I'm sorry, I don't understand what you are asking for. Could you please clarify?

[GRstudent]

Hi all,

I have been trying to solve the Einstein Field Equations for its (0,0) component. So I have got that (c=1)

Einstein Tensor (upper,0,0)=8*pi*G*T(upper,0,0)

Now, let's see what T (0,0) really is. It is energy density, right? So According to famous E=mc^2 the energy density is the same as mass density, assuming the speed of light to be equal to one. Therefore, the (0,0) components of the Stress-Energy Tensor is just the mass density. So we have that

Einstein Tensor (upper,0,0)=8*pi*G*ρ

Now, let's see what Einstein Tensor G(0,0) really is

Ricci(0,0) - 1/2*g(upper 0,0)*Ricci scalar

Ricci scalar is obtained by contracting it with metric tensor so we have that

Ricci (upper,0,0) - 1/2*g(upper 0,0)*Ricci (upper0,0)*g(lower, 0,0)=8*pi*G*ρ

So g(upper 0,0)*g(lower, 0,0) is 1 so we have that

Ricci (upper 0,0) - 1/2*Ricci(upper 0,0)=8*pi*G*ρ

1/2 Ricci (upper 0,0)=8*pi*G*ρRicci(0,0)=4*pi*G*ρNow look carefully to the Right Hand Side of the Equation. It is the same from the Poisson's Equation where

Set of second partial derivatives of the Gravitational potential=4*pi*G*ρ

Therefore, is it true that the zero-zero component of the Einstein Tensor, and subsequently the Ricci tensor, is just the [Set of second partial derivatives of the Gravitational potential] or 4*pi*G*ρ

Thanks!P.S I apologize for not using MathCodes--never used them before and would appreciate if someone will show me how to use them.

 

[Mentz114]

With c=G=1
[tex]
\begin{align}
R^\mu_\mu - (1/2)g^\mu_\mu R &= 8\pi T^\mu_\mu\\
R &= -8\pi \rho
\end{align}
[/tex]
for a pressureless, static matter ( dust) solution.

The Riemann and Ricci tensors contain first and second derivatives of the metric.
[tex]
R^{r}_{msq}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
[/tex]

If you right click on a formula, you get a menu. Select 'Tex' to see the format.

 

[GRstudent]

^
You did with one lower and one upper indexes. That's not what I meant. When I wrote (upper 0,0) it meant that BOTH zero and zero are upper indexes.

Also, I would appreciate you solving Ricci tensor zero-zero (both upper!) with 4*pi*G*density on the Right Hand Sight.

And why did you think that both indexes are equal? The are mu and nu; not mu and mu.

And, I didn't ask about the Riemann tensor at all. There was also no point making G=1

In general, I would like to see the proof that set of partial derivatives of the gravitational potential=ricci tensor with zero-zero

 

[Dale]

^
You did with one lower and one upper indexes. That's not what I meant. When I wrote (upper 0,0) it meant that BOTH zero and zero are upper indexes.

Also, I would appreciate you solving Ricci tensor zero-zero (both upper!) with 4*pi*G*density on the Right Hand Sight.

And why did you think that both indexes are equal? The are mu and nu; not mu and mu.

And, I didn't ask about the Riemann tensor at all. There was also no point making G=1

If you are going to be this picky then you really need to make the effort to learn LaTeX.

 

[GRstudent]

^
Absolutely true! I just cannot find where I can learn it.

 

[Dale]

Please start here:
https://www.physicsforums.com/showthread.php?p=3977517&posted=1#post3977517

Then once you know some of the basics you can just use Google to find many other sites.

 

[GRstudent]

OK, my question is how to solve this equation, assuming c=1

[itex]\mathrm R_{}{}^{00}[/itex]=[itex]\nabla^2 \phi[/itex]

where [itex]\phi[/itex] is a gravitational potential from Poisson's equation.

