Calculating G' for an Orthogonal Coordinate System

In summary: I'm not sure if I explained it well but in summary, the first method calculates G' using the inverse transformation formula, and the second method calculates G' by multiplying the primed basic vectors a1 and a2. I don't know if you're supposed to fill in the basis vectors with the value 1 or just leave them like that, but either way the result will be the same. I think you should try doing it that way to see if you get the same result.
  • #1
PhysicsDude1
8
0

Homework Statement



For the orthonormal coordinate system (X,Y) the metric is

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Calculate G' in 2 ways.

1) G'= M[itex]^{T}[/itex]*G*M
2) g[itex]\acute{}[/itex][itex]_{ij}[/itex] = [itex]\overline{a}\acute{}_{i}[/itex] . [itex]\overline{a}\acute{}_{j}[/itex]

Homework Equations



\begin{pmatrix} \overline{a}\acute{}_{1} \\ \overline{a}\acute{}_{2} \end{pmatrix}
= \begin{pmatrix} -cos(\phi).\overline{a}\acute{}_{1} -\overline{a}\acute{}_{2} \\
cos(\phi).\overline{a}\acute{}_{2}\end{pmatrix}

M= \begin{pmatrix} -cos(\phi) & 0 \\ -1 & cos(\phi) \end{pmatrix}
M[itex]^{T}[/itex] = \begin{pmatrix} -cos(\phi) & -1 \\ 0 & cos(\phi) \end{pmatrix}
G=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
=> G' = \begin{pmatrix} cos²(\phi) +1 & -cos(\phi) \\ -cos(\phi) & cos²(\phi) \end{pmatrix}


The Attempt at a Solution




So I'm having problems with the 2nd method i.e. g[itex]\acute{}[/itex][itex]_{ij}[/itex] = [itex]\overline{a}\acute{}_{i}[/itex] . [itex]\overline{a}\acute{}_{j}[/itex]

g[itex]\acute{}[/itex][itex]_{11}[/itex] = (-cos[itex](\phi)[/itex] . [itex]\overline{a}_{1}[/itex] -[itex]\overline{a}_{2}[/itex]) . (-cos[itex](\phi)[/itex] . [itex]\overline{a}_{1}[/itex] -[itex]\overline{a}_{2}[/itex]) = ?

What are the values for [itex]\overline{a}_{1}[/itex] and [itex]\overline{a}_{2}[/itex] ?
I think they're both 1 because they're both unit vectors of length 1 but I'm not sure.
Also, this is the first time ever I have used LaTeX so sorry if it's a bit sloppy.
 
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  • #2
Also, I can't fall asleep because of this. I REALLY want to know the answer :p
 
  • #3
If there's something confusing, just ask :)
 
  • #4
PhysicsDude1 said:
If there's something confusing, just ask :)

I don't know what problem you are trying solve. I know what "orthogonal coordinates" means and I know what "orthonormal vector fields" are. I not sure I know what "orthonormal coordinates" are. I can guess but could you explain that? Might be best to start with the definitions before you plow into trying to prove things. That might make it a lot easier than you think. And what IS this G' you are trying to calculate anyway?
 
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  • #5
Dick said:
I don't know what problem you are trying solve. I know what "orthogonal coordinates" means and I know what "orthonormal vector fields" are. I not sure I know what "orthonormal coordinates" are. I can guess but could you explain that? Might be best to start with the definitions before you plow into trying to prove things. That might make it a lot easier than you think. And what IS this G' you are trying to calculate anyway?


I'm sorry, I meant orthogonal coordinate system :)
Basically I'm trying to find G' which is the metric of the coordinate system (X', Y').
There are 2 ways to find this :
1) G' = M^T * G * M (M=inverse transformationformula // M^T = the transposed matrix of M // G = metric of the coordinate system (X,Y) )

2) Via the multiplication of the primed basic vectors a1 and a2 (see first matrix) and that's where I'm stuck. I don't know if I'm supposed to fill in those basis vectors with the value 1 or just leave them like that BUT that would mean that the 2nd method has a different result from the 1st method which can't be possible. I think I have to fill it in.

Sorry for all the confusion!
 

Related to Calculating G' for an Orthogonal Coordinate System

1. What is a metric orthonormal basis?

A metric orthonormal basis, also known as an orthonormal basis for a metric space, is a set of vectors that are orthogonal (perpendicular) to each other and have a unit length, with respect to a given metric. This means that the inner product (or dot product) of any two vectors in the basis is equal to 0, and the norm (or length) of each vector is equal to 1.

2. What is the importance of a metric orthonormal basis?

A metric orthonormal basis plays a crucial role in many mathematical and scientific fields, such as linear algebra, functional analysis, and quantum mechanics. It provides a convenient way to represent vectors in a metric space, and allows for easier calculations involving inner products, norms, and projections.

3. How is a metric orthonormal basis constructed?

A metric orthonormal basis can be constructed using the Gram-Schmidt process, which is a method for finding an orthogonal basis from a set of linearly independent vectors. This process involves repeatedly subtracting the projections of previously constructed vectors from the original set, until the resulting set is orthogonal. Then, the vectors are normalized to have a unit length to form an orthonormal basis.

4. Can a metric orthonormal basis exist in any metric space?

No, a metric orthonormal basis can only exist in a metric space that satisfies certain conditions. For example, the metric space must have a defined inner product, and the vectors in the basis must be finite in number and linearly independent. Additionally, the metric space must also satisfy the axioms of a normed vector space.

5. How is a metric orthonormal basis different from a standard orthonormal basis?

A metric orthonormal basis is similar to a standard orthonormal basis in that both sets of vectors are orthogonal (perpendicular) and have a unit length. However, a metric orthonormal basis is specific to a particular metric space, while a standard orthonormal basis is typically associated with a Euclidean space. Additionally, the inner product in a metric orthonormal basis may not necessarily be the dot product, as in the case of a non-Euclidean metric space.

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