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Transformation of Christoffel Symbol

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:

  1. I cannot figure out how the last term in (3.16) can be converted to that in (3.17). Where does the minus sign come from?
  2. I am skeptical about the last term in (3.17) because I reached the same conclusion with this post that this term vanishes. The author actually uses equation (3.17) in p.68 of the textbook, so I’m really confused.

I appreciate any help.

One Answer

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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