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

[JMessenger]

In fact, neither have I! The only term I have come across is no-where vanishing, so a tensor that is not no-where vanishing, is "not no-where vanishing"!

 

 

Ha! I had to read that a couple times

 

Consider just R³, which is flat by any definition you want. You can define a non-vanishing one form dz - x dy, in (global) coordinates (x,y,z). That one form is the standard contact form on R³.

 

So flat manifolds can come with no-where vanishing tensors.

 

Hmmmm...

 

Then the same should apply for R⁰, R¹ and R², no? Is there a rule concerning no-where vanishing tensors remaining in non-flat manifolds of R⁴ and higher? As in R⁰ remaining in R⁴, R¹ remaining in flat R⁵, etc...as in R^n-4 for [math]n\geq 4[/math]?

Edited December 11, 2012 by JMessenger

[JMessenger]

You had mentioned earlier about the theory having an additional scalar field, seemingly non-dynamical. I do need to look at the action of the theory later on and develop that, but going to expound on the stress energy tensor of a perfect fluid for a moment:

In this tensor

[math] T^{\mu\nu}_{normal}=(\rho+p)\mu^{\mu} \mu^{\nu}-p\eta^{\mu\nu} [/math]

 

there are two scalar values of [math]\rho[/math] and p, which are scalar values of density and pressure. The problem I have with this tensor is that it was borrowed from hydrodynamical theory based on Newtonian concepts. What isn't stated is that the scalar values have an implied "+ 0" with them, as in "[math]\rho + 0[/math]" and "p + 0". What is just as valid with these scalars is to measure density and pressure from a higher magnitude, as in "[math]C_{\rho}-\rho[/math]" and "C_p-p"

 

so that we get

[math] T^{\mu\nu}_{alternate}=(C_{\rho}-\rho+C_p-p)\mu^{\mu} \mu^{\nu}-(C_p-p)\eta^{\mu\nu} [/math].

 

Although the constant scalars in this tensor (the "C") may not change value, they are not independent of the metric (same as the Cosmological Constant), so the alternate tensor should predict different answers (although that may only occur at large radii). The alternate stress energy tensor seems to instead describe a constant reduction in the density of a medium, that perhaps travels as a wave. Note that for [math]C_{\rho}-\rho=0[/math] (and the same for pressure), there is no stress in the medium and that for [math]\Lambda g_{\mu\nu}-G_{\mu\nu}=0[/math] there is no curvature.

 

Edit: As an aside, I am aware of how the normal tensor is derived into 4-momentum and the "energy density" we are familiar with. I do have some questions concerning the definition of "flux" as it concerns "energy density".

Edited December 11, 2012 by JMessenger

[ajb]

Then the same should apply for R⁰, R¹ and R², no?

You can only have contact structures on odd dimensional manifolds, but the idea of no-where vanishing tensors is universal. (On classical manifolds anyway, maybe on supermanifolds it is not so clear, but that is another story.)

Is there a rule concerning no-where vanishing tensors remaining in non-flat manifolds of R⁴ and higher? As in R⁰ remaining in R⁴, R¹ remaining in flat R⁵, etc...as in R^n-4 for [math]n\geq 4?

You want some embedding theorems?

 

There are various theorems about embedding non-vanishing vectors, I have no idea about more general tensors. You would have to be very specific here to come up with theorems here.

[JMessenger]

You can only have contact structures on odd dimensional manifolds, but the idea of no-where vanishing tensors is universal. (On classical manifolds anyway, maybe on supermanifolds it is not so clear, but that is another story.)

 

You want some embedding theorems?

 

There are various theorems about embedding non-vanishing vectors, I have no idea about more general tensors. You would have to be very specific here to come up with theorems here.

 

 

 

Heh, I imagine that route has quite a bit of "mathematical fertility" but might not be the most direct route for some of my questions, so I will bow out of asking further along those lines (can I take a raincheck?). I am interested in how the metric is related to the cosmological constant, so it revolves around the physical understanding and derivation of a constant scalar with various metric values (including unimodular where the determinant magnitude of the metric is 1).

 

What is your view of the relationship between the metric and the cosmological constant?

[ajb]

What is your view of the relationship between the metric and the cosmological constant?

The relationship is given by the field equations. Not that that helps much.

 

 

[JMessenger]

The relationship is given by the field equations. Not that that helps much.

 

 

True, but as you are a mathematician and forgetting all physical models, doesn't it bother you that a constant scalar (with the metric) appears in a geometric theory but there are no foundational proofs of its purpose?

[ajb]

True, but as you are a mathematician and forgetting all physical models, doesn't it bother you that a constant scalar (with the metric) appears in a geometric theory but there are no foundational proofs of its purpose?

 

It does seem rather ad hoc from a pure geometry point of view. This is key, classical general relativity gives us no indication as to what the cosmological constant should be.

 

[math]R_{\mu \nu} + \frac{1}{2}R g_{\mu \nu} + \Lambda g_{\mu \nu} = \kappa T_{\mu \nu}[/math]

 

The left-hand side of the above is the most general local, coordinate-invariant, divergenceless, symmetric, two-index tensor one can construct using only the metric as well as its first and second derivatives. One could and should ask: why do we consider only up to second order in derivatives? (Physically, a quantum theory has no ghosts if it contains only up to second order derivatives. Any higher and one needs mechanisms to remove these ghosts, and people do think about higher order gravity. I don't know if there is some clear geometric reason why we should stop at second order derivatives.)

 

 

However, when we take quantum theory into account, the cosmological constant is a measure of the energy density of the vacuum. This gives us some hope of calculating it, but all attempts so far have been pretty bad.

 

In short, I think the cosmological constant is physical in origin rather than geometrical, in the sense outlined above.

[JMessenger]

 

It does seem rather ad hoc from a pure geometry point of view. This is key, classical general relativity gives us no indication as to what the cosmological constant should be.

 

[math]R_{\mu \nu} + \frac{1}{2}R g_{\mu \nu} + \Lambda g_{\mu \nu} = \kappa T_{\mu \nu}[/math]

 

The left-hand side of the above is the most general local, coordinate-invariant, divergenceless, symmetric, two-index tensor one can construct using only the metric as well as its first and second derivatives. One could and should ask: why do we consider only up to second order in derivatives? (Physically, a quantum theory has no ghosts if it contains only up to second order derivatives. Any higher and one needs mechanisms to remove these ghosts, and people do think about higher order gravity. I don't know if there is some clear geometric reason why we should stop at second order derivatives.)

