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

[JMessenger]

R_uv -(1/2) Rg_uv=G_uv given true statement

 

R_uv -(1/2) Rg_uv+Lambda g_uv=G_uv

G_uv+Lambda g_uv=G_uv false statement

 

Going to be giving a web lecture on this next month to those interested in alternate theories, what is required to make the false statement true and what it means physically. Questions more than welcome here.

[ajb]

 

G_uv+Lambda g_uv=G_uv false statement

 

Unless [math]\Lambda=0[/math].

 

Anyway, it looks to me like all you have is a redefinition of the Einstein tensor to include the cosmological constant.

 

Going to be giving a web lecture on this next month to those interested in alternate theories, what is required to make the false statement true and what it means physically. Questions more than welcome here.

 

What alternative theories are you interested in and why?

[JMessenger]

Unless [math]\Lambda=0[/math].

 

True.

 

 

Anyway, it looks to me like all you have is a redefinition of the Einstein tensor to include the cosmological constant.

True, but it isn't mine. MTW state the redefinition as [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]. They consider that with the cosmological constant the theory is no longer geometric.

 

 

What alternative theories are you interested in and why?

 

Alternative geometric theories of gravity in order to understand how to derive the magnitude of the cosmological constant. It seems to me that there is a simple missing case concerning parallel transport of vectors where the equation can still be considered as "geometric" even with a constant multiple of the metric included.

[JMessenger]

The first fundamental theorem of calculus states that, if is continuous on the closed interval and is the indefinite integral of on , then

 

[math]\int\limits_{b}^{a} f(x)dx=F(a)-F(b)[/math]

 

 

or we could write

[math]\int f(x)dx=F(x)+C[/math]

 

 

where C is a constant of integration, which means that there are an infinite number of antiderivatives of .

We can certainly simply choose C=0 but we need to also keep in mind that the theorem makes no assumptions on the magnitudes of or C.

 

By this I mean consider the following two plots:

 

 

Since both plots have equivalent areas and if [math]f'_1=(C-f_2)'[/math]

then neither the Poisson equation nor Gauss' theorem gives different answers between the two plots. By extension, I postulate that this general equivalency should also apply to differentiable manifolds and that the "Einstein tensor" is no more required to vanish for flat space-time with a multiple of the metric included than f₂ is required with C. This may require unimodular assumptions but even should this turn out to not be correct (but if not, why?), it should provide some interesting comparisons with classical General Relativity and why we set vacuum as the lowest energy state =0.

[ajb]

True, but it isn't mine. MTW state the redefinition as [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]. They consider that with the cosmological constant the theory is no longer geometric.

 

I think the point is that the Einstein tensor, without the cosmological constant, is built out of geometric information only. That is, once we know the metric we know the Einstein tensor. Moreover the Einstein tensor does have a geometric interpretation, though it is not as clear as the other curvature tensors. have a look at http://math.ucr.edu/home/baez/einstein/

 

The problem is that, as far as I know, no classical geometric argument can give us the cosmological constant. We know it is small and most likely an artefact of quantum theory, but this is not a resolved problem by far.

 

 

Alternative geometric theories of gravity in order to understand how to derive the magnitude of the cosmological constant.

 

A I said, I cannot imagine it is as simple as classical differential geometry.

 

 

It seems to me that there is a simple missing case concerning parallel transport of vectors where the equation can still be considered as "geometric" even with a constant multiple of the metric included.

 

I don't quite follow what you are saying.

 

By extension, I postulate that this general equivalency should also apply to differentiable manifolds and that the "Einstein tensor" is no more required to vanish for flat space-time with a multiple of the metric included than f₂ is required with C. This may require unimodular assumptions but even should this turn out to not be correct (but if not, why?), it should provide some interesting comparisons with classical General Relativity and why we set vacuum as the lowest energy state =0.

 

I know that people have looked into the idea that the cosmological constant is some kind of integration constant. However, this will not predict the value as it will be arbitrary; unless you input suitable boundary conditions. Though it would make it clear that the cosmological constant is non-dynamical.

 

Details here are outside of my expertise.

