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- Thread starter meteor
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16. Look at http://www.etsu.edu/physics/plntrm/relat/general.htm if you're not sure what's going in that one). About 3/4 of the way down they get to the field equations.

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Thanks.Then I suppose that this page:

http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Field-Equations.htm [Broken]

an this other:

http://csep10.phys.utk.edu/astr162/lect/cosmology/solutions.html

are wrong. Bad information in the net!!!

http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Field-Equations.htm [Broken]

an this other:

http://csep10.phys.utk.edu/astr162/lect/cosmology/solutions.html

are wrong. Bad information in the net!!!

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instanton

Einstein Tensor = Energy-Momentum tensor

Both tensors are symmetric. In 4 dimension symmetric matrices has 10 independent components. Indeed, among these 10 eqns. only 6 of them has dynamical information. The other 4 is constraints on initial data.

Instanton

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marcus

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the page cited is from the East Tennessee State UniversityOriginally posted by Zefram

16. Look at http://www.etsu.edu/physics/plntrm/relat/general.htm if you're not sure what's going in that one). About 3/4 of the way down they get to the field equations.

http://www.etsu.edu/physics/

and the specific page cited is

http://www.etsu.edu/physics/plntrm/relat/general.htm

the form of the Einstein equation cited on this page is "R=0" (with mu, nu subscripts which I dont want to have to write) and this is not the version that one usually sees

Usually one sees it arranged this way:

G = 8pi T (with mu, nu subscripts)

this is the form shown in the U. Winnipeg page cited by meteor,

where they say 10 equations.

Probably instanton represents the majority view (10 eqs.)

namely what shows up at meteor's winnipeg site

http://scholar.uwinnipeg.ca/courses...d-Equations.htm [Broken]

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Sure?I have seen that the formula isNo. It is 10. Field eqn is

Einstein Tensor = Energy-Momentum tensor

Einstein tensor=k*Energy-Momentum tensor

being k a constant. There is no agreement in what is the value of this constant, 8*pi or 8*pi*G. Anybody knows?

Is the Einstein tensor a variant or a contravariant tensor?

- #8

instanton

R = 0 is a special case of Einstein eqn, which is G_a_b = 8* pi*G* T_a_b. (here G is a Newton's gravitational constant. The factor 8*pi* G is for matcing Newtonian theory of gravity for slow motion - or weak field limit.)

G_a_b = R_a_b - (1/2)*(g_a_b)*R where g_a_b is a metric tensor, R_a_b is a Ricci tensor, and R is a Ricci scalar. If you contract G_a_b you will get -R in 4 dimensional spacetime. R = 0 is true when T = 0, usually for vacuum spacetime.

For meteor,

I've already answer to your first question. Einstein tensor is usually defined as a second rank covariant tensor.

Instanton

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I trust in your word, but believe me that there are certain pages where it appears like a contravariant tensor:I've already answer to your first question. Einstein tensor is usually defined as a second rank covariant tensor.

http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec10.html [Broken]

www.pa.uky.edu/~cvj/as500_lec17/as500_lec17.html[/URL]

or like a mixed tensor:

[PLAIN]http://folk.uio.no/kkarlsen/docu/gr1/node22.html [Broken]

I suppose that the Ricci tensor and the metric tensor are covariant tensors too

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Tom Mattson

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You can always convert from covariant-->contravariant by using the metric tensor.Originally posted by meteor

I trust in your word, but believe me that there are certain pages where it appears like a contravariant tensor:

A

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chroot

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There are, of course, 16 equations in any 4x4 tensor equation. 6 of them disappear when you impose symmetry, and there are thus only 10 unique equations.

meteor, the reason the proportionality constant is either 8 pi G or 8 pi in the Einstein equation is simply because some people choose to work in natural units (G = 1) while others do not.

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another question: how to multiply a tensor with an scalar? Exists any web that explain this?

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instanton

Einstein equation is 10 coupled non-linear second order partial diffrential equation of metric components. So, for given boundary condition and matter distribution you are solving for metric. Of course there are subtleties. To being able to define energy-momentum tensor of matter we need information on background metric. So, usually you fix backround metric and matter distribution, then perturbatively calculate it's solution. Of course there are few examples of exact solutions, but they are usually rare.Originally posted by meteor

another question: how to multiply a tensor with an scalar? Exists any web that explain this?

Multiplyng tensor with scalar is exactly like multiplying vector with scalar, if that is what you are asking.

Instanton

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