# Why is the spatial term for contravariant 4-gradient negative, whereas for other 4-vectors it is the covariant part that is negative spatially?

The contravariant 4-displacement is:

$${x}^{\alpha} = (ct,\mathbf{r})$$

$${\partial}^{\alpha} = (\frac{1}{c}\frac{\partial}{\partial{t}},-\nabla)$$

From what I can gather so far, other contravariant 4-vectors tend to follow the same pattern as for the displacement - only with the 4-gradient does the negative spatial part appear in the contravariant form. Can someone please explain why the negative is in this one, and not the covariant 4-gradient instead?

• What is the signature convention used for $\eta_{\alpha\beta}$? – Everiana Mar 13 '16 at 0:08
• Just because in the presence of a metric, we can convert vectors into dual vectors (or "contravariant vectors" into "covariant vectors", if you insist) and vice versa, some objects have more natural meaning in certain "variance" than others. For example the electromagnetic tensor $F_{\mu\nu}$ is naturally covariant. The flat space derivative operator $\partial_\mu$ naturally exists like this, even without a metric, so the "contravariant" derivative $\partial^\mu$ is the one that will have "unnatural" signs. – Bence Racskó Mar 13 '16 at 18:54