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







share|cite|improve this question











$endgroup$

















    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 Jan 27 at 21:57









      Qmechanic♦

      104k121871191




      104k121871191










      asked Jan 27 at 20:46









      Žarko TomičićŽarko Tomičić

      903511




      903511






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          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
            Jan 27 at 21:45










          • $begingroup$
            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
            Jan 27 at 22:12





















          3












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Yes, I know, but the key word is "believe" . Why would anyone choose that action?
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 12:59










          • $begingroup$
            The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
            $endgroup$
            – Qmechanic♦
            Jan 28 at 13:13












          • $begingroup$
            Yes yes...I see. I was just, you know, sayin..
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 13:15










          • $begingroup$
            That is a great way to think of it.
            $endgroup$
            – R. Rankin
            Jan 28 at 19:15











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          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
            Jan 27 at 21:45










          • $begingroup$
            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
            Jan 27 at 22:12


















          5












          $begingroup$

          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
            Jan 27 at 21:45










          • $begingroup$
            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
            Jan 27 at 22:12
















          5












          5








          5





          $begingroup$

          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 Jan 27 at 23:02

























          answered Jan 27 at 21:13









          Frederic ThomasFrederic Thomas

          1,5861115




          1,5861115












          • $begingroup$
            Great! Isnt this then better? From geometrical point of view?
            $endgroup$
            – Žarko Tomičić
            Jan 27 at 21:45










          • $begingroup$
            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
            Jan 27 at 22:12




















          • $begingroup$
            Great! Isnt this then better? From geometrical point of view?
            $endgroup$
            – Žarko Tomičić
            Jan 27 at 21:45










          • $begingroup$
            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
            Jan 27 at 22:12


















          $begingroup$
          Great! Isnt this then better? From geometrical point of view?
          $endgroup$
          – Žarko Tomičić
          Jan 27 at 21:45




          $begingroup$
          Great! Isnt this then better? From geometrical point of view?
          $endgroup$
          – Žarko Tomičić
          Jan 27 at 21:45












          $begingroup$
          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
          Jan 27 at 22:12






          $begingroup$
          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
          Jan 27 at 22:12













          3












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Yes, I know, but the key word is "believe" . Why would anyone choose that action?
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 12:59










          • $begingroup$
            The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
            $endgroup$
            – Qmechanic♦
            Jan 28 at 13:13












          • $begingroup$
            Yes yes...I see. I was just, you know, sayin..
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 13:15










          • $begingroup$
            That is a great way to think of it.
            $endgroup$
            – R. Rankin
            Jan 28 at 19:15
















          3












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Yes, I know, but the key word is "believe" . Why would anyone choose that action?
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 12:59










          • $begingroup$
            The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
            $endgroup$
            – Qmechanic♦
            Jan 28 at 13:13












          • $begingroup$
            Yes yes...I see. I was just, you know, sayin..
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 13:15










          • $begingroup$
            That is a great way to think of it.
            $endgroup$
            – R. Rankin
            Jan 28 at 19:15














          3












          3








          3





          $begingroup$

          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 Jan 27 at 22:28









          Qmechanic♦Qmechanic

          104k121871191




          104k121871191












          • $begingroup$
            Yes, I know, but the key word is "believe" . Why would anyone choose that action?
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 12:59










          • $begingroup$
            The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
            $endgroup$
            – Qmechanic♦
            Jan 28 at 13:13












          • $begingroup$
            Yes yes...I see. I was just, you know, sayin..
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 13:15










          • $begingroup$
            That is a great way to think of it.
            $endgroup$
            – R. Rankin
            Jan 28 at 19:15


















          • $begingroup$
            Yes, I know, but the key word is "believe" . Why would anyone choose that action?
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 12:59










          • $begingroup$
            The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
            $endgroup$
            – Qmechanic♦
            Jan 28 at 13:13












          • $begingroup$
            Yes yes...I see. I was just, you know, sayin..
            $endgroup$
            – Žarko Tomičić
            Jan 28 at 13:15










          • $begingroup$
            That is a great way to think of it.
            $endgroup$
            – R. Rankin
            Jan 28 at 19:15
















          $begingroup$
          Yes, I know, but the key word is "believe" . Why would anyone choose that action?
          $endgroup$
          – Žarko Tomičić
          Jan 28 at 12:59




          $begingroup$
          Yes, I know, but the key word is "believe" . Why would anyone choose that action?
          $endgroup$
          – Žarko Tomičić
          Jan 28 at 12:59












          $begingroup$
          The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
          $endgroup$
          – Qmechanic♦
          Jan 28 at 13:13






          $begingroup$
          The point is whether an action exists or not. Its specific form is irrelevant for the argument presented here. Actions play an important role in modern physics.
          $endgroup$
          – Qmechanic♦
          Jan 28 at 13:13














          $begingroup$
          Yes yes...I see. I was just, you know, sayin..
          $endgroup$
          – Žarko Tomičić
          Jan 28 at 13:15




          $begingroup$
          Yes yes...I see. I was just, you know, sayin..
          $endgroup$
          – Žarko Tomičić
          Jan 28 at 13:15












          $begingroup$
          That is a great way to think of it.
          $endgroup$
          – R. Rankin
          Jan 28 at 19:15




          $begingroup$
          That is a great way to think of it.
          $endgroup$
          – R. Rankin
          Jan 28 at 19:15


















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