Curvilinear coordinate system: Determine the standardized base vectors

In summary, the position vector is always \mathbf{r}(u,v,w) = (x(u,v,w),y(u,v,w),z(u,v,w)). You don't have to invert it to get (x,y,z) in terms of (u,v,w). The basis vector points in the direction of increasing u_i and is normal to surfaces of constant u_i. If the \nabla u_i are orthogonal, then \nabla u_i and \frac{\partial \mathbf{r}}{\partial u_i} are parallel.
  • #1
Karl Karlsson
104
12
Homework Statement
Determine the standardized base vectors in the curvilinear coordinate system $$\begin{cases} u_1 = x^2-y^2 \\ u_2 = xy \\ u_3 = z\end{cases}$$
Relevant Equations
$$\begin{cases} u_1 = x^2-y^2 \\ u_2 = xy \\ u_3 = z\end{cases}$$
How I would have guessed you were supposed to solve it:
IMG_0716.jpg


What you are supposed to do is just take the gradients of all the u:s and divide by the absolute value of the gradient? But what formula is that why is the way I did not the correct way to do it?

Thanks in advance!
 
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  • #2
The position vector is always [itex]\mathbf{r}(u,v,w) = (x(u,v,w),y(u,v,w),z(u,v,w))[/itex]. Here you are given [itex]u(x,y,z)[/itex] etc., and you would have to invert this to get [itex](x,y,z)[/itex] in terms of [itex](u,v,w)[/itex], but you don't have to.

The basis vector [itex]\mathbf{e}_{u_i}[/itex] points in the direction of increasing [itex]u_i[/itex] and is normal to surfaces of constant [itex]u_i[/itex]. The vector [itex]\nabla u_i[/itex] does exactly this.

Now it turns out that if the [itex]\nabla u_i[/itex] are orthogonal then [itex]\nabla u_i[/itex] and [itex]\frac{\partial \mathbf{r}}{\partial u_i}[/itex] are parallel.
 
  • #3
pasmith said:
The position vector is always [itex]\mathbf{r}(u,v,w) = (x(u,v,w),y(u,v,w),z(u,v,w))[/itex]. Here you are given [itex]u(x,y,z)[/itex] etc., and you would have to invert this to get [itex](x,y,z)[/itex] in terms of [itex](u,v,w)[/itex], but you don't have to.

The basis vector [itex]\mathbf{e}_{u_i}[/itex] points in the direction of increasing [itex]u_i[/itex] and is normal to surfaces of constant [itex]u_i[/itex]. The vector [itex]\nabla u_i[/itex] does exactly this.

Now it turns out that if the [itex]\nabla u_i[/itex] are orthogonal then [itex]\nabla u_i[/itex] and [itex]\frac{\partial \mathbf{r}}{\partial u_i}[/itex] are parallel.
Is it because they have taken ##grad φ = \sum_{i} \frac {1} {h_i} \frac {\partial φ} {\partial u_i}## and ##φ=u_i##?
 

Related to Curvilinear coordinate system: Determine the standardized base vectors

1. What is a curvilinear coordinate system?

A curvilinear coordinate system is a coordinate system that uses curved lines to define the position of a point in space. It is different from a Cartesian coordinate system, which uses straight lines.

2. What are standardized base vectors in a curvilinear coordinate system?

Standardized base vectors in a curvilinear coordinate system are unit vectors that are used to define the direction of the coordinate axes. They are typically perpendicular to each other and are often denoted by e1, e2, e3, etc.

3. How do you determine the standardized base vectors in a curvilinear coordinate system?

The standardized base vectors in a curvilinear coordinate system can be determined by taking the partial derivatives of the position vector with respect to each coordinate. These partial derivatives are then normalized to obtain the unit vectors.

4. What is the importance of standardized base vectors in a curvilinear coordinate system?

The standardized base vectors are important because they define the orientation of the coordinate system and are used to express vectors and tensors in terms of the curvilinear coordinates. They also allow for the calculation of derivatives and other operations in the coordinate system.

5. Can standardized base vectors change in a curvilinear coordinate system?

Yes, the standardized base vectors can change in a curvilinear coordinate system. This is because the coordinate axes may vary in direction and magnitude at different points in the space, resulting in different base vectors. However, they still follow the rules of being perpendicular to each other and are normalized to have a magnitude of 1.

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