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Einstein Field Equations

  1. May 17, 2003 #1
    Looking for Einstein Field Equations, in certain places put that they are 10 and in other places put that they are 16. Which is the correct number?
  2. jcsd
  3. May 17, 2003 #2
    Last edited by a moderator: Apr 20, 2017
  4. May 17, 2003 #3
    Thanks.Then I suppose that this page:
    http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Field-Equations.htm [Broken]
    an this other:
    are wrong. Bad information in the net!!!
    Last edited by a moderator: May 1, 2017
  5. May 17, 2003 #4
    BTW, I would like to give a look to them, but in all the webs appear in compact form. Does exist any web where I can find the 16 EFE?
  6. May 19, 2003 #5
    No. It is 10. Field eqn is

    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.

  7. May 19, 2003 #6


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    the page cited is from the East Tennessee State University

    and the specific page cited is

    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]
    Last edited by a moderator: May 1, 2017
  8. May 19, 2003 #7
    Sure?I have seen that the formula is
    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?
  9. May 19, 2003 #8
    First, for Marcus.

    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.

  10. May 19, 2003 #9
    I trust in your word, but believe me that there are certain pages where it appears like a contravariant tensor:
    http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec10.html [Broken]

    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
    Last edited by a moderator: May 1, 2017
  11. May 20, 2003 #10

    Tom Mattson

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

  12. May 20, 2003 #11


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    The answer is 10.

    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.

    - Warren
  13. May 22, 2003 #12
    In the EFE, in the metric tensor, you have to put the tensor of the metric that you are using? For example if you are using the Minkowski metric you have to put the Minkowski metric tensor, or if you are using the euclidean metric, you have to put the euclidean metric tensor?
    another question: how to multiply a tensor with an scalar? Exists any web that explain this?
  14. May 24, 2003 #13
    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.

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

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