Lorentz transformation $$y_\mu=\Lambda^{\nu}_\mu x_\nu$$, so $$\frac{\partial}{\partial y_\mu}=\frac{\partial x_\nu}{\partial y_\mu} \frac{\partial}{\partial x_\nu}=(\Lambda^{-1})^{\mu}_{\nu}\frac{\partial}{\partial x_\nu}$$ which is a contravariant transformation.
A more abstract and general way to look at it is the following: take $$m \in \mathcal{M}$$ a generic point on a manifold. The tangent space at the point $$m$$ is defined as the vector space spanned by all directional derivatives of the type $$\left.\frac{d}{dt} f(x+tv)\right|_{t=0}$$
however you choose a smooth function$$f$$. The value of such expression in the point $$m$$ is a scalar, namely it is left invariant by a chart transformation: this means that if the coordinate representation $$x=\varphi(m)\in \mathbb{R}^N$$ transforms in a certain way, the partial directional derivatives $$\partial_{\mu}$$ must transform in the exact opposite way. Then you can choose which one you want to call covariant or contra-variant.