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

[PeterDonis]

$R_{\mu \nu} = g_{\mu \nu}$

Find $G_{\mu \nu}$

What is this supposed to mean? If you are using $R_{\mu \nu}$ and $g_{\mu \nu}$ with their standard definitions (Ricci tensor and metric, respectively), then your equation is simply false.

 

[George Jones]

$R_{\mu \nu} = g_{\mu \nu}$

Find $G_{\mu \nu}$

Looks like pure dark energy to me.

$$\begin{align} \kappa T_{\mu \nu} &= G_{\mu \nu} \\ &= R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} \\ &= g_{\mu \nu} - \frac{1}{2} g^\alpha_\alpha g_{\mu \nu} \\ &= g_{\mu \nu} - 2 g_{\mu \nu} \\ \kappa T_{\mu \nu} &= -g_{\mu \nu} \end{align}$$

 

[PeterDonis]

Looks like pure dark energy to me.

But is it even possible to have the Ricci tensor equal the metric in the first place?

 

[George Jones]

But is it even possible to have the Ricci tensor equal the metric in the first place?

Yes, this is de Sitter space.

Here is another way to look at it. Start with the stress energy tensor for dark energy $\kappa T_{\mu \nu} = - g_{\mu \nu}$. This, together with Einstein's equation and some manipulations, leads to $R_{\mu \nu} = g_{\mu \nu}$.

 

[PeterDonis]

Yes, this is de Sitter space.

With a particular value of the cosmological constant, correct?

Googling also uncovered a general term, "Einstein manifold", for the general class of solutions which have the Ricci tensor proportional to the metric:

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

 

[GRstudent]

You guys seem to misunderstand what I was trying to say. I meant 2d Sphere. So on a surface of a sphere, the metric tensor is just the Einstein tensor.

$R_{\theta \theta}=1$
$R_{\theta \phi} = 0$
$R_{\phi \phi} = sin^2(\theta)$

Now, I am trying to calculate the Einstein tensor using these Ricci components. So I got that $G_{\theta \theta}=0$ and $G_{\phi \phi}= sin^2(\theta)-\dfrac{1}{sin^2(\theta)}$. Please check my answers! Thank you.

 

[GRstudent]

Anyone?

 

[Dale]

You guys seem to misunderstand what I was trying to say. I meant 2d Sphere. So on a surface of a sphere, the metric tensor is just the Einstein tensor.

I got that the metric tensor was equal to the Ricci curvature tensor, and that the Einstein tensor was 0.

 

[GRstudent]

^ yeah that's true. Why Einstein Tensor is zero on a 2d sphere (not in mathematical view)?

$R_{\theta \theta}=1$
$R_{\phi \phi}=sin^2(\theta)$

$G_{\theta \theta} = R_{\theta \theta} - \dfrac{1}{2} g_{\theta \theta} R$

$G_{\theta \theta} = 1 - \dfrac{1}{2} 2$

$G_{\theta \theta} = 0$

and

$G_{\phi \phi}= R_{\phi \phi} - \dfrac{1}{2} g_{\phi \phi} R$

$G_{\phi \phi} = sin^2(\theta) - \dfrac{1}{2} sin^2(\theta) 2$

$G_{\phi \phi} = 0$

WHY?

 

[George Jones]

There aren't enough degrees of freedom. The Einstein tensor is zero for any 2-dimensional manifold, not just for S^2.

 

[GRstudent]

^
That is as clear as day. Again, WHY? W H Y ?

There aren't enough degrees of freedom.

What are you talking about? What degrees? Fahrenheit or Celcius?

 

[ApplePion]

Isn't the integrand for the Gauss Bonnet quantity in 2 dimensions the square root of negative g times the Riemann Scalar?

If so, then any variation in that quantity will be zero because the topology does not change. But that same quantity is the Lagrangian density for General Relativity--its variation gives the Einstein tensor. So the Einstein tensor in two dimensions must vanish.

 

[HallsofIvy]

^
That is as clear as day. Again, WHY? W H Y ?


