Derive the relationship between Ricci scalar and Gauss Curvature

[jag]

Homework Statement

Derive the relationship between Ricci scalar and Gauss curvature in 2-surface R = 2K

Relevant Equations

Ricci scalar and Gauss curvature in 2-surface, $R=2K$, where $K \equiv \frac {R_{1212}} {g}$; given the Ricci tensor $R_{\alpha \beta} \equiv R^\lambda_{\, \alpha \lambda \beta}$ and Ricci scalar $R \equiv R^{\alpha}_{\, \alpha}$.

Hi,

I'm self-learning some physics topics and came across an exercise to derive the relationship between Ricci scalar and Gauss curvature in 2-surface, $R=2K$, where $K \equiv \frac {R_{1212}} {g}$; given the Ricci tensor $R_{\alpha \beta} \equiv R^\lambda_{\, \alpha \lambda \beta}$ and Ricci scalar $R \equiv R^{\alpha}_{\, \alpha}$.

My attempt:

1. $R = R^{\alpha}_{\, \alpha}$. The left-hand side of the equation remains the same in the next steps.
2. $\frac {R^{\alpha}_{\, \alpha} g_{\alpha \alpha}} {g_{\alpha \alpha}}$
3. $\frac {R_{\alpha \alpha}} {g_{\alpha \alpha}}$
4. $\frac {R^\lambda_{\, \alpha \lambda \alpha}} {g_{\alpha \alpha}}$ => using Ricci tensor definition
5. $\frac {R^\lambda_{\, \alpha \lambda \alpha} g_{\lambda \lambda}} {g_{\lambda \lambda}g_{\alpha \alpha}}$
6. $\frac {R_{\lambda \alpha \lambda \alpha}} {g_{\lambda \lambda}g_{\alpha \alpha}}$

After that, I just have the following idea but I'm not sure whether it makes sense. I'm thinking $\lambda = 1$ and $\alpha = 2$, yielding $\frac {R_{1212}} {g_{11}g_{22}}$. Then, $g_{11}g_{22} = \frac {g} {2}$, which will yield $\frac {2R_{1212}} {g} = 2K$. I'm also not sure whether repeating the index labels in the metric is correct.

Please kindly let me know if my thought process is incorrect. Looking for any kind of help. Thank you.

 

[Orodruin]

Step 2 have way too many $\alpha$ indices.

 

[jag]

Thank you for your response @Orodruin .

This is my second attempt:

1. $R = R^{\alpha}_{\, \alpha}$ The left-hand side of the equation remains the same in the next steps.
2. $\frac {R^{\alpha}_{\, \alpha} g_{\beta \alpha}} {g_{\beta \alpha}}$
3. $\frac {R_{\beta \alpha}} {g_{\beta \alpha}}$
4. $\frac {R_{\alpha \beta}} {g_{\alpha \beta}}$
5. $\frac {R^\lambda_{\, \alpha \lambda \beta}} {g_{\alpha \beta}}$ => using Ricci tensor definition
6. $\frac {R^\lambda_{\, \alpha \lambda \beta} g_{\mu \lambda}} {g_{\mu \lambda}g_{\alpha \beta}}$
7. $\frac {R_{\mu \alpha \lambda \beta}} {g_{\mu \lambda}g_{\alpha \beta}}$
8. Taking two cases
   1. $\mu = \lambda = 1$ and $\alpha = \beta = 2$ yields $\frac {R_{1212}} {g_{11}g_{22}}$
   2. $\mu = \lambda = 2$ and $\alpha = \beta = 1$ yields $\frac {R_{2121}} {g_{22}g_{11}}$ which by using symmetry of Riemann tensor becomes $\frac {R_{1212}} {g_{11}g_{22}}$
9. My assumption: The cross-terms $g_{12}g_{21}$ in metric determinant, $g$, of 2-surface is $0$ yielding $g = g_{11}g_{22} - g_{12}g_{21} = g_{11}g_{22} - 0 = g_{11}g_{22}$
10. Hence,
    1. $\frac {R_{1212} + R_{1212}} {g_{11}g_{22}}$
    2. $\frac {2R_{1212}} {g}$
    3. $2K$

Please kindly correct me if my thought process is incorrect. Thank you.

 

[Orodruin]

Thank you for your response @Orodruin .

This is my second attempt:

1. $R = R^{\alpha}_{\, \alpha}$ The left-hand side of the equation remains the same in the next steps.
2. $\frac {R^{\alpha}_{\, \alpha} g_{\beta \alpha}} {g_{\beta \alpha}}$

Still too many $\alpha$. You cannot create a new tensor equation by multiplying and dividing with a tensor component. You need to use the appropriate definitions.

You might benefit from reading this: https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

 

[renormalize]

Please kindly correct me if my thought process is incorrect. Thank you.

Why not just make use of the simple formula below for the Riemann tensor of a manifold of constant curvature?

(https://en.wikipedia.org/wiki/Riemannian_manifold)
Then you only have to contract twice with the inverse metric to get what you want.

EDIT: As correctly pointed out by @Orodruin, my comment is only valid for constant curvature (e.g., a 2-sphere) so it is inapplicable for a general 2d curved manifold; i.e., never mind!

