- #1

timeant

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The sign of Einstein equations is minus "-" , Eq. (1-236).

However, the sign of Riemann and Ricci tensor are the same as MTW's book.

The sign of Einstein equations in MTW's book are "+"!

Is there a error?

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- Thread starter timeant
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- #1

timeant

- 16

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The sign of Einstein equations is minus "-" , Eq. (1-236).

However, the sign of Riemann and Ricci tensor are the same as MTW's book.

The sign of Einstein equations in MTW's book are "+"!

Is there a error?

- #2

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- #3

PeterDonis

Mentor

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No, just a different choice of sign conventions. If you have a copy of MTW, look at the page titled "SIGN CONVENTIONS", just after the copyright page. You will see three different possible choices of sign convention described. On the next page is a table of the sign conventions used in various textbooks published up to the time MTW was published in 1973. Chandrasekhar's book was published later so it isn't listed in the table, but you should be able to figure out which conventions he is using from the information given in MTW.Is there a error?

- #4

timeant

- 16

- 2

However, metric sign do not affect the sign of Ricci tensor, (1,3) Riemann tensor and Einstein equation. I think so.

Chandrasekhar use the metric(+---) for Einstein equations. However, his electromagnetic stress-energy tensor T_{\mu\nu} is base on the the metric(-+++). May be I make mistakes?

- #5

PeterDonis

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The sign convention for the metric tensor is one of only three sign conventions involved. See the page in MTW that I referenced.Chandrasekhar use the metric(+---) for Einstein equations. However, his electromagnetic stress-energy tensor T_{\mu\nu} is base on the the metric(-+++). May be I make mistakes?

- #6

timeant

- 16

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Chandrasekhar's book is the same as Landau's book(The Classical Theory of Fields, 4th edn) except for the sign of Einstein equations.

Is Landau wrong? I do not think so.

In another words, the time-time component of stress-energy tensor T_{ab} is negative, i.e. T_{00} <0, in Chandrasekhar's book.

It was an unusual choice!

Is Landau wrong? I do not think so.

In another words, the time-time component of stress-energy tensor T_{ab} is negative, i.e. T_{00} <0, in Chandrasekhar's book.

It was an unusual choice!

Last edited:

- #7

PeterDonis

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Yes, so that just means Chandrasekhar has made the opposite choice for Landau for one of the three sign conventions that MTW describes.Chandrasekhar's book is the same as Landau's book(The Classical Theory of Fields, 4th edn) except for the sign of Einstein equations.

Why would Landau have to be wrong? Sign conventions are conventions. You can make either choice.Is Landau wrong? I do not think so.

- #8

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The signature of the metric is west-coast: ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## for flat spacetime wrt. a Lorentzian basis.

The Riemann curvature tensor is defined by

$$\nabla_{\mu} \nabla_{\nu} A^{\rho}-\nabla_{\nu} \nabla_{\mu} A^{\rho}=A^{\sigma} {R^{\rho}}_{\sigma \mu \nu}$$

and the Ricci tensor by

$$R_{\mu \nu} = g^{\rho \sigma} R_{\rho \mu \sigma \nu}.$$

The Einstein field equations read

$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=\frac{8 \pi G}{c^4} T_{\mu \nu}.$$

The signature of the metric is west-coast: ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## for flat spacetime wrt. a Lorentzian basis.

The Riemann curvature tensor is defined by

$$\nabla_{\mu} \nabla_{\nu} A^{\rho}-\nabla_{\nu} \nabla_{\mu} A^{\rho}=-A^{\sigma} {R^{\rho}}_{\sigma \mu \nu}$$

and the Ricci tensor by

$$R_{\mu \nu} = g^{\rho \sigma} R_{\rho \mu \sigma \nu}.$$

The Einstein field equations read

$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=-\frac{8 \pi G}{c^4} T_{\mu \nu}.$$

So the difference in sign is due to the definition of the Riemann curvature tensor (note that the Riemann tensor is antisymmtric wrt. to the 1st and 2nd as well the 3rd and 4th index), and the EFEs are the same in both books.

I'd be very surprised, if there were different sign conventions for the energy-momentum tensor of matter and radiation, because usually one defines energy densities as positive, ##T^{00}>0##.

