Mathematical principle for decomposing metric fluctuation

What's the underlying principle to demonposite metrict fluctuation as scalar ,vector and tensor?
Is this decomposition complete?unique?
for scalar mode,
[tex]\begin{equation}
\delta g_{\mu \nu}=a^{2} \left( \begin{array}{cccc}
2\phi & -B,_{i} \\
-B,_{i} & 2(\psi \delta_{ij}-E,_{i,j})
\end{array}\right)
\end{equation}[/tex]
why should 00 term to be a "scalar"?but it is not a lorentz scalar?and why shoud 0i terms to look like a 3 vector? thank you.
 

Chalnoth

Science Advisor
6,192
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What's the underlying principle to demonposite metrict fluctuation as scalar ,vector and tensor?
Is this decomposition complete?unique?
for scalar mode,
[tex]\begin{equation}
\delta g_{\mu \nu}=a^{2} \left( \begin{array}{cccc}
2\phi & -B,_{i} \\
-B,_{i} & 2(\psi \delta_{ij}-E,_{i,j})
\end{array}\right)
\end{equation}[/tex]
why should 00 term to be a "scalar"?but it is not a lorentz scalar?and why shoud 0i terms to look like a 3 vector? thank you.
The '0' index represents time. The 'i' index goes from 1-3 and represents the three spatial directions. Does that help?
 
The '0' index represents time. The 'i' index goes from 1-3 and represents the three spatial directions. Does that help?
thank you! But I know that already ,of course.
00 term of metric fluctuation is a SO(3) scalar,but when a boost happens to change the whole metric tensor,00 term is not a scalar. when you infer a quantity or a set of quantitites as scalar , vector or tensor,you should specify a coordinate thansformation to make the definition valid.what kind of coordinate transformation here is used to define the "scalar","vector",and "tensor" fluctuations?
 

Chalnoth

Science Advisor
6,192
441
thank you! But I know that already ,of course.
00 term of metric fluctuation is a SO(3) scalar,but when a boost happens to change the whole metric tensor,00 term is not a scalar. when you infer a quantity or a set of quantitites as scalar , vector or tensor,you should specify a coordinate thansformation to make the definition valid.what kind of coordinate transformation here is used to define the "scalar","vector",and "tensor" fluctuations?
Well, these are just useful divisions for us to make sense of what's going on here.

In the stress-energy tensor, for instance, the (00) term is the energy density, the (0i) terms are the linear momentum density, and the (ii) terms are pressure and stress.

I don't think this is an actual coordinate transformation, just a grouping of components of the original tensor into components within a particular coordinate system that allow for easier interpretation of the results.
 

cristo

Staff Emeritus
Science Advisor
8,050
72
Quantities in the perturbed metric tensor are decomposed into scalar, vector and tensor quantities depending upon how they transform under a spatial coordinate transformation in the background spacetime. See, for example, Phys. Rev. D 22, 1882 - 1905 (1980). It is useful to decompose the perturbations in this way since, to linear order, the different types decouple.
 
Well, these are just useful divisions for us to make sense of what's going on here.

In the stress-energy tensor, for instance, the (00) term is the energy density, the (0i) terms are the linear momentum density, and the (ii) terms are pressure and stress.

I don't think this is an actual coordinate transformation, just a grouping of components of the original tensor into components within a particular coordinate system that allow for easier interpretation of the results.
O I almost got there! Yes! it is SO(3) transformation! I was reading Brandenberger's lecture notes and run into the problem. I checked another textbook by Mukhanov and find out that it is SO(3), plus spatial translation!
Now the underlying principle is the cosmic principle to constrain metric to what it looks like according to the text book.The homegeneity requires the metric to be invarient under spatial translation, and isotropic requires metric to be ,wait, here is another problem.isotropic requires the metric function to be same after a rotation.but it is not! it just behaves like this: 0i terms behaves like SO(3) vector ,ij term SO(3) tensor.
yes you attacked the problem from another side of einstein equation ,the energy momentum tensor.I want to point out that energy momentum tensor is not necessarily stress energy tensor.Because only a sub set of observers in whose eyes the system is isotropic can call energy momentum tensor as stress energy tensor.In the eye of another observer who is moving with respect to the isotropic observer, the system is not isotropic and sterams of momentun (i,j terms) do not vanish.
 

Chalnoth

Science Advisor
6,192
441
O I almost got there! Yes! it is SO(3) transformation! I was reading Brandenberger's lecture notes and run into the problem. I checked another textbook by Mukhanov and find out that it is SO(3), plus spatial translation!
Now the underlying principle is the cosmic principle to constrain metric to what it looks like according to the text book.The homegeneity requires the metric to be invarient under spatial translation, and isotropic requires metric to be ,wait, here is another problem.isotropic requires the metric function to be same after a rotation.but it is not! it just behaves like this: 0i terms behaves like SO(3) vector ,ij term SO(3) tensor.
yes you attacked the problem from another side of einstein equation ,the energy momentum tensor.I want to point out that energy momentum tensor is not necessarily stress energy tensor.Because only a sub set of observers in whose eyes the system is isotropic can call energy momentum tensor as stress energy tensor.In the eye of another observer who is moving with respect to the isotropic observer, the system is not isotropic and sterams of momentun (i,j terms) do not vanish.
Right, so the cosmological principle only applies for a specific subset of observers. So more accurately stated, it is the statement that there exist some observers for whom the universe appears homogeneous and isotropic. A general observer will certainly not see this.
 

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