 

[Orodruin]

Why not just make use of the simple formula below for the Riemann tensor of a manifold of constant curvature?

I don’t think we were to assume constant curvature.

 

[jag]

Still too many $\alpha$. You cannot create a new tensor equation by multiplying and dividing with a tensor component. You need to use the appropriate definitions.

You might benefit from reading this: https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

Thank you for your feedback and the reading you gave me. It was very helpful!

This is my third attempt:

1. $R = R^{\alpha}_{\, \alpha}$ The left-hand side of the equation remains the same in the next steps.
2. $g^{\alpha \beta}R_{\beta \alpha}$
3. $g^{\alpha \beta}R_{\alpha \beta}$
4. $g^{\alpha \beta}R^\lambda_{\, \alpha \lambda \beta}$ => using Ricci tensor definition
5. $g^{\alpha \beta}g^{\lambda \mu}R_{\mu \alpha \lambda \beta}$
6. $\frac {R_{\mu \alpha \lambda \beta}} {g_{\alpha \beta}g_{\lambda \mu}}$
7. Taking two cases
   1. $\mu = \lambda = 1$ and $\alpha = \beta = 2$ yields $\frac {R_{1212}} {g_{11}g_{22}}$
   2. $\mu = \lambda = 2$ and $\alpha = \beta = 1$ yields $\frac {R_{2121}} {g_{22}g_{11}}$ which by using symmetry of Riemann tensor becomes $\frac {R_{1212}} {g_{11}g_{22}}$
8. My assumption: The cross-terms $g_{12}g_{21}$ in metric determinant, $g$, of 2-surface is $0$ yielding $g = g_{11}g_{22} - g_{12}g_{21} = g_{11}g_{22} - 0 = g_{11}g_{22}$
9. Hence, adding case $1$ and $2$
   1. $\frac {R_{1212} + R_{1212}} {g_{11}g_{22}}$
   2. $\frac {2R_{1212}} {g}$
   3. $2K$

Please correct me if I'm wrong. Thank you for your patience in guiding me. :)

 

[Orodruin]

6 is now incorrect. Multiplying by the components of the inverse is not the same as dividing by the components

 

[jag]

6 is now incorrect. Multiplying by the components of the inverse is not the same as dividing by the components

This is my fourth attempt:

1. $R = R^{\alpha}_{\, \alpha}$ The left-hand side of the equation remains the same in the next steps.
2. $g^{\alpha \beta}R_{\beta \alpha}$
3. $g^{\alpha \beta}R_{\alpha \beta}$
4. $g^{\alpha \beta}R^\lambda_{\, \alpha \lambda \beta}$ => using Ricci tensor definition
5. $g^{\alpha \beta}g^{\lambda \mu}R_{\mu \alpha \lambda \beta}$
6. Gauss curvature definition $K \equiv \frac {R_{\mu \alpha \lambda \beta}} {g}$ yields $R_{\mu \alpha \lambda \beta} = Kg = K(g_{\alpha \beta}g_{\lambda \mu} - g_{\alpha \lambda}g_{\beta \mu})$
7. Substituting $6$ into $5$ yields $g^{\alpha \beta}g^{\lambda \mu}(g_{\alpha \beta}g_{\lambda \mu} - g_{\alpha \lambda}g_{\beta \mu})K$
8. Contracting $g^{\lambda \mu}$ yields $g^{\alpha \beta}(n \cdot g_{\alpha \beta} - {\delta^{\mu}}_{\alpha}g_{\beta \mu})K$
9. $g^{\alpha \beta}(n \cdot g_{\alpha \beta} - g_{\alpha \beta})K$
10. Contracting $g^{\alpha \beta}$ yields $(n-1)g^{\alpha \beta}g_{\alpha \beta}K = (n-1) \cdot n \cdot K$
11. Hence, for 2-surface where $n=2$ yields $(2-1)2K = 2K$ as required

Please correct me if I'm wrong. :)

 

[Orodruin]

6 doesn’t make any sense because of index mismatch. What you need to do is to continue from 5 by writing out the sums. (Only four terms will be non-zero due to the symmetries of the curvature tensor)

 

[jag]

6 doesn’t make any sense because of index mismatch. What you need to do is to continue from 5 by writing out the sums. (Only four terms will be non-zero due to the symmetries of the curvature tensor)

Sorry, I know I need to type the equations in Latex. It just got too long when I expanded the sums. Hence, I captured my workings in the image below. Unfortunately, the picture quality seems to decrease when I uploaded. You could zoom in. Please kindly let me know if I'm wrong. Thank you.

 

[Orodruin]

Right. I would have just skipped writing out any terms that vanish because of the Riemann tensor symmetries, but you can of course write them out as well even if they vanish.

 

[jag]

Right. I would have just skipped writing out any terms that vanish because of the Riemann tensor symmetries, but you can of course write them out as well even if they vanish.

Thank you @Orodruin . I learned a lot from this exercise; fixing my misconceptions. I don't have much formal math and physics education, but I always wanted to learn more in-depth than reading popular science books. Hence, I embarked on a journey to self-learn math and physics. Thank you again. Have a pleasant week, and Happy New Year!