What are you talking about? What degrees? Fahrenheit or Celcius?

If you don't know what "degrees of freedom" means then you are probably in over your head.

 

[Mentz114]

I don't really understand what are you trying to explain yet I came up with another idea. If Einstein Tensor on a 2d Sphere is zero then the energy density on the sphere surface is zero as surface doesn't have any energy. Yet it is strange somewhat because the source of gravity is not only mass but also momentum, so when we take 2d sphere on the Earth, we shouldn't get 0 because Earth is in orbit 30km/sec. So this is unclear.

Nonsense.

I am not interested in your "probabilities" and "conclusions".

Rude, arrogant and ignorant. Are you a child ? Someone should teach you some manners.

 

[GRstudent]

If you don't wish to help me--fine. Ignore this thread.

 

[Dale]

^
That is as clear as day. Again, WHY? W H Y ?

What are you talking about? What degrees? Fahrenheit or Celcius?

Degrees of freedom are ways in which something can change. So, in 4D a t vector can deviate in the x direction while going around a yz loop, or it can deviate in the z direction while going in a xz loop, or ... there are lots of different ways. Not so in 2D.

If you just go through the definitions and calculate the Riemann curvature tensor of an arbitrary metric in 2D and then contract it down to the Einstein tensor then you will see that everything will cancel.

Btw, you are being a little impolite. Please make an extra effort at civility.

 

[George Jones]

If you don't wish to help me--fine. Ignore this thread.

But when I chose to do just this, i.e., I when I decided that I no longer wanted to try and help, and that I wanted to ignore the thread, you went back and edited an earlier post in order to direct sarcasm towards me.

In 4-dimensional spacetime, the Riemann curvature tensor has 20 independent components, the Ricci tensor has 10 independent components, the Ricci curvature scalar has 1. This makes sense, because Ricci is a contraction of Riemann, and the curvature scalar is a contraction of Ricci. In a 2-dimenional space, however, the Riemann curvature tensor, the Ricci tensor, and the Ricci curvature scalar all have just one independent component.

Consequently, it seems plausible to me that

$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}$$
vanishes for 2-dimensiona spaces. Plausible, but I can't just "see" it, and I need to use math to verify it. You wrote

Why Einstein Tensor is zero on a 2d sphere (not in mathematical view)?

to which I don't have an answer. I have to use math. This is (just) one reason that I chose to ignore the thread.

For 2-dimensional spaces,

$$R_{\mu \nu} = K g_{\mu \nu},$$
where $K$ is a function of position. Thus, in two dimensions,

$$R = R^\mu_\mu = K R^\mu_\mu = 2K.$$
Using the above two equations in the Einstein tensor gives zero. This is a general result for 2-dimensional space, and not something special about 2-dimensional spheres.

Do you really want to look at Einstein's equation for 2-dimensional spacetimes, or do you want to look at Einstein's equation's equation for a 4-dimensional spacetime and project the results on a 2-dimensional surface? These are two quite different things?

 

[GRstudent]

Degrees of freedom are ways in which something can change. So, in 4D a t vector can deviate in the x direction while going around a yz loop, or it can deviate in the z direction while going in a xz loop, or ... there are lots of different ways. Not so in 2D.

This idea seems to make some sense to me.

Do you really want to look at Einstein's equation for 2-dimensional spacetimes, or do you want to look at Einstein's equation's equation for a 4-dimensional spacetime and project the results on a 2-dimensional surface? These are two quite different things?

This is not about dimensions. I was just looking for a concrete example of EFE in use. I mean, if EFE is such a big topic in physics, there must be some real application of this.

The point of me asking for 2d was only because in 2d, Ricci has non-zero component so in this case, it was more interesting for me to understand this.

Besides, FRW metric (which is extremely complicated) I would like to get a standard case, when $G_{\mu \nu}$ has non-zero components.

 

[George Jones]

This is not about dimensions. I was just looking for a concrete example of EFE in use. I mean, if EFE is such a big topic in physics, there must be some real application of this.

