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Why does a degree of freedom vanish from 3D to 2D in that tensor construction?

Let's assume an arbitrary tensor in 3D coordinates: $g_{ij} $ with $i, j$ in $[1,3]$. It shall be arbitrary, meaning not symmetric. It has 9 entries which equals 9 degrees of freedom (dof). Now, I ...
MartyMcFly's user avatar
0 votes
0 answers
259 views

How many independent degrees of freedom does the metric tensor have in vacuum (at every point)?

A field of metric tensors fully characterises the curvature of a vacuum space-time. (For example, the spacetime between some single point masses which are themself not part of the manifold) The metric ...
Scibo's user avatar
  • 61
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1 answer
384 views

What are the properties of metric tensor? [duplicate]

It's frequently said that graviton has spin-2, so its wave function should have $5$ independent components. The metric tensor has $n^2=16$ components, but it obeys the following property: \begin{...
JavaGamesJAR's user avatar
2 votes
1 answer
961 views

Number of Independent Components of Levi-Civita Christoffel Symbol

Can anybody explain why Levi-Civita Christoffel symbol in general $N$ dimensional space have $\frac{N^2(N+1)}{2}$ independent components? I have read that in $N$-dimensional space, metric tensor has ...
Math noob's user avatar
3 votes
1 answer
2k views

How to count degrees of freedom in a symmetric $N \times N$ matrix?

I am reading Wayne Hu's short lecture on cosmology mathematical infrastructure (https://arxiv.org/abs/astro-ph/0402060), and have several questions. Some background for us lazy people that don't want ...
BeastRaban's user avatar
1 vote
1 answer
189 views

How to count degrees of freedom of metric in Newmann-Penrose formalism?

Usually a metric has 10 degrees of freedom. How to show the same in Newmann-Penrose formalism?
Dr. user44690's user avatar
3 votes
1 answer
387 views

Is there only one vacuum solution of the Einstein equations?

I am thinking about this: A vacuum solution means vanishing Ricci tensor. The Ricci tensor is a contraction of the Riemann, which itself involves contains second derivatives of the metric. Thus they ...
JoeGlas's user avatar
  • 33
1 vote
0 answers
249 views

Notion of 'functional degrees of freedom' for the metric function in GR?

I have read through the numerous questions on 'degrees of freedom' in the metric tensor, and won't list them all here. However none of them address my question on 'functional' degrees of freedom in ...
Meep's user avatar
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2 votes
1 answer
3k views

The 10 independent components in the Einstein's Field equations

I'm trying to understand the Einstein field equations, but there is something that is costing me to understand. From the Wikipedia: https://en.wikipedia.org/wiki/Einstein_field_equations, it says that ...
Alex Climent's user avatar
0 votes
0 answers
162 views

How many degrees of freedom to diagonalize the metric?

In A. Zee's Einstein Gravity in a Nutshell, he starts with the following expansion of the metric at some point $P$ of a Riemannian manifold, with coordinates $x^\mu$ that have the origin at $P$: $$ ...
Bass's user avatar
  • 1,507
2 votes
0 answers
1k views

What is the degrees of freedom of metric tensor?

As $g_{\mu\nu}$ can be taken to be symmetric, it contains 10 functions of spacetime in 4 dimensions. But, why we call these 10 functions as the degrees of freedom of the metric while they are the ...
Archimedes's user avatar
12 votes
2 answers
9k views

Variation of the metric with respect to the metric

For a variation of the metric $g^{\mu\nu}$ with respect to $g^{\alpha\beta}$ you might expect the result (at least I did): \begin{equation} \frac{\delta g_{\mu\nu}}{\delta g_{\alpha\beta}}= \delta_\mu^...
Sam's user avatar
  • 318