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Understanding dual basis & scale factors

  1. Jul 7, 2015 #1
    I'm confused by the following passage in our book (translated).
    An alternative too choosing the normed tangent vectors ##\vec e_i = \frac{1}{h_1}\frac{\partial \vec r}{\partial u_i}## with scale factors ##h_i = \left| \frac{\partial \vec r }{\partial u_i} \right| ## is to choose the normal vector ##\nabla u_i##.
    If the system is ortogonal the vectors ##\nabla u_i## and ##\frac{\partial \vec r}{\partial u_i}## point in the same direction so we can write ##\vec e_i = h_i \nabla u_i##.

    ##\{ u_i \}_{i=1}^3## is supposed to be curvilinear coordinates with a transformation describing the position ##\vec r = \vec r (u_1,u_2,u_3)##.
    What does it mean to take ##\nabla u_i##? am I supposed to express ##u_i## in cartesian coordinates and then take the gradient? And why multiply with the scale factors instead of dividing?
    Another question I'm wondering about is if it's always true that the jacobian ##J## is equal too ##h_1h_2h_3##.
     
  2. jcsd
  3. Jul 7, 2015 #2

    Orodruin

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    That would be one way of doing it yes. The coordinates ##u_i## are functions and therefore you can take the gradient of them.

    Because this is how you would obtain normalised base vectors. The norm of the dual basis is the reciprocal of the norm of the tangent vector basis.

    It is only true in orthogonal coordinate systems (which are the only ones for which it makes sense to deal with the scale factors rather than the full metric tensor).
     
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