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Meaning of tensor invariant, covariant differentiation

  1. Jan 14, 2015 #1
    E.g - considering co variant differentiation,

    The issue with the normal differentiation is it varies with coordinate system change.
    Covariant differentiation fixes this as it is in tensor form and so is invariant under coordinate transformations.


    'If a tensor is zero in one coordinate system, it is zero in all coordinate systems.'This is probably a stupid question, but following this, by 'invariant' do we mean you get the same value/matrix components? What's the significance of the form being unchanged?

    Would you get different matrix components but scaled suitable i.e. according to the coordinate tranistion laws which are such that the computed tensor will have the same physical meaning in all coordinate systems , and a tensor that is zero, will remain zero, and a non-zero tensor will remain non-zero.

    I'm not sure I understand the concepts...Are these thoughts correct?

    Thanks in advance.
     
  2. jcsd
  3. Jan 14, 2015 #2

    Matterwave

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    This is basically it. The components of a tensor will change under coordinate transformations, but they will change in a specific way so as to keep their physical meaning invariant. Physical meaning is obtained from tensors by contracting them with various 4-vectors. Because tensor and 4-vector components change in the correct way, their contraction, and therefore their physical meaning, will not change.

    First, let's look at the tensor transformation law:
    $$T^{a'b'...}_{~~~~~c'd'...}=\frac{\partial x^{a'}}{\partial x^e}\frac{\partial x^{b'}}{\partial x^f}...\frac{\partial x^g}{\partial x^{c'}}\frac{\partial x^h}{\partial x^{d'}}...T^{ef...}_{~~~~~gh...}$$
    This is how components of a tensor will change under coordinate transformations. It is then, not hard to show that an expression like ##E=u^a T_{ab} u^b## does not change under a coordinate transformation. In this case, ##E## is a scalar quantity and is the energy (density) measured by observers with 4-velocity ##u^a## where the stress-energy distribution is given by the stress-energy tensor ##T_{ab}##
     
  4. Jan 15, 2015 #3
    But it doesn't have to be a scalar quantity ? Looking at the Riemann tensor, contract it to the ricci vector and than to the ricci scalar, the ricci scalar having the physical meaning of curvature must be given by the same value regardless of coordinate choice. Can you directly compare the ricci vectors from two coordinate frames in a similar way? thanks.
     
  5. Jan 15, 2015 #4

    Matterwave

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    Comparing the same tensor in two different frames requires the tensor transformation law as I gave in my post #2. The Ricci tensor (not vector) is a rank 2 tensor, and its components transforms as: $$R_{\mu'\nu'}=\frac{\partial x^\rho}{\partial x^{\mu'}}\frac{\partial x^\tau}{\partial x^{\nu'}} R_{\rho\tau}$$
     
  6. Jan 15, 2015 #5
    On second thoughts, you can't really compare the matrix components of the ricci vector as given in 2 different coordinate systems as to get the scalar we contract with the metric, which takes different forms in the coordinate systems and so the components must differ to get the same scalar value?
     
  7. Jan 15, 2015 #6

    Matterwave

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    What? I could not understand this question. Both the metric tensor's and the Ricci scalar's components will change in the appropriate way as to make the Ricci scalar a scalar. Are you confused on explicitly how this happens? This happens not just for the Metric tensor and the Ricci tensor, but for any totally contracted tensor equation. Taking as an example: $$g^{\mu'\nu'}=\frac{\partial x^{\mu'}}{\partial x^\alpha}\frac{\partial x^{\nu'}}{\partial x^\beta} g^{\alpha\beta}$$ and the transformation I wrote down in post #4, we find: $$R_{\mu'\nu'}g^{\mu'\nu'}=\frac{\partial x^\rho}{\partial x^{\mu'}}\frac{\partial x^\tau}{\partial x^{\nu'}}R_{\rho\tau}\frac{\partial x^{\mu'}}{\partial x^\alpha}\frac{\partial x^{\nu'}}{\partial x^\beta}g^{\alpha\beta}$$ Which gives us, with a little bit of rearranging (and use of the inverse function theorem): $$R_{\mu'\nu'}g^{\mu'\nu'}=\delta^\rho_{~\alpha}\delta^\tau_{~\beta}R_{\rho\tau}g^{\alpha\beta}=R_{\alpha\beta}g^{\alpha\beta}$$ Therefore, we have showed that the Ricci scalar is in fact a scalar, and its value does not change based on a coordinate change.
     
  8. Jan 15, 2015 #7
    And does it have to give the same value in different frames? if it's 6, say, in one frame, is it 6 in every frame?
     
  9. Jan 15, 2015 #8

    Matterwave

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    Yes, but notice though that these scalars we have defined so far are actually scalar fields. So if you asked the question "at the point ##P##, what is the value of the scalar field?" The answer must be the same no matter who you asked, and no matter which coordinate system that person used. However, realize that the "point ##P##" will have different coordinate values in different coordinate systems. But this should not be surprising.
     
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