Edited December 1, 2012 by ajb

[JMessenger]

I think the point is that the Einstein tensor, without the cosmological constant, is built out of geometric information only. That is, once we know the metric we know the Einstein tensor. Moreover the Einstein tensor does have a geometric interpretation, though it is not as clear as the other curvature tensors. have a look at http://math.ucr.edu/.../baez/einstein/

 

The problem is that, as far as I know, no classical geometric argument can give us the cosmological constant. We know it is small and most likely an artefact of quantum theory, but this is not a resolved problem by far.

True on the unresolved problem. Baez also goes into that on his cosmological constant page.

 

 

A I said, I cannot imagine it is as simple as classical differential geometry.

 

I doubt it is as simple as that either, but the cosmological constant argument was introduced via differential geometry. Einstein was aware that it showed up in the full field equation, but his actual initial argument for a static universe in his paper Cosmological Considerations on the General Theory of Relativity was based on a modification of the Poisson equation for Newtonian gravity. Slight modifications of this are still used to illustrate the bizarre effects the constant would seem to have in opposing gravity. 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]

I don't quite follow what you are saying.

Let me bypass that for now, and just point out that the reason MTW "redefined" the Einstein tensor such that [math]G_{\mu\nu}^{new}=G_{\mu\nu}^{old}+\Lambda g_{\mu\nu}[/math] is because it violates the summation properties of tensors if [math]\Lambda \neq 0 [/math]. I see how this makes it easier to transfer the physical assumptions concerning the old Einstein tensor to the new, but I don't see how this rises to the same rigor that was used previously. I agree that the two tensors on each side of the equation can't be the same, but this research is based on what the proposed effects would be if we followed the equation through explicitly. By this I mean, changing the symbols so as not to confuse ourselves, if we let [math]G_{\mu\nu}^{new}=-L_{\mu\nu}[/math] and [math]\Lambda g_{\mu\nu}=-\Omega g_{\mu\nu}[/math] to get [math]G_{\mu\nu}^{old}=\Omega g_{\mu\nu}-L_{\mu\nu}[/math]. It would require a better understanding of what we can say definitively when a tensor "vanishes".

 

 

 

I know that people have looked into the idea that the cosmological constant is some king of integration constant. However, this will not predict the value as it will be arbitrary; unless you input suitable boundary conditions. Though it would make it clear that the cosmological constant is non-dynamical.

 

Details here are outside of my expertise.

 

Anderson and Finkelstein wrote a paper on this. I tend to favor the concept they propose: Cosmological Constant and the Fundamental Length

[JMessenger]

Going to continue here a bit in case I can interest a couple people in the web conference...

 

Take for example a two dimensional scalar field (note that this isn't a matrix, it is field representing scalar values at points):

 

 

The first field represents a rank 0 tensor at each point, and the second field is a rank 1 tensor at each point. Note that the cosmological constant tensor Baez shows here is a four dimensional rank 0 tensor. Normally it is partial derivatives (this is where we get "fluxes" from) that occupy each component but in the case of a rank 0 tensor it is a constant, i.e. a scalar. There are no partial derivatives of a constant, so the constant scalar field results in no directional derivatives. However, we could model this as orthogonal vectors with a sum of zero, regardless of the magnitude of the scalar values. This is where the "pressure" analogy comes from that Baez speaks of:

 

 

Note that the sum of the two top tensors results in a rank 1 tensor (there are more specifics I believe under what are called Nuemann boundary conditions) but you can see that even should the second tensor scalar values settle to all zeroes or all constant values, the rank 1 tensor "vanishes" and becomes solely a rank 0 tensor. There is a problem on uniqueness of how we view that sum of these tensors in the following:

 

 

This is what I believe the full Einstein field equation is showing us. Geometrically, the only thing the normal Einstein tensor (a second rank tensor) seems to rely upon are the presence of partial derivatives present as components, but even should the Einstein tensor devolve to a rank 0 tensor with arbitrary values, the constant multiple of the metric (cosmological constant) would also allow cancellation of these, still resulting in no curvature.

 

Why is this important? The argument for Newtonian gravity is that for flat space-time and slow speeds, we only need to take into consideration g₀₀, which would simply be a -1 (Minkowski space). The theory is that in the weak field limit, a small perturbation from this -1 will come to represent, through the equivalence principle of inertial and gravitational mass, the Newtonian [math]\Phi[/math].

