Proof that classical General Relativity with CC is an incomplete theory of gravity

[elfmotat]

 

There is nothing magical here, and I do agree that it is simple.Taking the derivative of a function may be equivalent to taking the derivative of a function plus a constant, but the same is not true in reverse. The anti-derivative does not give you the function. You could argue it via Occam's Razor, but that is just lazy and not rigorous. You cannot take the derivative of the following functions and determine which was the original function [math]f'_1[/math] or [math](C-f_2)'[/math]

 

gausstheoremequalityver6.jpg

 

If the other images didn't help explain it, then not sure that one will help either, though.

 

Though your image is confusing I know exactly what you're talking about, and I don't see the problem. Antiderivatives, like differential equations, yield a family of solutions. The specific solution that fits the physical situation is determined by boundary conditions. I'm not quite sure what about this you find problematic.

 

 

 

 

Perhaps I can use this and relate it to what I was stating in the linked paper. I will combine what you stated above and the formula they give and see what happens here.

 

Let's take equation 1.2 [math]\alpha \mu + \beta\frac{\partial \mu}{\partial n}=g[/math] This was supposed to be for a normalized vector on a surface so we will change this to [math]\alpha \mu + \beta\frac{\partial \mu}{\partial x_i}=g[/math]

 

I'm not sure what the Poisson equation has to do with scalar/vector field definitions.

 

 

Combining the notation you gave, for example [math]f(x^u)=\alpha\mu_1=k_1^*[/math] and [math]a^v(x^u)=\beta\left(\frac{\partial \mu}{\partial x_1},\frac{\partial \mu}{\partial x_2},\frac{\partial \mu}{\partial x_3}\right)=(k_1,k_2,k_3)[/math]

 

The problem with what you wrote is that the scalar [math]k_1^*[/math] is most certainly not the same type of scalar as [math]k_1[/math]. The first is a true scalar (like the absolute temperature at a point) and the other is the magnitude of a vector (such as the rate of change of a temperature in a basis direction).

 

First of all, I don't see what your point is.

 

Second, k₁ is a component, not a magnitude. Magnitudes are invariant under diffeomorphisms because the components change in a precise way.

[JMessenger]

 

Though your image is confusing I know exactly what you're talking about, and I don't see the problem. Antiderivatives, like differential equations, yield a family of solutions. The specific solution that fits the physical situation is determined by boundary conditions. I'm not quite sure what about this you find problematic.

 

I'm not sure what the Poisson equation has to do with scalar/vector field definitions.

 

First of all, I don't see what your point is.

 

Second, k₁ is a component, not a magnitude. Magnitudes are invariant under diffeomorphisms because the components change in a precise way.

 

I think we are starting to talk past each other, although I have found some of your comments very useful. I am approaching tensors from a very basic level, and with GR being so well developed there is a gap that I would have to close in order to be able to effectively communicate each and every difference.

 

One of my points is that even the equation [math]R=4\Lambda[/math] doesn't inform me precisely of the mathematical purpose of [math]\Lambda[/math], nor its value. The only explanations of it I have found are some conjectures that it may be related to QFT and that it appears to oppose gravity. As far as I can tell, most of the predictions from these don't lead to anything we physically observe. After all this time and effort by some of the brightest mathematicians, that just seems strange for such a simple term. I was familiar with simpler forms of tensors before I started this, and it occurred to me that with higher rank tensors you could lose information if you weren't careful. I was curious how classical GR handled this and as far as I can tell, it doesn't. I could rewrite the stress-energy tensor of a perfect fluid again to illustrate my point, but this doesn't seem to have helped previously in our discussion.

Edited December 17, 2012 by JMessenger

[elfmotat]

 

I think we are starting to talk past each other, although I have found some of your comments very useful. I am approaching tensors from a very basic level, and with GR being so well developed there is a gap that I would have to close in order to be able to effectively communicate each and every difference.

 

One of my points is that even the equation [math]R=4\Lambda[/math] doesn't inform me precisely of the mathematical purpose of [math]\Lambda[/math], nor its value. The only explanations of it I have found are some conjectures that it may be related to QFT and that it appears to oppose gravity. As far as I can tell, most of the predictions from these don't lead to anything we physically observe. After all this time and effort by some of the brightest mathematicians, that just seems strange for such a simple term. I was familiar with simpler forms of tensors before I started this, and it occurred to me that with higher rank tensors you could lose information if you weren't careful. I was curious how classical GR handled this and as far as I can tell, it doesn't. I could rewrite the stress-energy tensor of a perfect fluid again to illustrate my point, but this doesn't seem to have helped previously in our discussion.

 

Ah, I see. Well, the properties of the cosmological constant are derived mostly from the FLRW metric which is an exact solution to the field equations which describes the expanding universe:

 

[math]ds^2=-dt^2+a(t)^2d\Sigma^2[/math]

 

 

where [math]d\Sigma^2=\frac{dr^2}{1-kr^2}+r^2(d\theta ^2+sin^2\theta d\phi ^2)[/math], a(t) is a function called the "scale factor" of space, and k is a measure of the overall curvature (shape) of the universe. From observations, we know that the universe is probably flat (k=0).

 

If we assume the stress-energy tensor is approximately a perfect fluid, from the field equations we get:

 

[math]{\dot \rho} = - 3 \frac{\dot a}{a}\left(\rho+\frac{p}{c^{2}}\right)[/math]

 

[math]\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}[/math]

 

 

As you can see from the second equation, the pressure and energy density cause a deceleration in the expansion of the universe whereas the cosmological constant causes an acceleration in the expansion of the universe.

 

We see that if we assume the cosmological constant term to be part of the energy density and pressure, we can eliminate the CC term from the above equations:

 

[math]\rho_{\Lambda}=\frac{\Lambda }{8\pi G}[/math]

 

[math]p_{\Lambda}=-\frac{\Lambda }{8\pi G}[/math]

 

[math]\rho =\rho_T+\rho_{\Lambda }[/math]

 

[math]p =p_T+p_{\Lambda }[/math]

 

 

where [math]\rho_T[/math] and [math]p_T[/math] are the energy density and pressure contributed by the stress-energy tensor.

 

So, we can see that the cosmological constant can be treated as a form of energy density (what you could call "dark energy") with negative pressure:

 

[math]p_\Lambda =-\rho _\Lambda [/math]

Edited December 17, 2012 by elfmotat

[JMessenger]

 

Ah, I see. Well, the properties of the cosmological constant are derived mostly from the FLRW metric which is an exact solution to the field equations which describes the expanding universe:

 

[math]ds^2=-dt^2+a(t)^2d\Sigma^2[/math]

 

 

where [math]d\Sigma^2=\frac{dr^2}{1-kr^2}+r^2(d\theta ^2+sin^2\theta d\phi ^2)[/math], a(t) is a function called the "scale factor" of space, and k is a measure of the overall curvature (shape) of the universe. From observations, we know that the universe is probably flat (k=0).

 

If we assume the stress-energy tensor is approximately a perfect fluid, from the field equations we get:

 

[math]{\dot \rho} = - 3 \frac{\dot a}{a}\left(\rho+\frac{p}{c^{2}}\right)[/math]

 

[math]\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}[/math]

 

 

As you can see from the second equation, the pressure and energy density cause a deceleration in the expansion of the universe whereas the cosmological constant causes an acceleration in the expansion of the universe.

 

We see that if we assume the cosmological constant term to be part of the energy density and pressure, we can eliminate the CC term from the above equations:

 

[math]\rho_{\Lambda}=\frac{\Lambda }{8\pi G}[/math]

 

[math]p_{\Lambda}=-\frac{\Lambda }{8\pi G}[/math]

 

[math]\rho =\rho_T+\rho_{\Lambda }[/math]

 

[math]p =p_T+p_{\Lambda }[/math]

 

 

where [math]\rho_T[/math] and [math]p_T[/math] are the energy density and pressure contributed by the stress-energy tensor.

 

So, we can see that the cosmological constant can be treated as a form of energy density (what you could call "dark energy") with negative pressure:

 

[math]p_\Lambda =-\rho _\Lambda [/math]

 

I have read this before but I don't understand how this helps move us forward as a physical model on how to determine the structure of vacuum energy or how it relates to the standard model of particle physics. The [math]\rho[/math] of matter and energy is dynamical ( I can have particles with different masses and energy that dynamically change the stress-energy value at a point) and hence it is associated with terms that have indices ([math]\mu\nu[/math] of [math]G_{\mu\nu}\equiv \kappa T_{\mu\nu}[/math]). The [math]\rho_\Lambda[/math] is associated with a constant term that has no indices ([math]\Lambda[/math] and thus does not have the ability to vary in these indices. It does have the metric [math]g_{\mu\nu}[/math] associated with it and I am still working on understanding precisely what this means. I have seen it explained that the metric is a property of spacetime.

 

There are a couple of things I would like to understand from your point of view and perhaps you can also expound on them from the point of view of Lagrangian mechanics. The Ricci tensor [math]R_{\mu\nu}[/math] appears to me to be a dynamic term (hence the [math]\mu\nu[/math] indices), whereas the Ricci scalar [math]R[/math] appears to be a non-dynamical potential term. Would this be true?

 

Can you give the derivation of exactly when [math]\Lambda[/math] shows up in the derivation of [math]G_{\mu\nu}[/math]? It perhaps is in MTW, but as the text is almost 40 years old it would be nice to see a fresh take on this.

 

 

My motivation is whether I can formulate a complete theory of gravity based on multiplying both sides of [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=G_{\mu\nu}-\Lambda g_{\mu\nu}[/math] by -1 and changing (in order to differentiate the differences) [math]G_{\mu\nu}\equiv L_{\mu\nu}[/math] and [math]\Lambda g_{\mu\nu}=\Omega g_{\mu\nu}[/math]. So for my definition of flat spacetime (no matter or energy) I would get

 

 

 

[math]\frac{1}{2}R g_{\mu\nu}-R_{\mu\nu}=\Omega g_{\mu\nu}-L_{\mu\nu}=0[/math] with [math]L_{\mu\nu}=\Omega g_{\mu\nu}[/math] and [math]g_{\mu\nu}=diag[-1,1,1,1][/math].

Edited December 17, 2012 by JMessenger

[elfmotat]

I have read this before but I don't understand how this helps move us forward as a physical model on how to determine the structure of vacuum energy or how it relates to the standard model of particle physics.

 

You can't use GR to describe quantum phenomena nor to predict the value of the cosmological constant.

 

 

 

There are a couple of things I would like to understand from your point of view and perhaps you can also expound on them from the point of view of Lagrangian mechanics. The Ricci tensor [math]R_{\mu\nu}[/math] appears to me to be a dynamic term (hence the [math]\mu\nu[/math] indices), whereas the Ricci scalar [math]R[/math] appears to be a non-dynamical potential term. Would this be true?

 

Since the Ricci scalar is just a contraction of the Ricci tensor, if the Ricci tensor is dynamic then so is the Ricci scalar:

 

[math]\frac{dR}{dt}=\frac{d}{dt}[g_{\mu \nu}R^{\mu \nu}][/math]

 

 

 

Can you give the derivation of exactly when [math]\Lambda[/math] shows up in the derivation of [math]G_{\mu\nu}[/math]? It perhaps is in MTW, but as the text is almost 40 years old it would be nice to see a fresh take on this.

 

It doesn't. The Einstein tensor is just [math]R^{\mu \nu}-\frac{1}{2}g_{\mu \nu }R[/math]. The CC comes from considering the following action for gravity:

 

[math]S=\int \left [ \frac{1}{2\kappa}(R-2\Lambda) +\mathcal{L}_{M} \right ]\sqrt{-g}d^4x[/math]

 

where [math]\mathcal{L}_{M}[/math] is the Lagrangian density for the matter distribution in spacetime, and [math]g=det(g_{\mu \nu})[/math]. Varying this action W.R.T the metric yields the field equations.

 

 

My motivation is whether I can formulate a complete theory of gravity based on multiplying both sides of [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=G_{\mu\nu}-\Lambda g_{\mu\nu}[/math] by -1 and changing (in order to differentiate the differences) [math]G_{\mu\nu}\equiv L_{\mu\nu}[/math] and [math]\Lambda g_{\mu\nu}=\Omega g_{\mu\nu}[/math]. So for my definition of flat spacetime (no matter or energy) I would get

 

 

 

[math]\frac{1}{2}R g_{\mu\nu}-R_{\mu\nu}=\Omega g_{\mu\nu}-L_{\mu\nu}=0[/math] with [math]L_{\mu\nu}=\Omega g_{\mu\nu}[/math] and [math]g_{\mu\nu}=diag[-1,1,1,1][/math].

 

But that's exactly the same thing as general relativity. Those are exactly the same field equations; all you did was change the symbols. If you want a new classical metric theory of gravity, you need to consider:

• Scalar-tensor or scalar-vector-tensor theories
• Additional spacial dimensions
• Higher-order terms in the action

 

Examples of the first would be Brans-Dicke Theory and MOG. Brans-Dicke in particular is a popular competitor with GR. In these theories, additional scalar and vector fields contribute to gravitational interaction.

 

Another type of variation on gravity is the addition of extra spacial dimensions. Theories of this kind are called Kaluza-Klein theories, and have some interesting consequences. For example, (4+1)-D KK theory unifies classical electromagnetism with GR - its field equations separate into the Einstein field equations and Maxwell's equations.

 

If you consider powers, derivatives, different contractions, etc. of the curvature in the action then you can come up with infinitely many different theories, each with different field equations. Einstein-Hilbert action (which GR is derived from) is the simplest possible metric gravity theory because the action is just:

 

[math]S=\int \alpha R \sqrt{-g}d^4x[/math]

 

where [math]\alpha[/math] is some constant. A simple example of a higher-order gravity theory would be:

 

[math]S=\int \left [ \alpha_{1} R+\alpha_{2} R^2 \right ]\sqrt{-g}d^4x[/math]

 

Indeed, you can come up with infinitely many theories this way:

 

 

where [math].~.~.[/math] represents the infinitely many other scalars we can form from the curvature tensor. This type of GR alternative is interesting because with good choice of higher-order terms, you can render gravity renormalizable which is good for quantization. Unfortunately, these theories also generally feature ghosts.

Edited December 17, 2012 by elfmotat

[JMessenger]

But that's exactly the same thing as general relativity. Those are exactly the same field equations; all you did was change the symbols.

 

I am not sure I understand how they are the same. Take for example the [math]\kappa T_{00}[/math] for slow velocities as described by GR.

As [math]\rho[/math] increases [math]\uparrow[/math], the stress energy increases and gravity increases.

Taking my example [math]\kappa T_{00}[/math] of a perfect fluid, [math]\Omega g_{\mu\nu}[/math] would represent the density [math]C_{\rho}[/math] in a neighborhood about a point. [math]\L_{\mu\nu}[/math] would represent the density of a the fluid at a point [math]\rho_{res}[/math]. If the point density is equal to the surrounding density, there is no stress-energy [math]C_{\rho}-\rho_{res}=0[/math], but the more the density [math]\rho_{res}[/math] decreases [math]\downarrow[/math] with respect to the surrounding [math]C_{\rho}[/math], the higher the stress-energy [math](C_{\rho})\leftrightarrow-(\rho_{res})\downarrow=\uparrow[/math]. In the description by GR, the "energy density" [math]\rho[/math] increases from zero (with respect to what, the vacuum?), whereas I am proposing that the density decreases with respect to its neighborhood. I see how these might produce similar answers, but I don't see how they are equivalent.

 

I will have to take some time to review the other things you posted. I do appreciate it.

Edited December 17, 2012 by JMessenger

[elfmotat]

 

I am not sure I understand how they are the same. Take for example the [math]\kappa T_{00}[/math] for slow velocities as described by GR.

As [math]\rho[/math] increases [math]\uparrow[/math], the stress energy increases and gravity increases.

Taking my example [math]\kappa T_{00}[/math] of a perfect fluid, [math]\Omega g_{\mu\nu}[/math] would represent the density [math]C_{\rho}[/math] in a neighborhood about a point. [math]\L_{\mu\nu}[/math] would represent the density of a the fluid at a point [math]\rho_{res}[/math]. If the point density is equal to the surrounding density, there is no stress-energy [math]C_{\rho}-\rho_{res}=0[/math], but the more the density [math]\rho_{res}[/math] decreases [math]\downarrow[/math] with respect to the surrounding [math]C_{\rho}[/math], the higher the stress-energy [math](C_{\rho})\leftrightarrow-(\rho_{res})\downarrow=\uparrow[/math]. In the description by GR, the "energy density" [math]\rho[/math] increases from zero (with respect to what, the vacuum?), whereas I am proposing that the density decreases with respect to its neighborhood. I see how these might produce similar answers, but I don't see how they are equivalent.

 

They're the same because all you did was change the symbol for the Einstein tensor to L and the symbol for the CC to omega. That doesn't change anything.

[JMessenger]

 

They're the same because all you did was change the symbol for the Einstein tensor to L and the symbol for the CC to omega. That doesn't change anything.

 

It gives me one extra option for flatness. In GR when [math]\rho[/math] vanishes, so does [math]G_{\mu\nu}[/math] and the curvature.

 

In what I am proposing, when [math]\rho_{res}[/math] matches the background [math]C_{\rho}[/math], then the difference between them vanishes along with the curvature. Neither [math]L_{\mu\nu}[/math] vanishes nor [math]\Omega g_{\mu\nu}[/math].

 

I do not think this relationship holds true for non-flatness, but with [math]\Omega g_{\mu\nu}-L_{\mu\nu}=G_{\mu\nu}[/math], if I set [math]\Omega g_{\mu\nu}=0[/math] then L is nothing more than the negative of the Einstein tensor.

Edited December 18, 2012 by JMessenger

[elfmotat]

Neither [math]L_{\mu\nu}[/math] vanishes nor [math]\Omega g_{\mu\nu}[/math].

 

I do not think this relationship holds true for non-flatness, but with [math]\Omega g_{\mu\nu}-L_{\mu\nu}=G_{\mu\nu}[/math], if I set [math]\Omega g_{\mu\nu}=0[/math] then L is nothing more than the negative of the Einstein tensor.

 

You just defined [math]L_{\mu \nu}\equiv G_{\mu \nu}[/math]. Now you're saying they're equal only when [math]\Omega=0[/math]. Which is it?

Edited December 18, 2012 by elfmotat

[JMessenger]

 

You just defined [math]L_{\mu \nu}\equiv G_{\mu \nu}[/math]. Now you're saying they're equal only when [math]\Omega=0[/math]. Which is it?

 

Heh, I was wondering if you were actually reading these posts. The [math]L_{\mu \nu}\equiv G_{\mu \nu}[/math] is incorrect, but it will also take some time for me to fully understand [math]\Omega g_{\mu\nu}-L_{\mu\nu}=G_{\mu\nu}[/math]. If they truly are equal, for any value of the metric, then why would they predict anything differently? That is the reason I was curious to see your version of including the CC with the derivation of Newtonian gravity. I can come up with a specific radius for when gravity switches over to repulsive but I need to make sure I understand the assumptions in each step. I don't just want rote memorization, I would like to understand the logic that goes behind each. Takes time and learning as I go.

[JMessenger]

An interesting aspect of looking at it this way, is that for a perfect fluid stress-energy tensor, when the density at a point decreases (from flat), the pressure would appear to increase in order to follow [-1,1,1,1] and [1,-1,-1,-1].

 

[math]\Omega g_{\mu\nu}-L_{\mu\nu}=\left(\begin{array}{cccc} \Omega-L_{00} & 0 & 0 & 0 \\ 0 & L_{11}-\Omega & 0 & 0 \\ 0 & 0 & L_{22}-\Omega & 0 \\ 0 & 0 & 0 & L_{33}-\Omega \end{array}\right)[/math]

 

[math]T_{\mu\nu}=\left(\begin{array}{cccc} C_{\rho}-\kappa\rho^{res}_{00} & 0 & 0 & 0 \\ 0 & \kappa p^{res}_{11}-C_p & 0 & 0 \\ 0 & 0 & \kappa p^{res}_{22}-C_p & 0 \\ 0 & 0 & 0 & \kappa p^{res}_{33}-C_p \end{array}\right)[/math]

Edited December 19, 2012 by JMessenger

[JMessenger]

In GR, a Ricci flat manifold is a special case of an Einstein manifold with a "vanishing" cosmological constant. Here, [math]G_{\mu\nu}[/math] is theorized to have "vanished" since it was defined as the stress-energy tensor of matter.

[math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=\Lambda g_{\mu\nu}[/math] Einstein manifold (arbitrary multiple of the metric remaining, no accepted physical theory, when viewed this way is measured to have a very small value)

[math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=\Lambda g_{\mu\nu}=0[/math] Ricci flat manifold (zero curvature)(classical GR, but doesn't fit with current observations due to appearance of positive accelerating expansion)

 

 

In this theory, the equation is viewed mathematically the same but physically different:

 

[math]\frac{1}{2}R g_{\mu\nu}-R_{\mu\nu}=\Omega g_{\mu\nu}-L_{\mu\nu}=0[/math]

This is viewed from the mathematical context that [math]\Omega g_{\mu\nu}[/math] is a Rank 0 tensor ([math]g_{\mu\nu}=[-1,1,1,1][/math], meaning [math]\Omega g_{\mu\nu}-L_{\mu\nu}=0[/math] (curvature vanishes) holds true for any arbitrary value of [math]\Omega[/math], including zero. In this theory, the only vacuum solution would be that [math]\Omega g_{\mu\nu}-L_{\mu\nu}=0[/math] with [math]\Omega\neq 0[/math] and perhaps based on Planck length.

 

The major difference is that in GR, [math]G_{\mu\nu}[/math] is restricted to being a Rank 2 tensor, and must "vanish" for flat spacetime, whereas [math]L_{\mu\nu}[/math] is not restricted to being a Rank 2 tensor and can become a Rank 0 tensor for flat spacetime. In the special case when [math]\Omega g_{\mu\nu}=0[/math], THEN [math]L_{\mu\nu}[/math] would reduce to classical GR and thus be required to "vanish" (instead of becoming a Rank 0 tensor).