# Rank-2 tensor: multiple definitions

1. Jul 7, 2013

### fairy._.queen

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 $\phi$, then this quantity, say A, transforms like
$A\mapsto Ae^{i2\phi}$

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
$T'_{ij}=\frac{\partial x^{'a}}{\partial x^i}\frac{\partial x^{'b}}{\partial x^j}T_{ab}$

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

2. Jul 7, 2013

### Bill_K

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.

3. Jul 7, 2013

### fairy._.queen

Why is the tensor corresponding to the spin-2 object traceless?

4. Jul 7, 2013

### Bill_K

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.

5. Jul 7, 2013

### fairy._.queen

There is a thing I don't get: if $T_{00}\mapsto T_{00}$ 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 $T_{++}$ component of a rank-2 tensor... where am I wrong?

6. Jul 7, 2013

### Bill_K

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.

7. Jul 7, 2013

### fairy._.queen

I'm afraid I didn't really understand how an object of spin $s$ transforms, then.
Do the $m=1$ component of a spin-1 and spin-2 objects both transform as
$C\mapsto Ce^{i\phi}$?

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

Sorry if the question is trivial!

8. Jul 7, 2013

### Bill_K

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.

9. Jul 7, 2013

### fairy._.queen

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 $m=2$ rule, another 2 are called vectors, because they transform with an $m=1$ rule and 1 is called scalar, $m=0$.
From what you say, it seems to me that, strictly speaking, you shouldn't say that the $m=0,1$ parts of the perturbation are scalars or vectors, in that they could be the $m=0,1$ components of a spin-2 object. If you confirm that this terminology is not 100% correct, then I think I got the point.