Motivation behind vacuum Einstein equation

  • #1
Hi,

I am using Hartle to study GR and at one point, there is a leap that I don't understand. He finds the result for the geodesic deviation equation and introduces the Riemann curvature tensor.

Then, we are told that there is an object called the Ricci curvature tensor which is a contracted version of the Riemann curvature tensor and that in a vacuum, the components of the Ricci tensor go to zero.

Can anyone tell me what the motivation behind this is? Hartle's explanation is not very clear at all. After getting the Riemann curvature tensor, how would one logically go to the Ricci tensor and equate it to 0 for vacuum?

Thank you!
 

Answers and Replies

  • #2
George Jones
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Let's work through this.

First, what is Einstein's equation?
 
  • #3
zonde
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that in a vacuum, the components of the Ricci tensor go to zero.

Can anyone tell me what the motivation behind this is? Hartle's explanation is not very clear at all. After getting the Riemann curvature tensor, how would one logically go to the Ricci tensor and equate it to 0 for vacuum?
Isn't the statement that Ricci tensor goes to zero equivalent to Einstein's equivalence principle?
It means that locally spacetime is Euclidean, right?

If that would be so then we just have to sort out why Ricci tensor would be non zero in presence of mass/energy.
 
  • #4
George Jones
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Isn't the statement that Ricci tensor goes to zero equivalent to Einstein's equivalence principle?
It means that locally spacetime is Euclidean, right?
In general, it is not a good idea to use the the equivalence principle as a substitute for a calculation. At any event in spacetime, it is always possible to find an orthonormal basis, even in the presence of matter. Also, even if the Ricci tensor is zero, the curvature tensor does not have to be zero.

The way to proceed is to start with Einstein's equation in its usual form, do a little calculation that transforms Einstein's equation to a slightly less usual form in which is obvious that Ricci is zero for a vacuum.
My take on the equivalence principle is as follows. The equivalence principle, in the hands of Albert Einstein, was an amazing conceptual insight that led from special relativity to general relativity. Mere mortals like us who want to know what the now mature theory of relativity says about a given situation should grab hold of the metric appropriate for the situation and calculate. Use of the equivalence principle is not a substitute for calculation and can even be misleading. See comments 11, 12, 18 and 20 in

http://blogs.discovermagazine.com/cosmicvariance/2009/06/02/susskind-lectures-on-general-relativity/.

Sean is Sean Carroll.
 
  • #5
We haven't got to the Einstein equation yet. It was introduced first using [itex]R_{\alpha\beta}=0[/itex] in a vacuum. This is section 21.4 of Hartle, in case anyone has the book too.

Hartle justifies this as a generalization of the Newtonian version [itex]\nabla^{2}\phi=0[/itex] where [itex]\phi[/itex] is the graviational potential. His line of reasoning, if I understand correctly is as follows. Let's call [itex]\chi[/itex] the separation vector which measures the separation between two nearby particles as they fall freely. The Newtonian calculation yields
[tex]\frac{d^{2}\chi_{i}}{dt^{2}}=-\frac{\partial^{2}\phi}{\partial x_{i}\partial x_{k}}\chi_{k}[/tex]

We also have the relation [itex]\nabla^{2}\phi=4\pi G\mu[/itex] where [itex]\mu[/itex] is the mass distribution. In a vacuum, this is just [itex]\nabla^{2}\phi=0[/itex].

Then, there is the separation vector for nearby geodesics which obeys a similar relation as the Newtonian one but this time, the right hand side has the Riemann curvature tensor instead of the partial derivative of [itex]\phi[/itex]. The relationship is
[tex](\nabla_{\mu}\nabla_{\mu}\chi)^{\alpha}=-R^{\alpha}_{\beta\theta\delta}u^{\beta}u^{\delta} \chi^{\theta}[/tex]

From here, he somehow makes the conclusion that we must have the Ricci curvature zero to get the analog of the Newtonian case. But, I'm not sure how it came about. How does one even think of the contracting the Riemann tensor in such a way to get the Ricci curvature tensor?

I understand this is the sort of thing that can't be derived but I am looking for some sort of insight. I feel like I am missing something obvious because Hartle generally does proceed very logically. Thank you for your help!
 
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  • #6
George Jones
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When, I first responded, I did not have Hartle at hand; now I do.

Compare equation (21.27) to (21.13).
 
  • #7
Oh I see it now! Must have missed that earlier.

Thank you, George Jones!
 
  • #8
Bill_K
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Compare equation (21.27) to (21.13).
Could someone who has the book reveal the answer to those of us who don't? Thanks!
 
  • #9
Sure, let me try to explain. I'm only sort of sure so if someone finds something wrong, please do correct it. Hartle takes the following equation in a local inertial frame.
[tex](\nabla_{\mu}\nabla_{\mu}\chi)^{\alpha}=-R^{\alpha}_{\beta\theta\delta}u^{\beta}u^{\delta} \chi^{\theta}[/tex]

This yields,
[tex]\frac{d^{2}\chi^{\alpha}}{d\tau^{2}}=R^{\alpha}_{ \beta \theta\delta}\chi^{\beta}[/tex]

The R here is the Riemann curvature tensor in the LIF.

Now, he uses the static weak field metric to work out this Riemann tensor and contracts it to get the Ricci tensor. This yields (approximately, since it is a weak perturbation of the flat space metric)

[tex]\frac{d^{2}\chi^{i}}{dt^{2}}=R^{i}_{ t j t}\chi^{j}[/tex]

[tex]R_{ij}=\frac{\partial^{2}\phi}{\partial x^{i}\partial x^{j}}[/tex]

Comparing with the Newtonian case, the Ricci tensor must be zero in a vacuum.
 
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