How Are Basis Vectors Defined in Different Coordinate Systems?

[olgerm]

hi.

1. if I know how to convert coordinates from a system to cartesian system, then how can I find basevectors of that coordinatesystem?
2. Is it possible that basevectors are different in different points(with different coordinates)?
3. What is most general definition of basevectors? I tought it would be $\vec{e_i}(\vec{v})=\frac{\partial \vec{v}}{\partial x_i}$, because then basevector would show in which direction the point, that is described by coordinates moves when corresponding coordinate increases. $\vec{e_i}(\vec{v})$ is i'th basevector in point $\vec{v}$ and $x_i$ is i'th coordinate of $\vec{v}$. I am not sure if this is correct definition.

 

[Orodruin]

if I know how to convert coordinates from a system to cartesian system, then how can I find basevectors of that coordinatesystem?

That you mention a Cartesian coordinate system implies that you are working in Euclidean space. The answer to your question depends on what you mean by "base vectors of that coordinate system". Given a coordinate system with coordinates $x^a$ (note that the typical convention is to use super indices rather than sub indices on coordinates), there is a holonomic basis of tangent vectors to the coordinate lines defined by
$$
\vec E_a = \frac{\partial \vec x}{\partial x^a},
$$
where $\vec x$ is the position vector. This set of vectors forms a basis at every point. However, there is also a dual basis defined by
$$
\vec E^a = \nabla x^a,
$$
i.e., the gradient of the coordinate functions. These basis vectors are normal to the coordinate hypersurfaces, i.e., the hypersurfaces along which $x^a$ is constant, and also form a basis at every point. Note that generally $\vec E^a \cdot \vec E_b = \delta^a_b$.

Furthermore, in orthogonal coordinate systems, it is quite common to define a set of normalised basis vectors $\vec e_a$, which are just the $\vec E_a$ rescaled to have unit length. All of these sets of basis vectors can to some extent be said to "belong to" the coordinate system. Also note that a given set of vector fields that form a basis at each point not necessarily must be either of these for some coordinate system.

Is it possible that basevectors are different in different points(with different coordinates)?

Yes. Consider any curvilinear coordinate system such as polar coordinates.

What is most general definition of basevectors?

See above. It is not necessary for a set of basis vector fields to be any of the above described bases for some coordinate system. Any set of vector fields that are linearly independent everywhere will do fine.

 

[Cryo]

I very much agree with @Orodruin. Here is the same thing said in a slightly different way:

if I know how to convert coordinates from a system to cartesian system, then how can I find basevectors of that coordinatesystem?

Start by writing what is a squared difference between two points separated by small step. For example in 2D, Eucledian

Cartesian coordinates: $dr^2=dx^2+dy^2$
Cyllindrical coordinates: $dr^2=d\rho^2+\rho^2d\phi^2$ (so you go $dx=\partial_\rho x d\rho + \partial_\phi x d\phi$ etc, substitute and simplify)

Then note that the difference in a function $f$ between these two points is:

$df=f\left(\vec{r}+d\vec{r}\right)-f\left(\vec{r}\right)=\partial_\rho f d\rho +\partial_\phi f d\phi$ (e.g. for cyllindrical coordinates)

We can also define a vector:

$d\vec{r}=\vec{e}_\rho d\rho+\vec{e}_\phi d\phi$

and then: $df=d\vec{r}.\vec{\nabla}f$ by the definition of gradient. But $d\vec{r}.d\vec{r}=d\rho^2+\rho^2d\phi^2$ , so $\vec{e}_\rho.\vec{e}_\rho=1$, $\vec{e}_\phi.\vec{e}_\phi=\rho^2$ and $\vec{e}_\rho.\vec{e}_\phi=0$

It then follows that $\vec{\nabla}=\vec{e}_\rho \partial_\rho+\frac{1}{\rho}\vec{e}_\phi \partial_\phi$

Finally $\vec{e}_\rho=\vec{\nabla}\rho$ and $\vec{e}_\phi=\rho\vec{\nabla}\phi$. Why is this a good thing? Because gradient is also defined in Cartesian coordinates, so (e.g.):

