- #1
Shirish
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I'm reading 'Core Principles of Special and General Relativity' by Luscombe, specifically the introductory section on problems with defining usual notion of differentiation for tensor fields. I'll quote the relevant part:
$$\bigg(\frac{\partial T^i}{\partial x^j}\bigg)_p=\partial_j(T^i\circ x^{-1})(x(p))=\lim_{h\to 0}\frac{(T^i\circ x^{-1})(x(p)+[0,\ldots,h,\ldots,0])-(T^i\circ x^{-1})(x(p))}{h}$$ where ##[0,\ldots,h,\ldots,0]\in\mathbb{R}^n## has ##h## as its ##j##-th coordinate.
Is my above interpretation correct? If so, what's the issue with defining the derivative of a vector field component in this way?
Since the equation above is a notational mess, here's my attempt to interpret it:The second way (to see whether the partial derivative of a tensor is a tensor) is to look at the definition of derivative, $$\frac{\partial T^i}{\partial x^j}=\lim_{dx^j\to 0}\frac{T^i(x+dx^j)-T^i(x)}{dx^j}$$ The numerator is not in general a vector! We're comparing (subtracting) vectors from different points, yet the transformation properties of tensors are defined at a point.
$$\bigg(\frac{\partial T^i}{\partial x^j}\bigg)_p=\partial_j(T^i\circ x^{-1})(x(p))=\lim_{h\to 0}\frac{(T^i\circ x^{-1})(x(p)+[0,\ldots,h,\ldots,0])-(T^i\circ x^{-1})(x(p))}{h}$$ where ##[0,\ldots,h,\ldots,0]\in\mathbb{R}^n## has ##h## as its ##j##-th coordinate.
Is my above interpretation correct? If so, what's the issue with defining the derivative of a vector field component in this way?