Einstein and Riemann curvature tensor












2












$begingroup$


Riemann curvature tensor is dirrectly related to a path dependence of parallel transport. I read that Einstein first thought of this tensor to be the one that goes into his field equation but it didn't fit. Now, I know why this was problematic. There is a clear explanation of that. What I am asking is, is there a formulation of theory of gravity where equations give us directly $R$ and not $G$? $G$ is Einsteins tensor and $R$ Riemanns. Because, non-relativistic equation was used and directly translated, with energy-momentum tensor. Is there some other tensor related to the physics which would give $R$ instead of $G$? $G$ itself is not curvature...but from $G$ I guess we could calculate $R$ so there should be some equation that gives $R$ directly?






general-relativity curvature stress-energy-momentum-tensor







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$endgroup$

















    2












    $begingroup$


    Riemann curvature tensor is dirrectly related to a path dependence of parallel transport. I read that Einstein first thought of this tensor to be the one that goes into his field equation but it didn't fit. Now, I know why this was problematic. There is a clear explanation of that. What I am asking is, is there a formulation of theory of gravity where equations give us directly $R$ and not $G$? $G$ is Einsteins tensor and $R$ Riemanns. Because, non-relativistic equation was used and directly translated, with energy-momentum tensor. Is there some other tensor related to the physics which would give $R$ instead of $G$? $G$ itself is not curvature...but from $G$ I guess we could calculate $R$ so there should be some equation that gives $R$ directly?






    general-relativity curvature stress-energy-momentum-tensor







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Riemann curvature tensor is dirrectly related to a path dependence of parallel transport. I read that Einstein first thought of this tensor to be the one that goes into his field equation but it didn't fit. Now, I know why this was problematic. There is a clear explanation of that. What I am asking is, is there a formulation of theory of gravity where equations give us directly $R$ and not $G$? $G$ is Einsteins tensor and $R$ Riemanns. Because, non-relativistic equation was used and directly translated, with energy-momentum tensor. Is there some other tensor related to the physics which would give $R$ instead of $G$? $G$ itself is not curvature...but from $G$ I guess we could calculate $R$ so there should be some equation that gives $R$ directly?






      general-relativity curvature stress-energy-momentum-tensor







      share|cite|improve this question











      $endgroup$




      Riemann curvature tensor is dirrectly related to a path dependence of parallel transport. I read that Einstein first thought of this tensor to be the one that goes into his field equation but it didn't fit. Now, I know why this was problematic. There is a clear explanation of that. What I am asking is, is there a formulation of theory of gravity where equations give us directly $R$ and not $G$? $G$ is Einsteins tensor and $R$ Riemanns. Because, non-relativistic equation was used and directly translated, with energy-momentum tensor. Is there some other tensor related to the physics which would give $R$ instead of $G$? $G$ itself is not curvature...but from $G$ I guess we could calculate $R$ so there should be some equation that gives $R$ directly?






      general-relativity curvature stress-energy-momentum-tensor




      general-relativity curvature stress-energy-momentum-tensor






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Qmechanic♦

      103k121851180




      103k121851180










      asked 2 hours ago









      Žarko TomičićŽarko Tomičić

      898511




      898511






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          The formulation you are looking for exists. Starting off from Einstein's field equations: ($R_{ik}$ Ricci tensor, $R$ curvature scalar and $T_{ik}$ the stress tensor and $g_{ik}$ the metric tensor, $kappa=frac{8pi G}{c^4}$ with $G$ gravitational constant)



          $$R_{ik} -frac{1}{2}g_{ik} R = kappa T_{ik}$$



          One can alternatively move one index up in all tensors:



          $$R_i^k -frac{1}{2}delta_i^k R = kappa T_i^k$$



          and then taking the trace (we put the indices $i$ and $k$ equal and sum over it according to Einstein's summation convention):



          $$ R = - kappa T$$



          with $T = T_i^iequivsum^3_{i=0} T_i^i$. We substitute this expression for $R$ in Einstein's original field equations and bring the term with $T$ on the right side:



          $$ R_{ik} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$



          and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $R_{ik}:=g^{lm} R_{limk}$



          $$ g^{lm} R_{limk} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Great! Isnt this then better? From geometrical point of view?
            $endgroup$
            – Žarko Tomičić
            1 hour ago










