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I Finding Covariant and Contravariant components of Cylin Coor

  1. Nov 2, 2017 #1
    I'm going through Introduction_to_Tensor_Calculus by Wiskundige_Ingenieurstechnieken.

    I want to find the covariant and contravariant components of the cylindrical cooridnates.

    Ingerniurstechnieken sets tangent vectors as basis (E1, E2, E3). He also sets the normal vectors as another basis (E1, E2, E3)

    The coorindates relationships are:

    x=r*cosθ
    y=r*sinθ
    z=z

    r=sqrt(x2+y2)
    θ=arctan(y/x)
    z=z

    I'm setting R= vector r

    R=[r*cosθ, r*sinθ, 1]

    I found:

    E1 = [cosθ, sinθ, 0]
    E2 = [-r*sinθ, r*cosθ, 0]
    E3 = [0, 0, 1]

    He says to take gradient of r, θ, and z for normal basis (E1, E2, E3). What do I take the gradient of to get this basis?

    Is it ∇r=(1, 0, 0)
    ∇θ=(0, 1, 0)
    ∇z=(0, 0, 1)?
     
  2. jcsd
  3. Nov 3, 2017 #2

    vanhees71

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    Your calculations of the basis vectors are correct. Now the most simple way to get the dual basis is to evaluate the metric components
    $$g_{jk}=\vec{E}_j \vec{E}_k.$$
    The nonvanishing components are
    $$g_{11}=\vec{E}_1 \cdot \vec{E}_1=1, \quad g_{22}=\vec{E}_2 \cdot \vec{E}_2=r^2, \quad g_{33}=\vec{E}_3 \cdot \vec{E}_3.$$
    It's more conveniently summarized in writing the length of an arbitrary line element,
    $$\mathrm{d} s^2=\mathrm{d} r^2 + r^2 \mathrm{d} \theta^2 + \mathrm{d} z^2.$$
    You should make a little drawing and understand this line element geometrically.

    To evaluate the dual basis, you need the contravariant metric components. They are defined by the inverse of the matrix of the covariant metric components. Here it's very easy since this matrix is diagonal. You get for the non-zero components
    $$g^{11}=\frac{1}{g_{11}}=1, \quad g^{22}=\frac{1}{g_{22}}=\frac{1}{r^2}, \quad g^{33}=\frac{1}{g_{33}}=1.$$
    Thus the cobasis is
    $$\vec{E}^j=\vec{E}_i g^{ij},$$
    i.e.,
    $$\vec{E}^1=\vec{E}_1, \quad \vec{E}^2=\frac{1}{r^2} \vec{E}_2, \quad \vec{E}^3=\vec{E}_3.$$
    It's easy to check that indeed
    $$\vec{E}^j \cdot \vec{E}_k=\delta_{k}^{j}$$
    as it must be.

    Obviously you have a misconception concerning the gradient. For an arbitrary scalar field ##\Phi## it's defined as
    $$\vec{\nabla} \Phi =\vec{E}^j \partial_j \Phi.$$
    You should also note that here we deal with the holonomes basis and cobasis of cylinder coordinates.

    Usually for orthogonal curvilinear coordinates one rather uses local orthornormal (Cartesian) basis. You can easily calculate everything by the definitions
    $$\vec{e}_j=\frac{1}{\sqrt{g_{jj}}} \vec{E}_j.$$
    If you understand German, you find everything in my lecture notes,

    https://th.physik.uni-frankfurt.de/~hees/publ/matherg1.pdf
     
  4. Nov 3, 2017 #3
    How do you type equations? It looks much better than what I'm writing.

    Thank you. That helps a lot.

    Using your formulation, I found the two basis:

    E1=[cosθ, sinθ, 0]
    E2=[-r*sinθ, r*cosθ, 0]
    E3=[0, 0, 1]

    E1=[cosθ, sinθ, 0]
    E2=[-sinθ/r, cosθ/r, 0]
    E3=[0, 0, 1]

    I would like to find A components.

    Writing out R=AmEm and R=AnEn.

    I got Am=An= [r, 0, z]
    Is this right?

    I did the same for spherical coordinates.

    E1=[cosθ*sin∅, sinθ*sin∅, cos∅]
    E2=[-r*sinθ*sin∅, r*cosθ*sin∅, 0]
    E3=[r*cosθ*cos∅, r*sinθ*cos∅, -r*sin∅]

    E1=[cosθ*sin∅, sinθ*sin∅, cos∅]
    E2=[-sinθ/(r*sin∅), cosθ/(r*sin∅), 0]
    E3=[cosθ*cos∅/r, sinθ*cos∅/r, -sin∅/r]

    Finding Am=An=[r, 0, 0]

    I hope the A's are correct. I do get the x, y, z components when they are paired with their basis.
     
  5. Nov 4, 2017 #4

    Orodruin

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    https://www.physicsforums.com/help/latexhelp/

    Your components look fine. However, I strongly suggest you get rid of the [...] notation in anything but Cartesian coordinates.

    Also, φ is the Greek letter, ∅ is the empty set symbol. :rolleyes:
     
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