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

[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,

<br /> 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),<br />

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

<br /> \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},<br />

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,

<br /> \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}}}<br />.

Running, the numbers, I get

<br /> \frac{d\tau_{centre}}{d\tau_{surf}} = 1 - 3.5 \times 10^{-10}.<br />

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.

 

[ApplePion]

<<It will if g_rr is fixed>>

The g_tt metric component looks wrong, too. It is a bit difficult to understand what he is saying because he made a typing mistake. Note in the second radical for the g_tt formula he has a closed parenthesis without an open parenthesis. Regardless, it does not appear that you get the required g_tt = 1 - 2M/R, for r=R. (I'm setting G and c equal to 1, for simplicity)

 

[PeterDonis]

The g_tt metric component looks wrong, too. It is a bit difficult to understand what he is saying because he made a typing mistake. Note in the second radical for the g_tt formula he has a closed parenthesis without an open parenthesis.

The open parenthesis is at the beginning of the entire thing, before the first radical. The whole expression adding both radicals together is inside the parentheses; then that entire expression gets squared to give the final g_tt coefficient. It's the same as what George Jones quoted in his post.

Regardless, it does not appear that you get the required g_tt = 1 - 2M/R, for r=R. (I'm setting G and c equal to 1, for simplicity)

It does when read correctly; see above. You have 3/2 times sqrt(1 - 2M/R) minus 1/2 times sqrt(1 - 2M/R), which gives sqrt(1 - 2M/R); then you square that, which gives the correct g_tt at r = R to match up with the exterior vacuum metric.

 

[ApplePion]

OK, thanks Peter. I did not notice the open parenthesis.

 

[GRstudent]

OK, here I modified it (please confirm that it is correct so I can go and calculate Einstein Tensor):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^2}{R^3c^2})}

g_{\theta \theta}=r^2

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

 

[PeterDonis]

OK, here I modified it (please confirm that it is correct so I can go and calculate Einstein Tensor

This looks correct to me.

 

[Mark M]

Yes, it's correct.

EDIT: Oh, Peter beat me to it.

 

[PeterDonis]

This looks correct to me.

Oops, there is one other thing: either there needs to be a minus sign in front of g_tt, or there need to be minus signs in front of g_rr, g_theta_theta, and g_phi_phi. The former sign convention is more common and is probably easier to work with.

 

[GRstudent]

what's \dfrac{\partial g_{rr}}{\partial r}?

 

[GRstudent]

^^

Right.

R_{tt}=-\dfrac{\partial^2_{r} g_{tt}}{2g_{rr}}+\dfrac{\partial_r g_{tt}}{4g_{rr}}(\dfrac{\partial_{r} g_{tt}}{g_{tt}}+\dfrac{\partial_{r}g_{rr}}{g_{rr}}) - \dfrac{\partial_{r}g_{tt}}{r g_{rr}}

Can someone show me how to solve above equation? I have some trouble with differentiating those messy terms.

Can anyone calculate this? I am getting some crazy mess on my Ti-89.

 

[ApplePion]

GRstudent, why are you interested in knowing R_tt explicitly?

 

[GRstudent]

^
All I want that someone puts my metric into Mathematica and calculates Gammas, Riemanns, Ricci, and Einstein Tensor. I have mathematica but I don't know why GR package doesn't work on my PC.

 

[PeterDonis]

what's \dfrac{\partial g_{rr}}{\partial r}?

If you're not comfortable taking the derivative of that g_rr expression with respect to r, I would suggest that you need a calculus review before tackling GR. Andy by "taking the derivative" I mean doing it by hand, not with computer assistance.

Can someone show me how to solve above equation?

How are you coming up with that equation for R_tt?

 

[GRstudent]

^
When I wanted to calculate the derivative of g_rr with respect to r, my Ti-89 gave me some crazy answers.

I need someone to solve this with Mathematica and then, I will analyze the solution. Please! Help HS student!

 

[PeterDonis]

When I wanted to calculate the derivative of g_rr with respect to r, my Ti-89 gave me some crazy answers.

Did you read the part where I said you should be able to calculate that derivative yourself, without computer assistance? If you aren't comfortable doing that with your expression for g_rr, which is reasonably simple as such expressions go, then you are probably not ready to dig into the math of GR. You need to learn the requisite calculus first.

I need someone to solve this with Mathematica and then, I will analyze the solution.

If you are not capable of deriving the solution yourself, I don't think you are capable of "analyzing" it. Certainly not without the math skills I referred to above. The solution is just going to be more math expressions that you will need to be able to understand and manipulate.

If you are just trying to get some kind of basic physical understanding of what the Einstein Field Equation says, then I would not recommend trying to do so by just looking at equations and trying to "analyze" them. You might try reading this discussion by John Baez:

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

 

[GRstudent]

^
Oh, not g_{rr}. I meant g_{tt} with respect to {r}. If my Ti-89 gives the messiest equation that I have ever seen in my life how would you expect me to solve it? I have taken Calculus courses and successfully competed Differentiation chapter.

 

[GRstudent]

Unless I am presented with some contradictory evidence, I think it is correct:

R_{tt}=-\dfrac{\partial^2_{r} g_{tt}}{2g_{rr}}+\dfrac{\partial_r g_{tt}}{4g_{rr}}(\dfrac{\partial_{r} g_{tt}}{g_{tt}}+\dfrac{\partial_{r}g_{rr}}{g_{rr}} ) - \dfrac{\partial_{r}g_{tt}}{r g_{rr}}

 

[PeterDonis]

Oh, not g_{rr}. I meant g_{tt} with respect to {r}.

