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SVT decomposition

  1. Apr 6, 2014 #1
    Hi there, Im reading about inflation, perturbations and so on, and every book I read take the SVT decomposition as granted. Something like "every perturbation can be decomposed in Scalar, Vector and Tensor perturbations". I have 2 questions:

    1) What is the definition of a Scalar, Vector and Tensor perturbation? is it just a perturbation that can be written with just 1 function (Scalar) 4 functions (Vector) and 16 functions (Tensor)? or it has to behave in some why when changing coordinates? Are the numbers I wrote right or they are 1, 3 and 9 (just the space components)?

    2) Why every perturbation can be wrote as a sum of Scalar, Vector and Tensor perturbations?

  2. jcsd
  3. Apr 6, 2014 #2
    scalar, vector and tensor are all mathematical as well as physics terms.

    -scalar is a change in magnitude only, although in physics its, a quantity that is independent of specific classes of coordinate systems
    -vector is magnitude and direction, however their are numerous forms of vectors some include coordinate position
    -tensors are geometric objects that describe linear relations between vectors, scalars, and other tensors, "In differential geometry an intrinsic geometric statement may be described by a tensor field on a manifold, and then doesn't need to make reference to coordinates at all. The same is true in general relativity, of tensor fields describing a physical property. The component-free approach is also used heavily in abstract algebra and homological algebra, where tensors arise naturally." cut and paste from last post wiki page

  4. Apr 7, 2014 #3


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    Wikipedia has an article that goes over the details of how this works:

    Here's my description of the above article.

    Perturbations in cosmology are done by taking an "average" metric (generally the FLRW metric) and adding a general symmetric metric tensor to it. But a general tensor is hard to work with: it's got 10 independent components. You can get rid of four of these by choosing the right coordinates, but that still leaves 6 components to deal with, which is difficult to understand.

    The trick, then, is writing these in a way that makes sense. What was discovered back in the 40's was that one particular way of decomposing the tensor results in three separate components with different physical behavior: two scalar components, a vector component, and a tensor component. The trick is that each component has mathematical properties such that it disappears when you calculate some quantities, splitting the three components nicely in different situations.

    Does that help?
  5. Apr 20, 2014 #4
    I have been reading your answers, wikipedia and so on. However, Im not sure if I understood it well. Let me ask you a simple question (probably it has a simple answer and it allows me to go on). In several places (ie the Tasi lectures) it says "lets write the more general perturbation of the metric" and then he writes a metric wher the 4 terms of the diagonal depend on only two scalar functions (psi for g(0,0) and only fi for g(1,1), g(2,2), g(3,3)). If it is "the more general perturbatin" shouldnt the diagonal depend on 4 scalar (or a vector with 4 elements)? Another way to make the question is, if the metric is defined by 10 functions, shouldnt the "morge general perturbation" be defined by 10 functions?

    Thanks for your answers and thanks in advance for your help!

    Ps: Let me guess the answer and if it is right you just have to say yes and that´s all. We can write g(1,1), g(2,2), g(3,3) with only fi because adding other two functions only changes the scales in the metric -it is the same physical system but measured with other rules-. Am I right? If this is the case, we have only 8 "important" components? How many degrees of freedom are in the metric? 10?, 8? (I think Ive read somewhere that the real degrees of freedom is 6)

    Thanks again
    Last edited: Apr 21, 2014
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