- #1
Pencilvester
- 184
- 42
Hello PF,
I was reading through “A First Course in General Relativity” by Schutz and I got to the part where he derives the divergence formula for a vector:$$V^α { } _{;α} = \frac {1} {\sqrt{-g}} ( \sqrt{-g} V^α )_{,α}$$I’m having trouble with a couple of the steps he made. So we start with the standard covariant derivative formula:$$V^α { } _{;α} = V^α { } _{,α} + Γ^α_{μα} V^μ$$We use the formula for ##Γ## that uses the metric and its derivatives:$$Γ^α_{μα} = \frac {1} {2} g^{αβ} (g_{βα,μ} + g_{βμ,α} - g_{μα,β})$$then we rearrange to get$$Γ^α_{μα} = \frac {1} {2} g^{αβ} (g_{βμ,α} - g_{μα,β}) + \frac {1} {2} g^{αβ} g_{βα,μ}$$and we notice that ##\frac {1} {2} (g_{βμ,α} - g_{μα,β})## is antisymmetric on ##α## and ##β## and therefore contracting with ##g^{αβ}## (which is symmetric) gets you a 0. So we’re left with$$Γ^α_{μα} = \frac {1} {2} g^{αβ} g_{βα,μ}$$I can follow this so far just fine. This next part is where I start having issues. He tells me the derivative of the determinant of the metric (##g##) is this:$$g_{,μ} = g g^{αβ} g_{βα,μ}$$and I can’t for the life of me work this out on my own. It’s probably due to the fact that I have no experience manipulating determinants even algebraically, let alone taking their derivatives. I would assume you somehow use the Laplace expansion way of defining the determinant, but again, I have no experience manipulating the minor of a matrix. So my first question is how do you derive this?
Moving on, it looks to me that if we just divide the previous equation by ##g##, we can sub back into the equation for ##Γ## and get$$Γ^α_{μα} = \frac {g_{,μ}} {2g}$$but in the book, without any explanation, he gets$$Γ^α_{μα} = \frac {(\sqrt {-g})_{,μ}} {\sqrt {-g}}$$and again, I have no idea where this came from. The rest of it I can follow, but any help on either of my two issues here would be much appreciated. Thank you!
I was reading through “A First Course in General Relativity” by Schutz and I got to the part where he derives the divergence formula for a vector:$$V^α { } _{;α} = \frac {1} {\sqrt{-g}} ( \sqrt{-g} V^α )_{,α}$$I’m having trouble with a couple of the steps he made. So we start with the standard covariant derivative formula:$$V^α { } _{;α} = V^α { } _{,α} + Γ^α_{μα} V^μ$$We use the formula for ##Γ## that uses the metric and its derivatives:$$Γ^α_{μα} = \frac {1} {2} g^{αβ} (g_{βα,μ} + g_{βμ,α} - g_{μα,β})$$then we rearrange to get$$Γ^α_{μα} = \frac {1} {2} g^{αβ} (g_{βμ,α} - g_{μα,β}) + \frac {1} {2} g^{αβ} g_{βα,μ}$$and we notice that ##\frac {1} {2} (g_{βμ,α} - g_{μα,β})## is antisymmetric on ##α## and ##β## and therefore contracting with ##g^{αβ}## (which is symmetric) gets you a 0. So we’re left with$$Γ^α_{μα} = \frac {1} {2} g^{αβ} g_{βα,μ}$$I can follow this so far just fine. This next part is where I start having issues. He tells me the derivative of the determinant of the metric (##g##) is this:$$g_{,μ} = g g^{αβ} g_{βα,μ}$$and I can’t for the life of me work this out on my own. It’s probably due to the fact that I have no experience manipulating determinants even algebraically, let alone taking their derivatives. I would assume you somehow use the Laplace expansion way of defining the determinant, but again, I have no experience manipulating the minor of a matrix. So my first question is how do you derive this?
Moving on, it looks to me that if we just divide the previous equation by ##g##, we can sub back into the equation for ##Γ## and get$$Γ^α_{μα} = \frac {g_{,μ}} {2g}$$but in the book, without any explanation, he gets$$Γ^α_{μα} = \frac {(\sqrt {-g})_{,μ}} {\sqrt {-g}}$$and again, I have no idea where this came from. The rest of it I can follow, but any help on either of my two issues here would be much appreciated. Thank you!