- #1
ChrisVer
Gold Member
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I have one question, which I don't know if I should post here again, but I found it in GR...
When you have a metric tensor with components:
[itex] g_{\mu \nu} = \eta _{\mu \nu} + h_{\mu \nu}, ~~ |h|<<1 [/itex] (weak field approximation).
Then the inverse is:
[itex] g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu} [/itex] right? However that doesn't give exactly that [itex] g^{\mu \rho}g_{\rho \nu} = \delta^{\mu}_{\nu} [/itex] because of the existence of the [itex]- h^{\mu \rho}h_{\rho \nu} [/itex] which is of course small but it's not zero... Can the inverse matrix be defined approximately?
Also I don't understand why should the derivatives of [itex]h[/itex] behave as [itex]h[/itex] itself? I mean they take the terms like [itex] h \partial h, ~~ \partial h \partial h [/itex] to be of order [itex]\mathcal{O}(h^2)[/itex]... why?
When you have a metric tensor with components:
[itex] g_{\mu \nu} = \eta _{\mu \nu} + h_{\mu \nu}, ~~ |h|<<1 [/itex] (weak field approximation).
Then the inverse is:
[itex] g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu} [/itex] right? However that doesn't give exactly that [itex] g^{\mu \rho}g_{\rho \nu} = \delta^{\mu}_{\nu} [/itex] because of the existence of the [itex]- h^{\mu \rho}h_{\rho \nu} [/itex] which is of course small but it's not zero... Can the inverse matrix be defined approximately?
Also I don't understand why should the derivatives of [itex]h[/itex] behave as [itex]h[/itex] itself? I mean they take the terms like [itex] h \partial h, ~~ \partial h \partial h [/itex] to be of order [itex]\mathcal{O}(h^2)[/itex]... why?
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