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Rank-2 tensor: multiple definitions

  1. Jul 7, 2013 #1
    Hi all!
    In a paper they say that a certain quantity is a rank-2 tensor because it transforms like a spin-2 object under rotations, that is: if the basis vectors undergo a rotation of angle [itex]\phi[/itex], then this quantity, say A, transforms like
    [itex]A\mapsto Ae^{i2\phi}[/itex]

    As far as I knew, a rank-2 tensor is a 2-linear functions or vectors or one forms. In the case it is, for example, completely covariant, it will transform like
    [itex]T'_{ij}=\frac{\partial x^{'a}}{\partial x^i}\frac{\partial x^{'b}}{\partial x^j}T_{ab}[/itex]

    What is the link between the two definitions? Why should a spin-2 object be a tensor?

    Thanks in advance!
  2. jcsd
  3. Jul 7, 2013 #2


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    A spin-2 object corresponds to a symmetric traceless rank-2 tensor.

    In Cartesian coordinates the components are Txx, Txy = Tyx, Tyy, etc, with five independent components. To relate this to spin 2, we go to the spherical basis,

    e+ = (ex + iey)/√2,
    e0 = ez,
    e- = (ex - iey)/√2.

    The five independent components are then T++, T+0 = T0+, T00, T-0 = T0-, T--, and these transform under rotations like the |j=2, m> components of a spin 2 object.
  4. Jul 7, 2013 #3
    Thank you for your reply!
    Why is the tensor corresponding to the spin-2 object traceless?
  5. Jul 7, 2013 #4


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    To get a definite spin we need an object that is irreducible. But in general a symmetric tensor can be reduced. That is, it can be split into parts that behave differently under rotations.

    The trace T = ∑Tii is a scalar, meaning it's invariant under 3-D rotations. So to look at how the rest of Tab behaves, we split the trace off and consider what's left - the traceless part, which is Tab - δabT/3.

    This leaves us with five independent components rather than six. And under 3-D rotations these five components transform only into linear combinations of themselves - the trace is no longer involved. In the spherical basis the five components correspond to the five values of m allowed for spin 2, namely T++ is m = 2, T+0 is m = 1, T00 is m = 0, etc. Under a rotation about the z-axis, T++ ~ T++ e2iφ, and so on.
  6. Jul 7, 2013 #5
    Thank you again for replying!

    There is a thing I don't get: if [itex]T_{00}\mapsto T_{00}[/itex] under rotations, doesn't that mean it is a scalar, rather than a component of a spin-2 object?

    From what you said in your last post, it seems to me that a spin-2 object corresponds to the [itex]T_{++}[/itex] component of a rank-2 tensor... where am I wrong?
  7. Jul 7, 2013 #6


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    This happens only under one particular rotation - the rotation about the z-axis. Under other rotations, the five components get mixed together.
    T++ is only the state in which the spin is spin-up (m = 2). Of course the spin of a spin-2 object can have any other orientation besides.
  8. Jul 7, 2013 #7
    I'm afraid I didn't really understand how an object of spin [itex]s[/itex] transforms, then.
    Do the [itex]m=1[/itex] component of a spin-1 and spin-2 objects both transform as
    [itex]C\mapsto Ce^{i\phi}[/itex]?

    I thought all the components of a spin-2 object would transform as
    [itex]C\mapsto Ce^{i2\phi}[/itex]

    Sorry if the question is trivial!
  9. Jul 7, 2013 #8


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    You're thinking of "spin" as meaning the spin projection, m = 2. A spin-2 object meaning total spin, j = 2 can be in different states |j=2, m> with any spin projection m between ±2. Under a rotation about the z-axis, each of these five states does something different:

    |j=2, m=+2> → |j=2, m=+2> e2iφ
    |j=2, m=+1> → |j=2, m=+1> e
    |j=2, m=0> → |j=2, m=0>
    |j=2, m=-1> → |j=2, m=-1> e-iφ
    |j=2, m=-1> → |j=2, m=-2> e-2iφ

    And under rotations about some other axis, the states will mix together. But forgive me, for spin 2 it gets rather messy, so let me do it instead for spin 1, where the three spin-1 states transform like the three components of a vector.

    Under a rotation through an angle θ about the y-axis, the Cartesian components of a vector or spin-1 object do this:

    Vx → Vx cos θ - Vz sin θ
    Vy → Vy
    Vz → Vz cos θ + Vx sin θ [Eq 1]

    Now replace the Cartesian components with the spherical components (see way up above). I'll just write out one of them, [Eq 1]:

    V0 → V0 cos θ + (V+ + V-)/√2 sin θ

    So there they are mixing together as promised.
  10. Jul 7, 2013 #9
    Thank you for your thorough reply!
    In the paper I am reading, the anisotropies of the CMB which are due to a Bianchi background geometry (not the usual statistical perturbations to the FRW Universe) can be split in 5 according to their behaviour under rotation around some preferred axis which reflects some symmetry. 2 of them are called rank-2 tensors, because they transform with an [itex]m=2[/itex] rule, another 2 are called vectors, because they transform with an [itex]m=1[/itex] rule and 1 is called scalar, [itex]m=0[/itex].
    From what you say, it seems to me that, strictly speaking, you shouldn't say that the [itex]m=0,1[/itex] parts of the perturbation are scalars or vectors, in that they could be the [itex]m=0,1[/itex] components of a spin-2 object. If you confirm that this terminology is not 100% correct, then I think I got the point.
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