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Physics
Special and General Relativity
Deriving Contravariant Form of Levi-Civita Tensor
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[QUOTE="AndersF, post: 6531318, member: 677147"] [B]TL;DR Summary:[/B] If the covariant form for the Levi-Civita is defined as ##\varepsilon_{i,j,k}:=\sqrt{g}\epsilon_{i,j,k}##, how could be shown from this definition that it's contravariant form is given by ##\varepsilon^{i,j,k}=\frac{1}{\sqrt{g}}\epsilon^{i,j,k}##? The covariant form for the Levi-Civita is defined as ##\varepsilon_{i,j,k}:=\sqrt{g}\epsilon_{i,j,k}##. I want to show from this definition that it's contravariant form is given by ##\varepsilon^{i,j,k}=\frac{1}{\sqrt{g}}\epsilon^{i,j,k}##.[B]My attempt[/B]What I have tried is to express this tensor ##\varepsilon^{i j k}## through the contraction with the metric tensor of ##\varepsilon_{i j k}## the contravariant form, and then to replace the definition of ##\varepsilon_{i j k}##:##\varepsilon^{i j k}=g^{i p} g^{j q} g^{k r} \varepsilon_{p q r}=g^{i p} g^{j q} g^{k r} \sqrt{g} \epsilon_{p q r}## This expression reminds me of the the expression of the determinant of the dual metric tensor, ##g^{-1}=\det (g^{ij})##, through the Levi-Civita symbol: ##g^{-1}=g^{1 p} g^{2 q} g^{3 r} \epsilon_{p q r}## But I'm stuck here, as I don't know how to match these expressions... Would there be any way to achieve this? Would this be a good way to prove the theorem? [/QUOTE]
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Special and General Relativity
Deriving Contravariant Form of Levi-Civita Tensor
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