About the Einstein tensor.

[grayson]

So, I have been trying to understand the Einstein field equations to better understand general relativity. I figured out the stress-energy tensor but the einstein tensor puzzles me. There are no fun little diagrams showing me what the different parts of the tensor represent. Another thing is I cannot find the value of the cosmological constant. Can anybody help me?

[Genady]

13 minutes ago, grayson said:

the einstein tensor puzzles me

Which part?

 

14 minutes ago, grayson said:

the value of the cosmological constant

[astro-ph/0102033] Value of the Cosmological Constant: Theory versus Experiment (arxiv.org)

[grayson]

Just now, Genady said:

Which part?

 

[astro-ph/0102033] Value of the Cosmological Constant: Theory versus Experiment (arxiv.org)

I mean all of the parts of the Einstein tensor. Like is there any diagrams to show me what they represent?

[joigus]

It is the only rank-2 tensor that has identically zero covariant derivative. Einstein wanted something as analogous as is possible to get to Maxwell's equations.

I know of piecewise ways to understand its different pieces, but not the thing itself.

IOW, if you had a conversation with a pure differential-geometry person and you asked them:

"What is this G = Ricci-(1/2)R(metric) tensor to you?"

They would probably go,

"well, it's the way to obtain a 2-rank tensor from the Riemann tensor that is covariantly constant, if you want to, for some reason."

You, 

"But does it codify anything about the manifold, holes, winding numbers, other topological features? Can I see it coming from something more basic"

They,

"Well, the R that you mention codifies the genus (number of holes), but not the G. Not that I know of, why you ask?"

You,

"You don't happen to know of any way to quantise it, do you?"

Them,

"Our time is up."

[studiot]

1 hour ago, grayson said:

So, I have been trying to understand the Einstein field equations to better understand general relativity. I figured out the stress-energy tensor but the einstein tensor puzzles me. There are no fun little diagrams showing me what the different parts of the tensor represent. Another thing is I cannot find the value of the cosmological constant. Can anybody help me?

 

1 hour ago, grayson said:

I mean all of the parts of the Einstein tensor. Like is there any diagrams to show me what they represent?

 

This pdf article may well address what you are asking. It doesn't get any simpler but there is no heavy maths in it.

Here is the first page

Quote

https://math.ucr.edu/home/baez/einstein/einstein.pdf

 

The Meaning of Einstein’s Equation
John C. Baez∗ and Emory F. Bunn†
January 4, 2006
Abstract
This is a brief introduction to general relativity, designed for both students
and teachers of the subject. While there are many excellent expositions of
general relativity, few adequately explain the geometrical meaning of the
basic equation of the theory: Einstein’s equation. Here we give a simple
formulation of this equation in terms of the motion of freely falling test
particles. We also sketch some of the consequences of this formulation and
explain how it is equivalent to the usual one in terms of tensors. Finally, we
include an annotated bibliography of books, articles and websites suitable
for the student of relativity.
1 Introduction
General relativity explains gravity as the curvature of spacetime. It’s all about
geometry. The basic equation of general relativity is called Einstein’s equation.
In units where c = 8πG = 1, it says
Gαβ = Tαβ . (1)
It looks simple, but what does it mean? Unfortunately, the beautiful geometri-
cal meaning of this equation is a bit hard to find in most treatments of relativity.
There are many nice popularizations that explain the philosophy behind rela-
tivity and the idea of curved spacetime, but most of them don’t get around to
explaining Einstein’s equation and showing how to work out its consequences.
There are also more technical introductions which explain Einstein’s equation
in detail — but here the geometry is often hidden under piles of tensor calculus.
This is a pity, because in fact there is an easy way to express the whole
content of Einstein’s equation in plain English. In fact, after a suitable prelude,
one can summarize it in a single sentence! One needs a lot of mathematics to
derive all the consequences of this sentence, but it is still worth seeing — and
we can work out some of its consequences quite easily.
∗Department of Mathematics, University of California, Riverside, California 92521, USA.
email: baez@math.ucr.edu
†Physics Department, University of Richmond, Richmond, VA 23173, USA. email:
ebunn@richmond.edu

 

[Markus Hanke]

12 hours ago, grayson said:

There are no fun little diagrams showing me what the different parts of the tensor represent.

It’s best to look at the Einstein tensor as a kind of machine with two slots - if you feed into both slots a unit vector that points into the future time direction, it will give you as result a real number that represents the average Gaussian curvature in a small spatial region around the event you are working at:

\[G_{\mu\nu}t^{\mu}t^{\nu}=k\]

So the Einstein tensor is a kind of average curvature around some event. I don’t think it is helpful to try and find specific “meanings” for specific components.

[KJW]

15 hours ago, joigus said:

It is the only rank-2 tensor that has identically zero covariant derivative.

Actually, it's not. In fact, for every scalar function of the metric tensor and its partial derivatives of any order, there is a rank-2 tensor that has identically zero covariant divergence. I call this the general relativity version of Noether's theorem. For the Ricci scalar as the scalar function, the corresponding rank-2 tensor that has identically zero covariant divergence is the Einstein tensor. Two other independent scalar functions are the square of the magnitude of the trace-free part of the Ricci tensor, and the square of the magnitude of the Weyl conformal tensor. There are also various covariant derivatives of curvature tensors from which scalar functions can be derived as the square of the magnitudes. However, the Ricci scalar is probably the only scalar function that is not derived from the square of a magnitude.

 

[joigus]

5 minutes ago, KJW said:

Actually, it's not. In fact, for every scalar function of the metric tensor and its partial derivatives of any order, there is a rank-2 tensor that has identically zero covariant divergence. I call this the general relativity version of Noether's theorem. For the Ricci scalar as the scalar function, the corresponding rank-2 tensor that has identically zero covariant divergence is the Einstein tensor. Two other independent scalar functions are the square of the magnitude of the trace-free part of the Ricci tensor, and the square of the magnitude of the Weyl conformal tensor. There are also various covariant derivatives of curvature tensors from which scalar functions can be derived as the square of the magnitudes. However, the Ricci scalar is probably the only scalar function that is not derived from the square of a magnitude.

 

You mean a scalar multiple of it?

"Essentially". I forgot to say "essentially". It's essentially the only tensor...

You can introduce a scalar that can be gauged away in virtually every theory we have. So...

[KJW]

50 minutes ago, joigus said:

You mean a scalar multiple of it?

"Essentially". I forgot to say "essentially". It's essentially the only tensor...

You can introduce a scalar that can be gauged away in virtually every theory we have. So...

No. I'm talking about scalars that are distinct from the Ricci scalar and obtained from the Riemann tensor or its covariant derivatives of any order.

 

I should also remark that one such scalar:

R_ij^kl R_pq^rs δ^ijpq_klrs

is the integrand of a topological invariant in four dimensions, and therefore not all the second-degree scalars are independent.

 

 

Edited December 17, 2023 by KJW

[Markus Hanke]

27 minutes ago, KJW said:

No. I'm talking about scalars that are distinct from the Ricci scalar and obtained from the Riemann tensor or its covariant derivatives of any order.

Yes, but then you’d end up with a different theory of gravity. In GR, the tensor we’re looking for can only contain first and second derivatives of the metric, must be linear in the latter, symmetric, and divergence-free. Lovelock’s theorem guarantees that the only tensor that fulfills these criteria is the Einstein tensor.

[KJW]

33 minutes ago, Markus Hanke said:

Yes, but then you’d end up with a different theory of gravity. In GR, the tensor we’re looking for can only contain first and second derivatives of the metric, must be linear in the latter, symmetric, and divergence-free. Lovelock’s theorem guarantees that the only tensor that fulfills these criteria is the Einstein tensor.

Yes, as I alluded to in my first post, the Einstein tensor is the only tensor that is linear in the second-order derivatives of the metric tensor.

One of the questions that I had considered for a long time was what justified the use of the Ricci scalar in the Einstein-Hilbert action. It turned out that the discovery that every scalar function of the metric tensor and its partial derivatives of any order leads to a conserved quantity was not the answer I was expecting. But the Ricci scalar is the only scalar function (that I'm aware of) that is linear in the second-order derivatives of the metric tensor. That doesn't quite answer my question but at least it is something that distinguishes the Ricci scalar from other scalars.

 

 

33 minutes ago, Markus Hanke said:

Yes, but then you’d end up with a different theory of gravity.

For me, there is the question of why the theory of gravity we have is the one that matches physics. I believe that GR needs to not only conform to physical reality, but also be logical as well.

 

 

Edited December 17, 2023 by KJW

[joigus]

The Ricci scalar is also of special interest because its integral to the whole manifold gives you the genus of your manifold (number of holes).

So yes, physical information is coded in it. What not so clear to me is how to visualise the tensor itself.

For some reason, people keep talking about "seeing" tensors.

[KJW]

6 minutes ago, joigus said:

The Ricci scalar is also of special interest because its integral to the whole manifold gives you the genus of your manifold (number of holes).

That's only true in two dimensions. The corresponding quantity in four dimensions (ignoring the numeric factor) was given above:

R_ij^kl R_pq^rs δ^ijpq_klrs

For even dimensions in general, the quantity is:

R_ij^kl R_pq^rs ... R_uv^wx δ^ijpq...uv_klrs...wx

I am not aware of any corresponding quantity for odd dimensions.

 

[joigus]

8 minutes ago, KJW said:

That's only true in two dimensions. The corresponding quantity in four dimensions (ignoring the numeric factor) was given above:

R_ij^kl R_pq^rs δ^ijpq_klrs

For even dimensions in general, the quantity is:

R_ij^kl R_pq^rs ... R_uv^wx δ^ijpq...uv_klrs...wx

I am not aware of any corresponding quantity for odd dimensions.

 

Thank you. Yes, that's right. I was trying to recall ways in which the different pieces of the Einstein tensor might have some kind of meaning easy to visualise.

[Genady]

2 hours ago, joigus said:

people keep talking about "seeing" tensors

Of course, I "see" multilinear machines taking in vectors and producing numbers.

I also "see" equivalence classes of indexed collections of numbers with certain transformations being the equivalence relations.

Edited December 17, 2023 by Genady

[Markus Hanke]

18 hours ago, KJW said:

For me, there is the question of why the theory of gravity we have is the one that matches physics. I believe that GR needs to not only conform to physical reality, but also be logical as well.

Are you familiar with the topological arguments put forward in MTW with regards to why the EFE has the form it does? Not sure if one can call it a “derivation” exactly, but it’s nonetheless very interesting.

[joigus]

On 12/17/2023 at 1:20 PM, KJW said:

That's only true in two dimensions. The corresponding quantity in four dimensions (ignoring the numeric factor) was given above:

R_ij^kl R_pq^rs δ^ijpq_klrs

For even dimensions in general, the quantity is:

R_ij^kl R_pq^rs ... R_uv^wx δ^ijpq...uv_klrs...wx

I am not aware of any corresponding quantity for odd dimensions.

 

Being a little fast and loose with your logic allows you to detect people who aren't.  

Welcome to the forums, if I didn't say it before.

[KJW]

On 12/18/2023 at 4:00 PM, Markus Hanke said:

Are you familiar with the topological arguments put forward in MTW with regards to why the EFE has the form it does? Not sure if one can call it a “derivation” exactly, but it’s nonetheless very interesting.

I don't recall seeing it.

It did occur to me that my question could be addressed by considering a principle that, given that reality is based on a spacetime with a non-flat metric, every quantity that emerges from Ricci Calculus corresponds to some quantity in the physical world. So, for example, the Einstein tensor corresponds to some conserved quantity in the physical world. That it corresponds to the energy-momentum tensor is largely a matter of physics rather than mathematics. However, which physical quantity a given mathematical quantity corresponds to does depend on the precise matching of the mathematical properties of the two quantities. Nevertheless, I'm still curious about the rationale behind choosing the Ricci scalar for the Einstein-Hilbert action.

 

3 hours ago, joigus said:

Welcome to the forums, if I didn't say it before.

Thank you. 🙂

 

[Markus Hanke]

13 hours ago, KJW said:

Nevertheless, I'm still curious about the rationale behind choosing the Ricci scalar for the Einstein-Hilbert action.

I think this discussion on Physicsforums, particularly post #12, might shed some light, though I need to go through it more carefully myself.