That one is more complicated, yes, but you should still be able to differentiate it by hand if you've completed courses in calculus including differentiation. Hint: a lot of the symbols represent constants; to make the differentiation easier you might want to roll as many of the constants as possible into single symbols. When you do that you get an expression that looks like this:

g_{tt} = \left( \frac{3}{2} \sqrt{A} - \frac{1}{2} \sqrt{1 - B r^{2}} \right)^{2}

where A and B are constants. That should help in figuring out how to take its derivative.

If my Ti-89 gives the messiest equation that I have ever seen in my life how would you expect me to solve it?

The old-fashioned way, as I've said repeatedly. You should not be relying on your calculator to figure things out that you can't figure out without it. The calculator can help with speed and accuracy for computations that you already understand how to do; but if you don't understand how to do it without the calculator, you won't understand what the calculator is telling you.

Unless I am presented with some contradictory evidence, I think it is correct

How did you obtain this equation for R_tt?

 

[GRstudent]

The only way to compute these messy equations is to simplify them--establish new variables:

A=\dfrac{2MG}{Rc^2}

B=\dfrac{2MG}{R^3c^2}

g_{\mu \nu}=\left [ \begin{matrix} <br /> (1.5\sqrt{1-A}-0.5\sqrt{1-Br^2})^2&amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; \dfrac{1}{-(1-Br^2)}&amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; -r^2 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; -r^2 sin^2 \theta <br /> \end{matrix} \right ]

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2)) \dfrac{\partial g_{rr}}{\partial r}

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2)) \dfrac{2rB}{r^2 B -1}

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-\dfrac{2MG}{R^3c^2} r^2)) \dfrac{2r\dfrac{2MG}{R^3c^2}}{r^2 \dfrac{2MG}{R^3c^2} -1}

Please correct me if I am wrong!

Others:

\Gamma^{\theta}_{\theta r }=\dfrac{1}{r}

\Gamma^{\phi}_{\phi r}=\dfrac{1}{r}

\Gamma^{\phi}_{\phi \theta} = \dfrac{1}{\tan \theta}

 

[PeterDonis]

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2)) \dfrac{\partial g_{rr}}{\partial r}

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2)) \dfrac{2rB}{r^2 B -1}

This doesn't look right. g_{rr} is of the form

\frac{1}{F(r)}

where F(r) is a function of r; so its derivative should be

\frac{-1}{F(r)^{2}} \frac{dF}{dr}

Edit: Corrected typo in the last formula, both F's are capital F's.

 

[GRstudent]

^
Don't understand what you wrote? Can you express your equation in terms of A and B?

 

[PeterDonis]

^
Don't understand what you wrote? Can you express your equation in terms of A and B?

g_{rr} = \frac{1}{F(r)} = \frac{1}{- \left(1 - Br^{2} \right)}

 

[GRstudent]

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2))\dfrac{1}{- \left(1 - Br^{2} \right)}

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-\dfrac{2MG}{R^3c^2}r^2))\dfrac{1}{- \left(1 - \dfrac{2MG}{R^3c^2}r^{2} \right)}

Is this correct? What can I do with it further?

Wait...What?

Gamma is 1/2? With c=G=1 and r=R it should reduce to \dfrac{M}{2MR-R^2}

Something is wrong...

 

[PeterDonis]

Is this correct?

No. The gammas involve the *derivative* of g_rr, not g_rr itself. I was pointing out that you took the derivative of g_rr wrong; you need to go back and correct that before going further.

 

[GRstudent]

^
What do you mean? You seem very confused:

\Gamma^{r}_{rr}=\dfrac{1}{2}g^{rr}(\dfrac{\partial g_{rr}}{\partial r} +\dfrac{\partial g_{rr}}{\partial r}-\dfrac{\partial g_{rr}}{\partial r})

After cancelations:

\Gamma^{r}_{rr}=\dfrac{1}{2}g^{rr}(\dfrac{\partial g_{rr}}{\partial r})

g^{rr} IS INVERSE OF g_{rr}. You said that derivative of g_{rr} what you wrote in terms A and B.

 

[PeterDonis]

I said that you computed

\dfrac{\partial g_{rr}}{\partial r}

wrong. I didn't say anything about g^{rr}.

 

[GRstudent]

^
Can you RE-write the derivative of g_{rr} in terms of A and B correctly?

 

[PeterDonis]

Can you RE-write the derivative of g_{rr} in terms of A and B correctly?

I could, but I'd prefer that you did it yourself, using the hints I gave you in posts #89 and #91.

 

[GRstudent]

Is this one correct? I re-did my earlier calculations:

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-Br^2)) \dfrac{-2rB}{(Br^2 -1)^2}

\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-\dfrac{2MG}{R^3c^2}r^2)) \dfrac{-2r\dfrac{2MG}{R^3c^2}}{(\dfrac{2MG}{R^3c^2}r^2 -1)^2}

 

[PeterDonis]

Is this one correct?

Looks OK to me.

 

[GRstudent]

^
Nope, there is a mistake in factor somewhere. When reduced to ordinary Schwarzschild metric (c=G=1 and r=R) it should be

\dfrac{M}{2MR-R^2}

Instead I got \dfrac{-2M}{2MR-R^2}

Factor (-2) is somewhere unnecessary, where?