Derivatives of contravariant and covariant vectors

[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.

 

[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.

 

[bcrowell]

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.

 

[vanhees71]

Sorry, I thought the question is about special relativity.

 

[nigelscott]

Thanks for your responses. I think my question should really have asked about the 4-gradient in SR.

∂_μ = ∂/x^μ = [∂/∂t, ∇]

and

∂^μ = ∂/x_μ = [∂/∂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.