          • $begingroup$
            When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
            $endgroup$
            – Frederic Thomas
            58 mins ago





















          2












          $begingroup$

          Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $G_{munu}$ is preferred over the Ricci curvature tensor $R_{munu}$, as only the former has the form of a functional derivative $frac{1}{sqrt{|g|}}frac{delta S}{delta g^{munu}}$ of some action functional $S$. (For the Einstein tensor $G_{munu}$ this is the Einstein-Hilbert action.)






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            The formulation you are looking for exists. Starting off from Einstein's field equations: ($R_{ik}$ Ricci tensor, $R$ curvature scalar and $T_{ik}$ the stress tensor and $g_{ik}$ the metric tensor, $kappa=frac{8pi G}{c^4}$ with $G$ gravitational constant)



            $$R_{ik} -frac{1}{2}g_{ik} R = kappa T_{ik}$$



            One can alternatively move one index up in all tensors:



            $$R_i^k -frac{1}{2}delta_i^k R = kappa T_i^k$$



            and then taking the trace (we put the indices $i$ and $k$ equal and sum over it according to Einstein's summation convention):



            $$ R = - kappa T$$



            with $T = T_i^iequivsum^3_{i=0} T_i^i$. We substitute this expression for $R$ in Einstein's original field equations and bring the term with $T$ on the right side:



            $$ R_{ik} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$



            and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $R_{ik}:=g^{lm} R_{limk}$



            $$ g^{lm} R_{limk} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Great! Isnt this then better? From geometrical point of view?
              $endgroup$
              – Žarko Tomičić
              1 hour ago










            • $begingroup$
              When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
              $endgroup$
              – Frederic Thomas
              58 mins ago


















            3












            $begingroup$

            The formulation you are looking for exists. Starting off from Einstein's field equations: ($R_{ik}$ Ricci tensor, $R$ curvature scalar and $T_{ik}$ the stress tensor and $g_{ik}$ the metric tensor, $kappa=frac{8pi G}{c^4}$ with $G$ gravitational constant)



            $$R_{ik} -frac{1}{2}g_{ik} R = kappa T_{ik}$$



            One can alternatively move one index up in all tensors:



            $$R_i^k -frac{1}{2}delta_i^k R = kappa T_i^k$$



            and then taking the trace (we put the indices $i$ and $k$ equal and sum over it according to Einstein's summation convention):



            $$ R = - kappa T$$



            with $T = T_i^iequivsum^3_{i=0} T_i^i$. We substitute this expression for $R$ in Einstein's original field equations and bring the term with $T$ on the right side:



            $$ R_{ik} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$



            and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $R_{ik}:=g^{lm} R_{limk}$



            $$ g^{lm} R_{limk} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Great! Isnt this then better? From geometrical point of view?
              $endgroup$
              – Žarko Tomičić
              1 hour ago










            • $begingroup$
              When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
              $endgroup$
              – Frederic Thomas
              58 mins ago
















            3












            3








            3





            $begingroup$

            The formulation you are looking for exists. Starting off from Einstein's field equations: ($R_{ik}$ Ricci tensor, $R$ curvature scalar and $T_{ik}$ the stress tensor and $g_{ik}$ the metric tensor, $kappa=frac{8pi G}{c^4}$ with $G$ gravitational constant)



            $$R_{ik} -frac{1}{2}g_{ik} R = kappa T_{ik}$$



            One can alternatively move one index up in all tensors:



            $$R_i^k -frac{1}{2}delta_i^k R = kappa T_i^k$$



            and then taking the trace (we put the indices $i$ and $k$ equal and sum over it according to Einstein's summation convention):



            $$ R = - kappa T$$



            with $T = T_i^iequivsum^3_{i=0} T_i^i$. We substitute this expression for $R$ in Einstein's original field equations and bring the term with $T$ on the right side:



            $$ R_{ik} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$



            and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $R_{ik}:=g^{lm} R_{limk}$