The point of me asking for 2d was only because in 2d, Ricci has non-zero component so in this case, it was more interesting for me to understand this.

But the 2-dimesional case is not interesting physically precisely because of the lack of degrees of freedom. For this case physicists sometimes add extra degrees of freedom, like a dilaton field.

Besides, FRW metric (which is extremely complicated) I would like to get a standard case, when $G_{\mu \nu}$ has non-zero components.

I don't know of a simple example that has non-zero T. Other interesting cases that have non-zero T are the Schwarzschild constant density spherical "star" (different from the "standard" Schwarzschild solution), Oppenheimer-Snyder collapse, and the Vaidya solution.

 

[GRstudent]

Schwarzschild constant density spherical "star"

Sounds interesting to me. Where can I find more details about this? Can you write down its components (T or G)?

 

[George Jones]

Sounds interesting to me. Where can I find more details about this? Can you write down its components (T or G)?

Here is the metric, which is treated in many relativity texts (e.g., texts by Schutz, by Hobson, Efstathiou, Lasenby, and by Misner, Thorne, Wheeler),:

If the Earth is modeled as a constant density, non-rotating sphere, then Schwarzschild's interior solution can be used. When $G=c=1$,

$$d\tau^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right),$$

where $R$ is the $r$ coordinate at the surface of the Earth.

If an observer on the Earth's surface uses a telescope to look down a tunnel to a clock at the Earth's centre, he will see his clock running faster than the clock at the Earth's centre.

Consider two dentical clocks, one moving around the Earth once a day on the Earth's surface at the equator ($\theta = \pi/2$) and one at the Earth's centre. Both clocks have constant $r$ values, so $dr=0$ for both clocks, and, after factoring out a $dt^2$, the above equation becomes

$$\left( \frac{d\tau }{dt}\right) ^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}-v^{2},$$

where $v=rd\phi/dt$ is, approximately, the speed of something moving along a circular path. At the centre, $v=r=0$, and, on the surface, $r = R$ and $v = 1.544 \times 10^{-6}$, which is one Earth circumference in one day.

Then, with $G$ and $c$ restored,

$$\frac{d\tau_{centre}}{d\tau_{surf}}=\left( \frac{d\tau_{centre}}{dt}\right) \left( \frac{d\tau_{surface}}{dt}\right)^{-1} =\frac{\frac{3}{2}\sqrt{1-\frac{2GM}{c^{2}R}}-\frac{1}{2}}{\sqrt{1-\frac{2GM}{c^{2}R}-v^{2}}}$$.

Running, the numbers, I get

$$\frac{d\tau_{centre}}{d\tau_{surf}} = 1 - 3.5 \times 10^{-10}.$$

Lots of places errors could have crept in, though.

 

[GRstudent]

^
I don't understand this. Where is Stress-Energy Tensor or Einstein Tensor?

 

[Mark M]

^
I don't understand this. Where is Stress-Energy Tensor or Einstein Tensor?

George Jones showed you the Schwarzschild metric. It's a solution of the Einstein field equations for a perfectly spherical non-rotating body. You don't need either of those things to make calculations with it, since they were used to derive it.

 

[George Jones]

The Einstein tensor can be calculated from the metric (R is constant); I am not going to type this into a post.

This can be inverted. Einstein's equation, together with symmetry and physical reasonableness, give rise to relativistic equations of structure for stars.

In my opinion, there are not internet substitutes for some of the excellent relativity texts.

 

[PeterDonis]

I was just looking for a concrete example of EFE in use. I mean, if EFE is such a big topic in physics, there must be some real application of this.

*Every* spacetime used in GR is a solution of the EFE, so every application of GR is an application of the EFE. If you check out the Wikipedia page on solutions of the EFE that I linked to many posts ago, you will see many examples.

If you're looking for the actual mechanics of how the EFE is solved in specific cases, much of the heavy lifting nowadays is done by computers, particularly in "real applications" where the system being studied is more complicated than the simple ones for which we can write down analytic solutions.

