Understanding dual basis & scale factors

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SUMMARY

This discussion clarifies the relationship between dual basis vectors and scale factors in curvilinear coordinates. The tangent vectors are defined as ##\vec e_i = h_i \nabla u_i##, where ##h_i = \left| \frac{\partial \vec r }{\partial u_i} \right|##. The gradient ##\nabla u_i## can be taken directly from the curvilinear coordinates, and the Jacobian ##J## equals the product of the scale factors ##h_1h_2h_3## only in orthogonal coordinate systems. This understanding is crucial for normalizing basis vectors in differential geometry.

PREREQUISITES
  • Understanding of curvilinear coordinates and their transformations
  • Familiarity with gradient operations in vector calculus
  • Knowledge of Jacobians and their significance in coordinate transformations
  • Basic concepts of differential geometry and basis vectors
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  • Study the properties of orthogonal coordinate systems in differential geometry
  • Learn about the derivation and application of Jacobians in transformations
  • Explore the concept of normalized basis vectors in vector calculus
  • Investigate the relationship between scale factors and metric tensors
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Incand
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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##.
 
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Incand said:
##\{ 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?

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

Incand said:
And why multiply with the scale factors instead of dividing?

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.

Incand said:
Another question I'm wondering about is if it's always true that the jacobian ##J## is equal too ##h_1h_2h_3##.
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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