$\vec{e}_\phi=\rho\left(x,\,y\right)\left(\frac{\partial\phi}{\partial x}\vec{\hat{x}}+\frac{\partial\phi}{\partial y} \vec{\hat{y}}\right)$

etc

 

[kent davidge]

there is a holonomic basis of tangent vectors to the coordinate lines defined by
$$
\vec E_a = \frac{\partial \vec x}{\partial x^a},
$$

That makes me wonder why one usually defines a vector as $V = V^a \partial / \partial x^a$, i.e., the basis vectors are $\partial / \partial x^a$ instead of $\partial \vec x / \partial x^a$.

 

[Orodruin]

That makes me wonder why one usually defines a vector as $V = V^a \partial / \partial x^a$, i.e., the basis vectors are $\partial / \partial x^a$ instead of $\partial \vec x / \partial x^a$.

Because the only thing that actually matters is the derivative (there is a bijection between tangent vectors at a point and directional derivatives in that point) and in a general manifold there is no position vector. However, the OP is explicitly working in a Euclidean space where it does exist.

 

[olgerm]

The answer to your question depends on what you mean by "base vectors of that coordinate system".

in orthogonal coordinate systems, it is quite common to define a set of normalised basis vectors $\vec e_a$, which are just the $\vec E_a$ rescaled to have unit length. All of these sets of basis vectors can to some extent be said to "belong to" the coordinate system.

I am quite confused with term "base vectors of a coordinate system". I thought that it is strictly defined so that we can calculate the "base vectors of a coordinate system" in every point of space. But if there are many possible values to "base vectors of a coordinate system" in some points, then I misunderstood the term.

How to call the vectors of a coordinatesystem that must be orthogonal and have length 1 if the coordinatesystem is orthonormalcoordinatesystem?

there is a holonomic basis of tangent vectors to the coordinate lines defined by
$$\vec E_a = \frac{\partial \vec x}{\partial x^a},$$
where $\vec x$ is the position vector.

Are these tangent vectors to the coordinate lines?

However, there is also a dual basis defined by $$\vec E^a = \nabla x^a,$$ i.e., the gradient of the coordinate functions.

Are vectors of that base covariant vectors?

 

[Orodruin]

I thought that it is strictly defined so that we can calculate the "base vectors of a coordinate system" in every point of space. But if there are many possible values to "base vectors of a coordinate system" in some points, then I misunderstood the term.

Your lecturer may be using one particular choice and referring to that particular choice as "base vectors of a coordinate system". However, it is worth keeping in mind that there are several choices of basis that to some extent "belong" to a given coordinate system.

How to call the vectors of a coordinatesystem that must be orthogonal and have length 1 if the coordinatesystem is orthonormalcoordinatesystem?

I would just call it a normalised basis of the orthogonal coordinate system, e.g., the normalised basis of spherical coordinates. Note that a coordinate system by itself can be orthogonal (meaning that coordinate lines intersect at right angles), but it usually does not make sense to call a coordinate system "orthonormal". However, you can normalise the basis of any orthogonal coordinate system to obtain an orthonormal basis.

Are these tangent vectors to the coordinate lines?

Yes.

Are vectors of that base covariant vectors?

In Euclidean space, there really is no distinction between the vectors as such. These basis vectors transform contravariantly so the components of any vector expressed in this basis are the covariant vector components of that vector. In a general manifold, the corresponding basis would be $dx^\alpha$, which are one-forms that would span the dual space at each point - unlike the partial derivatives $\partial_\alpha$, which span the tangent space at each point.

Geometrically, these vectors are the surface normals of the coordinate level surfaces.

 

[olgerm]

I would just call it a normalised basis of the orthogonal coordinate system, e.g., the normalised basis of spherical coordinates. Note that a coordinate system by itself can be orthogonal (meaning that coordinate lines intersect at right angles), but it usually does not make sense to call a coordinate system "orthonormal". However, you can normalise the basis of any orthogonal coordinate system to obtain an orthonormal basis.