- #9

timeant

- 16

- 2

Landau Lifshitz (4th edition)

The signature of the metric is west-coast: ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## for flat spacetime wrt. a Lorentzian basis.

The Riemann curvature tensor is defined by

$$\nabla_{\mu} \nabla_{\nu} A^{\rho}-\nabla_{\nu} \nabla_{\mu} A^{\rho}=A^{\sigma} {R^{\rho}}_{\sigma \mu \nu}$$

and the Ricci tensor by

$$R_{\mu \nu} = g^{\rho \sigma} R_{\rho \mu \sigma \nu}.$$

The Einstein field equations read

$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=\frac{8 \pi G}{c^4} T_{\mu \nu}.$$

Chandrasekhar, The mathematical theory of black holes (1983)

The signature of the metric is west-coast: ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## for flat spacetime wrt. a Lorentzian basis.

The Riemann curvature tensor is defined by

$$\nabla_{\mu} \nabla_{\nu} A^{\rho}-\nabla_{\nu} \nabla_{\mu} A^{\rho}=-A^{\sigma} {R^{\rho}}_{\sigma \mu \nu}$$

and the Ricci tensor by

$$R_{\mu \nu} = g^{\rho \sigma} R_{\rho \mu \sigma \nu}.$$

The Einstein field equations read

$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=-\frac{8 \pi G}{c^4} T_{\mu \nu}.$$

So the difference in sign is due to the definition of the Riemann curvature tensor (note that the Riemann tensor is antisymmtric wrt. to the 1st and 2nd as well the 3rd and 4th index), and the EFEs are the same in both books.

I'd be very surprised, if there were different sign conventions for the energy-momentum tensor of matter and radiation, because usually one defines energy densities as positive, ##T^{00}>0##.

I will give the component equations.

The Riemann curvature tensor is defined by Eq 1-154

$$R^j_{\cdot lnm}= \Gamma^j_{lm,n}- \Gamma^j_{ln,m}+ \Gamma^j_{pn}\Gamma^p_{lm}-\Gamma^j_{pm}\Gamma^p_{ln}$$

The Ricci curvature tensor is defined by Eq 1-172

$$R_{lm}=R^n_{\cdot lnm}= \Gamma^n_{lm,n}- \Gamma^n_{ln,m}+ \Gamma^n_{pn}\Gamma^p_{lm}-\Gamma^n_{pm}\Gamma^p_{ln}$$

The Riemann curvature tensor is defined by Eq 91.4

$$R^i_{\cdot klm}= \Gamma^i_{km,l}- \Gamma^i_{kl,m}+ \Gamma^i_{pl}\Gamma^p_{km}-\Gamma^i_{pm}\Gamma^p_{kl}$$

The Ricci curvature tensor is defined by Eq 92.7

$$R_{km}=R^l_{\cdot klm}= \Gamma^l_{km,l}- \Gamma^l_{kl,m}+ \Gamma^l_{pl}\Gamma^p_{km}-\Gamma^l_{pm}\Gamma^p_{kl}$$

Are they different?

I think they are the same. right?

- #10

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For Chandrasekhar I inferred it from Eq. (231) and made use of the antisymmetry of the Riemann tensor in interchanging the first two components. For Landau Lifshitz I used Eq. (91.7) and made use of the antisymmetry in interchanging the 3rd and 4th index.

- #11

timeant

- 16

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Otherwise, it will be different from Landau's book.

- #12

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- #13

timeant

- 16

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No puzzles!

MTW said: All authors agree positive energy density, ##T_{00}>0##!

Chandrasekhar said: No, I am an exception!

All equations are correct in Chandrasekhar's book. It's just that the energy density is negative.

- #14

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Landau: (91.6)

$$A_{i;k;l}-A_{i;l;k}={R^m}_{ikl} A_m=\nabla_l \nabla k A_i-\nabla_k \nabla_l A_i.$$

Now I see that indeed Chandrasekhar (231) is indeed the same. Somehow I was obviously confused from Eq. (91.7) in LL (having overlooked that there the contraction was wrt. the second rather than the first index, while in (91.6) the contraction is over the first index of the Riemann tensor).

Sorry for that blunder :-(((!

But to make energies negative is really very unconventional. What a confusion!

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