 

 

However, when we take quantum theory into account, the cosmological constant is a measure of the energy density of the vacuum. This gives us some hope of calculating it, but all attempts so far have been pretty bad.

 

In short, I think the cosmological constant is physical in origin rather than geometrical, in the sense outlined above.

 

Good point about stopping at only second order derivatives. But ignoring quantum theory of the vacuum, let's still take all physical interpretations out of the equation, meaning the stress-energy of matter also:

[math]R_{\mu \nu} + \frac{1}{2}R g_{\mu \nu} + \Lambda g_{\mu \nu} = G_{\mu \nu}[/math]

 

Don't you find it odd, that if the metric is [math] g_{\mu\nu}=diag [-1,1,1,1][/math], then whatever you decide you want [math]\Lambda[/math] to represent, it is Euclidean? When the metric changes away from flatness, how is it that this term no longer has some fundamental role to play in this equation if it comes up in the derivation?

[JMessenger]

To put it in more basic terms, if we currently measure curvature away from the flat metric of [math]g_{\mu\nu}=\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)[/math]

 

what is the difference of measuring curvature from the reference point of an arbitrary flat [math]\Lambda[/math]? Aren't we arbitrarily setting [math]\Lambda[/math] to 1 when we get the Newtonian approximation?

[elfmotat]

Aren't we arbitrarily setting [math]\Lambda[/math] to 1 when we get the Newtonian approximation?

 

No, we're arbitrarily setting it to zero. Setting it to 1 would result in some funky vacuum geometry.

[JMessenger]

 

No, we're arbitrarily setting it to zero. Setting it to 1 would result in some funky vacuum geometry.

 

In every derivation for the Newtonian [math]\phi[/math] that I have seen, it is arrived at as a perturbation away from "1", not zero.

[elfmotat]

 

In every derivation for the Newtonian [math]\phi[/math] that I have seen, it is arrived at as a perturbation away from "1", not zero.

 

The metric is a perturbation of the Minkowski metric, yes. That has nothing to do with the cosmological constant. A nonzero CC means nonzero vacuum Ricci curvature. A quick contraction of the vacuum equations yields R=4Λ, so another form of the vacuum equations would be:

 

[math]R_{\mu \nu}=g_{\mu \nu}\Lambda[/math]

 

With such a large value for Λ, you're going to get some funky geometry.

 

 

When you're deriving the low-energy metric, you assume Λ=0. What you do is define the metric as [math]g_{\mu \nu}=\eta_{\mu \nu} + h_{\mu \nu}[/math], where [math]|h_{\mu \nu}|\ll 1[/math] because we're assuming the field is weak.

 

The geodesic equation, [math]\frac{d^2 x^\mu}{d\tau^2}=-\Gamma^{\mu}_{\lambda \sigma} \frac{dx^\lambda}{d\tau} \frac{dx^\sigma}{d\tau}[/math], can be simplified under the condition that we're considering velocities much less than that of light, i.e. [math]\frac{dx^i}{d\tau} \ll \frac{dt}{d\tau }[/math]. This reduces it to:

 

[math]\frac{d^2 x^\mu }{d\tau^2}=-\Gamma^{\mu}_{00}\left ( \frac{dt}{d\tau } \right )^2[/math]

 

We can expand the remaining Christoffel symbols as the following if we assume the metric is static:

 

[math]\Gamma^{\mu}_{00}=-\frac{1}{2}g^{\mu \lambda}\partial_{\lambda}g_{00}=-\frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00}[/math]

 

The geodesic equation then becomes:

 

[math]\frac{d^2 x^\mu }{d\tau^2}= \frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00} \left ( \frac{dt}{d\tau } \right )^2[/math]

 

The 0-component obviously gives us [math]d^2t/d\tau^2=0[/math]. Applying the chain rule for the remaining components gives us:

 

[math]\frac{dx^i}{dt^2}=\frac{1}{2}\partial_ih_{00}[/math].

 

We see here that if we set [math]h_{00}=-2\Phi [/math] where [math]\Phi[/math] is the Newtonian potential then we have reproduced Newtonian gravity. Since the remain components of [math]h_{\mu \nu }[/math] don't affect our differential equation, it is sufficient to set them equal to zero. So the way we represent Newtonian gravity with curved spacetime is with the following metric:

 

[math]ds^2=-(1+2\Phi )dt^2+dr^2+r^2d\Omega ^2[/math]

 

where [math]d\Omega^2=d\theta^2+sin^2\theta d\phi ^2[/math].

Edited December 14, 2012 by elfmotat

[JMessenger]

 

The metric is a perturbation of the Minkowski metric, yes. That has nothing to do with the cosmological constant. A nonzero CC means nonzero vacuum Ricci curvature. A quick contraction of the vacuum equations yields R=4Λ, so another form of the vacuum equations would be:

 

[math]R_{\mu \nu}=g_{\mu \nu}\Lambda[/math]

 

With such a large value for Λ, you're going to get some funky geometry.

 

 

When you're deriving the low-energy metric, you assume Λ=0. What you do is define the metric as [math]g_{\mu \nu}=\eta_{\mu \nu} + h_{\mu \nu}[/math], where [math]|h_{\mu \nu}|\ll 1[/math] because we're assuming the field is weak.

 

The geodesic equation, [math]\frac{d^2 x^\mu}{d\tau^2}=-\Gamma^{\mu}_{\lambda \sigma} \frac{dx^\lambda}{d\tau} \frac{dx^\sigma}{d\tau}[/math], can be simplified under the condition that we're considering velocities much less than that of light, i.e. [math]\frac{dx^i}{d\tau} \ll \frac{dt}{d\tau }[/math]. This reduces it to:

 

[math]\frac{d^2 x^\mu }{d\tau^2}=-\Gamma^{\mu}_{00}\left ( \frac{dt}{d\tau } \right )^2[/math]

 

We can expand the remaining Christoffel symbols as the following if we assume the metric is static:

 

[math]\Gamma^{\mu}_{00}=-\frac{1}{2}g^{\mu \lambda}\partial_{\lambda}g_{00}=-\frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00}[/math]

 

The geodesic equation then becomes:

 

[math]\frac{d^2 x^\mu }{d\tau^2}= \frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00} \left ( \frac{dt}{d\tau } \right )^2[/math]

 

The 0-component obviously gives us [math]d^2t/d\tau^2=0[/math]. Applying the chain rule for the remaining components gives us:

 

[math]\frac{dx^i}{dt^2}=\frac{1}{2}\partial_ih_{00}[/math].

