Derivative of metric along curve
$begingroup$
Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
$$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).
I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
$$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
but
$$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$
I do not see how these two expressions are equal.
Is there something wrong? Is there something I am missing?
Thanks!
differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
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Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
$$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).
I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
$$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
but
$$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$
I do not see how these two expressions are equal.
Is there something wrong? Is there something I am missing?
Thanks!
differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
$$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).
I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
$$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
but
$$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$
I do not see how these two expressions are equal.
Is there something wrong? Is there something I am missing?
Thanks!
differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $D$. Let $alpha : [a,b] rightarrow M$ be a smooth curve on $M$, and let $frac{D}{dt}$ be the induced covariant derivative on $alpha$. I want to prove that for all $X,Y$ smooth vector fields on $alpha$,
$$frac{d}{dt}big{(}g(X,Y)big{)} = g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) $$
(Proposition 3.18(4) from O'Neill's Semi-Riemannian Geometry).
I've tried using substituting coordinates (for a chart with coordinate functions $x^1,...,x^n$), which is what the book suggests: $g(X,Y) = g_{ij}X^iY^j$, and $frac{D}{dt}X = left( frac{dX^i}{dt} + Gamma^i_{jk}frac{d(x^jcircalpha)}{dt}X^kright)partial_i$. But I am stuck:
$$frac{d}{dt}big{(}g(X,Y)big{)} = frac{dg_{ij}}{dt}X^iY^j + g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt}$$
but
$$ g(frac{D}{dt}X,Y) + g(X,frac{D}{dt}Y) = g_{ij}frac{dX^i}{dt}Y^j + g_{ij}X^ifrac{dY^j}{dt} + g_{ij}(Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^lY^j + Gamma^i_{kl}frac{d(x^kcircalpha)}{dt}X^jY^l) $$
I do not see how these two expressions are equal.
Is there something wrong? Is there something I am missing?
Thanks!
differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry
differential-geometry riemannian-geometry general-relativity semi-riemannian-geometry
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
asked Jan 7 at 2:08
Quim LlorensQuim Llorens
132
132
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Quim Llorens is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$
Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.
answered Jan 7 at 7:24
kobekobe
34.8k22247
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$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$
Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.
answered Jan 7 at 7:24
kobekobe
34.8k22247
$endgroup$
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
add a comment |
$begingroup$
From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$
Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.
answered Jan 7 at 7:24
kobekobe
34.8k22247
$endgroup$
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
add a comment |
$begingroup$
From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$
Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.
answered Jan 7 at 7:24
kobekobe
34.8k22247
$endgroup$
From the last equation, let's manipulate the last group of terms on the right-hand side. We can rewrite it as
$$g_{ij}Gamma^{i}_{kl}dot{x}^k X^l Y^j + g_{il}Gamma^{i}_{kj}dot{x}^k X^lY^j = (g_{ij}Gamma^i_{kl} + g_{il}Gamma^i_{kj})dot{x}^kX^lY^j$$
Since $g_{ij}Gamma^{i}_{kl} = dfrac{1}{2}(partial_k g_{lj} + partial_l g_{jk} - partial_j g_{kl})$ and $g_{il}Gamma^{i}_{kj} = dfrac{1}{2}(partial_k g_{jl} + partial_j g_{lk} - partial_l g_{kj})$, then $g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj} = partial_k g_{lj}$. By the chain rule, $partial_k g_{lj}, dot{x}^k = dfrac{dg_{lj}}{dt}$. Hence,
$$(g_{ij}Gamma^{i}_{kl} + g_{il}Gamma^{i}_{kj})dot{x}^k X^l Y^j = frac{d g_{lj}}{dt}X^l Y^j = frac{d g_{ij}}{dt} X^i Y^j$$ as was needed.
answered Jan 7 at 7:24
kobekobe
34.8k22247
answered Jan 7 at 7:24
kobekobe
34.8k22247
answered Jan 7 at 7:24
kobekobe
34.8k22247
answered Jan 7 at 7:24
kobekobe
34.8k22247
34.8k22247
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
add a comment |
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
$begingroup$
Lovely, thank you!!!! :)
$endgroup$
– Quim Llorens
Jan 7 at 12:38
add a comment |
Quim Llorens is a new contributor. Be nice, and check out our Code of Conduct.
Quim Llorens is a new contributor. Be nice, and check out our Code of Conduct.
Quim Llorens is a new contributor. Be nice, and check out our Code of Conduct.
Quim Llorens is a new contributor. Be nice, and check out our Code of Conduct.
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