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Why use Tensor in GR?

  1. Mar 17, 2014 #1
    I think that is a fundamental question of why we need Tensor when dealing with GR?
    Quoting from the text book (Relativity, Gravitation and Cosmology: A Basic Introduction)

    Tensors are mathematical object having definite transformation properties under coordinate transformations. The simplest examples are scalars and vector components. The principle of relativity says that physics equations should be covariant under coordinate transformation.
    To ensure that this principle is automatically satisfied, all one need s to do is to write physics equation in terms of tensors.

    So, what is, if I say don't have definite transformation properties?
    Any example about if I am not going to choose Tensors as my tool, what trouble will I get?

  2. jcsd
  3. Mar 17, 2014 #2


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    Your theory likely won't be covariant if you don't use tensors - which means that its physical predictions won't be independent of the choice of observer and/or coordinate system.

    You can express a covariant theory without tensors, but as the text says, tensors make it automatically covariant.
  4. Mar 17, 2014 #3


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    tensors are in a sense an organizing tool even if you decided to work things out using different methods you'd be doing the same math as is done by the tensors.

    As a comparison, linear algebra is an organizing tool that makes it easier to solve systems of equations. You can solve for x in one equation and substitute into the next one and eventually work out a solution or you can use the rules of linear algebra to get there faster.
  5. Mar 17, 2014 #4


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    In SR, there are arguably quite a few cases where three-dimensional language is the better tool for the job. In Rindler's SR book, he has a preface where he expresses this point of view and explains that the initial part of the book will be in three-dimensional language, and only later will he introduce four-vectors.

    But for GR...just the thought of trying to do it with non-tensorial tools makes my head hurt. We have rules in GR for how to parallel-transport a tensor. We don't have rules for how to parallel-transport a non-tensorial quantity.
  6. Mar 18, 2014 #5


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    This doesn't only apply to relativity. Trying to do classical continuum mechanics of solids or fluids without tensors isn't fun, either, for the same reason: the behavior of the physical system is independent of the mathematical coordinate system you use to model it.

    Try working with an arbitrary anisotropic material whose orientation varies at different points in space, without a tensor equation like ##\sigma_{ij} = C_{ijkl}\epsilon_{kl}## (plus some symmetry conditions on the three tensors) to connect stresses and strains. If you prefer to write down 21 separate equations to describe how to transform the 21 independent coefficients in the 81 terms in ##C## from one coordinate system to another, go right ahead....

    Of course you can just about survive without tensors for isotropic materials, by writing the independent terms in ##\sigma## and ##\epsilon## as vectors, and ##C## as a 6x6 matrix. But even in that simple case the "rules" for transforming ##\sigma## and ##\epsilon## between different coordinate systems are a mess, and engineers give themselves headaches even in 2D, trying to remember how to do it geometrically with Mohr's circle, etc.
    Last edited: Mar 18, 2014
  7. Mar 18, 2014 #6
    It seems to me this question is analogous to why use vectors in Newtonian mechanics or electromagnetism? You may eventually need to pick a coordinate system to calculate something, but up to that point being tied to a coordinate system just obscures the physics.
  8. Mar 18, 2014 #7


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  9. Mar 18, 2014 #8
    I remember reading that up til 1911, Einstein was still using scalars for the gravitational field, but scalars proved insufficient for the model, so he moved to tensors.

    Correct me if I am wrong.
  10. Mar 18, 2014 #9


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    Consider the math entity defined as V=(1,0) in every 2-dimensional vector space. This is a perfectly valid mathematical entity. It can not be a physical entity because (1,0) in inches and (1,0) in miles can not be the same physical thing. In other words, its definition does not allow it to transform correctly under coordinate transformations. Tensors can represent physical entities that exist, whether there is a human to define coordinate systems and units or not. When you know that something physically exists like force and acceleration, you want a way to represent it so that the equations you right down, like F = mA, will make sense without regard to choice of coordinate system. That is why Tensors have to have "definite transformation properties under coordinate transformations." They are mathematical constructs that can represent physical entities.
    Last edited: Mar 18, 2014
  11. Mar 18, 2014 #10
    I don’t disagree with this (side note, MTW has a nice exercise showing why relativistic scalar (precession of Mercury’s perihelion wrong) and vector theories don’t work as theories of gravity, these could still be considered tensor theories of rank-0 and rank-1 tensors). I just took the question to mean something different, I took it to mean something like why use tensor methods.
  12. Mar 21, 2014 #11
    Thx a lot. I understood something in here.
    And thanks all of you.

    I will keep on trying to learn more from the text books and go back here to check my understanding.

    All of the text books using Tensor from the beginning and tell you that Tensor is a wonderful tool in general relativity. But none of them, even though my professor cannot not show me the difference about using or not using.

    Like Remainder Theorem

    f(x)=x^101+x^91 what is the remainder when divided by (x-1)
    In here, it is convenience that just substitute x=1 to get the remainder.
    If I don't to that, I will get a horrible things to be divided.
    Hence, you know that Remainder Theorem is powerful through this example.

    I hope I can get an example like that.

    Once again, thx all of you.
  13. Mar 21, 2014 #12


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    In a same manner someone can ask about vectors... Vectors have some specific transformation properties and that's why we use them to describe some physical quantities. If you'd try to describe these quantities by scalars instead of vectors, you'd eventually get really weird results (if you get any result at all). For example, a vector under rotations of the axis will have a transformation law like:
    [itex] \vec{w} \rightarrow \vec{w'}= R \vec{w} [/itex]
    with [itex]R[/itex] being a rotation matrix. In general you have that [itex]|\vec{w}|^{2}=|\vec{w'}|^{2}[/itex]
    Things that in nature transform like this, are recognized as vectors and we use them as such.

    In the same manner, in Riemannian geometry, you have to introduce tensors because they follow some certain transformation laws, which keep some quantities (eg the 1st fundamental form, or else put [itex]ds^{2}[/itex]) invariant under these transformations. You'll eventually have to use tensors.
    You don't in general use tensors in GR...for example the Christofel Symbols (or connections) are not tensors, but they are used in several GR formalism (mainly in creating tensors, like the Riemann tensor). One really interesting fact for tensors, is that if they are 0 in one reference frame, they'll be zero in all others (due to transformation laws). So in fact you can always choose a ref.frame which will be much easier to compute the quantities involved.
  14. Mar 21, 2014 #13


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    This isn't as fundamental a question as you might think.

    I can easily just ask you: "why use vector calculus to formulate Maxwell's equations? Why not just write them all out explicitly in terms of partials like Maxwell originally did? Hell why not just write out Einstein's equation explicitly?" This is all simply a matter of calculational fluidity and, perhaps to a lesser extent, aesthetics.

    Do you want to go the pedestrian, masochistic route and prove that ##\xi^2 \xi^{\nu}\nabla_{\nu}\psi^{\mu} = (\xi^{\delta}\psi_{\delta})\xi^{\nu}\nabla_{\nu}\xi^{\mu} \Leftrightarrow \xi_{[\delta}\nabla_{\mu}\xi_{\nu]} = 0## for time-like and axial Killing fields ##\xi^{\mu},\psi^{\mu}## by specializing to some coordinate system and writing out a mess of terms involving coordinate dependent quantities (Christoffel symbols, partials etc.) or do you want to take the more elegant, efficient route and just prove the above using the abstract index formalism?

    The latter is always faithful whereas the former is only pragmatic when synchronous or locally inertial coordinate systems can be utilized and even then you're still making use of the relationship between general covariance and tensor fields, and more generally spinor fields.
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