0
$\begingroup$

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).

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    May 27 '19 at 14:37
  • $\begingroup$ 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. $\endgroup$
    – tparker
    May 27 '19 at 14:57
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy