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Einstein Tensor

  1. Oct 30, 2012 #1
    Hello,

    Please excuse me if the following questions are very mundane:

    (1) The Einstein tensor in indices form: Guv=Ruv-1/2guvR

    Does uv means the indices i.e. just as we define row column in a matrix.?

    (2) The Einstein tensor is symmetric. Does that mean in matrix as we mean aij=aji? Does it carry in physical meaning? I mean to say other than mathematical pt.of view is there any substantial physical implication of it?

    (3) Like stress energy tensor is divergence-less. What does it mean?

    Thanks,

    -- Shounak
     
  2. jcsd
  3. Oct 30, 2012 #2
    (1): Yes, an easy way to interpret the ##\mu, \nu## are as row and column specifications for a matrix. Because we often deal with objects that have more than two indices, though, it's handy to be able to think about such tensors outside of their matrix representations. To get an initial grasp on things though, it's sound.

    (2): Yes, this symmetry is precisely like symmetry in matrices. Regarding the last question of this part and (3), usually we talk about the physical meaning of the symmetries of the stress-energy tensor. The symmetry of T can be read as "the flux of the #\mu# component of four-momentum across a surface perpendicular to the ##\nu## vector is the same as if the ##\mu, \nu## were switched." I'm sure this has an analogous meaning for the Einstein tensor, though I don't know it off the top of my head.

    That the stress-energy tensor is divergence-free, and usually this is taken as an analogue of the conservation of momentum and energy. Note that in GR, this is the covariant divergence (i.e. associated with the covariant derivative).
     
  4. Oct 30, 2012 #3
    Thank you Muphrid. I try to verify that whatever I am learning is right or not.

    Well, I have one more question. The properties of different matrices that we have, say, symmetric matrix, identity matrix, diagonal matrix and others, I understand they have mathematical properties. We calculate certain values......etc.etc.

    Physically, I mean to say, say energy, density, or flux whatever there is; if we get tensors (from matrices) which have the properties of matrices mentioned above; does it at all create difference, I mean to say say a matrix is symmetric as you have mentioned in case of Einstein tensor is analogous to the flux component , perpendicular to vector v, is the same as mu,v are swapped.

    Similarly tensors which have certain properties carrying from matrices do they physically signify anything? Do they help in calculating any physical property?

    -- Shounak
     
  5. Oct 31, 2012 #4

    bcrowell

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    A matrix is a 2-index tensor. Tensors can have 0, 1, 2, 3, ... indices.

    The real point of tensors is that they're a way of writing equations in a form in which we can tell that they're valid regardless of the coordinate system.
     
  6. Oct 31, 2012 #5
    Hello Ben,

    Do you mean they remain the same in co-ordinate transformation or they are co-ordinate independent?
    Can you please give me an example with a 2x2 tensor?
     
  7. Oct 31, 2012 #6
    There is a definite transformation law for tensors.They are not co-ordinate independent but some properties of it can be derived without having resort to special coordinate system.Like a scalar product,which does not depend on any specific co-ordinate system.
     
  8. Oct 31, 2012 #7

    haushofer

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    This can be confusing. The tensor itself is absolutely coordinate independent. Tensors can be defined in a coordinate-free way. The components however depend on the choice of coordinates, just as the choice of basis. These choices are such that the expansion of the tensor in the specific basis is invariant under a coordinate transformation. That's how one can understand the concept of "covariance" and "contravariance" in the first place.

    The way to loosely visualize this for e.g. a vector (a rank-1 tensor) in two dimensions is to see the vector being attached to a point, and that under a coordinate transformation the coordinate grid just transforms, leaving the vector untouched.

    If you want to understand differential geometry in a sensible way, then I think it's important not to confuse tensors with their components. It's like saying that the object (a,b,c) is a vector without further specification.
     
    Last edited: Oct 31, 2012
  9. Oct 31, 2012 #8

    haushofer

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    A tensor with two lower indices (never mind the dimension, that's not important) can be expanded in the basis [itex]e^{(\mu)}[/itex] as

    [tex]
    T = T_{\mu\nu}\, e^{(\mu)} \otimes e^{(\nu)}
    [/tex]


    The round brackets () are there to indicate that the indices on the basis vectors label whole vectors, not just their components! Now under a coordinate transformation

    [tex]
    x^{\rho} \rightarrow x^{'\rho}
    [/tex]

    the basis vectors and components transform in the opposite direction such that T itself is left unchanged. I leave it up to you to check this explicitly; the answer can be found in any GR book.
     
  10. Oct 31, 2012 #9
    Yes,that is what is meant by it.The same applies to vectors but whenever one wants to deal with it,he most probably go on with their components.It is the power of tensor calculus that the eqn can be written in a compact form without resorting to any special coordinate system.
     
