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Questions on tensors in GR and the Newtonian limit

  1. Aug 27, 2015 #1

    dyn

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    Hi. I am self-studying GR and have many questions. Here are a few. If anyone can help me with any of them I would be grateful.
    1 - What is the difference between Tu v and Tvu ?

    2 - I have read that the order of indices matters in tensors but when transforming tensors from one coordinate system to another it involves products of partial derivatives but the order of these doesn't matter so why does the position of indices matter ?

    3 - When looking at the derivation of the Newtonian limit using gab = nab+hab could anyone explain how to get from
    Γa 00 = -1/2 gad ∂g00/ ∂xd to the following equation
    Γa 00 = -1/2 nad ∂h00 / ∂xd
    I don't understand what has happened to the had term

    4 - I have been looking at the model solution to a question involving the Schwarzchild solution and timelike geodesics for radial motion only which states that the coordinate r does not correspond to physical distance from the gravitational source. What does it correspond to then ?

    5 - In SR almost everything seems to boil down to comparisons between 2 inertial frames ie. the unprimed and primed frames. In the GR I have done so far I see no mention of any other frames. There are no primed frames anywhere. Is basic GR only concerned with one frame ?
     
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  3. Aug 27, 2015 #2

    Geofleur

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    I will try to answer question 1. When there is a metric available (and usually there is), you can use it to "raise" and "lower" indices of a vector, or more generally, a tensor. Consider a vector ## v = v^i e_i ##. I have used the summation convention here, where an index is summed over if it appears exactly once in an upper position and exactly once in a lower position. The ## e_i ## may not be orthogonal to one another in general, but we can still define a "cobasis" ## e^j ## so that ## e^j \cdot e_i = \delta^j_i ##. Then, the same vector can be written in terms of the cobasis as ## v = v_i e^i ##. Using the metric ## g ##, one can show that ## v_i = g_{ij} v^j ##. There is a single object, the vector ## v ##, and it can be represented by either ## v^i ## (when using the original basis ## e_i ##) or by ## v_i ## (when using the cobasis ## e^i ##). A tensor is the same way! There is one object, ## T ##, and we can write it as ## T_{ij} e^i \otimes e^j = T^i{_ j} e_i \otimes e^j = T^{ij} e_i \otimes e_j = T{_i}^j e^i \otimes e_j ##. We need tensor products of basis and cobasis vectors now because the tensor is rank two (as opposed to the vector, which has rank one). All of these different ways of writing the tensor components refer to a single object, ## T ##, but expressed in terms of different bases.
     
  4. Aug 27, 2015 #3

    dyn

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    For a rank 2 tensor ie. a matrix how does Tuv differ from Tvu in terms of rows and columns of the matrix ?
     
  5. Aug 27, 2015 #4

    PeterDonis

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    Do you understand the difference between upper and lower indexes? If you do, the difference between the two tensors you've written should be obvious, since the index order is reversed.

    The order of indexes matters because each index corresponds to a "slot" in the tensor that can be contracted with either a vector or a covector (depending on whether the index is lower or upper). Reordering the indexes means reordering the slots, which means reordering the set of vectors and covectors that the tensor is contracted with. That changes the physical meaning of the contraction, and since all actual observables are expressed as contractions with no free indexes (i.e., scalars), changing the order of indexes means changing which actual observable you are talking about.

    Note that none of this has anything to do with how the tensor transforms between coordinate systems.

    It got dropped because it's second order (because it has a factor of ##h^{ad}## and a factor of ##h_{00}##), and the Newtonian limit only includes first order terms.
     
    Last edited: Aug 28, 2015
  6. Aug 27, 2015 #5

    PeterDonis

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    The second tensor would be the transpose of the first if both were viewed as matrices. But a matrix representation doesn't really capture the distinction between upper and lower indexes.
     
  7. Aug 27, 2015 #6

    dyn

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    In terms of a matrix I still don't understand the difference between Tab and Tba and Tab.
    as regards the Newtonian limit the had term is multiplied by the derivative of h00. Is this classed as a 2nd order term. Just because had is small doesn't necessarily mean that its derivative is small ?
     
  8. Aug 27, 2015 #7

    PeterDonis

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    There isn't really; that's why I said the matrix representation doesn't really capture the difference between upper and lower indexes. To see the difference, it helps to look at the simplest example: a vector ##v^a## as compared to a covector ##v_a##.

    A vector ##v^a## is a set of numbers that transform in a particular way when you change coordinates; if we change from coordinates ##x^u## to new coordinates ##x'^u##, the vector ##v^a## transforms like this:

    $$
    v'^b = \frac{\partial x'^b}{\partial x^a} v^a
    $$

    A covector ##v_a## (also called a "1-form") is a set of linear maps from vectors to numbers. The transformation law for covectors is this:

    $$
    v'_b = \frac{\partial x^a}{\partial x'^b} v_a
    $$

    Notice that the coefficients here are the inverses of the coefficients in the vector transformation law above.

    The following Wikipedia page has a decent discussion:

    https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors

    Yes, it does; if the derivative of ##h^{ad}## isn't small, then ##h^{ad}## itself, even if it starts out small, won't stay small for long.

    In any case, that's not the comparison being made. The comparison being made is between the product of ##n^{ad}## and the derivative of ##h_{00}##, and the product of ##h^{ad}## and the derivative of ##h_{00}##. The latter is much smaller than the former.
     
