# Showing isotropy of tensor

1. Jul 26, 2014

### erogard

Hi, I am trying to show explicitly the isotropy of the stress energy tensor for a scalar field Phi.

By varying the corresponding action with respect to a metric g, I obtain:

$$T_{\mu \nu} = \frac{1}{2} g_{\mu \nu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\nu \Phi$$

Assuming further that Phi has not spatial dependence (only function of time) and using the Friedmann Robertson Walker metric, which is diagonal, I end up with a diagonal energy tensor:

$$T_{\mu \mu} = \frac{1}{2} g_{\mu \mu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\mu \Phi$$

where
$$g_{\mu \mu} = Diag \left( 1 , -R(t)^2, -s^2 R(t)^2, -\sin^2(\theta) s^2 R(t)^2 \right)$$

R(t) being the scale factor.

I then use
$$T_{\mu \nu}' = R_{\mu i} R_{\nu j} T_ij$$
where R is a 4x4 rotation matrix, with the (1,1) entry being equal to one, and the remaining 3x3 matrix being the usual rotation matrix (here about the z axis).

However, I am unable to recover the original tensor upon performing the above product. For example, I get
$$T_{22}' = \cos^2(\theta')T_{22} + \sin^2(\theta')T_{33}$$

where I use theta prime (the angle of rotation) to differentiate between theta in the FRW metric, and since T_22 is not equal to T_33 (unless sin theta is one), I don't quite get the same result. Haven't worked out the other diagonal entries yet.

I suspect that I am using a wrong form for the RW metric (maybe wrong coordinate?). Any help would be appreciated. Thanks.

Last edited: Jul 26, 2014
2. Jul 26, 2014

### ChrisVer

Try once to use cartesian coordinates, since you are doing a rotation around the z axis.
Also, are you sure about the form of your EM tensor? First of all, your last term doesn't vanish, it survives only for $\mu,\nu=0$ simultaneously.
The action is:

$S= \int d^{4}x \sqrt{|g|} L_{scalar}(\Phi, \partial \Phi)$
which will lead to:
http://catarina.udlap.mx/u_dl_a/tales/documentos/lfa/juarez_a_ba/capitulo2.pdf
(2.4)

Last edited: Jul 26, 2014
3. Jul 26, 2014

### George Jones

Staff Emeritus
But space in a FLRW spacetime is not necessarily flat, so Cartesian coordinates are not always possible for space.

erogard, ChrisVer has a good point, care must be used for curvilinear coordinates. Also, since space isn't flat, things are a little tricky. Yon need to translate along the integral curves for the Killing vectors that give spatial isotropy.

I think some of this is given in the book " An Introduction to General Relativity and Cosmology" by Plebanski and Krasinski, but I don't have it home with me.

4. Jul 26, 2014

### erogard

Thanks for the replies.

ChrisVer: You are right regarding the last term, corrected that.
If you look closely at (2.4), ignoring the addition term V[phi], this is really the same tensor up to an overall minus sign, I believe.

George Jones: so, just to make sure: is the rotation matrix I'm using correct? also, could you elaborate on what you mean by translating along the integral curves? since T contains g in it, isn't it taken care of so long as you use the FLRW metric for g? a bit confused here.

question to both of you I guess: by using cartesian coordinate you mean using the appropriate form of the above metric, correct? with terms like dx dy dz etc. instead of angular terms?

5. Jul 27, 2014

### ChrisVer

I don't know if there are difficulties in using cartesian coordinates, but I am pretty sure our prof had used them quite oftenly in class for the FRW metric.
I also don't understand why just a change of coordinates is not applicable in a curved spacetime. Going from spherical coords to cartesian is always possible...I guess...

As for the question, yes you go from spherical coords (rθφ) to cartesian coords (xyz) and in that case your metric becomes (for $k =0$:
$g_{00}=1$
$g_{ij}= -R^{2}(t) \delta_{ij}$
$g_{i0}=g_{0i}=0$

http://theory.physics.helsinki.fi/~cosmology/cosmo2013_03.pdf
(3.2)

But even if you used the whole expression ($k \ne 0$) then you'd get a similar result - all the $g_{ii}$ components are the same. That's also to be expected (for the metric to have this form) because the EM tensor has the form:
$T= diag( \rho, ~~ p, ~~p,~~p)$
So it has the same entries in its spatial components.

Last edited: Jul 27, 2014
6. Jul 27, 2014

### erogard

I think that works, since now T_22 = T_33 = T_44 so that
$$T_{22}' = \cos^2(\theta')T_{22} + \sin^2(\theta')T_{33} = \cos^2(\theta')T_{22} + \sin^2(\theta')T_{22} = T_{22}$$

and similarly for T'_33 and T'_44.

Didn't check the off diagonal entries but they'd have to be zero. Also since the first entry of T and R is 1, we're essentially multiplying the remaining 3x3 block matrices together, and since the T block is just a multiple of the 1 matrix (since T_22 = T_33 = T_44), it has to yield T back when we perform T' = R.Rtranspose.(constant*1 matrix) = T.

Thanks for the help.

7. Jul 27, 2014

### ChrisVer

the off diagonal entries are zero, and you have shown that in your first post, the 2nd equation you've written.

8. Jul 28, 2014

### George Jones

Staff Emeritus
Yes. For example Dodelson largely deals with the $k = 0$ case, and consequently can use Cartesian coordinates. Our universe is close to being spatially, so $k=0$ gives physically meaningful results. This simplifies the analysis. To check that oure universe is close to flat, it is important to compare non-flat universes to empirical data, so non-zero $k$ also are important

Yes, but they aren't Cartesian coordinates, and they aren't the coordinates used in the original post in this thread..

Yes, but for $k = 1$ and $k = -1$, these aren't Cartesian coordinates, they are called isotropic coordinates. This the third of the three "popular" coordinate systems used for FLRW spacetimes, but many texts don't mention them. With respect to isotropic coordinates, the metric is

$$g = dt^2 - R\left(t\right)^2 \frac{dx^2 + dy^2 +dz^2}{\left(1 +\frac{1}{4} k \left(x^2 + y^2 + z^2 \right) \right)^2 .}$$

When $k = 0$, these are Cartesian coordinates, but when $k$ is non-zero, they are not Cartesian coordinates.

Consider a one-parameter group of rotations, the rotations about the "z-axis", say. Letting all of these rotations operate on a point (event) $p$ in spacetime traces out a curve (line) in spacetime. Take another point $q$ in spacetime that is not on this curve and do the same. This traces out a second curve that does not intersect the first curve. The set of all such curves is called a flow (like water streamlines) for the rotations about the z-axis.

The rotations about the z-axis are generated by a 4-vector field $\mathbf{k}$ that, with respect to isotropic coordinates, has components $k^\mu = \left(0, -y, x, 0 \right)$. Similar to the situation in quantum mechanics, the exponential of $\theta \mathbf{k}$ gives a rotation of $\theta$ about the z-axis. At every point in spacetime, the vector field $\mathbf{k}$ is tangent to the flow (set of integral curves).

As a useful analog for visualization, think of electric filed lines for an electric field in space. Electric filed lines never cross, and the electric field vectors are tangent to the the field lines, i.e., the field lines are the integral curves for the electric field vector field.

Rotations about the z-axis is a symmetry for a tensor field $\mathbf{T}$ if, in an appropriate sense, $\mathbf{T}$ is constant along every integral curve. This suggests that some derivative of $\mathbf{T}$ is zero. This derivative is not a partial or covariant derivative, it is the Lie derivative in the direction of $\mathbf{k}$, the derivative with respect to flows.

The Lie derivative obeys a product rule, so, for your $\mathbf{T}$, this amounts to the vanishing of the Lie derivative of the metric tensor. So, yes, taken care of by g. If the metric tensor is constant along generated integral curves, then the generator is called a Killing vector and the group is called an isometry group. The components of the Lie derivative of the metric are given by (derived in many GR texts)

$$\left( \mathcal{L}_k g \right)_{\alpha \beta} = k^\mu \frac{\partial g_{\alpha \beta}}{\partial x ^\mu} + \frac{\partial k^\mu}{\partial x^\alpha} g_{\mu \beta} + \frac{\partial k^\mu}{\partial x^\beta} g_{\alpha \mu}$$

For example, using the components of k and g given above, a short calculation gives $\left( \mathcal{L}_k g \right)_{1 1} = 0$.

The whole apparatus that I have introduced above is perhaps over-kill for this example, but it generalizes in a straightforward fashion to arbitrary symmetries.

Last edited: Jul 28, 2014
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