Physics Asked on May 10, 2021
I’m studying General Relativity by Hobson and got stuck in understanding the derivation of the transformation law of Christoffel symbols (p. 64). Let $x^a$ and $x’^a$ be two coordinate systems on a manifold. In this book, the Christoffel symbol (or the affine connection) is defined by $$frac{partial vec{e}_a}{partial x^c} = Gamma^b_{,ac} vec{e}_b tag{3.12}$$ where $vec{e}_a$ denote the basis vectors corresponding to the $x_a$ coordinates. After some calculations, the book reaches the transformation law below which can be found in many other sources, e.g. equation (10) here:
$$Gamma’^a_{,,bc} = vec{e}’^a cdot frac{partial vec{e}’_b}{partial x’^c} = cdots = frac{partial x’^a}{partial x^d}frac{partial x^f}{partial x’^b}frac{partial x^g}{partial x’^c}Gamma^d_{,fg} + frac{partial x’^a}{partial x^d}frac{partial^2 x^d}{partial x’^c partial x’^b} ,. tag{3.16}$$
This is all very well, but I have a trouble with the next step. The book says that by swapping derivatives with respect to $x$ and $x’$ in the last term on the right-hand side of (3.16), we can arrive at an alternative but equivalent expression:
$$ Gamma’^a_{,,bc} = frac{partial x’^a}{partial x^d}frac{partial x^f}{partial x’^b}frac{partial x^g}{partial x’^c}Gamma^d_{,fg} – frac{partial x^d}{partial x’^b}frac{partial x^f}{partial x’^c}frac{partial^2 x’^a}{partial x^d partial x^f} ,. tag{3.17}$$
My question is:
I appreciate any help.
I was also blocking for a time on this point in the book, I came accross your question which was helpfull for me.
In fact, I think the intuitive way is using the derivative of the product of functions:
In 3.16, let assume $f$ to be the first term $ frac{partial x^{'a}}{partial x^{d}} $ and $g'$ to be the second $ frac{{partial}^2 x^{d}}{partial x^{'c}partial x^{'b}} $ ):
Then considering that $(fg)'=f'g+fg' rightarrow fg'=(fg)'-f'g$ (this is where the minus sign comes from.
So you then have: $ g=frac{partial x^{d}}{partial x^{'b}} $, $ f'= frac{{partial}^2 x^{'a}}{partial x^{d}partial x^{'c}}$
which yields that $fg = delta^a_b$ which derivative is $0$: $(fg)'=0$
and $ -f'g=-frac{{partial}^2 x^{'a}}{partial x^{d}partial x^{'c}} frac{partial x^{d}}{partial x^{'b}} = -frac{{partial}^2 x^{'a}}{partial x^{d}partial x^{f}} frac{partial x^{d}}{partial x^{'b}} frac{partial x^{f}}{partial x^{'c}} $, the second term in 3.17
Is all of this correct?
Answered by FrançoisF on May 10, 2021
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