# Einstein Tensor

1. Jan 25, 2010

### Orion1

Schwarzschild metric:
$$c^{2} d\tau^{2} = e^{\nu(r)} dt^{2} - e^{\lambda(r)} dr^{2} - r^2 d\theta^{2} - r^2 \sin^2 \theta d\phi^2$$

According to reference 1, the Maple 13 'tensor' package generated this solution for the $G_{11}[/tex] component: $$G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}$$ According to reference 2, the Mathematica 6 'Einsteintensor' package generated this solution for the [itex]G_{11}[/tex] component: $$G_{11} = \frac{e^{-\lambda} (-r \nu' + e^{\lambda} - 1)}{r^2}$$ According to reference 3 - eq. series 7, the solution for the [itex]G_{11}[/tex] component: $$G_{11} = \frac{\nu'}{r} - \frac{e^{\lambda}}{r^2} + \frac{1}{r^2}$$ According to reference 4 - eq. 4, the solution for the [itex]G_{11}$ component:
$$G_{11} = \frac{e^{-\lambda} (r \nu' - e^{\lambda} + 1)}{r^2}$$

Which $G_{11}$ component is the correct solution?

Reference:
https://www.physicsforums.com/showpost.php?p=2543074&postcount=1"
https://www.physicsforums.com/showpost.php?p=2547561&postcount=2"
http://www.bergshoeff.fmns.rug.nl/gr/form1.pdf" [Broken]

Last edited by a moderator: May 4, 2017
2. Jan 25, 2010

### Altabeh

Maple is always the master of all! I calculated G11 by hand and got exactly the same result as Maple!

Edit: The result from Mathematica has an extra factor $$e^{-\lambda}$$ which along with the computation obtained by Krori and Borgohain are worst. But the third calculation doesn't seem to be wrong as it is just the result obtained by Maple with a flipped sign: this comes from the fact that one uses, for example, D'inverno's definition of Riemann tensor, but the other uses Weinberg's!

AB

Last edited by a moderator: May 4, 2017
3. Jan 25, 2010

### nicksauce

Using my self-made GR package in Maple, I get the same result as Maple.

4. Jan 25, 2010

### bcrowell

Staff Emeritus
Are you sure this is an apples-apples comparison? E.g., in the ctensor package used with Maxima, which is what I'm familiar with, I believe it outputs $G^i_j$, not $G_{ij}$, and the order of indices on Christoffel symbols is also goofy. The order of indices on the Riemann tensor is not standardized at all in the literature.

5. Jan 26, 2010

### Orion1

a theoretical problem...

There is a theoretical problem, the Einstein_tensor used by Krori and Borgohain (1974) is the same Einstein_tensor used by J.R. Oppenheimer and G.M. Volkoff (1939) and R.C. Tolman (1939) and used to derive the Tolman-Oppenheimer-Volkoff (TOV) equation and is based upon an incorrect Einstein tensor.

Derivation of the Tolman-Oppenheimer-Volkoff (TOV) equation...

Einstein field equation:
$$\frac{8 \pi G}{c^4} T_{11} = G_{11}$$

Tolman Einstein tensor: (ref. 2 - eq. 4)
$$G_{11} = e^{-\lambda} \left(\frac{\nu '}{r} + \frac{1}{r^2} \right) - \frac{1}{r^2}$$

Integration via substitution:
$$\frac{8 \pi G}{c^4} T_{11} = e^{-\lambda} \left(\frac{\nu '}{r} + \frac{1}{r^2} \right) - \frac{1}{r^2}$$

Differential Equation of State for hydrostatic equilibrium: (ref. 2 - eq. 6)
$$\frac{dP(r)}{dr} = - \frac{(T_{11} + T_{00})}{2} \nu'$$

$$\nu ' = - \frac{dP(r)}{dr} \left( \frac{2}{T_{11} + T_{00}} \right) = \frac{1}{r} \left( \frac{8 \pi G T_{11} r^2 e^{\lambda}}{c^4} + e^{\lambda } - 1 \right)$$

Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \frac{\left( \left( e^{\lambda} - 1 \right) c^4 + e^{\lambda} 8 \pi G r^2 T_{11} \right) \left(T_{11} + T_{00} \right)}{2 c^4 r}$$

Metric identity:
$$e^{\lambda} = \left (1 - \frac{r_s}{r} \right)^{-1}$$

Metric identity:
$$\boxed{e^{\lambda} - 1 = \frac{r_s}{r - r_s}}$$

Equation of State for hydrostatic equilibrium:
$$\boxed{\frac{dP(r)}{dr} = -\frac{ \left(r_s c^4 + 8 \pi G r^3 T_{11} \right) \left( T_{11} + T_{00} \right)}{2 c^4 r \left(r - r_s \right)}}$$

