The Lagrangian function [tex]L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )[/tex] on a field theory's action-integral is a generally covariant scalar-density so the resulting field equations obtained by using the standard Lagrange's expresions will inevitably involve tensor-densities:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]I\equiv k.\int _{D}L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} ).d\Omega[/tex]

[tex]\delta I\equiv 0[/tex] [tex]\leftrightarrow[/tex] [tex]\partial _{k} \left(\frac{\partial L}{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial L}{\partial \mathop{q}\limits_{(x)} } =0[/tex] (tensor densitie's equations)

The following method avoids working with them: let's assume the Lagrangian density can be represented as the product of a reference density (which can never be null) and a scalar function both depending on the dynamic variables:

[tex]L(\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )\equiv \sigma (\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} ).\Lambda (\mathop{q}\limits_{(x)} ,\partial _{\bullet } \mathop{q}\limits_{(x)} )[/tex]

Based on them the Lagrange's equations can be modified for directly returning tensor expressions:

[tex]\[\partial _{k} \left(\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial \Lambda }{\partial \mathop{q}\limits_{(x)} } +\left(\partial _{k} (\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } )-\frac{\partial \psi }{\partial \mathop{q}\limits_{(x)} } +\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \psi \right).\Lambda +\frac{\partial \psi }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \Lambda +\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } .\partial _{k} \psi =0\][/tex]

/ [tex]\psi \equiv \ln (\sigma )[/tex] , [tex]\sigma \ne 0[/tex]

On a curved Riemann space (also on a flat one when working with curvilinear coordinates) there is already a reference density based on the metric tensor:

[tex]\sigma (g_{\bullet \bullet } )\equiv \sqrt{-\left|g_{\bullet \bullet }[/tex]

[tex]\psi \equiv \ln (\sqrt{-\left|g_{\bullet \bullet } \right|} )[/tex] [tex]\to[/tex] [tex]\frac{\partial \psi }{\partial g^{ij} } =-{\tfrac{1}{2}} .g_{ij}[/tex] , [tex]\partial _{i} \psi =\Gamma _{i} [/tex]

For a metric-affine dependent scalar function the equivalent Lagrange's equations become:

[tex]\Lambda \equiv \Lambda (g^{\bullet \bullet } ,{\Gamma }^{\bullet } _{\bullet \bullet } ,\partial _{\bullet } {\Gamma }^{\bullet } _{\bullet \bullet }) [/tex]

[tex]\to[/tex] [tex]\partial _{k} \left(\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } \right)-\frac{\partial \Lambda }{\partial \mathop{q}\limits_{(x)} } +\Gamma _{k} .\frac{\partial \Lambda }{\partial \partial _{k} \mathop{q}\limits_{(x)} } =0[/tex] / [tex]\mathop{q}\limits_{(x)} \ne g^{ij}[/tex]

[tex]\frac{\partial \Lambda }{\partial g^{ij} } -{\tfrac{1}{2}} .g_{ij} .\Lambda =0[/tex] / [tex]\mathop{q}\limits_{(x)} =g^{ij} [/tex]

This can be applied directly to General Relativity's action for obtaining the known Einstein's equations:

[tex]\[I_{G} \equiv \int _{D}\sqrt{-\left|g_{\bullet \bullet } \right|} .g^{kr} .R_{kr}. d\Omega \] [/tex]

[tex]\to[/tex] [tex]\nabla _{i} g^{jk} =0 [/tex]

[tex]\[R_{ij} -{\tfrac{1}{2}} .R.g_{ij} =0\] [/tex]

I hope this helps.

CGR_JAMA

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# Alternative Lagrange's Equations

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