Besides, FRW metric (which is extremely complicated) I would like to get a standard case, when $G_{\mu \nu}$ has non-zero components.

Just out of curiosity, why do you find the FRW metric "extremely complicated"? In some ways it's simpler than other solutions that are being discussed in this thread.

 

[Matterwave]

George gave you the metric. You can calculate the Einstein tensor from it using standard definitions.

EDIT: Whoops got triple sniped.

 

[George Jones]

Have a look at Chapter 25 from Blandford and Thorne's notes,

http://www.pma.caltech.edu/Courses/ph136/yr2011/,

but I agree with Peter, standard cosmology is easier. You might also want to look at other Chapters.

 

[GRstudent]

You can calculate the Einstein tensor from it using standard definitions

This sounds nice! I will try.

Just out of curiosity, why do you find the FRW metric "extremely complicated"?

It's not super hard yet it is harder than ordinary Schwarzschild metric.

********

Basically, Schwarzschild interior is gravity for inside Earth, right?

I am really excited!

 

[PeterDonis]

It's not super hard yet it is harder than ordinary Schwarzschild metric.

I would say that depends on what you're trying to do. As far as the components of the EFE are concerned, the FRW metric is actually simpler. But it's true that the FRW metric is time-dependent, while the Schwarzschild metric is static, which can make the latter seem simpler in some ways.

Basically, Schwarzschild interior is gravity for inside Earth, right?

If you are referring to the metric that George Jones posted, not exactly--it's for an idealized spherically symmetric body with constant density that is not rotating. The Earth is not quite spherically symmetric, it's rotating, and its density is certainly not constant; it increases with depth. As far as I know, nobody has written down an exact metric in closed form for the case of a rotating massive body with non-constant density; cases like that are solved numerically.

 

[GRstudent]

So here is the metric of Schwarzschild interior metric:

$g_{tt}=(\dfrac{3}{2}\sqrt{1-\dfrac{2MG}{Rc^2}}-\dfrac{1}{2}\sqrt{1-\dfrac{2MGr^{2}}{R^{3}c^2}})^{2}$

$g_{rr}=\dfrac{1}{(1-\dfrac{2MGr}{R^3c^2})}$

$g_{\theta \theta}=r^2$

$g_{\phi \phi}=r^2sin^2\theta$

Anyone who has Mathematica can calculate Ricci components of this metric. I am looking forward to see them!

Also, you can check whether I put $G$ and $c$ correctly.

What is the difference between R and r?

 

[Dale]

What is the difference between R and r?

r is the Schwarzschild radial coordinate. R is the Schwarzschild radial coordinate at the surface of the planet.

 

[GRstudent]

^
What about my metric?

 

[ApplePion]

GRstudent, a test of your metric's plausibility is to examine what happens when r=R. If you insert r= R into your formula for the metric components you will see that what you get does *not* correspond to the accepted Schwarzschild Solution for that spherical surface.

 

[GRstudent]

So which one is a correct one? What is the metric (written as a matrix) of Schwarzschild interior solution?

 

[PeterDonis]

So here is the metric of Schwarzschild interior metric:

You made a mistake in g_rr; there should be an $r^{2}$ in it, not an $r$. If you fix that the metric you wrote down will be correct for the interior of an idealized non-rotating planet with perfect spherical symmetry and constant density.

Anyone who has Mathematica can calculate Ricci components of this metric. I am looking forward to see them!

It would be better for you to work through the computation yourself. It's tedious, but straightforward.

Also, you can check whether I put $G$ and $c$ correctly.

Looks OK to me.

GRstudent, a test of your metric's plausibility is to examine what happens when r=R. If you insert r= R into your formula for the metric components you will see that what you get does *not* correspond to the accepted Schwarzschild Solution for that spherical surface.

It will if g_rr is fixed. But you're right, GRStudent should verify that explicitly to confirm that this metric makes sense.