Here is my logic:

[itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]g_{00}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]\frac{1}{2}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]4[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

Poisson's equation tells us that

[itex]\nabla^2 \phi[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

So it must be that
[itex]4[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]=[/itex][itex]\nabla^2 \phi[/itex]

So my question is what is its solution?

Also [itex]\phi[/itex][itex]=[/itex][itex]\frac{GM}{R}[/itex], right?

 

[TheEtherWind]

First off [itex]\frac{1}{2}R^{00}=8 \pi G \rho[/itex] is not equivalent to [itex]R^{00}=4 \pi G \rho[/itex]

It's equivalent to [itex]R^{00}=16 \pi G \rho[/itex]

And there is a limiting case somewhere to compare the metric with the Newtonian gravitational potential I believe. But I do not believe [itex]R^{00}= \nabla^2 \phi[/itex]

 

[GRstudent]

I am confused...

 

[PeterDonis]

[itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

[itex]\mathrm R_{}{}^{00}[/itex][itex]-[/itex][itex]\frac{1}{2}[/itex][itex]g_{}{}^{00}[/itex][itex]\mathrm R_{}{}^{00}[/itex][itex]g_{00}{}^{}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G[/itex][itex]\rho[/itex]

This step is not correct. The Ricci scalar [itex]R[/itex] involves all of the components of the Ricci tensor:

[tex]R = R^{ab} g_{ab} = R^{00} g_{00} + R^{11} g_{11} + R^{22} g_{22} + R^{33} g_{33}[/tex]

In fact, even the above is only true in a coordinate chart where the metric is diagonal, i.e., where the only nonzero components have a = b. In a more general coordinate chart you will need to include terms like [itex]R^{01} g_{01}[/itex], etc.

Also, EtherWind is correct that the equation you give relating [itex]R^{00}[/itex] to the Newtonian "potential", [itex]R^{00} = \nabla^{2} \phi[/itex], is not true in general. First, it is a limiting case where gravity is very weak; second, even that limiting case only applies in situations where the field is static, i.e., where gravity at a given radius does not change with time.

You might want to check out this page by John Baez on the meaning of Einstein's Equation:

http://math.ucr.edu/home/baez/einstein/einstein.html

 

[GRstudent]

I understand now...I will get back to my calculations and Reply to this thread in a couple of days. Because I am a first year undergrad, it is kinda difficult for me to grasp these concepts.

 

[WannabeNewton]

Well if [itex]\frac{1}{2}x = 8 [/itex] then [itex]x = 16[/itex] but anyways what you are thinking of is the limiting case of the field equations. What you are trying to say would be valid for the linearized field equations where you consider a small perturbation on a background flat space -time. Anyways in this limit you can recast the field equations, with a bit of plugging in and simplifying etc. and using that for small velocities [itex]T^{00}[/itex] will dominate and [itex]\partial _{t}[/itex] will be higher order, and write it as [itex]\triangledown ^{2}\bar{h^{00}} = -16\pi \rho [/itex] ([itex]h[/itex] is the perturbation). You can compare that to the usual Poisson's equation to extract a solution which will eventually give you the weak field Newtonian metric.

 

[GRstudent]

I reviewed my older calculations and I have that

[itex]\mathrm T_{}{}^{00}[/itex][itex]=[/itex][itex]\rho[/itex][itex]c_{}{}^{2}[/itex]

I mean that this fact is non-negotiable, right?

This doesn't mean that all other components are zero--we are just considering its zero-zero component. (Right Hand Sight of the EFE has c^-4 term in original form, so having c^2 up and c^-4 down gives c^-2)

[itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]8[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex][itex]c_{}{}^{-2}[/itex]

[itex]\frac{c_{}{}^{2}G_{}{}^{00}}{2}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex]

if we set c=1, we have that

[itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex]

So we have that

[itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]\nabla^2 \phi[/itex]

Please correct me if I am wrong.

Also, [itex]\mathrm T_{}{}^{00}[/itex] of the Earth is [itex]\rho[/itex][itex]c^2[/itex][itex]=[/itex][itex]5520[/itex][itex]c^2[/itex], right?

 

[Dale]

I reviewed my older calculations and I have that

[itex]\mathrm T_{}{}^{00}[/itex][itex]=[/itex][itex]\rho[/itex][itex]c_{}{}^{2}[/itex]

I mean that this fact is non-negotiable, right?

Well, it isn't generally true, but it certainly can be taken as a given for a specific coordinate system that you wish to consider.

[itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]4[/itex][itex]\pi[/itex][itex]\mathrm G_{}{}^{}[/itex][itex]\rho[/itex]

All fine so far.

So we have that

[itex]\frac{1}{2}[/itex][itex]\mathrm G_{}{}^{00}[/itex][itex]=[/itex][itex]\nabla^2 \phi[/itex]

This relies on the substitution: [itex]\nabla^2 \phi = 4 \pi \mathrm G \rho[/itex]. This substitution is generally true in Newtonian gravity, but it is only true in GR in the "weak field" limit.

When solving the EFE in GR, the usual goal is to find a metric, [itex]g_{\mu\nu}[/itex], which satisfies the EFE for a given stress energy tensor. So you usually need to expand the Einstein tensor in terms of the metric.

 

[GRstudent]

^
Thank you

 

[ApplePion]

You wrote:

"Ricci scalar is obtained by contracting it with metric tensor so we have that
Ricci (upper,0,0) - 1/2*g(upper 0,0)*Ricci (upper0,0)*g(lower, 0,0)=8*pi*G*ρ"You did the contraction for getting the R scalar incorrectly. You did R= (g00)(R00), thinking that R00 was the only non-vanishing part of the Ruv tensor when Tuv has T00 as its only non-vanishing component. Surprisingly it turns out different than what you expected.

I'll modify what you were trying to do to it correctly. (I might make careless arithmetic mistakes, so check for them)

Take Ruv - 1/2 guv R = 8 pi Tuv and do a scalar contraction. We get

R - 2R = 8 pi T

R = - 8 pi T where T is the contraction of the Tuv tensor.

Now insert this back into the Einstein Equarions

Ruv - (1/2)(- 8 pi guv T) = 8 pi Tuv

Ruv = 8 pi [Tuv - 1/2 guv T]

You wanted to look at the T00 equation. OK:

R00 = 8 pi [ T00 - 1/2 g00 T]

Note that it is not that only the T00 component of the stress energy tensor that is driving R00! The 1/2 g00 T term contains components in addition to T00!

So if we have two otherwise identical mass distributions except one has more pressure, the one with more pressure (but the same mass!) will generate a stronger gravitational field. This might at first seem unphysical, but it actually makes sense. Faster particles are more strongly deflected by the gravitational field of a static mass distribution than are slower particles--this is because the static object actually turns out to generate velocity squared forces. So it makes sense that there is a recipricosity-- an object with velocity squared characteristics, being more affected by other gravitational bodies, also affects them more strongly than if it were stationary.

Another place where the fact that the right-hand side of R00 = 8 pi [T00 - 1/2 g00 T] having the T term manifests itself is in the acceleration of the cosmological expansion. The [T00 - 1/2 g00 T] drives the acceleration of the expansion, and it contains pressure. Quantitatively both the mass density and the pressure in that term, if they are positive, produce deceleration of the expansion of the universe. The current view is that the observational evidence is that the Universe is expanding with positive acceleration. There is a reason (another relativistic cosmological equation) why the mass density appears to need to be positive...so this seems to leave us stuck with a negative pressure, a disturbing thought considering pressure goes as the square of the velocities. This is the "dark energy" problem.

OK, so we need to add the T term to your equation. As you claimed there then is a Poisson Equation appearance. But you seem not aware that the Poisson nature is only approximate--the General Relativity equations have non-linear terms not appearing in the Poisson Equation. Also the General Relativity Equation has time derivatives.

 

[GRstudent]

Thanks for your answer. Here I re-wrote what you tried to say in more clear way:

[itex]R_{\mu\nu}-\dfrac{1}{2} g_{\mu\nu} R =8 \pi T_{\mu\nu}[/itex]

[itex] R-2R=8\pi T[/itex] (how 1/2 turned into 2?)

[itex]R=-8 \pi T [/itex]

[itex]R_{\mu\nu}+\dfrac{1}{2} g_{\mu\nu}8 \pi T =8 \pi T_{\mu\nu}[/itex]

[itex]R_{\mu\nu}= 8 \pi (T_{\mu\nu}-\dfrac{1}{2} g_{\mu\nu} T) [/itex]

[itex]R_{00}= 8 \pi (T_{00}-\dfrac{1}{2} g_{00} T) [/itex]

In any way, the T_{00} is the energy density. It would be much clearer if you make an example of this formula in use.

 

[ApplePion]

GRstudent, I think you did not mean to type what you typed in your last post.

 

[GRstudent]

^
Just look above--I re-edited my message. Thanks.

 

[ApplePion]

"R−2R=8πT (how 1/2 turned into 2?)"

When we did the contraction of the metric tensor we did (g00) squared + (g11) squared + (g22) squared + (g33) squared. That is (1) squared + (-1) squared + (-1) squared + (-1) squared. So the contraction of the metric tensor is 4. So the contraction of the -(1/2) guv R piece is -2R.

 

[ApplePion]

"In any way, the T_{00} is the energy density. It would be much clearer if you make an example of this formula in use."

Let's look at a situation where we have a stable star with no time derivatives of the metric occurring. And we will assume the fields are weak, and so we can ignore non-linear terms.

R00 = 8 pi [ T00- 1/2 g00 T]

R00 in the linear situation specified is approximately del squared g00.

We can approximate the g00 on the right-hand side as one. (The induced errors from this approximating are non-linear, and we are ignoring non-linear stuff.)

Again approximating the metric on the right-hand side as Lorentzian we get that T = T00 + T11 + T22 + T33, where the indices are up, and the coordinate system is a Cartesian type, rather than a spherical coordinate type.

T11 would be the mass density times the square of the velocity in the x direction. Consider molecules of a gas hitting a container, and you will see that this termm is the density of a gas' pressure. Amd remember there are 3 such terms.

OK, putting this all together we get:

del squared g00 = 4 pi [ p + 3P]

where p is the mass density, and P is the pressure density.

So it is a Poisson equation with the sort of solution you were discussing...but there is a 3P term.

Also the g11, g22 and g33 terms are not simply negative one.

 

[GRstudent]

OK, I'll try to rewrite it here:
**************************************************
EXAMPLE 1:

A stable star with weak gravitational field is given. The mass and pressure density of the star are [itex]\rho[/itex] and [itex]P[/itex], respectively. Find all information about its gravitational field.

SOLUTION:

1) Because the field is weak, we have (ignoring non-linear terms) that:

[itex]R_{00}= 8 \pi (T_{00}-\dfrac{1}{2} g_{00} T)[/itex]

2) Therefore, the [itex]\nabla^2 g_{00}[/itex][itex]=[/itex][itex]R_{00}{}[/itex]

3) The Stress-Energy tensor would be

[itex]T^{\mu \nu}=\left [ \begin{matrix}
T^{00}=\rho & 0 & 0 & 0 \\
0 & T^{11}=\rho v^2_{x}=P & 0 & 0 \\
0 & 0 & T^{22}=\rho v^2_{y} & 0\\
0 & 0 & 0 & T^{33}=\rho v^2_{z}
\end{matrix} \right ] [/itex]

4) Finally, [itex]\nabla^2 g_{00}= 4\pi (\rho + 3P)[/itex]

***********************************************

Now, I have questions. What does [itex]g_{00}[/itex] exactly represent here? I guess it is proportional to the gravitational potential.

