- #1

- 6

- 1

- Homework Statement
- Following a book "Introducing Einstein's Relativity" by Ray d'Inverno, I came across this problem (6.16):

Show directly that $$\Gamma^a_{bc}=\frac{1}{2}g^{ad}(\partial_b g_{dc}+\partial_c g_{db}-\partial_d g_{bc})$$ transforms like a connection.

- Relevant Equations
- The definition of ##\Gamma^a_{bc}## given above, and the usual rules for transforming tensors, i. e. $$g'^{ab}=\frac{\partial x'^a}{\partial x^c}\frac{\partial x'^b}{\partial x^d}g^{cd}$$

A useful result is also

$$\partial'_b g'_{dc}=\frac{\partial x^e}{\partial x'^b} \frac{\partial x^f}{\partial x'^c} \frac{\partial x^g}{\partial x'^d}\partial_e g_{gf}+\frac{\partial x^e}{\partial x'^b}\frac{\partial}{\partial x^e}\left (\frac{\partial x^g}{\partial x'^d}\frac{\partial x^f}{\partial x'^c}\right ) g_{gf}$$

I simply just wrote down the definition of ##\Gamma'^a_{bc}##, and inserted the transformations of ##g'^{ad}##, ##g'_{dc,b}##, and the like terms. After some rearranging and cancelling out,

$$\Gamma'^a_{bc}=\frac{\partial x'^a}{\partial x^e}\frac{\partial x^f}{\partial x'^b}\frac{\partial x^g}{\partial x'^c}\Gamma^e_{fg}+\frac{\partial x'^a}{\partial x^e}\frac{\partial^2 x^e}{\partial x'^b\,\partial x'^c}+\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

The first two terms correspond to the usual transformation law for connections, hence the third term should vanish. If my calculations until this point are correct, the original problem reduces to proving that the following expression reduces to zero:

$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

However, I couldn't show this, despite making lots of attempts. One of more promising attempts is swapping the dummy indices ##g## and ##h## in the second term, and arriving at

$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x^g}{\partial x'^b}\frac{\partial^2 x^h}{\partial x'^c\, \partial x'^d}\left( \frac{\partial x'^d}{\partial x^h}g^{eh}-\frac{\partial x'^d}{\partial x^g}g^{eg}\right )$$

which would have to be proved to be equal to zero.

Unfortunately, I only know the basics of tensors, and don't know how such equation could be proved. As such, I am kindly asking you to post some insight on how such a problem could be tackled. Of course, I have almost certainly already made some errors, and I would be happy if you'd pointed that out as well.

$$\Gamma'^a_{bc}=\frac{\partial x'^a}{\partial x^e}\frac{\partial x^f}{\partial x'^b}\frac{\partial x^g}{\partial x'^c}\Gamma^e_{fg}+\frac{\partial x'^a}{\partial x^e}\frac{\partial^2 x^e}{\partial x'^b\,\partial x'^c}+\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

The first two terms correspond to the usual transformation law for connections, hence the third term should vanish. If my calculations until this point are correct, the original problem reduces to proving that the following expression reduces to zero:

$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

However, I couldn't show this, despite making lots of attempts. One of more promising attempts is swapping the dummy indices ##g## and ##h## in the second term, and arriving at

$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x^g}{\partial x'^b}\frac{\partial^2 x^h}{\partial x'^c\, \partial x'^d}\left( \frac{\partial x'^d}{\partial x^h}g^{eh}-\frac{\partial x'^d}{\partial x^g}g^{eg}\right )$$

which would have to be proved to be equal to zero.

Unfortunately, I only know the basics of tensors, and don't know how such equation could be proved. As such, I am kindly asking you to post some insight on how such a problem could be tackled. Of course, I have almost certainly already made some errors, and I would be happy if you'd pointed that out as well.