  11. Oct 31, 2012 #10
    Ok, haushofer. That means if I have a tensor T ij under co-ordinate transformation ij might change but T remain unchanged?

    One more thing, co-ordinate transformation means? From Cartesian to spherical, spherical to hyperbola?

    Thanks,

    -- Shounak
     
  12. Oct 31, 2012 #11
    I am a bit confused here:

    Guv=Ruv-1/2Rguv

    Here, G is the Einstein tensor or gravitational constant?
     
  13. Oct 31, 2012 #12

    Nabeshin

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    Einstein tensor. The gravitational constant is a scalar -- it has no indicies.
     
  14. Oct 31, 2012 #13
    One more thing as Einstein tensor is symmetric:

    Guv=Gvu

    that means can it be written

    G=GT (transpose of A)?
     
  15. Oct 31, 2012 #14
    Hello Muphrid,

    "The symmetry of T can be read as "the flux of the #\mu# component of four-momentum across a surface perpendicular to the ν vector is the same as if the μ,ν were switched."

    If you can explain a little bit further on that.

    -- Shounak
     
  16. Oct 31, 2012 #15

    haushofer

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    Yes. The Einstein tensor can be represented by a symmetric matrix. (By A you mean G, I suppose), just like e.g. the metric.
     
  17. Oct 31, 2012 #16

    haushofer

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    Yes, if you mean by "ij" the components T_{ij}.

    Whatever you want; it depends on the tensors you have at your disposal. If people say that T is a tensor, they often implicitly have a certain group of transformations in mind. To be complete they should mention this group. In GR one talks about general coordinate transformations, gct's. Such a transformation can be described as

    [tex]
    x^{\mu} \ \rightarrow \ \ x^{'\mu}(x^{\nu})
    [/tex]

    where the coordinate maps are general functions (but of course have to be invertible etc). You often hear the statement that the connection [itex]\Gamma^{\rho}_{\mu\nu}[/itex] is not a tensor. People mean by that that the connection is not a tensor under gcts. Under Poincaré transformations (Lorentz transformations and constant spacetime translations) the connection is an honest tensor, which you can check for yourself.

    Another nice example is the Newton potential, which is a tensor (a scalar, to be precise) under Galilei transformations. However, under accelerations

    [tex]
    x^i \ \rightarrow x^{'i} = x^i + \xi^i(t)
    [/tex]

    the Newton potential does not transform as a scalar anymore, but inhomogeneously (don't pin me down on the sign):

    [tex]
    \phi(x) \ \rightarrow \phi'(x') = \phi(x) + \delta_{ij }\ddot{\xi}^i x^j
    [/tex]

    For Galilei transformations [itex]\ddot{\xi}^i = 0 [/itex] and you see that phi transforms as a scalar again. So the proper thing to say is that the Newton potential is a Galilei-scalar.
     
    Last edited: Oct 31, 2012
  18. Nov 1, 2012 #17
    Hello,

    Can anybody please explain me this lowering and raising of indices. In matrix we write A ij to mention row,column. But raising the index to A ij, what does it mean? Can anybody please give me a simple example to demonstrate that?

    -- Shounak
     
  19. Nov 1, 2012 #18
    Raising and lowering indices are mainly done with the help of metric tensor.The main point is that the indices should be balanced on both side.It will be better for you to take a look here for this thing
    http://www.mathpages.com/rr/s5-02/5-02.htm
     
  20. Nov 1, 2012 #19

    haushofer

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    This involves the metric tensor, and is explained in every text on GR, like Wald or Carroll. The metric tensor takes you from vector spaces to dual vector spaces (and vice versa for the inverse). It allows you to define an inner product, and to define scalar traces.

    If I have a tensor [itex]A_{\mu\nu}[/itex] and I want to define the scalar (!) trace, I cannnot write down [itex] \Sigma_{\mu} \, A_{\mu\mu}[/itex], because this is not invariant under coordinate transformations as you can check. However, I can use the metric tensor to define e.g.

    [tex]
    A^{\mu}{}_{\nu} \equiv g^{\mu\lambda} A_{\lambda\nu}
    [/tex]

    and then sum over mu, [itex]A^{\mu}{}_{\mu}[/itex] (which is of course the same as [itex]g^{\mu\lambda} A_{\lambda\mu}[/itex]). This trace is invariant under coordinate transformations, as you should check.
     
  21. Nov 1, 2012 #20
    Ok, haushofer. I am just getting near to it, just a bit confusion please clear me up:

    We write:

    Auv=g.

    Why do we write λ?

    How do we sum over mu?

    For metric tensor can we write gxx or guu
     
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