 

We see here that if we set [math]h_{00}=-2\Phi [/math] where [math]\Phi[/math] is the Newtonian potential then we have reproduced Newtonian gravity. Since the remain components of [math]h_{\mu \nu }[/math] don't affect our differential equation, it is sufficient to set them equal to zero. So the way we represent Newtonian gravity with curved spacetime is with the following metric:

 

[math]ds^2=-(1+2\Phi )dt^2+dr^2+r^2d\Omega ^2[/math]

 

where [math]d\Omega^2=d\theta^2+sin^2\theta d\phi ^2[/math].

 

Excellent description, thank you. Can you also show how to derive down to Newtonian gravity with the cosmological constant included? Hobson et. al. General Relativity: An Introduction for Physicists come up with [math]\vec g=-\vec \nabla \Phi =-\frac{GM}{r^2}\hat{\vec r}+\frac{c^2 \Lambda r}{3}\hat{\vec r} [/math] but they do not show the assumptions made to arrive at this.

[elfmotat]

Excellent description, thank you. Can you also show how to derive down to Newtonian gravity with the cosmological constant included? Hobson et. al. General Relativity: An Introduction for Physicists come up with [math]\vec g=-\vec \nabla \Phi =-\frac{GM}{r^2}\hat{\vec r}+\frac{c^2 \Lambda r}{3}\hat{\vec r} [/math] but they do not show the assumptions made to arrive at this.

 

The form of the field equations with zero cosmological constant are:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}[/math]

 

Compare this with the analogous Poisson equation for Newtonian gravity:

 

[math]\nabla^2 \phi =4\pi G\rho [/math]

 

 

The field equations with nonzero CC are:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda[/math]

 

Because the metric is analogous to the Newtonian potential, this suggests a modified Poisson equation of the form:

 

[math]\nabla^2 \phi =4\pi G\rho-\phi \Lambda[/math]

 

 

However, if we assume the gravitational field is weak (which is the only time when Newtonian gravity is accurate anyway) and we linearize the field equations with [math]g_{\mu \nu }=\eta_{\mu \nu }+h_{\mu \nu }[/math] and [math]|h_{\mu \nu}|\ll 1[/math], then we arrive at the following field equations:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}-\eta_{\mu \nu}\Lambda[/math]

 

where [math]G_{\mu \nu}=-\frac{1}{2} \partial^{\alpha} \partial_{\alpha}(h_{\mu \nu}-\frac{1}{2}\eta_{\mu \nu} h^{\sigma}_{~\sigma })[/math], though the form of the Einstein Tensor isn't really important for our purposes. These field equations imply (because the relevant components of the Minkowski metric are of magnitude 1) a Newtonian Poisson equation of the following form:

 

[math]\nabla^2 \phi =4\pi G\rho-\Lambda[/math]

 

This is clearly much simpler than the previous form we considered. We can now apply Gauss' Law to a spherically symmetric mass to obtain a modified version of Newton's Law:

 

[math]\nabla^2 \phi=\nabla \cdot (\nabla\phi )=\nabla \cdot(-\mathbf{g})[/math]

 

[math]\int \nabla\cdot \mathbf{g}dV=\int \mathbf{g}\cdot d\mathbf{A}=\int (-4\pi G \rho +\Lambda )dV[/math]

 

[math]g\int dA=-4\pi G\int \rho dV + \int \Lambda dV[/math]

 

[math]4\pi r^2g=-4\pi GM + \frac{4}{3}\pi r^3 \Lambda [/math]

 

[math]g=-\frac{GM}{r^2} + \frac{\Lambda r}{3} [/math]

Edited December 15, 2012 by elfmotat

[JMessenger]

 

The form of the field equations with zero cosmological constant are:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}[/math]

 

Compare this with the analogous Poisson equation for Newtonian gravity:

 

[math]\nabla^2 \phi =4\pi G\rho [/math]

 

 

The field equations with nonzero CC are:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda[/math]

 

Because the metric is analogous to the Newtonian potential, this suggests a modified Poisson equation of the form:

 

[math]\nabla^2 \phi =4\pi G\rho-\phi \Lambda[/math]

 

 

However, if we assume the gravitational field is weak (which is the only time when Newtonian gravity is accurate anyway) and we linearize the field equations with [math]g_{\mu \nu }=\eta_{\mu \nu }+h_{\mu \nu }[/math] and [math]|h_{\mu \nu}|\ll 1[/math], then we arrive at the following field equations:

 

[math]G_{\mu \nu}=\kappa T_{\mu \nu}-\eta_{\mu \nu}\Lambda[/math]

 

where [math]G_{\mu \nu}=-\frac{1}{2} \partial^{\alpha} \partial_{\alpha}(h_{\mu \nu}-\frac{1}{2}\eta_{\mu \nu} h^{\sigma}_{~\sigma })[/math], though the form of the Einstein Tensor isn't really important for our purposes. These field equations imply (because the relevant components of the Minkowski metric are of magnitude 1) a Newtonian Poisson equation of the following form:

 

[math]\nabla^2 \phi =4\pi G\rho-\Lambda[/math]

 

This is clearly much simpler than the previous form we considered. We can now apply Gauss' Law to a spherically symmetric mass to obtain a modified version of Newton's Law:

 

[math]\nabla^2 \phi=\nabla \cdot (\nabla\phi )=\nabla \cdot(-\mathbf{g})[/math]

 

[math]\int \nabla\cdot \mathbf{g}dV=\int \mathbf{g}\cdot d\mathbf{A}=\int (-4\pi G \rho +\Lambda )dV[/math]

 

[math]g\int dA=-4\pi G\int \rho dV + \int \Lambda dV[/math]

 

[math]4\pi r^2g=-4\pi GM + \frac{4}{3}\pi r^3 \Lambda [/math]

 

[math]g=-\frac{GM}{r^2} + \frac{\Lambda r}{3} [/math]

 

Thanks, I need to compare this to something I tried before.

