Covariant and contravariant analysis

[ehrenfest]

Homework Statement

Can someone explain/help me prove the formulas
$$\vec{e_a}' = \frac{ \partial{x^b}}{\partial x'^a} \vec{e_b}$$

$$\vec{e^a}' = \frac{ \partial{x'^a}}{\partial x^b} \vec{e^b}$$

I do not understand why the partial derivative flip?

Homework Equations

The Attempt at a Solution

 

[dextercioby]

I'll give you 2 hints: 1. What is a coordinate basis ? 2. What is the chain rule for partial derivatives ?

 

[nrqed]

Homework Statement

Can someone explain/help me prove the formulas
$$\vec{e_a}' = \frac{ \partial{x^b}}{\partial x'^a} \vec{e_b}$$

$$\vec{e^a}' = \frac{ \partial{x'^a}}{\partial x^b} \vec{e^b}$$

I do not understand why the partial derivative flip?

Homework Equations

The Attempt at a Solution

Basically, this is essentially a definition of what we mean by covariant and contravariant indices (sometimes people start from the components of a covariant or contravariant vector instead of the basis vectors in which case the transformations are the opposite of the ones given here)

 

[ehrenfest]

My book defines the coordinate basis vector at a point P as follows:

$$\vec{e}_a = lim_{\delta x^a \to 0} \frac{ \delta \vec{s}}{\delta x^a}$$

where delta s is the infinitesimal vector displacement between P and a nearby point Q whose coordinate separation from P is delta x^a.

I think you must be referring to perhaps a more useful definition!

 

[javierR]

Homework Statement

Can someone explain/help me prove the formulas
$$\vec{e_a}' = \frac{ \partial{x^b}}{\partial x'^a} \vec{e_b}$$

$$\vec{e^a}' = \frac{ \partial{x'^a}}{\partial x^b} \vec{e^b}$$

I do not understand why the partial derivative flip?

A vector V is a geometrical object, that can be written in a coordinate basis as
$$V=v^{a}e_{a}$$
where $$e_{a}$$ are basis vectors (I'm giving them a lower index). The $$v^{a}$$ are the coefficient functions of the vector. As a geometrical object, V cannot care about what coordinate basis you have chosen, so under changes of bases it must be left unchanged. So if the coefficient functions transform as
$$v^{a}(x')=\frac{\partial x^{a}}{\partial x'^{b}}v^{b}(x)$$ the basis vectors must transform as
$$e_{a}(x')=\frac{\partial x'^{c}}{\partial x^{a}}e_{c}(x)$$
since
$$\frac{\partial x'^{c}}{\partial x^{a}}\frac{\partial x^{a}}{\partial x'^{b}}=\delta^{c}_{b}$$
implies that
$$V=v^{a}(x')e_{a}(x')=v^{b}(x)e_{b}(x)=V$$.

Similar arguments go for co-vectors $$W=w_{a}e^{a}$$
where $$e^{a}$$ are a basis of co-vectors (also known as 1-forms) contracted with their coefficient functions. The location of the index on all these objects determines the form of the transformation. Note: There is confusion in the literature regarding the names covariant and contravariant as applies to upper vs. lower index objects (e.g. there are generally opposite naming conventions between maths and physics).

 

[ehrenfest]

OK. I see why the basis case holds. But for the dual basis case, we then need that:

$$\frac{\partial x_a}{\partial x'_b} = \frac{\partial x'^b}{\partial x^a}$$

which is what I do not understand?