Are all vector fields invariants?

[Je m'appelle]

Are all vector fields invariants or is this a particular characteristic to some fields?

For example, suppose the vector field E = x^2 x + xy y.

If I write it in terms of covariant and contravariant basis through the polar coordinates I get the same results, or in other words

$$\vec{E} = \bar{E^i} \bar{e_i} = \bar{E_i} \bar{e^i}$$

Where

$$\bar{E^i} \bar{e_i} \equiv \frac{\partial \bar{x^i}}{\partial x^j} E^j \bar{e_i}$$

is the vector field in it's contravariant basis and$$\bar{E_i} \bar{e^i} \equiv \frac{\partial x^j}{\partial \bar{x^i}} E_j \bar{e^i}$$

is the vector field in it's covariant basis.

By evaluating the above I found out that

$$\bar{E^i} = \bar{E_i} = r^2 cos \theta$$

Or in other words

$$r^2 cos \theta \ \hat{r_n} = r^2 cos \theta \ \hat{r_t}$$

Where

$$\hat{r_n} \equiv \bar{e^i}$$

is the normal vector of the contravariant basis and

$$\hat{r_t} \equiv \bar{e_i}$$

is the tangent vector of the covariant basis.

Then, I picked another vector field F = x x + y y and wrote it in terms of it's contravariant and covariant basis through the polar coordinates system, and my results were that

$$\vec{F} = \bar{F^i} \bar{e_i} = \bar{F_i} \bar{e^i}$$

That is, F is just as invariant as E.

So my question is, are all vector fields invariants or is this just some coincidence that I picked two arbitrary vector fields and they came out to be invariants?

 

[lippyka]

The answer is no, not all vector fields are invariants. Some vector fields may be invariant under certain coordinate transformations, but not all. It is possible for a vector field to be invariant only under a particular set of transformations, and not all.