# Mathematical principle for decomposing metric fluctuation

• aristurtle
In summary: 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.
aristurtle
What's the underlying principle to demonposite metrict fluctuation as scalar ,vector and tensor?
Is this decomposition complete?unique?
for scalar mode,
$$\delta g_{\mu \nu}=a^{2} \left( \begin{array}{cccc} 2\phi & -B,_{i} \\ -B,_{i} & 2(\psi \delta_{ij}-E,_{i,j}) \end{array}\right)$$
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.

aristurtle said:
What's the underlying principle to demonposite metrict fluctuation as scalar ,vector and tensor?
Is this decomposition complete?unique?
for scalar mode,
$$\delta g_{\mu \nu}=a^{2} \left( \begin{array}{cccc} 2\phi & -B,_{i} \\ -B,_{i} & 2(\psi \delta_{ij}-E,_{i,j}) \end{array}\right)$$
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?

Chalnoth said:
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?

aristurtle said:
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.

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.

Chalnoth said:
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 textbook.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.

aristurtle said:
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 textbook.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.

## 1. What is the mathematical principle for decomposing metric fluctuation?

The mathematical principle for decomposing metric fluctuation is known as the Singular Value Decomposition (SVD). It is a method for breaking down a complex matrix into simpler, more manageable parts that can be analyzed and understood more easily.

## 2. How does the SVD work?

The SVD works by finding the eigenvectors and eigenvalues of a matrix. These are used to create a diagonal matrix, which represents the "simpler" parts of the original matrix. By using this diagonal matrix, the original matrix can be decomposed into two simpler matrices that can be analyzed separately.

## 3. What are some applications of using the SVD in decomposing metric fluctuation?

The SVD is commonly used in fields such as signal processing, image and sound compression, and data analysis. In terms of decomposing metric fluctuation, it can be used to identify patterns or trends in data that may not be apparent in the original form.

## 4. Are there any limitations to using the SVD?

While the SVD is a powerful tool for decomposing metric fluctuation, it does have some limitations. For example, it may not be well-suited for large datasets, as it can be computationally intensive. Additionally, it may not always accurately capture all the information in the original matrix.

## 5. How can the SVD be applied in other scientific fields?

The SVD has applications in a variety of scientific fields, including physics, chemistry, biology, and economics. In physics, it can be used to analyze quantum states and wave functions. In biology, it can be used to analyze genetic data and identify patterns in gene expression. In economics, it can be used to analyze financial data and forecast market trends.

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