EFE: Why is there a curvature tensor and curvature scalar?

In summary, the two terms for curvature in the Einstein-Hilbert action are the curvature tensor and the curvature scalar multiplied by the metric tensor. These terms arise when extremizing the action with respect to the inverse metric. The curvature scalar is simply the contraction of the Ricci tensor, while the curvature tensor is the tensor associated with the Riemann-Christoffel curvature. The constant in the Einstein Field Equations comes from matching the equations to the Newtonian equation for gravitation in the low speed, low gravity case.
  • #1
PerpStudent
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In the Einstein tensor equation for general relativity, why are there two terms for curvature: specifically the curvature tensor and the curvature scalar multiplied by the metric tensor?
 
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  • #2
PerpStudent said:
In the Einstein tensor equation for general relativity, why are there two terms for curvature: specifically the curvature tensor and the curvature scalar multiplied by the metric tensor?
When the Einstein-Hilbert action is extremized wrt the inverse metric, that is what emerges. See here http://en.wikipedia.org/wiki/Einstein–Hilbert_action.

Are you aware that the curvature scalar is the contraction of the Ricci tensor, so R = Rμμ.
 
  • #3
Hi PerpStudent! :smile:

Because the traceless symmetric part of Aij is Aij - 1/4 tr(A) gij.

Any tensor equation can be "traced" and "tracelesssed".

ie the trace of the equation is true, and the traceless symmetric part of the equation is true.

So in the Einstein field equations we expect …

R = constant*T

Rij - 1/4 R gij = constant* (Tij - 1/4 T gij)​

(and it turns out the constants have to be -8π and 8π, to give the Newtonian inverse-square law in the low-field limit)
 
  • #4
PerpStudent said:
In the Einstein tensor equation for general relativity, why are there two terms for curvature: specifically the curvature tensor and the curvature scalar multiplied by the metric tensor?

There's only one independent / fundamental curvature, namely the Riemann-Christoffel curvature tensor. The so-called Ricci curvature and the curvature scalar are simply contractions of the 4th rank tensor with respect with the metric once and twice, respectively. They are susequently derived concepts.

One can write down the EFE in terms of the Riemann-Christoffel curvature tensor only (in the absence of matter) as:

[tex] g^{\mu \alpha}R_{\mu \nu|\alpha \beta}-\frac{1}{2}g_{\nu \beta}g^{\mu \alpha}g^{\lambda \sigma}R_{\mu \lambda|\alpha \sigma} = 0 [/tex]

but it won't look pretty, that's why the Ricci curvature tensor and the Ricci scalar are put into GR.
 
  • #5
The "reason" the particular combination

[tex]G^{\mu\nu} \equiv R^{\mu\nu} - \frac12 R g^{\mu\nu}[/tex]
appears is because is this the unique combination of curvatures that satisfies

[tex]\nabla_\mu G^{\mu\nu} = 0.[/tex]
 
  • #6
Ben Niehoff said:
The "reason" the particular combination

[tex]G^{\mu\nu} \equiv R^{\mu\nu} - \frac12 R g^{\mu\nu}[/tex]
appears is because is this the unique combination of curvatures that satisfies

[tex]\nabla_\mu G^{\mu\nu} = 0.[/tex]

To expand on this , Einstein's equation says that G_uv = 8 pi T_uv, where T_uv is the stress-energy tensor.

Continuiity conditions on the stress-energy tensor, T_uv require that
[tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] i.e. that the tensor be divergence free.

So since T_uv, the rhs is divergence free, the lhs has to be divergence free as well.
 
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  • #7
pervect, ben, as a matter of interest, do you know any easy-to-understand reason why it's 8π ?

4π I'm more or less used to (and even 4π 10-7 :rolleyes:) …

but why 8 ? :confused:
 
  • #8
My 2ç. There's no <physical> reason, the "extra" 2 comes from the 1/2 of the Christoffel symbols which has to do with the metric being assumed torsionless and symmetric.
 
  • #9
tiny-tim said:
pervect, ben, as a matter of interest, do you know any easy-to-understand reason why it's 8π ?

4π I'm more or less used to (and even 4π 10-7 :rolleyes:) …

but why 8 ? :confused:

The constant comes from matching the Einstein Field Equations to the Newtonian equation for gravitation in the low speed, low gravity case (makes sure that General Relativity gives the same predictions in this case as Newtonian gravity).
 
  • #10
pervect said:
To expand on this , Einstein's equation says that G_uv = 8 pi T_uv, where T_uv is the stress-energy tensor.

Continuiity conditions on the stress-energy tensor, T_uv require that
[tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] i.e. that the tensor be divergence free.

So since T_uv, the rhs is divergence free, the lhs has to be divergence free as well.

Is the requirement that [tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] due to energy and momentum conservation?
 
  • #13
PerpStudent said:
Is the requirement that [tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] due to energy and momentum conservation?

No, see the reason in the my post above this one.
 
  • #14
PerpStudent said:
Is the requirement that [tex]\nabla_\mu T^{\mu\nu} = 0.[/tex] due to energy and momentum conservation?

It's very closely related.

You can think of it as being due to energy and momentum conservation in a local sense, i.e. at a point.

See for instance http://en.wikipedia.org/wiki/Continuity_equation

I'm pretty sure Wald and MTW discuss this with more rigor - I'd have to look stuff up to refresh my recollection to give any real detail, at least if I wanted to avoid misleading anyone.
 

1. What is the curvature tensor?

The curvature tensor, also known as the Riemann curvature tensor, is a mathematical object that describes the curvature of a manifold, which is a geometric space that can be curved. It is represented by a set of numbers that measure how much the space is curved at each point.

2. Why is there a curvature tensor?

The existence of a curvature tensor is a consequence of the theory of general relativity, which states that gravity is not a force between masses, but rather a curvature of spacetime caused by the presence of mass and energy. The curvature tensor is used to describe this curvature and its effects on the motion of objects.

3. What is the significance of the curvature tensor?

The curvature tensor is essential in understanding the behavior of gravity and the structure of the universe. It allows us to make predictions about the motion of objects, the bending of light, and the evolution of the universe. It also plays a crucial role in the mathematical formulation of general relativity.

4. What is the curvature scalar?

The curvature scalar, also known as the Ricci scalar, is a mathematical quantity that is derived from the curvature tensor. It is obtained by contracting (summing) the components of the curvature tensor and represents the overall curvature of a space at a given point. It is a crucial parameter in the equations of general relativity.

5. Why is there a curvature scalar?

The existence of the curvature scalar is a consequence of the mathematical properties of the curvature tensor. It provides a way to summarize the local curvature of a space in a single value, making it easier to analyze and compare different spaces. It is also used in the Einstein field equations to describe the relationship between the curvature of space and the distribution of matter and energy.

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