Metric identity:
$$r (r - r_s) = r^2 \left( 1 - \frac{r_s}{r} \right)$$

Stress-energy tensor:
$$T_{00} = \rho(r) c^2 \; \; \; T_{11} = P(r)$$

$$r_s = \frac{2 G M(r)}{c^2}$$

Equation of State for hydrostatic equilibrium:
$$\boxed{\frac{dP(r)}{dr} = -\frac{ \left( P(r) + \rho(r) c^2 \right) \left(r_s c^4 + 8 \pi G r^3 P(r) \right) \left( 1 - \frac{r_s}{r} \right)^{-1}}{2 c^4 r^2}}$$

Factoring out a $c^2$ from the first two numerator terms and a $2 G$ from one numerator term results in the TOV equation.

Tolman-Oppenheimer-Volkoff (TOV) equation:
$$\boxed{\frac{dP(r)}{dr} = - \frac{G}{r^2} \left[ \rho(r) + \frac{P(r)}{c^2} \right] \left[M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right] \left[1 - \frac{2GM(r)}{c^2r} \right]^{-1}}$$

Reference:
http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation" [Broken]

Last edited by a moderator: May 4, 2017
6. Jan 26, 2010

### Orion1

a theoretical problem...

Derivation of the Equation of State for hydrostatic equilibrium...

Einstein field equation:
$$\frac{8 \pi G}{c^4} T_{11} = G_{11}$$

Schwarzschild-Einstein tensor:
$$\boxed{G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}}$$

Integration via substitution:
$$\frac{8 \pi G}{c^4} T_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}$$

Differential Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \frac{(T_{11} + T_{00})}{2} \nu'$$

$$\nu ' = - \frac{dP(r)}{dr} \left( \frac{2}{T_{11} + T_{00}} \right) = \frac{1}{r} \left( - \frac{8 \pi G T_{11} r^2}{c^4} + e^{\lambda} - 1 \right)$$

Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = -\frac{\left((e^{\lambda} - 1)c^4 - 8 \pi G r^2 T_{11} \right) \left( T_{11} + T_{00} \right)}{2 c^4 r}$$

Metric identity:
$$e^{\lambda} = \left (1 - \frac{r_s}{r} \right)^{-1}$$

Metric identity:
$$\boxed{e^{\lambda} - 1 = \frac{r_s}{r - r_s}}$$

Equation of State for hydrostatic equilibrium:
$$\boxed{\frac{dP(r)}{dr} = - \frac{\left(\frac{c^4 r_s}{r - r_s} - 8 \pi G r^2 T_{11} \right)\left(T_{11} + T_{00} \right)}{2 c^4 r}}$$

Stress-energy tensor:
$$T_{00} = \rho(r) c^2 \; \; \; T_{11} = P(r)$$

$$r_s = \frac{2 G M(r)}{c^2}$$

Factoring out $2 G c^4$ from the numerator results in the Equation of State for hydrostatic equilibrium equation.

Equation of State for hydrostatic equilibrium:
$$\boxed{\frac{dP(r)}{dr} = - \frac{G}{r} \left( P(r) + \rho(r) c^2 \right) \left( \frac{M(r)}{c^2 r - 2 G M(r)} - 4 \pi r^2 \frac{P(r)}{c^4} \right)}$$

Reference:
http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation" [Broken]

Last edited by a moderator: May 4, 2017
7. Jan 26, 2010

### Orion1

correct equation?...

Another theoretical problem is all my scientific astrophysics papers downloaded from the internet are based upon the Equation of State of neutron star models, and are all based upon the Tolman-Oppenheimer-Volkoff (TOV) equation, which is based upon an incorrect Einstein tensor.

Tolman-Oppenheimer-Volkoff (TOV) equation:
$$\frac{dP(r)}{dr} = - \frac{G}{r^2} \left[ \rho(r) + \frac{P(r)}{c^2} \right] \left[M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right] \left[1 - \frac{2GM(r)}{c^2r} \right]^{-1}$$

Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \frac{G}{r} \left( P(r) + \rho(r) c^2 \right) \left( \frac{M(r)}{c^2 r - 2 G M(r)} - 4 \pi r^2 \frac{P(r)}{c^4} \right)$$

Which equation is correct?

Last edited: Jan 26, 2010
8. Jan 26, 2010

### Altabeh

Re: correct equation?...

