# Derivatives of contravariant and covariant vectors

1. Apr 25, 2014

### nigelscott

Can someone explain why the derivative with respect to a contravariant coordinate transforms as a
covariant 4-vector and the derivative with respect to a covariant coordinate transforms as a
contravariant 4-vector.

2. Apr 25, 2014

### vanhees71

Take a scalar field. Then its differential is
$$\mathrm{d} \phi=\mathrm{d} x^{\mu} \frac{\partial \phi}{\partial x^{\mu}}$$
is also a scalar. Thus, since $\mathrm{d} x^{\mu}$ transforms contravariantly the four-gradient must transform covariantly, i.e., the correct notation is
$$\mathrm{d} \phi=\mathrm{d} x^{\mu} \partial_{\mu} \phi.$$
In the same way you can show that deriving with respect to the covariant components leads to a contravariant object.

3. Apr 25, 2014

### bcrowell

Staff Emeritus
The title says "Derivatives of contravariant and covariant vectors," which would be stuff like $\nabla_a v_b$ versus $\nabla_a v^b$. But #1 seems to be talking about $\nabla_a v_b$ versus $\nabla^a v_b$ , and #2 seems to be talking about the gradient of a scalar, $\nabla_a\phi$ versus $\nabla^a\phi$. Which are we really talking about here?

Not to be too pedantic, but we also don't have contravariant coordinates and covariant coordinates. Coordinates are always upper-index, and an ntuple of coordinates is not a vector or covector (at least not in GR). An infinitesimal *change* in the coordinates is an upper-index vector.

Assuming that the question is really the one posed in #1, then an easy way to see this is in terms of scaling. For example, suppose you change your units from meters to centimeters. All of your coordinates (which are upper-index quantities) get bigger by a factor of 100. Now suppose you have a scalar such as the electrical potential, and you take a gradient in order to find the electric field. The electric field is *smaller* in units of V/cm than it is in units of V/m. So the coordinates transform in one way under scaling, while a gradient transforms in the opposite way. This is what we expect for covariant quantities compared to contravariant ones.

4. Apr 25, 2014

### vanhees71

Sorry, I thought the question is about special relativity.

5. Apr 26, 2014

### nigelscott

μ = ∂/xμ = [∂/∂t, ∇]

and

μ = ∂/xμ = [∂/∂t, -∇]

c = 1

So in these cases the indeces are just telling you that there is a change of sign in the spatial
coordinates. What I don't understand is how the process of taking the derivative of the contravariant components results in a covariant vector and vice versa.

Last edited: Apr 26, 2014