This example is being made with myriads of assumptions; and furthermore, it's a weak field. Can you come up with the strong field example? Like dwarf or a black hole?

Thanks!

 

[ApplePion]

<<Now, I have questions. What does g 00 exactly represent here? I guess it is proportional to the gravitational potential>>

Typically, numerically g00 is approximately 1 + 2 phi/(c^2) where phi is the Newtonian potential. That is the typical approximate numerical value. What it "is" is part of a quantity that tells you what the physical 4-dimensional distance between two nearby space-time points based on what the coordinate difference between those points.

"Can you come up with the strong field example?"

In most cases that are not extremely simple it is very hard to get exact solutions for the full non-linear equations.

 

[GRstudent]

Then why would you go such extreme lengths to study the differential geometry, the geometry of curved space, all those tensors, if you cannot come up with an example where all of this is actually in real life usage?

 

[Mentz114]

This example is being made with myriads of assumptions; and furthermore, it's a weak field. Can you come up with the strong field example? Like dwarf or a black hole?

The simplest exact fluid solution is probably the Einstein fluid. The metric is
[tex]
ds^2 = -dt^2 +(1-r^2/R^2)^{-1} dr^2 + r^2 d\theta^2 + r^2\sin(\theta)^2 d\phi^2
[/tex]
where R is a constant and r<R.

The Einstein tensor has components [itex]G_{00}= 3/R^2[/itex], [itex]G_{nn}= g_{nn}P[/itex] where n is a spatial index and P = -1/R².

Before you complain about this, read http://arxiv.org/pdf/gr-qc/9809013.pdf

 

[ApplePion]

"Then why would you go such extreme lengths to study the differential geometry, the geometry of curved space, all those tensors, if you cannot come up with an example where all of this is actually in real life usage?"

If you want to apply the theory to real life applications you do not need an exact solution if the approximate solution is highly numerically accurate.

For example, suppose that there was no exact solution for the Earth's gravitational field (there actually is, if you assume that the Earth is spherically symmetric, but ignore that), and you want to get Global Positioning Devices to work right. You do not need an exact solution.

Suppose you had less applied interests--gravitational waves are not implied by Newtonian gravity, but are implied by the linear approximation of General Relativity. So you get a new phenomenon in an approximate theory. Of course if you wanted to deal with waves with very high field strength you could not use the linear approximation.

But on a deeper level, even if you cannot use something for applications it is still of intellectual interest to know the actual laws.

I also point out to you that in the real world, you always need to make approximations. No one can know, for example, the exact gravitational field of the Earth, either Newtonianly of Einsteinanly, because the exact mass distribution of the Earth cannot possibly be known. So we can approximate the Earth as a sphere...or we can use better approximations if needed. But we cannot get it exact.

Indeed, even if we solved the General Relativity equations exactly we still would not have exactness--General Relativity itself is a classical approximation to some quantum gravity theory. (That theory, BTW, is likely currently unknown)

Here is something for you to consider. When you write out the equations for the force diagram of a pendulum, the sine of theta appears. It is standard if theta is not large to use the approximation that sine of theta is theta. When you do that you get an equation that while not exact is easily solved--without that approximation the differential equation is difficult--and you get to learn lots of important things about the behavior of pendulums.

 

[ApplePion]

Mentz114, the metric from that paper is not correct. It would take a while to give you a thorough explanation, but there is a simple way to see there is a bad problem.

Consider the Christoffel Symbol [1, 00]. (I hope my notation is understandable.) If you calculate that from the metric given, you will get zero--that metric does not have a spatial derivative of g00 (or a time derivative of g01). But that Chistoffel Symbol is the Newtonian gravitational field! So that metric implies that the Newtonian gravitational field inside a star is zero.

 

[GRstudent]

Can you give me an example from cosmology where Ricci and Einstein Tensors are not zero?

 

[GRstudent]

"Then why would you go such extreme lengths to study the differential geometry, the geometry of curved space, all those tensors, if you cannot come up with an example where all of this is actually in real life usage?"

