Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: saspinski on 28/10/2017 19:54:00

Title: How can the field equation be zero while the Ricci tensor is not?
Post by: saspinski on 28/10/2017 19:54:00
In the so called vacuum solutions, the Einstein tensor Gμν = Rμν  - 1/2*gμν*R = 0. I saw a proof that the former equation implies in R = 0 and Rμν = 0. It follows from multiplying by gμν, given R - 1/2*4*R = 0.

But when deriving the Schwarzschild solution, after calculating the Ricci tensor (Rμν) and the Ricci scalar (R), from the spherical symmetries of the problem, both are non zero. And the equation Rμν  - 1/2*gμν*R = 0 is used anyway to derive 4 differential equations that eventually result in the Schwarzschild metric.

How is it possible that the field equation are zero while Rμν and R are non zero?

Title: Re: How can the field equation be zero while the Ricci tensor is not?
Post by: PmbPhy on 28/10/2017 20:42:27
But when deriving the Schwarzschild solution, after calculating the Ricci tensor (Rμν) and the Ricci scalar (R), from the spherical symmetries of the problem, both are non zero.
If so then that's news to me. Can you find a source on the internet which shows that to be true? Clearly from Einstein's equations in vacuo R = 0 so I can't see how it can not be zero.

How is it possible that the field equation are zero while Rμν and R are non zero?
They can't be. It seems to me that there has to be problem with your assertion above where you state that the Ricci scalar and tensor as zero.

Note: caution should be used when using the "/" operator in textual form. The expression 1/ab is defined as 1/(ab). I say this because you wrote 1/2*4 where you meant to write (1/2)*4. It really means 1/(2*4).
Title: Re: How can the field equation be zero while the Ricci tensor is not?
Post by: saspinski on 29/10/2017 13:09:57
The web site is: https://web.stanford.edu/~oas/SI/SRGR/notes/SchwarzschildSolution.pdf

It is not written there really that the Ricci tensor and the Ricci scalar are different from zero. But, after getting an expression for R as a function of U, V, and its derivatives, (g00 = U and g11 = -V); instead of using for example R00 = 0 and R = 0 separately, the equation used was: R00 - (1/2)*g00*R = 0, suggesting (for me) that each term was not zero.

But after your answer, I tested the obtained value of U, V and its derivatives on R00 and it is really zero.

So, I think that R00 - (1/2)*g00*R = 0 was used because the expression was greatly simplified, leading to a differential equation only for V and V', that could be easily integrated.