It behooves one to do take into account that since something carries the names of famous physicists must always be correct is based upon the strong defective idea of "following wind leads certainly one to get the position" kind of stuff! Of course the TOV equation is incorrect because its G11 consists of two $$e^{\lambda}$$-terms which as discussed earlier, is not true!

AB

9. Jan 27, 2010

### Orion1

Einstein Tensor package source code...

I thought that both packages generated output as $$G^{Mathematica}_{ij} = G^{Maple}_{ij}$$.

I could not locate the source code for the Maple 'tensor' package.

I have posted the Mathematica 'Einstein Tensor' package source code at reference 1.

Reference:

Last edited by a moderator: Apr 24, 2017
10. Jan 27, 2010

### Altabeh

True! If we reckon that Mathematica gives $$G^i_j$$ like that, a simple calculation would be able to verify this hypothesis:

$$G^1_1= g^{1i}G_{1i}=g^{11}G_{11} =-e^{-\lambda}\frac{ r{\nu}' - e^{\lambda} + 1}{r^2}=\frac{e^{-\lambda} (-r{\nu}'+ e^{\lambda} - 1)}{r^2}$$.

Remember that Mathematica appears to have made use of the Ricci tensor $$R_{ab}=R^c_{acb}$$. Since the Riemann tensor is antisymmetric in c and b, then a contraction of the form $$R_{ac}=R^b_{acb}$$ would lead to the Ricci tensor used by Maple which has a different sign. To put another way, $$G^{Mathematica}_{ij} = - G^{Maple}_{ij}$$.

AB

Last edited: Jan 27, 2010
11. Jan 27, 2010

### Orion1

Mathematica Einstein tensor...

Mathematica Ricci tensor:
$$R_{ab} = R^c_{acb}$$

Maple Ricci tensor:
$$R_{ac} = R^b_{acb}$$

Package criteria:
$$R^{Mathematica} = -R^{Maple}$$

$$G^{Mathematica}_{ij} = - G^{Maple}_{ij}$$

The different signs are due to the use of different Ricci tensors.

Mathematica generated output:
$$G^1_1 = g^{1i}G_{1i}=g^{11}G_{11} = -e^{-\lambda}\frac{ r{\nu}' - e^{\lambda} + 1}{r^2} = \frac{e^{-\lambda} (-r{\nu}'+ e^{\lambda} - 1)}{r^2}$$

Maple generated output:
$$G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}$$

Reference:

Last edited by a moderator: Apr 24, 2017
12. Jan 28, 2010

### Altabeh

Re: Mathematica Einstein tensor...

No, it is not! From

$$g^{11}=-e^{- \lambda}$$ and
$$G^{Mathematica}_{ij} = -G^{Maple}_{ij}=-\frac{- r{\nu}'+ e^{\lambda} - 1}{r^2}=\frac{ r{\nu}'- e^{\lambda} + 1}{r^2},$$

one would get

$$(G^1_1)^{Mathematica}=\frac{e^{-\lambda} (-r{\nu}'+ e^{\lambda} - 1)}{r^2}$$

which shows exactly the same equation for Mathematica's G11 you wrote in your first post in this thread!

AB

Last edited by a moderator: Apr 24, 2017
13. Jan 28, 2010

### Orion1

$$G^1_1 = g^{1i}G_{1i} = g^{11} G^{Mathematica}_{11} = (- e^{- \lambda} ) \left( \frac{ r{\nu}' - e^{\lambda} + 1}{r^2} \right) = \frac{e^{- \lambda} (- r \nu' + e^{\lambda} - 1)}{r^2}$$

$$\boxed{G^1_1 = \frac{e^{- \lambda} (- r \nu' + e^{\lambda} - 1)}{r^2}}$$

Is this the correct solution for $$G^1_1$$?

According to reference 1 equation series 8, the $$G^1_1$$ is:
$$G^1_1 = r^{-1} e^{- \lambda} \nu' - r^{-2} e^{- \lambda} + r^{-2} = \frac{e^{- \lambda} r {\nu}' - e^{- \lambda} + 1}{r^2} = \frac{e^{- \lambda} (r {\nu}' + e^{\lambda} - 1)}{r^2}}$$

$$G^1_1 = \frac{e^{- \lambda} (r {\nu}' + e^{\lambda} - 1)}{r^2}}$$

Reference:
http://www.bergshoeff.fmns.rug.nl/gr/form1.pdf" [Broken]

Last edited by a moderator: May 4, 2017
14. Jan 28, 2010

### Altabeh

Both are correct! As I said before, this unimportant difference comes from the definition of Ricci tensor. One uses $$R_{ab}=R^c_{acb}$$ and other makes use of $$R_{ac}=R^b_{acb}$$. Remember that this affects R (the scalar curvature), too! So $$R^{Maple}=-R^{Mathematica}$$.