  9. Aug 27, 2015 #8

    Orodruin

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    These are all components of different tensors (if you have a metric they are related). In the first one, the first index has been raised and in the second the second index has been raised. In matrix notation it corresponds to multiplying with the metric from the left and right, respectively, although I would suggest not always trying to use matrix representation because it only works for rank 2 tensors and gives you the impression that tensors and matrices are one and the same..

    I would also make sure that I was fluent in how tensors are used in SR before going on to GR.

    Everything involving h is considered small and may be expanded in, h itself as well as its derivatives.
     
  10. Aug 28, 2015 #9

    Orodruin

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    Oh yes, as it has not been pointed out: If T is symmetric, i.e., if ##T_{ab} = T_{ba}##, then ##T^a_{\phantom a b} = T_b^{\phantom b a}##.
     
  11. Aug 28, 2015 #10

    bcrowell

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    It can be interpreted as the circumference divided by 2 pi. It wouldn't make sense to define it as a physical distance from the center because there is no measurement process that could measure such a distance.

    Frames are not fundamental in relativity and never necessary. They are sometimes a convenience in SR. In GR global frames don't exist. Einstein thought that frames needed to play a central role in relativity. This was one of the things that history proved Einstein wrong about.
     
  12. Aug 28, 2015 #11

    PeterDonis

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    What resources are you using?

    Distinguishing different frames by putting a prime on quantities in one of them is only one convention. In GR it's more usual to just use different letters to distinguish coordinates in different charts.
     
  13. Aug 28, 2015 #12

    dyn

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    Thanks for all your replies. For resources I am working through lecture notes from universities and using books such as Schutz , Hobson and Carroll.
    If Tuv is the transpose of Tvu what would Tuv be then ?
    I found the following in some notes - "the order of free indices is not important so Sab = Taccb + Wba
    Is this correct ? It seems wrong to me as order of indices is supposed to matter ?
     
  14. Aug 28, 2015 #13

    PeterDonis

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    As I have said several times now, the matrix interpretation of tensors does not really capture the difference between upper and lower indexes. If you want to understand the difference between ##T^u{}_v## and ##T_{uv}##, trying to interpret them as matrices will not help you.

    You need to understand the difference between vectors and covectors. Please read post #7 and the Wikipedia page I linked to there. (I don't have Schutz or Hobson, but I have read Carroll's online lecture notes; I don't think he spends a lot of time on the difference between vectors and covectors. Misner, Thorne, and Wheeler go into it in detail.)

    Reference please?

    The expression as written is weird, but I don't think it's wrong as it stands, because it's a relationship between different tensors. What would be wrong would be to say that, for example, ##S^a{}_b## and ##S_b{}^a## were the same tensor in the general case. This will only be true in the special case of ##S## being a symmetric tensor.
     
  15. Aug 29, 2015 #14

    stevendaryl

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    This is a little off-track from the original post, but I'm a little confused about your explanation. The way you are describing it, [itex]e^j[/itex] is a vector, just like [itex]e_j[/itex]. But isn't it a covector? I understand that when there is a metric, vectors and covectors can be interconverted, but they aren't the same thing.
     
  16. Aug 29, 2015 #15

    Geofleur

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    The ## e^i ## are in fact vectors. They are like the reciprocal lattice vectors in crystal diffraction. When there is a metric present, there is a one-to-one correspondence between one-forms and vectors and one can identify them. We look for the vector w such that ## dx^i(v) = \langle w, v \rangle ## for any vector v. w turns out to be exactly ## e^i ##. Instead of writing ## dx^i(v) = v^i ## we can then write ## e^i \cdot v = v^i ##. We can even write things like ## e^j = g^{ij} e_i##, even though the e's are vectors and not components.

    I was trying to avoid the use of one-forms because I thought that might be more confusing. This perspective of opting for reciprocal basis vectors instead of one-forms I got from Don Koks' book, Explorations in Mathematical Physics.
     
    Last edited: Aug 29, 2015
  17. Aug 29, 2015 #16

    stevendaryl

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    Well then I have been misreading equations involving [itex]e^i[/itex]. I thought that it was supposed to be a basis of one-forms. I guess as long as there is a metric, that confusion doesn't matter much.
     
  18. Aug 29, 2015 #17

    Geofleur

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    Right. When there is not a metric, vectors and one-forms are certainly different kinds of objects.
     
  19. Aug 29, 2015 #18

    Geofleur

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    I should probably point out that, even when there is a metric present, sometimes there are good reasons to use forms ##dx^i## instead of the ##e^i##. Just because one can identify them doesn't mean one always should. When integrating, forms are much more natural. In curvilinear coordinates, for example, it's nicer to deal with the exterior derivative of a one-form than it is to deal with the curl of a vector.
     
    Last edited: Aug 29, 2015
  20. Aug 29, 2015 #19

    bcrowell

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    Either I'm misunderstanding you or you're misunderstanding Koks, or I'm misunderstanding Koks. Amazon let me look at what seems to be the relevant portion of the book (p. 284, first page of section 8.4). Koks refers to the ##e_\alpha## as the "basis vectors," and the ##e^\alpha## as the "cobasis vectors," which makes it seem pretty clear to me that he is following the usual custom of using upper-index notation for components of vectors and the basis for the space of covectors, and lower indices for components of covectors and the basis for the space of vectors.
     
  21. Aug 29, 2015 #20

    Geofleur

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    He calls them cobasis vectors, but he is very clear that he is talking about vectors of the "arrow" type. I've read the whole book many times, and I'm pretty certain. He even devotes several paragraphs to how one-forms are not needed. In the beginning of the book he introduces the reciprocal lattice vectors, and then he says later on that the cobasis vectors are the same thing.
     
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