            $$ g^{lm} R_{limk} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$






            share|cite|improve this answer











            $endgroup$



            The formulation you are looking for exists. Starting off from Einstein's field equations: ($R_{ik}$ Ricci tensor, $R$ curvature scalar and $T_{ik}$ the stress tensor and $g_{ik}$ the metric tensor, $kappa=frac{8pi G}{c^4}$ with $G$ gravitational constant)



            $$R_{ik} -frac{1}{2}g_{ik} R = kappa T_{ik}$$



            One can alternatively move one index up in all tensors:



            $$R_i^k -frac{1}{2}delta_i^k R = kappa T_i^k$$



            and then taking the trace (we put the indices $i$ and $k$ equal and sum over it according to Einstein's summation convention):



            $$ R = - kappa T$$



            with $T = T_i^iequivsum^3_{i=0} T_i^i$. We substitute this expression for $R$ in Einstein's original field equations and bring the term with $T$ on the right side:



            $$ R_{ik} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$



            and if you like with the Riemann curvature tensor according to the definition of the Ricci tensor $R_{ik}:=g^{lm} R_{limk}$



            $$ g^{lm} R_{limk} = kappaleft(T_{ik} -frac{1}{2} g_{ik} Tright)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 8 mins ago

























            answered 1 hour ago









            Frederic ThomasFrederic Thomas

            1,5661115




            1,5661115












            • $begingroup$
              Great! Isnt this then better? From geometrical point of view?
              $endgroup$
              – Žarko Tomičić
              1 hour ago










            • $begingroup$
              When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
              $endgroup$
              – Frederic Thomas
              58 mins ago




















            • $begingroup$
              Great! Isnt this then better? From geometrical point of view?
              $endgroup$
              – Žarko Tomičić
              1 hour ago










            • $begingroup$
              When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
              $endgroup$
              – Frederic Thomas
              58 mins ago


















            $begingroup$
            Great! Isnt this then better? From geometrical point of view?
            $endgroup$
            – Žarko Tomičić
            1 hour ago




            $begingroup$
            Great! Isnt this then better? From geometrical point of view?
            $endgroup$
            – Žarko Tomičić
            1 hour ago












            $begingroup$
            When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
            $endgroup$
            – Frederic Thomas
            58 mins ago






            $begingroup$
            When one searches for a vacuum solution of the EFEs, one can immediately start off with $R_{ik}=0$ which is easier than $R_{ik}-0.5g_{ik}R=0$
            $endgroup$
            – Frederic Thomas
            58 mins ago













            2












            $begingroup$

            Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $G_{munu}$ is preferred over the Ricci curvature tensor $R_{munu}$, as only the former has the form of a functional derivative $frac{1}{sqrt{|g|}}frac{delta S}{delta g^{munu}}$ of some action functional $S$. (For the Einstein tensor $G_{munu}$ this is the Einstein-Hilbert action.)






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $G_{munu}$ is preferred over the Ricci curvature tensor $R_{munu}$, as only the former has the form of a functional derivative $frac{1}{sqrt{|g|}}frac{delta S}{delta g^{munu}}$ of some action functional $S$. (For the Einstein tensor $G_{munu}$ this is the Einstein-Hilbert action.)






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $G_{munu}$ is preferred over the Ricci curvature tensor $R_{munu}$, as only the former has the form of a functional derivative $frac{1}{sqrt{|g|}}frac{delta S}{delta g^{munu}}$ of some action functional $S$. (For the Einstein tensor $G_{munu}$ this is the Einstein-Hilbert action.)






                share|cite|improve this answer









                $endgroup$



                Here is one difference: If you believe that the EFEs should follow from a stationary action principle (as Hilbert did), then the Einstein tensor $G_{munu}$ is preferred over the Ricci curvature tensor $R_{munu}$, as only the former has the form of a functional derivative $frac{1}{sqrt{|g|}}frac{delta S}{delta g^{munu}}$ of some action functional $S$. (For the Einstein tensor $G_{munu}$ this is the Einstein-Hilbert action.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 42 mins ago









                Qmechanic♦Qmechanic

                103k121851180




                103k121851180






























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