AB

Last edited by a moderator: May 4, 2017
15. Jan 31, 2010

### Orion1

Oppenheimer-Volkoff equation....

Schwarzschild metric:
$$c^{2} d\tau^{2} = e^{\nu} c^2 dt^{2} - e^{\lambda} dr^{2} - r^2 d\theta^{2} - r^2 \sin^2 \theta d\phi^2$$

Derivation of the Oppenheimer-Volkoff equation (O-V)...

Stress-energy tensor Schwarzschild field hydrostatic density: (ref. 1 - pg. 255 - eq. 10.20)
$$T_{00} = \rho(r) c^2 g_{00} = \rho(r) e^{\nu} c^4$$

Stress-energy tensor Schwarzschild field hydrostatic pressure: (ref. 1 - pg. 255 - eq. 10.21)
$$T_{11} = P(r) g_{11} = - P(r) e^{\lambda}$$

Where $$\rho(r)$$ is the fluid density and $$P(r)$$ is the fluid pressure.

Schwarzschild-Einstein tensor: (ref. 1 - pg. 255 - eq. 10.15)
$$G_{11} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}$$

Einstein field equation:
$$\frac{8 \pi G}{c^4} T_{11} = G_{11}$$

Integration via substitution:
$$- \frac{8 \pi G}{c^4} P(r) e^{\lambda} = \frac{- r \nu' + e^{\lambda} - 1}{r^2}$$

Differential Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \left( \frac{T_{00} g^{00} + T_{11} g^{11}}{2} \right) \nu' = - \left( \frac{\rho(r) c^2 + P(r)}{2} \right) \nu'$$

$$\nu' = - \frac{dP(r)}{dr} \left( \frac{2}{\rho(r) c^2 + P(r)} \right) = \frac{1}{r} \left( \frac{8 \pi G r^2 e^{\lambda} P(r)}{c^4} + e^{\lambda} - 1 \right)$$

Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = -\frac{ \left( \rho(r) c^2 + P(r) \right) \left((e^{\lambda} - 1)c^4 + 8 \pi G r^2 e^{\lambda} P(r) \right)}{2 c^4 r}$$

Metric identity:
$$e^{\lambda} = \left (1 - \frac{r_s}{r} \right)^{-1}$$

Metric identity:
$$e^{\lambda} - 1 = \frac{r_s}{r - r_s}$$

Equation of State for hydrostatic equilibrium:
$$\frac{dP(r)}{dr} = - \frac{ \left( \rho(r) c^2 + P(r) \right) \left(c^4 r_s + 8 \pi G r^3 P(r) \right)}{2 c^4 r \left(r - r_s \right)}$$

$$r_s = \frac{2 G M(r)}{c^2}$$

Factoring out a $2 G c^2$ from one numerator term results in the Oppenheimer-Volkoff equation.

Oppenheimer-Volkoff equation:
$$\frac{dP(r)}{dr} = - \frac{G \left( \rho(r) c^2 + P(r) \right) \left(M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right)}{r \left(c^2 r - 2 G M(r) \right)}$$

Oppenheimer-Volkoff equation (O-V): (ref. 1 - pg. 255 - eq. 10.21)
$$\frac{dP}{dr} = - \frac{( \rho + P)(m + 4 \pi r^3 P)}{r(r - 2m)}$$

Metric identity:
$$r (r - r_s) = r^2 \left( 1 - \frac{r_s}{r} \right)$$

Factoring out a $c^2$ from the numerator and integrating the metric identity via substitution results in the Tolman-Oppenheimer-Volkoff equation.

Tolman-Oppenheimer-Volkoff equation:
$$\frac{dP(r)}{dr} = - \frac{G}{r^2} \left( \rho(r) + \frac{P(r)}{c^2} \right) \left(M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right) \left( 1 - \frac{2 G M(r)}{c^2 r} \right)^{-1}$$

Tolman-Oppenheimer-Volkoff (TOV) equation: (ref. 2)
$$\frac{dP(r)}{dr} = - \frac{G}{r^2} \left[ \rho(r) + \frac{P(r)}{c^2} \right] \left[M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right] \left[1 - \frac{2GM(r)}{c^2r} \right]^{-1}$$

Are these equations correct?

Reference: