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A Unification: Classical field theory with GR


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Due to the mathcode still not operational, I will post the codes of my equations but I'll link at the end to a paper which displays my work. I am the sole author of this work, all rights are reserved to me. 


Elastic Action: A Wedding of Quantum Field Theory with the General Relativiatic Action

This tackles a question on how Sakharov's ground state field for virtual particles enters the gravitational action. In it I conclude that maybe the cosmological constant is in fact a renormalization constant which is only set to zero for flat Euclidean spacetime. The jury is still out, but most respected astrophysicists tend to agree that while spacetime looks quite flat, it probably isn't exactly flat, it's just a very good approximation and his equations on a cosmological scale would predict a small curve like we expect.

Let's identify variables

A - action

g - metric

x - variable spatial coordinate

c - speed of light

\mathbf{R} - invariant Ricci curvature tensor

\hbar - Planck constant, reduced

G - Newtons constant

k - wave number

* We will use \mathbf{k} as a constant \frac{8πG}{c^4} which is the upper value of the gravitational constant

In the style of Sakharov, we'd like to write a Langrangian of the ground state fluctuations which has a contribution of geometry by off-shell virtual particles.

It's a rare paper to find, but his original ideas can be found here:


His equation features like:

\mathcal{L} = \mathbf{R}\ \hbar c\ k\ \int dk + \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k} + C

Where C is a constant and

\frac{dk}{k} = d\log_e k \approx 137

His original ideas can be taken as a precursor to Bogoliubov transformations used to describe how gravity jiggles these off-shell particles at the horizon of supermassive Black holes, owing to their name as Hawking radiation.

So how do we do this? How does Sakharov's equation enter the action? It turns out there are a number of different ways we can do it and all are equally interesting.

Varied Action

Using the formulation set above, we now can apply the varied action.

\delta A = \frac{\int d^4x}{\mathbf{k}}\ \left[\delta\mathcal{L}\sqrt{-g} + \mathcal{L}\delta(\sqrt{-g})\right]

We can simplify the second term using the variation of the determinant of the metric tensor:

\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}

substitution gives

\delta A = \frac{\int d^4x}{\mathbf{k}} \left[\delta\mathcal{L}\sqrt{-g} + \mathcal{L}\left(\frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}\right)\right]

And yes, the Langrangian is split up into parts such that when we write out the Sakharov Langrangian. For instance, let us now use

\mathcal{L} = \mathbf{R}\ \hbar c\ k\ \int dk + \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k}

Plugging the Langrangian density in we get:

\delta A = \frac{\int d^4x}{\mathbf{k}}[\delta( \mathbf{R}\hbar c\ k\int dk + \mathbf{R}^2\hbar c\ \int \frac{dk}{k}) \sqrt{-g}

+ \frac{1}{2}(\mathbf{R}\hbar c\ k \int dk + \mathbf{R}^2\hbar c\ \int \frac{dk}{k}) \sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}]\frac{16 \pi G}{c^4}

We just want to focus on how this altered the dimensions and how we must fix those dimensions. Focusing on this,

\mathbf{R}\ \hbar c\ k dk

We know what R is as it's dimensions are still inverse length squared. The k is called the wave number and has inverse unit of length. All-in-all, we have dimensions of charge squared divided by a length, giving an energy, further with another inverse length cubed, giving the appropriate dimensions of energy density. What we "put in" those brackets, must be undone, and there's a straight-forward way to do it. We don't need to "undo" what we have in since it already features in the action, but we will concentrate on

\hbar c\ k\ \int dk = \frac{\hbar c}{\ell^2}

We understand that the following dimensions must hold true:

\frac{G}{c^2} \equiv \frac{\ell}{m}

Since dimensionally-speaking

\frac{\hbar c}{\ell^2} = \frac{Gm^2}{\ell^2}

Then we can decompose it in the following way:

\frac{Gm}{\ell^2} = \frac{Gm}{\ell}\frac{c^2}{G} = \frac{m}{\ell} \cdot c^2 = \frac{c^4}{G}

Interesting isn't it? It seems then the solution has been found. In order to "undo" what we did, it requires a correction coefficient of the upper limit of gravity as \frac{c^4}{G} (by taking its inverse). Why its inverse? Simply because if

\frac{\hbar c}{\ell^2} = \frac{Gm^2}{\ell^2} = \frac{c^4}{G}

Then we must invert to remove these unwanted dimensions.

Dimensions Check

\mathbf{k}=\frac{8\pi G}{c^4}=2.08\times 10^{-43}\text{N}^{-1}

So \frac{1}{\mathbf{k}} has the dimension of a force.

Considering this you get the dimension of

dx\ dy\ dz\ dt \frac{1}{\mathbf{k}}\mathbf{R}


\text{m}^3\cdot\text{s}\cdot\text{N} \cdot\text{m}^{-2}=\text{J}\cdot\text{s}

which is the dimension of an action as it should be, which is force times length times time, or energy times time.

Stress Energy

Sakharov intended to speak about fluctuations as a conteibution to the background spacetime. Loosely-speaking that means his Langrangian density must have to translate into the stress energy tensor. We can express this as:

If you define T^{\mu\nu} explicitly, by writing

\delta S = -\int d^dx \sqrt{g}\, T^{\mu\nu} \delta g_{\mu\nu}


T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g_{\mu\nu}}

where S_m is the matter action

S_m =\int d^4x\sqrt{-g}\mathcal{L}_m

and \mathcal{L}_m is the matter Lagrangian-density. It's important that we include the definition of the stress energy so that we can build a healthy picture of the stress energy contribution of the fluctuations to the background geometry. We shall do this now. A proportional way to speak about the stress energy tensor is the following

T^{\mu\nu}= -\frac{2}{\sqrt{-g}}\frac{\delta \mathcal{L}\sqrt{g}}{\delta g_{\mu\nu}}

Rearranging we get

T^{\mu\nu}\sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{g}

It will be this form from which we derive the following results. The factor of 2 looks a bit peculiar, but for arguments sake it could be dropped at any time.

Second Varied Action

Earlier I argued that the stress energy must be encorporates somehow into Sakharovs vision. We see now how this directly comes about since in uts contracted form the relationship is simply \mathbf{G} = \mathbf{k} \mathbf{T}, such that we have:

\delta A = \frac{\int d^4x}{\mathbf{k}}[\delta( \mathbf{R}\hbar c\ k\int dk + \mathbf{R}^2 \hbar c\ \int \frac{dk}{k}) \sqrt{-g}

+ \frac{1}{2}(\mathbf{R}\hbar c\ k\int dk + \mathbf{R}^2\hbar c\ \int \frac{dk}{k}) \sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}]\frac{16 \pi G}{c^4}

= \int d^4x\ [\mathbf{k} \mathbf{T}^{\mu \nu}\delta g_{\mu\nu}\sqrt{-g}]

See, originally, I had a picture in mind:

A = \int\ d^4x\ \sqrt{-g}[\frac{\mathbf{R}}{\mathbf{k}} + \mathcal{L}]

I asked “what if the Ricci curvature is broken into two parts just like in Sakharovs equation?” There turned out to be two theoretical ways to assume this sort of set up. We will cover one of them now another later. For instance, we bring the constant out of the square brackets and we get:

A = \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R} +…]

Also there is nothing stopping us writing it as,

A = \int\ d^4x\ \frac{1}{\mathbf{k}}[\sqrt{-g}\mathbf{R} +…]

Where we pull k out and distribute the determinant of the metric… Which means when we apply the product rule of variations we can also have

\delta A = \int\ d^4x\ \frac{1}{\mathbf{k}}[\delta \mathbf{R}\sqrt{-g} + \mathbf{R}\delta\sqrt{-g}]

The variation of the determinant of the metric tensor \delta\sqrt{-g} can be related to the variation of the metric tensor \delta g_{\mu\nu} using

\delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}

Substituting this expression into

\delta A, we obtain

\delta A = \int\ \frac{d^4x}{\mathbf{k}}\ [\delta \mathbf{R}\sqrt{-g} - \frac{1}{2}\mathbf{R}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}]

We can then use the fact that

\mathbf{R}^{\mu\nu} - \frac{1}{2}\mathbf{R} g^{\mu\nu} = \mathbf{k} \mathbf{T}^{\mu\nu}

Which retrieves the Einstein tensor which is a mixture of both scalar and tensor curvatures.

But This Doesn't Stop The Following Speculation

So could Sakharov's Langrangian

\mathcal{L} = \mathbf{R}\ \hbar c\ k\ \int dk + \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k}

Be incorporated into the action

A = \int\ d^4x\ \sqrt{-g}[\frac{\mathbf{R}}{\mathbf{k}} + ...]

Like so?

A = \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R}\ \hbar c\ k \int dk + \mathbf{R}^2 \hbar c\ \int \frac{dk}{k}]\frac{16 \pi G}{c^4}

Where \frac{16 \pi G}{c^4} simply corrects the dimensions added in. This was the original set-up I considered and I still find it an interesting equation. If thus truly is also a valid line of research for investigation, then we can say something about higher power corrections.

Higher Powers

Sakharov concludes the higher powers are taken like so:

\int \frac{dk}{k}(\mathbf{B}\ \mathbf{R}^2 + \mathbf{C}\ \mathbf{R}^{ik}\mathbf{R}_{ik} +\mathbf{D}\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} … + higher\ powers)

Where \int \frac{dk}{k} \approx 137

These higher powers on the equation we presented looks like:

A = \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk + \int \frac{dk}{k}(\mathbf{B}\ \mathbf{R}^2 + \mathbf{C}\ \mathbf{R}^{ik}\mathbf{R}_{ik}+\mathbf{D}\mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{16\pi G}{c^4}

Notice that this hasn't been subjected to the calculus of variations.

Higher Powers of Fluctuations

\mathbf{R}\ \hbar c\ k \int dk, taking higher powers of \hbar c requires that the dimensions are scaled appropriately… Say the higher powers don't just affect the curvature, but affects higher powers of \hbar c = (\mathbf{A}, \mathbf{B},\mathbf{C},\mathbf{D}) \approx 1

And using the rule

\mathbf{R}^2 \hbar^2 c^2(\frac{1}{Gm^2}) k \int dk = \mathbf{R}^2 \hbar c\ \alpha^{-1}_G k \int dk

So taking higher powers of \hbar c in

A = \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk + \int \frac{dk}{k}(\mathbf{B}\ \mathbf{R}^2 + \mathbf{C}\ \mathbf{R}^{ik}\mathbf{R}_{ik}+\mathbf{D}\mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{16\pi G}{c^4}


A = \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk

+ \int \frac{dk}{k}(\mathbf{B}^2\ \alpha_G^{-1} \mathbf{R}^2

+ \mathbf{C}^3\ \alpha^{-2}_G \mathbf{R}^{ik}\mathbf{R}_{ik}

+\mathbf{D}^4 \alpha^{-3}_G\mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{16\pi G}{c^4}

Where we use the gravitational fine structure to normalise the higher dimensions, for example. While the fine structure constant and Newton's gravitational constant are both fundamental constants of nature, they describe completely different physical phenomena. Therefore, it is not accurate to say that higher powers of the fine structure constant are equivalent to correcting gravity at higher powers.

However, there are some theories, such as string theory, that suggest a connection between the values of fundamental constants and the properties of space-time, including the strength of gravity. In these theories, it is possible that changes in the fine structure constant could lead to changes in the properties of space-time and therefore, to corrections in the theory of gravity. But this is a highly speculative area of research and is not yet well understood.

Taking higher powers of the fine structure constant can reveal higher order corrections to physical phenomena that cannot be accounted for by classical or first-order quantum mechanical calculations. In particular, higher-order quantum corrections are important in understanding the behavior of subatomic particles and the interactions between them.

For example, the anomalous magnetic moment of the electron can be calculated with higher and higher accuracy by taking into account higher order QED corrections involving higher powers of the fine structure constant. The electron's anomalous magnetic moment has been measured experimentally to extremely high precision, and the agreement between theory and experiment is a remarkable demonstration of the power of higher-order quantum corrections.

Similarly, in quantum chromodynamics (QCD), the theory that describes the strong nuclear force, higher order corrections involving higher powers of the strong coupling constant (the analog of the fine structure constant for the strong force) are important for understanding the properties of hadrons (particles made of quarks), and for calculating the scattering amplitudes of quarks and gluons.

In general, higher order corrections involving higher powers of dimensionless coupling constants are important for understanding the behavior of quantum field theories at energies far beyond the scales of current experiments. Such corrections can also provide clues to the existence of new physics beyond the Standard Model of particle physics.



Misc. Notes:

\delta(\sqrt{-g}) are the variations of the Lagrangian density and the determinant of the metric tensor, respectively, under a small variation of the metric tensor g_{\mu\nu}.

This expression follows from the fact that the variation of the determinant of a matrix is given by:

\delta\det(A) = \det(A) \operatorname{tr}(A^{-1} \delta A)

where \delta\det(A) denotes the variation of the determinant of the matrix A, \det(A) is the determinant of A, \operatorname{tr}(A^{-1} \delta A) is the trace of the matrix product A^{-1} \delta A and \delta A is an arbitrary matrix of the same size as A.

Okay second try. So far I know that the first two equations are true:

\delta A = \int d^4x\ \left[\delta\mathcal{L}\sqrt{-g} + \mathcal{L}\delta(\sqrt{-g})\right]

We can simplify the second term using the variation of the determinant of the metric tensor:

\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}

The substitution should have been

\delta A = \int d^4x\ \left[\delta\mathcal{L}\sqrt{-g} + \mathcal{L}\left(\frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}\right)\right]

And yes, the Langrangian is split up into parts such that when we writ out the Sakharov Langrangian parts, the L will be replaced by the Ricci scalar. For instance, let us now use

A useful identity to always keep in mind is


Stress Tensor

Tensors and Lagrangian Density

We refer back to

T^{\mu\nu}= -\frac{2}{\sqrt{-g}}\frac{\delta \mathcal{L}\sqrt{g}}{\delta g_{\mu\nu}}

Where we rearranged to get:

T^{\mu\nu}\sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{-g}

And we shall use

\delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}

Plugging the third equation into the second on the LHS gives

T^{\mu\nu} \sqrt{-g} \left(-\frac{1}{2}\sqrt{-g}\ g^{\alpha\beta}\delta g_{\alpha\beta}\right)

= -2\delta \frac{\mathcal{L}\sqrt{g}}{\delta g_{\mu\nu}}

Simplifying the left-hand side, we have:

We can simplify this expression by using the identity g^{\alpha\beta}g_{\alpha\beta}=4 and some algebraic manipulation.

Starting with the left-hand side:

T^{\mu\nu} \sqrt{-g} \left(-\frac{1}{2}\sqrt{-g}\ g^{\alpha\beta}\delta g_{\alpha\beta}\right)

Expanding the first factor and simplifying the second factor:

T^{\mu\nu} \sqrt{-g}\left(-\frac{1}{2}\sqrt{-g}g^{\alpha\beta}\delta g_{\alpha\beta}\right)

= -\frac{1}{2}T^{\mu\nu}g^{\alpha\beta}\sqrt{-g}\delta g_{\alpha\beta}

Now substituting

T^{\mu\nu} \sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{-g}

-\frac{1}{2}T^{\mu\nu}g^{\alpha\beta}\sqrt{-g}\delta g_{\alpha\beta}=-2\delta\mathcal{L}^{\mu\nu}\sqrt{g}

Using the chain rule of functional derivatives, we can write

\delta\mathcal{L}\sqrt{g} = \frac{\delta\mathcal{L}}{\delta g_{\mu\nu}} \delta g_{\mu\nu}\sqrt{g}

-2\delta\mathcal{L}^{\mu\nu}\sqrt{g} = -2\frac{\delta\mathcal{L}}{\delta g_{\mu\nu}} \delta g_{\mu\nu}\sqrt{g}

Finally, multiplying both sides by g_{\mu\nu}

and using


we get:

-T^{\mu\nu} \delta g_{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta \mathcal{L}\sqrt{g}}{\delta g_{\mu\nu}} g_{\mu\nu}

So we have simplified the original expression to the more compact form

-T^{\mu\nu} \delta g_{\mu\nu} = 2\frac{\delta \mathcal{L}}{\delta g_{\mu\nu}} \frac{g_{\mu\nu}}{\sqrt{-g}}

This time plugging the third equation into the second equation on the RHS gives

T^{\mu\nu}\sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{-g} - \mathcal{L}\frac{\delta(\sqrt{-g})}{\sqrt{-g}}g^{\mu\nu}\delta g_{\mu\nu}

To get this, we start from

T^{\mu\nu}\sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{g}

and use the identity

\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}

but with a slight modification. We note that

\delta(\sqrt{-g}) = \frac{1}{2\sqrt{-g}}\delta(\sqrt{-g})\sqrt{-g} = \frac{1}{2}\frac{\delta(\sqrt{-g})}{\sqrt{-g}}\sqrt{-g}

And substitute this into the above equation to get:

T^{\mu\nu}\sqrt{-g}\ \delta g_{\mu\nu} = -2\delta \mathcal{L}\sqrt{g} - \frac{1}{2}\mathcal{L}\delta(\sqrt{-g})g^{\mu\nu}\delta g_{\mu\nu}

Useful Operations

Variation of Metric Tensor

$$\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}$$

This formula can be derived using the following steps:

  1. Use the chain rule to express the variation of the determinant of the metric tensor in terms of the variation of the metric tensor components:

$$\delta(\sqrt{-g}) = \frac{\partial(\sqrt{-g})}{\partial g_{\mu\nu}}\delta g_{\mu\nu}$$

  1. Use the formula for the derivative of the determinant of a matrix, which is:

$$\frac{\partial}{\partial g_{\mu\nu}}(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}$$

2. Substitute the formula for the derivative of the determinant into the expression for the variation of the determinant, and simplify:

$$\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}$$

Therefore, the expression $\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}$ is a valid formula relating the variation of the determinant of the metric tensor to the variation of the metric tensor components.


Reference to displayed equations:


Edited by Dubbelosix
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Also related, the only one thing I disagree with Sakharov on, is the application of the Singularity at the initial condition of the universe. I was also attacked concerning my work "not being new" so I point out some fallacies in this poorly constructed disagreement. 


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