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 February 24, 2005, 11:30 Gradients. #1 J. Ca Guest   Posts: n/a Hi all, Can anyone tell me how to calculate the angle between the gradient of a scalar field and a vector field in general curvilinear cordinates. Cheers.

 February 24, 2005, 15:06 Re: Gradients. #2 Runge_Kutta Guest   Posts: n/a Well, first you have to decide how you have represented the two vectors; as contravariant or covariant vectors. Also, you need your metric tensor. So, for n^{i} = nabla(scalar) v^{j} = vector n^{i}g_{ij}v^{j} = Sqrt(n^{i}g_{ij}n^{j})*Sqrt(n^{i}g_{ij}n^{j}) Cos(theta) or n_{i} = nabla(scalar) v_{j} = vector n_{i}g^{ij}v_{j} = Sqrt(n_{i}g^{ij}n_{j})*Sqrt(n_{i}g^{ij}n_{j}) Cos(theta) You could also use physical components.

 February 25, 2005, 07:27 Re: Gradients. #3 J. Ca Guest   Posts: n/a Thank you Runge_Kutta. What is the g^{ij} you use in your notation above? Also, can you give me a simple intepretation of what the terms covariant and contravariant mean, particularly as applied to vectors. I am having difficulty understanding books when they try to explain this.

 February 25, 2005, 14:04 Re: Gradients. #4 Runge_Kutta Guest   Posts: n/a g^{ij} is the inverse of g_{ij} - the metric tensor. A year ago, I could have said something intellegent about covariant and contravariant quantities but I'm not going to post my fuzzy recollections on this board. Sorry!

 February 25, 2005, 15:11 Re: Gradients. #5 J. Ca Guest   Posts: n/a Thank you. Yes, I am empatize with you!

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