If you want to apply the theory to real life applications you do not need an exact solution if the approximate solution is highly numerically accurate.

For example, suppose that there was no exact solution for the Earth's gravitational field (there actually is, if you assume that the Earth is spherically symmetric, but ignore that), and you want to get Global Positioning Devices to work right. You do not need an exact solution.

Suppose you had less applied interests--gravitational waves are not implied by Newtonian gravity, but are implied by the linear approximation of General Relativity. So you get a new phenomenon in an approximate theory. Of course if you wanted to deal with waves with very high field strength you could not use the linear approximation.

But on a deeper level, even if you cannot use something for applications it is still of intellectual interest to know the actual laws.

I also point out to you that in the real world, you always need to make approximations. No one can know, for example, the exact gravitational field of the Earth, either Newtonianly of Einsteinanly, because the exact mass distribution of the Earth cannot possibly be known. So we can approximate the Earth as a sphere...or we can use better approximations if needed. But we cannot get it exact.

Indeed, even if we solved the General Relativity equations exactly we still would not have exactness--General Relativity itself is a classical approximation to some quantum gravity theory. (That theory, BTW, is likely currently unknown)

Here is something for you to consider. When you write out the equations for the force diagram of a pendulum, the sine of theta appears. It is standard if theta is not large to use the approximation that sine of theta is theta. When you do that you get an equation that while not exact is easily solved--without that approximation the differential equation is difficult--and you get to learn lots of important things about the behavior of pendulums.

If good old Schwarzschild metric (which is a good approximation) is all we need, why do we spent some much time on understanding the differential geometry and curvature tensor?

 

[ApplePion]

"Can you give me an example from cosmology where Ricci and Einstein Tensors are not zero?"

They are not zero in regions where there is matter. This follows directly for the Einstein tensor--it is equal to the stress energy tensor. For the Ricci tensor, you can derive it from from the Equation Ruv = 8 pi [Tuv- 1/2 guv T] that I derived for you yesterday.

 

[ApplePion]

"If good old Schwarzschild metric (which is a good approximation)"

The Schwarzschild Solution is an exact solution.

" is all we need, why do we spent some much time on understanding the differential geometry and curvature tensor?"

To understand the fundamental character of Nature.

 

[PeterDonis]

This example is being made with myriads of assumptions; and furthermore, it's a weak field. Can you come up with the strong field example? Like dwarf or a black hole?

The SET you wrote down for a fluid is valid for *any* fluid, not just a fluid with a weak field; the only restriction is that it must be a perfect fluid, i.e., zero viscosity and a perfect fluid equation of state. So it would describe a white dwarf or neutron star.

However, there is one other key restriction: the SET components you wrote down are in the rest frame of the fluid--i.e., *every* small fluid element is at rest in these coordinates. That is what makes the SET look so simple; in other coordinates it would not.

This SET does *not* describe a black hole--a black hole is a vacuum solution to the Einstein Field Equation (as I think has been mentioned before). See below.

If good old Schwarzschild metric (which is a good approximation) is all we need, why do we spent some much time on understanding the differential geometry and curvature tensor?

The "good old Schwarzschild metric", by itself, only describes one particular solution of the Einstein Field Equation--a black hole. The SET is zero everywhere. There are lots of other solutions that describe other kinds of spacetimes.

Check out this Wikipedia page and the pages it links to:

http://en.wikipedia.org/wiki/Solutions_of_the_Einstein_field_equations

 

[PeterDonis]

Can you give me an example from cosmology where Ricci and Einstein Tensors are not zero?

The standard FRW spacetimes all have nonzero Ricci and Einstein tensors:

http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

 

[Dale]

I hope my notation is understandable.

It would really help if you would take a few minutes to learn how to use the forum LaTeX feature.

 

[GRstudent]

[itex]R_{\mu \nu} = g_{\mu \nu} [/itex]

Find [itex] G_{\mu \nu}[/itex]