My lecturer said that coordinatesystem is orthonormal if it is normalized and orthogonal, but I do not understand, which conditions a coordinatesystem must satisfy to be normalized.
I guess it is that length of tangent vectors of all coordinate lines must be 1 in all points of space.$$\forall_a(|\frac{\partial \vec x}{\partial x^a}|=1)$$

 

[Orodruin]

My lecturer said that coordinatesystem is orthonormal if it is normalized and orthogonal, but I do not understand, which conditions a coordinatesystem must satisfy to be normalized.

This is not an appropriate way to describe a coordinate system. A set of coordinates being orthogonal means that its coordinate lines intersect at right angles. A set of vecors can be orthonormal and this is the case for the normalised basis vectors of orthogonal coordinates.

If you require the coordinate lines to have orthonormal tangent vectors, then you are restricting yourself to coordinate systems that preserve the form of the metric to be the Kronecker delta. This restricts coordinate transformations to translations and rotations and we already have a name for such coordinates: Cartesian coordinates.

 

[olgerm]

This is not an appropriate way to describe a coordinate system.

Why? Because cartesian coordinatesystem is only orthonormal coordinatesystem?

If you require the coordinate lines to have orthonormal tangent vectors, then you are restricting yourself to coordinate systems that preserve the form of the metric to be the Kronecker delta. This restricts coordinate transformations to translations and rotations and we already have a name for such coordinates: Cartesian coordinates.

Are you sure Cartesian coordinatesystem is only coordinatesystem, which tangent vectors to coordinatelines are orthonormal? Note that tangent vectors being orthonormal does not implicate that coordinatelines must not be bent nor that tangent vectors must be same in every point in space.

 

[Orodruin]

Are you sure Cartesian coordinatesystem is only coordinatesystem, which tangent vectors to coordinatelines are orthonormal? Note that tangent vectors being orthonormal does not implicate that coordinatelines must not be bent nor that tangent vectors must be same in every point in space.

Yes it does, by the very argument that I have already given you, the preservation of the metric tensor as the Kronecker delta.

Assume that $\vec E_i \cdot \vec E_j = \delta_{ij}$, then the Christoffel symbols are equal to zero because all the metric derivatives are zero. By definition, the Christoffel symbols are $\Gamma_{ij}^k = \vec E^k \cdot \partial_i \vec E_j$ and since this is zero for all $i,j,k$, it must hold that $\partial_i \vec E_j = 0$ and thus the $\vec E_j$ are the same everywhere.

 

[olgerm]

By definition, the Christoffel symbols are $\Gamma_{ij}^k = \vec E^k \cdot \partial_i \vec E_j$ and since this is zero for all $i,j,k$

Is this 0 if i=j?

Can explain more easily why cartesian coordinatesystem is only coordinatesystem, which tangent vectors to coordinatelines are orthonormal?

 

[Orodruin]

Is this 0 if i=j?

Yes. Again, all Christoffel symbols are equal to zero.

 

[olgerm]

I understand, that if $g=\delta$ ,then ${e}_i\cdot {e}_k=0$ (for all i and k if i≠k), but do not understand why ${\frac {\partial \mathbf {e} _{i}}{\partial x^{j}}}\cdot \mathbf {e} ^{k}$ (for all i, j and k).

 

[Orodruin]

I understand, that if $g=\delta$ ,then ${e}_i\cdot {e}_k=0$ (for all i and k if i≠k), but do not understand why ${\frac {\partial \mathbf {e} _{i}}{\partial x^{j}}}\cdot \mathbf {e} ^{k}$ (for all i, j and k).

Because those are the Christoffel symbols per definition. Expressed in the metric, the Christoffel symbols take the form
$$
\vec E^k \cdot \partial_i \vec E_j = \frac 12 g^{k\ell} (\partial_i g_{\ell j} + \partial_j g_{i \ell} - \partial_\ell g_{ij})
$$
and since all of the metric components are constant, this is necessarily zero - which means that all $\partial_i \vec E_j$ are identically equal to zero.