Understanding Coordinate Transforms: Partial Derivatives

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Transforming coordinates using partial derivatives is primarily applicable for local transformations or when adjusting differentials in integrals, such as applying the Jacobian. The Jacobian matrix plays a crucial role in these transformations, allowing for the conversion of volume elements between coordinate systems. Understanding covariant and contravariant vectors is essential for grasping how these transformations function. The discussion emphasizes the importance of the Jacobian in calculating changes in coordinates rather than using partial derivatives directly for the transformation itself. Clarifying these concepts can significantly aid in understanding coordinate transformations.
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Hey, I am having trouble understanding how you can transform one set of coordinates into another using partial derivatives i just don't get the whole thought process behind it. I came across this while reading about covariant and contrivariant vectors. If anyone can help clear up how this works i would really appreciate it. Also if i need to be more specific about what i mean or if you would like to see an example of what I am talking about just ask and i would be happy to post it up.
 
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Hi Storm Butler! :wink:
Storm Butler said:
Hey, I am having trouble understanding how you can transform one set of coordinates into another using partial derivatives …

We don't use partial derivatives for the transformation itself, we can only use them for a very local transformation, or for transforming the d part in an integral: ∫∫∫dx dy dz = ∫∫∫ (jacobian) da db dc.

see http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant" for some details. :smile:
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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