 

The problem I have with this derivation is that [math]G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda[/math] is based on a "redefintion" of the Einstein tensor. Meaning we are changing the equation in order to satisfy our physical understanding instead of following the equation through rigorously in order to see what it predicts. General Relativity interprets the field equation

[math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=G_{\mu\nu}-g_{\mu \nu}\Lambda=0[/math] when [math]g_{\mu \nu}=diag[-1,1,1,1][/math] AND [math]\Lambda=0[/math]

 

when in actuality it states

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

[math]g_{\mu \nu}=diag[-1,1,1,1][/math] for [math]-\infty\leq\Lambda\leq\infty[/math] (including [math]\Lambda=0[/math]. Vacuum solutions, deSitter space etc. all appear to be artifacts based on a redefinition.

 

One can arrive at a Newtonian approximation either way, but only one seems to provide a definite radius when gravity no longer appears to be attractive.

[elfmotat]

 

Thanks, I need to compare this to something I tried before.

 

The problem I have with this derivation is that [math]G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda[/math] is based on a "redefintion" of the Einstein tensor. Meaning we are changing the equation in order to satisfy our physical understanding instead of following the equation through rigorously in order to see what it predicts.

 

I'm not really sure what you mean. I've only ever seen one definition of the Einstein tensor: [math]G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R[/math].

 

 

 

 

General Relativity interprets the field equation

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

 

Those aren't the field equations, and that equation is wrong.

 

 

 

 

when [math]g_{\mu \nu}=diag[-1,1,1,1][/math] AND [math]\Lambda=0[/math]

 

What do you mean? Why are you assuming the metric is the Minkowski metric? The purpose of the field equations is that it's relation between the metric and the energy-momentum distribution of spacetime (the stress-energy tensor), analogous to how Poisson's equation gives a relation between the potential and the mass distribution in space.

 

Also, why are you including the CC term in your equation if you're letting it equal zero anyway?

 

 

 

when in actuality it states

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

[math]g_{\mu \nu}=diag[-1,1,1,1][/math] for [math]-\infty\leq\Lambda\leq\infty[/math] (including [math]\Lambda=0[/math].

 

That's wrong and doesn't make any sense.

 

 

 

 

Vacuum solutions, deSitter space etc. all appear to be artifacts based on a redefinition.

 

One can arrive at a Newtonian approximation either way, but only one seems to provide a definite radius when gravity no longer appears to be attractive.

 

Again, I have no idea what you're talking about. What "redefinition?" I've never heard of anything like this in all of the textbooks and papers that I've read.

[JMessenger]

Again, I have no idea what you're talking about. What "redefinition?" I've never heard of anything like this in all of the textbooks and papers that I've read.

 

We should probably tackle this first. I am fairly sure you have, it just seems to get glossed over. Misner, Thorne and Wheeler Gravitation page 410 :

 

Denote a new, modified lefthand side by "G", with quotation marks to avoid confusion with the standard Einstein tensor. To abandon [math]\nabla \cdot "G" \equiv 0[/math] is impossible on dynamic grounds (see 17.2). To change the symmetry or rank of "G" is impossible on mathematical grounds, since "G" must be equated to T. To let "G" be nonlinear in the Riemann would vastly complicate the theory. To construct "G" from anything except Riemann and g would make "G" no longer a measure of spacetime geometry and would thus violate the spirit of the theory. After much anguish, one concludes that the assumption which one might drop with the least damage to the beauty and spirit of the theory is assumption (1), that "G" vanish when spacetime is flat. But even dropping this assumption is painful; (1) although "G" might still be in some sense a measure of geometry, it can no longer be a measure of curvature; and (2) flat, empty spacetime will no longer be compatible with the geometrodynamic law ([math]"G"\neq 0[/math] in flat, empty space, where T=0). Nevertheless, these consequences were less painful to Einstein than a dynamic universe.

The only tensor that satisfies conditions (2) and (3) [with (1) abandoned] is the Einstein tensor plus a multiple of the metric:[math]``G_{\alpha\beta}"=R_{\alpha\beta}-\frac{1}{2}Rg_{\alpha\beta}+\Lambda g_{\alpha\beta}=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math].

The problem I have with this is that, in spite of everything stated here, (1)[math]``G_{\alpha\beta}"=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math].fundamentally violates the summation properties of tensors (and even matrices for that matter) and (2) [math]R_{\alpha\beta}-\frac{1}{2}Rg_{\alpha\beta}=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math] is, as ajb points out in post #32,

 

the most general local, coordinate-invariant, divergenceless, symmetric, two-index tensor one can construct using only the metric as well as its first and second derivatives.

(substituing the Einstein tensor back in). Note that Einstein did not "invent" the cosmological constant, he just proposed a physical use for it (based on the Poisson equation).

While the Einstein tensor and geometrodynamics may be complicated, the cosmological constant with a metric of [-1,1,1,1] is the simplest tensor possible (a simple Rank 0 tensor) and very easy to understand.

 

I have posted quite a bit in earlier posts, but don't mind reposting if it helps to illustrate my point.

[elfmotat]

 

We should probably tackle this first. I am fairly sure you have, it just seems to get glossed over. Misner, Thorne and Wheeler Gravitation page 410 :

 

I'm not really sure what your point is. MTW defines a new tensor which is equal to the Einstein tensor plus the CC term. So what? This doesn't change any physics like you claimed in your last post, it just changes the notation a bit. The form of the field equations remains the same.

 

 

 

 

 

The problem I have with this is that, in spite of everything stated here, (1)[math]``G_{\alpha\beta}"=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math].fundamentally violates the summation properties of tensors (and even matrices for that matter)

 

 

No, it certainly doesn't.

 

 

and (2) [math]R_{\alpha\beta}-\frac{1}{2}Rg_{\alpha\beta}=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math] is, as ajb points out in post #32,

 

This equation is only correct if you are using MTW's new "G" tensor in place of the usual Einstein tensor.

 

 

 

(substituing the Einstein tensor back in). Note that Einstein did not "invent" the cosmological constant, he just proposed a physical use for it (based on the Poisson equation).

While the Einstein tensor and geometrodynamics may be complicated, the cosmological constant with a metric of [-1,1,1,1] is the simplest tensor possible (a simple Rank 0 tensor) and very easy to understand.

 

It's quite useless (and needlessly non-general) to declare the metric to be something specific in a general tensor equation. It's also impossible for the metric to be globally flat (which you are declaring it to be) with nonzero CC.

[JMessenger]

 

I'm not really sure what your point is. MTW defines a new tensor which is equal to the Einstein tensor plus the CC term. So what? This doesn't change any physics like you claimed in your last post, it just changes the notation a bit. The form of the field equations remains the same.

 

I agree that the form stays the same, but I disagree it doesn't change the physics. I have no problem with the full field equation, but there is a systemic error in the derivation of GR. It all relies on what your definition of a vector is.

 

 

 

No, it certainly doesn't.

 

Can you show me in mathematical detail how it does not?

 

 

 

This equation is only correct if you are using MTW's new "G" tensor in place of the usual Einstein tensor.

 

Now we are getting somewhere, as this seems to be a falsifiable. Perhaps the sources I am reading where this is stated are incorrect Can you point me to a full historically accepted derivation of the equation? I take my source as quoted below from "The Collected Papers of Albert Einstein" The Foundations of General Theory of Relativity vol. 6

 

 

It must be pointed out that there is only a minimum of arbitrariness in the choice of these equations. For besides G_uv there is no tensor of second rank which is formed from the g_uv and its derivatives, contains no derivations higher than second, and is linear in these derivatives. *

(The * refers to a footnote which reads)

Properly speaking, this can only be affirmed of the tensor [math]G_{\mu\nu} + \lambda g_{\mu\nu}g^{\alpha\beta}G_{\alpha\beta}[/math], where [math]\lambda[/math] is a constant. If, however, we set this tensor =0, we come back again to the equation G_uv=0.

 

I know that Einstein made errors in his papers and that these have been corrected over the years. Can you show me where his footnote was refuted?

 

 

 

It's quite useless (and needlessly non-general) to declare the metric to be something specific in a general tensor equation. It's also impossible for the metric to be globally flat (which you are declaring it to be) with nonzero CC.

 

We shall see.

[elfmotat]

I agree that the form stays the same, but I disagree it doesn't change the physics.

 

 

How could it possibly change the physics? The CC term is simply absorbed into a new tensor.

 

[math]"G_{\mu \nu}"=G_{\mu \nu }+g_{\mu \nu}\Lambda =\kappa T_{\mu \nu}[/math]

 

There's nothing special here, it's just simple algebra.

 

 

I have no problem with the full field equation, but there is a systemic error in the derivation of GR. It all relies on what your definition of a vector is.

 

 

What are you talking about? The definition of a vector in the context of differential geometry is standard. There's absolutely nothing wrong with the derivation of GR - all you have to do is vary the Hilbert action with respect to the metric. Obviously GR is incomplete because it doesn't get along with QFT, but that doesn't mean there's anything "wrong" with it. It's a mathematically consistent theory.

 

 

Can you show me in mathematical detail how it does not?

 

Can you show me in detail how it does? Because I have no idea what you're talking about. How could a tensor "violate the summation properties of tensors?" I don't even know what that means.

 

 

Now we are getting somewhere, as this seems to be a falsifiable. Perhaps the sources I am reading where this is stated are incorrect Can you point me to a full historically accepted derivation of the equation? I take my source as quoted below from "The Collected Papers of Albert Einstein" The Foundations of General Theory of Relativity vol. 6

 

Again, this is just basic algebra. The Einstein tensor is defined as [math]G_{\mu \nu }\equiv R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R[/math]. So we see that the only way for your equation could possibly be correct is if:

1. The cosmological constant is set to zero.
2. [math]G_{\mu \nu }[/math] is something other than the Einstein tensor. The "G" tensor defined by MTW would work here, so long as the sign of the CC is reversed.

I know that Einstein made errors in his papers and that these have been corrected over the years. Can you show me where his footnote was refuted?

 

It wasn't, and I don't know why it would need to be. This is just math here; there's nothing to refute. I'm also unsure how it's relevant.

 

 

We shall see.

 

Um, no. It's not an opinion, you're needlessly restricting the equations to the point where they're useless. Since you made the metric flat, you've already defined all of the information you could possibly extract from the field equations. With a flat metric, the curvature (Riemann, Ricci, and Einstein tensors) is identically zero, the stress-energy tensor is zero, and the cosmological constant has to be zero (which is physically wrong anyway).

 

Nothing about this is interesting or useful.

Edited December 15, 2012 by elfmotat

[JMessenger]

How could it possibly change the physics? The CC term is simply absorbed into a new tensor.

 

[math]"G_{\mu \nu}"=G_{\mu \nu }+g_{\mu \nu}\Lambda =\kappa T_{\mu \nu}[/math]

 

There's nothing special here, it's just simple algebra.

 

What are you talking about? The definition of a vector in the context of differential geometry is standard. There's absolutely nothing wrong with the derivation of GR - all you have to do is vary the Hilbert action with respect to the metric. Obviously GR is incomplete because it doesn't get along with QFT, but that doesn't mean there's anything "wrong" with it. It's a mathematically consistent theory.

 

The standard definition of any vector I have found in differential geometry basically boils down to [math]f'_1[/math]. Since it is known that [math](f_1+C)'[/math] or [math](f_1-C)'[/math] make no difference then it is justified that there is no reason to include a constant of integration in the standard definition. The problem is that [math]f'_1=(C-f_2)'[/math] both give exactly equivalent vectors. Both also give exactly the same answer to the Poisson equation and Gauss' theorem.

 

 

Can you show me in detail how it does? Because I have no idea what you're talking about. How could a tensor "violate the summation properties of tensors?" I don't even know what that means.

 

In classical GR, the Einstein tensor "vanishes" with no mass or energy present so let's assume that means all components are zero_. We should have [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=G_{\mu\nu}=diag[0,0,0,0][/math]. No?

Now let us "add" the cosmological constant on, but here is what I don't understand. The metric for flat spacetime is proven to be [math]g_{\mu\nu}=diag[-1,1,1,1][/math] (or similar pattern such as +,-,-,-) so when we insert the cosmological constant we are getting [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=diag[0,0,0,0]=G_{\mu\nu}-g_{\mu\nu}\Lambda=diag[0,0,0,0]-diag[-\Lambda,\Lambda,\Lambda,\Lambda][/math]

 

It would seem that the redefinition that MTW has shown is since we want to simply allow G_uv to "vanish" (and by definition the stress energy tensor [math]\kappa T_{\mu\nu}[/math]) with some value of the cosmological constant present, even if [math]0\neq0-\Lambda[/math]. In other words, we want to be able to allow matter and energy to "vanish" but keep the cosmological constant present so we can use it for other purposes.

 

 

 

 

 

Again, this is just basic algebra. The Einstein tensor is defined as . So we see that the only way for your equation could possibly be correct is if:

1. The cosmological constant is set to zero.

1. is something other than the Einstein tensor. The "G" tensor defined by MTW would work here, so long as the sign of the CC is reversed.

 

It appears that he "defined" the field equation after​ setting the cosmological constant to zero. I see where he also "defined" is [math]G_{\mu\nu} \equiv \kappa T_{\mu\nu}[/math].

 

I agree that [math]G_{\mu\nu}[/math] would have to be something other than the Einstein tensor. I see no reason that it cannot become a Rank 0 tensor for flat spacetime, just like the cosmological constant, instead of "vanishing".

 

 

 

It wasn't, and I don't know why it would need to be. This is just math here; there's nothing to refute. I'm also unsure how it's relevant.

 

Einstein was stating that even though G_uv vanished for no mass or energy, the choice of the equation also isn't arbitrary and he mentioned that one must set the cosmological constant to zero (meaning that the cosmological constant is already present in the field equation and then is set to zero). I stated the same thing previously, which you did not seem to accept.

 

 

 

Um, no. It's not an opinion, you're needlessly restricting the equations to the point where they're useless. Since you made the metric flat, you've already defined all of the information you could possibly extract from the field equations. With a flat metric, the curvature (Riemann, Ricci, and Einstein tensors) is identically zero, the stress-energy tensor is zero, and the cosmological constant has to be zero (which is physically wrong anyway).

Who said I was restricting them to flat? We haven't gotten into what happens when they aren't. My point is that for a flat metric it cannot be definitively stated that the Einstein tensor "vanishes". It could just as well become a Rank 0 tensor.

 

 

 

Nothing about this is interesting or useful.

 

That is your opinion now but allow me to attempt to sway it. The problems that surround the cosmological constant aren't small ones. I am of the opinion that they have a common thread running through them. 10¹²⁰ orders of magnitude off of QFT predictions, seems to predict a vacuum energy that repulses attractive gravity from normal matter and energy. These problems almost make it seem we are measuring energy not only differently than QFT, but perhaps even from the wrong end. When you set the standard definition of a vector to [math] f'_1 [/math] without also allowing for the possibility of [math](C-f_2)'[/math], you are a priori defining your function relative to a constant energy level of zero.

12 years after Einstein passed, Zel'dovich came up with the comparison of the CC being some type of "vacuum energy" and dark energy wasn't discovered until 1998. I don't understand why people believe that the great thinkers in science must be saddled with their interpretations and wouldn't drastically re-evaluate their assumptions were they given the opportunity to know about these observations.

Edited December 16, 2012 by JMessenger

[elfmotat]

 

The standard definition of any vector I have found in differential geometry basically boils down to [math]f'_1[/math]. Since it is known that [math](f_1+C)'[/math] or [math](f_1-C)'[/math] make no difference then it is justified that there is no reason to include a constant of integration in the standard definition.

 

A vector is defined as something that transforms like a vector:

 

[math]a^{A}=\frac{\partial x^A}{\partial x^a}a^a[/math]

 

Of course adding or subtracting a constant vector is going to make a difference. What gave you the idea that it didn't? For example:

 

 

 

They are both constant vectors, but the decision of whether to add or subtract them makes a huge difference.

 

 

 

The problem is that [math]f'_1=(C-f_2)'[/math] both give exactly equivalent vectors. Both also give exactly the same answer to the Poisson equation and Gauss' theorem.

 

In what way could you possibly consider the above vectors equivalent? There are no vectors in the Poisson equation to begin with, so I'm not sure what you're talking about there. As far a Gauss' theorem goes, that's just a gauge invariance. Are you equally troubled by the fact that in classical mechanics the potential [math]V[/math] yields the the same equations of motion as the potential [math]V+C[/math]? What about in classical electrodynamics, where the magnetic vector potential and electric scalar potential yield the same equations of motion under [math]\mathbf{A}\rightarrow \mathbf{A}+\nabla f[/math] and [math]\phi \rightarrow \phi -\frac{\partial f}{\partial t}[/math] ? Or the fact a Lagrangian [math]L[/math] produces the same equations of motion as a function of that Lagrangian: [math]f(L)[/math]? I could go on, but I think you get the idea.

 

If you really have a problem with gauge invariance then you have a lot of work ahead of you because you're going to have to rewrite the entirety of mathematics and physics.

 

 

 

In classical GR, the Einstein tensor "vanishes" with no mass or energy present so let's assume that means all components are zero_.

 

As long as the cosmological constant is set to zero, yes.

 

 

 

We should have [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=G_{\mu\nu}=diag[0,0,0,0][/math]. No?

 

No. All of the components are zero, not just the diagonal ones: [math]G_{\mu \nu}=0[/math]

 

 

 

Now let us "add" the cosmological constant on, but here is what I don't understand. The metric for flat spacetime is proven to be [math]g_{\mu\nu}=diag[-1,1,1,1][/math] (or similar pattern such as +,-,-,-) so when we insert the cosmological constant we are getting [math]R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}=diag[0,0,0,0]=G_{\mu\nu}-g_{\mu\nu}\Lambda=diag[0,0,0,0]-diag[-\Lambda,\Lambda,\Lambda,\Lambda][/math]

 

I'm starting to think you have no idea what the field equations are, nor how they work. By making the metric flat you already assumed that the cosmological constant was zero! As I already showed you in a previous post, if the CC is nonzero then the vacuum field equations reduce to [math]R_{\mu \nu}=g_{\mu \nu}\Lambda [/math]. Even in a vacuum the spacetime is curved, so the metric is obviously not Minkowski. You can't have it both ways: either the metric is flat and the CC is zero, or the CC is nonzero and you have a curved metric.

 

 

 

I see where he also "defined" is [math]G_{\mu\nu} \equiv \kappa T_{\mu\nu}[/math].

 

Those are the field equations, not a definition of the Einstein tensor.

 

 

 

I see no reason that it cannot become a Rank 0 tensor for flat spacetime, just like the cosmological constant, instead of "vanishing".

 

I don't even know what that means. How can a rank-2 tensor "become" a scalar without contractions?

 

 

 

Einstein was stating that even though G_uv vanished for no mass or energy, the choice of the equation also isn't arbitrary and he mentioned that one must set the cosmological constant to zero (meaning that the cosmological constant is already present in the field equation and then is set to zero). I stated the same thing previously, which you did not seem to accept.

 

That Einstein quote has absolutely nothing to do with the cosmological constant. His footnote refers to:

 

"It must be pointed out that there is only a minimum of arbitrariness in the choice of these equations. For besides G_uv there is no tensor of second rank which is formed from the g_uv and its derivatives, contains no derivations higher than second, and is linear in these derivatives."

 

What his footnote means is that his above description is technically valid for any tensor [math]R_{\mu \nu}+\lambda Rg_{\mu \nu}[/math] for arbitrary constant [math]\lambda[/math]. The reason [math]\lambda=-1/2[/math] is chosen is because of the Bianchi identities. If any other value for [math]\lambda[/math] is used then the equation [math]\nabla_{\nu}G^{\mu \nu}[/math] wouldn't be valid, and so neither would [math]\nabla_{\nu}T^{\mu \nu}[/math] and then we wouldn't have energy/momentum conservation.

 

 

 

Who said I was restricting them to flat?

 

You did, when you made the metric Minkowski.

 

 

 

My point is that for a flat metric it cannot be definitively stated that the Einstein tensor "vanishes".

 

It does by definition. The Einstein tensor is a measure of curvature. If spacetime is flat then there is, by definition, no curvature.

 

 

 

It could just as well become a Rank 0 tensor.

 

Again, I have no idea what that means.

 

 

That is your opinion now but allow me to attempt to sway it. The problems that surround the cosmological constant aren't small ones. I am of the opinion that they have a common thread running through them. 10¹²⁰ orders of magnitude off of QFT predictions, seems to predict a vacuum energy that repulses attractive gravity from normal matter and energy. These problems almost make it seem we are measuring energy not only differently than QFT, but perhaps even from the wrong end. When you set the standard definition of a vector to [math] f'_1 [/math] without also allowing for the possibility of [math](C-f_2)'[/math], you are a priori defining your function relative to a constant energy level of zero.

12 years after Einstein passed, Zel'dovich came up with the comparison of the CC being some type of "vacuum energy" and dark energy wasn't discovered until 1998. I don't understand why people believe that the great thinkers in science must be saddled with their interpretations and wouldn't drastically re-evaluate their assumptions were they given the opportunity to know about these observations.

 

I'm not really sure what your responding to here. I was simply saying that the field equations when the metric is defined as Minkowski are not useful or interesting.

Edited December 16, 2012 by elfmotat

[JMessenger]

Trying to post, but it won't allow me to post any images (states "Sorry, but you have posted more images than you are allowed to")

 

 

In what way could you possibly consider the above vectors equivalent?

 

 

Let me try to find some common ground with a simple example of answering your question of how [math]f'_1=(C-f_2)'[/math] could possibly give equivalent vectors.

 

Take a simple two dimensional scalar field (note that this isn't a matrix)

 

 

 

at each point is a Rank 0 tensor and no higher.

 

If we sum this with another field, one that has not only a Rank 0 tensor present, but also a Rank 1 tensor at each point, we get

 

 

 

 

note that the sum of these two scalar fields do not change the value of the Rank 1 tensor, and should the second scalar field level out to all equivalent scalar values, the Rank 1 tensor has "vanished".

As an aside:

I don't even know what that means. How can a rank-2 tensor "become" a scalar without contractions?

 

a four dimensional scalar field that had varying scalar values would become a higher rank tensor. If the scalar values settled to all equivalent values then it can be represented by only a Rank 0 tensor at all points.

back to the discussion...

 

Let's call the summation of these two [math](C_1+f_1)'[/math]

 

Now let's take (C_2-f_2)' but for the first Rank 0 tensors, lets model them as orthogonal vectors with the sum of zero (kind of like a pressure):

 

 

 

The derivative of the summation of [math](C_2-f_2)'[/math] gives the exactly equivalent answer as [math](C_1+f_1)'[/math]. If all I have is the vector result, how do I tell which it came from?

 

 

I could take every vector in the diagram you posted and replace each

 

 

 

with

 

 

 

 

and there is no way of determining whether the original function was f₁ or f_2.

 

 

There are no vectors in the Poisson equation to begin with, so I'm not sure what you're talking about there.

 

I am not purposely trying to pick and choose which statements to respond to, but I do not know what to make of this statement.

The first equation in http://eprints.ma.man.ac.uk/894/02/0-19-852868-X.pdf

is [math]-\nabla^2\mu=f[/math] which is

[math]-\nabla\cdot\nabla\mu=f[/math].

 

In the simplest case [math]\nabla \mu[/math] is defined as the gradient of [math]\mu[/math] which, by definition, is in the direction of the greatest increase, meaning a vector. I don't understand your statement.

 

 

 

That Einstein quote has absolutely

nothing

to do with the cosmological constant. His footnote refers to:

 

 

 

 

I will have to re-read and consider this more in depth. I could be wrong in what I thought his meaning was.

(interestingly, when I quoted you in full in that segment, it seemed to create errors in the posting system)

 

 

Are you equally troubled by the fact that in classical mechanics the potential

yields the the same equations of motion as the potential ...(causes errors)

 

 

Yes, I get the idea and it intrigues me. It also interests me that the Einstein tensor is a dynamical term but the cosmological constant isn't. Similarly, it also intrigues me if I have a constant [math]C_{\rho}[/math] density and pressure that are reduced down to a dynamic residual [math]\rho_{res}[/math] density and pressure, how would I tell the difference between the standard stress-energy tensor of a perfect fluid

[math]T_{\mu\nu}=(\rho + p)u_{\mu} u_{\nu}+ p g_{\mu\nu}[/math]

and this alternative

[math]T_{\mu\nu}=((C_{\rho}-\rho_{res}) + (C_p-p_{res})) u_{\mu}u_{\nu}+(C_p-p_{res})g_{\mu\nu}[/math].

[elfmotat]

Trying to post, but it won't allow me to post any images (states "Sorry, but you have posted more images than you are allowed to")

 

Let me try to find some common ground with a simple example of answering your question of how [math]f'_1=(C-f_2)'[/math] could possibly give equivalent vectors.

 

Take a simple two dimensional scalar field (note that this isn't a matrix)

 

 

 

at each point is a Rank 0 tensor and no higher.

 

If we sum this with another field, one that has not only a Rank 0 tensor present, but also a Rank 1 tensor at each point, we get

 

 

 

 

note that the sum of these two scalar fields do not change the value of the Rank 1 tensor, and should the second scalar field level out to all equivalent scalar values, the Rank 1 tensor has "vanished".

 

back to the discussion...

 

Let's call the summation of these two [math](C_1+f_1)'[/math]

 

Now let's take (C_2-f_2)' but for the first Rank 0 tensors, lets model them as orthogonal vectors with the sum of zero (kind of like a pressure):

 

 

 

The derivative of the summation of [math](C_2-f_2)'[/math] gives the exactly equivalent answer as [math](C_1+f_1)'[/math]. If all I have is the vector result, how do I tell which it came from?

 

 

I could take every vector in the diagram you posted and replace each

 

 

 

with

 

 

 

 

and there is no way of determining whether the original function was f₁ or f_2.

 

 

Yes, I get the idea and it intrigues me. It also interests me that the Einstein tensor is a dynamical term but the cosmological constant isn't. Similarly, it also intrigues me if I have a constant [math]C_{\rho}[/math] density and pressure that are reduced down to a dynamic residual [math]\rho_{res}[/math] density and pressure, how would I tell the difference between the standard stress-energy tensor of a perfect fluid

[math]T_{\mu\nu}=(\rho + p)u_{\mu} u_{\nu}+ p g_{\mu\nu}[/math]

and this alternative

[math]T_{\mu\nu}=((C_{\rho}-\rho_{res}) + (C_p-p_{res})) u_{\mu}u_{\nu}+(C_p-p_{res})g_{\mu\nu}[/math].

 

I can't make heads nor tails of any of your pictures. I have no idea what they're supposed to represent.

 

 

Again, from what I could tell, you seem to have a problem with gauge invariance. It's a very simple concept really: the derivative of a constant is zero. So taking the derivative of a function is equivalent to taking the derivative of a function plus a constant. In other words, the function f'(x) is invariant under the gauge transformation f(x)→f(x)+C. There's nothing magical, inconsistent, or otherwise here.

 

 

 

 

As an aside:

a four dimensional scalar field that had varying scalar values would become a higher rank tensor. If the scalar values settled to all equivalent values then it can be represented by only a Rank 0 tensor at all points.

 

 

 

That's simply not true. A scalar field can have infinite degrees of freedom. It can have a different value at every point in space. The key idea is that if you plug in a point x^u the scalar function (we'll call f) returns a single number f(x^u)=k₁. If you were to plug in some other point, say y^u, then you would get another number f(y^u)=k₂. It is not necessarily the case that k₁=k₂. An example of a scalar field is a temperature field, where each point in space has a different temperature.

 

In order to have a rank-1 tensor (vector or one-form) field, when you plug a point x^u into a vector a^v, you get n numbers back where n is the dimension of the space. In three dimensions for example: a^v(x^u)=(k₁,k₂,k₃).

 

In order to have a rank-2 tensor field, when you plug x^u into the field you get back n² numbers. I.e. in 3D: b^av(x^u) = ((k₁₁,k₁₂,k₁₃), (k₂₁,k₂₂,k₂₃), (k₃₁,k₃₂,k₃₃)).

 

 

A rank-2 tensor field can't magically turn into a scalar field; that makes literally no sense.

 

 

 

 

The first equation in http://eprints.ma.man.ac.uk/894/02/0-19-852868-X.pdf

is [math]-\nabla^2\mu=f[/math] which is

[math]-\nabla\cdot\nabla\mu=f[/math].

 

In the simplest case [math]\nabla \mu[/math] is defined as the gradient of [math]\mu[/math] which, by definition, is in the direction of the greatest increase, meaning a vector. I don't understand your statement.

 

 

What I mean is that, without changing the form of the equation to something that resembles Gauss' Law, the Poisson equation is not a vector equation. It states that the Laplacian of some scalar function is equal to some other function. No vectors here.

[JMessenger]

 

I can't make heads nor tails of any of your pictures. I have no idea what they're supposed to represent.

 

That is a bummer.

 

 

 

Again, from what I could tell, you seem to have a problem with gauge invariance. It's a very simple concept really: the derivative of a constant is zero. So taking the derivative of a function is equivalent to taking the derivative of a function plus a constant. In other words, the function f'(x) is invariant under the gauge transformation f(x)→f(x)+C. There's nothing magical, inconsistent, or otherwise here.

 

 

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]

 

 

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

 

That's simply not true. A scalar field can have infinite degrees of freedom. It can have a different value at every point in space. The key idea is that if you plug in a point x^u the scalar function (we'll call f) returns a single number f(x^u)=k₁. If you were to plug in some other point, say y^u, then you would get another number f(y^u)=k₂. It is not necessarily the case that k₁=k₂. An example of a scalar field is a temperature field, where each point in space has a different temperature.

 

Agreed.

 

In order to have a rank-1 tensor (vector or one-form) field, when you plug a point x^u into a vector a^v, you get n numbers back where n is the dimension of the space. In three dimensions for example: a^v(x^u)=(k₁,k₂,k₃).

 

In order to have a rank-2 tensor field, when you plug x^u into the field you get back n² numbers. I.e. in 3D: b^av(x^u) = ((k₁₁,k₁₂,k₁₃), (k₂₁,k₂₂,k₂₃), (k₃₁,k₃₂,k₃₃)).

 

 

A rank-2 tensor field can't magically turn into a scalar field; that makes literally no sense.

 

 

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]

 

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).

 

 

 

 

 

What I mean is that, without changing the form of the equation to something that resembles Gauss' Law, the Poisson equation is not a vector equation. It states that the Laplacian of some scalar function is equal to some other function. No vectors here.