 

If we instead consider that this small perturbation would be equal to (vacuum) stress energy tensors equated to [math]\Omega g_{00}-L_{00}[/math], and taking in what I previously stated concerning the Poisson equation and Gauss' theorem, I will proceed to make some arguments concerning why we are looking for a repulsive vacuum energy that opposes attractive gravity. If you are into Lawrence Krauss lectures, starting at

he goes into that there does not seem to be any natural way to cancel the value of the cosmological constant. I will be arguing with that proposition in the web conference..

[JMessenger]

Wikipedia isn't good for a cited reference, but can be adequate for introduction to subjects.

 

If you are wondering about the vectors shown above and how they relate to manifolds, in this theory they are the vector space which you can introduce yourself to here: Vector Space

 

These vectors are what we currently understand to be forces, which is why a constant scalar field with orthogonal vectors would seem to be a "pressure".

 

Another thing to keep in mind is that what someone may define as a scalar would actually seem to be a magnitude in the tangent space. Take the examples given here. where it is stated "since the LHS is a vector and the RHS a scalar." As understood from the above post, the scalars |x_i| shown as components are actually magnitudes of partial derivatives of the curvature of the manifold. These scalar magnitudes are different from the absolute scalars shown in the above two dimensional field (which is why the Cosmological Constant in flat Minkowski space doesn't seem to make mathematical sense otherwise). If the vectors have no magnitudes then there is no force so it isn't something normally stated when introduced to the history of field theory. These absolute scalars in this theory could be previously argued as unobservable, however I will argue that evidence is building that they are required to exist in order to make sense of dark energy and the cosmological constant.

[JMessenger]

Attempting to disprove this theory, if you are able to help and catch any mistakes, much appreciated...

 

 

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

 

Einstein tensor vanishes if and only if the Ricci tensor vanishes, only for Minkowski space [math]g_{\mu\nu}=diag[-1,1,1,1][/math] and only if [math]Ricci=\Lambda g_{\mu\nu}[/math]:

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

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

[math]L_{\mu\nu}=\Omega g_{\mu\nu}[/math]

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

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

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

 

 

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

[ajb]

What is Omega exactly? This reminds me of an Einstein Manifold.

[JMessenger]

What is Omega exactly? This reminds me of an Einstein Manifold.

Yea, sorry about that. It isn't anything other than a constant multiple of the metric, same as Lambda. I just used a different symbol since most people that understand Lambda already have preconceived notions of its magnitude. That empirical magnitude is in reference to the magnitude of the Einstein tensor. Since magnitude is one of the cosmological constant problems, I am trying to take this back to just tensors, scalars and first principles.

 

 

In a nutshell, all I am stating is that for flat spacetime,

 

no cosmological constant, the Einstein tensor vanishes, G_uv->0

 

with cosmological constant, G_uv->Lambda g_uv

 

 

I just used different symbols above to attempt a proof.

 

 

 

 

(What happened to the LaTex?)

[math]LaTex[/math] (Edit: Nevermind, this one works)

[tex]LaTex[/tex]

[itex]LaTex[/itex]

 

What is Omega exactly? This reminds me of an Einstein Manifold.

 

I guess I would be stating that Ricci-flat manifolds and Einstein manifolds are actually the exact same thing.

 

Let me retract that last statement. I am proposing Ricci-flat manifolds and Einstein manifolds are the exact same thing for flat spacetime, but not curved.

 

(Wow, a "Reply to this topic" is added as an edit to the last post automatically)

Edited December 4, 2012 by JMessenger

[ajb]

So in vacco, the field equations can be written as

 

[math]R_{\mu \nu} = \frac{2 \Lambda}{n-2 } g_{\mu \nu}[/math].

 

Vaccum solutions to the field equation with vanishing cosmological constant are Ricci flat and speical cases of Einstein manifolds. If the cosmological constant is non-zero then the solution will not be Ricci flat, but just an Einstein manifold.

 

Look up de Sitter and anti de Sitter spaces as classical examples.

[JMessenger]

So in vacco, the field equations can be written as

 

[math]R_{\mu \nu} = \frac{2 \Lambda}{n-2 } g_{\mu \nu}[/math].

 

Vaccum solutions to the field equation with vanishing cosmological constant are Ricci flat and speical cases of Einstein manifolds. If the cosmological constant is non-zero then the solution will not be Ricci flat, but just an Einstein manifold.

 

Look up de Sitter and anti de Sitter spaces as classical examples.

 

Thanks, I have looked into it a bit but takes time to separate out the terminology from where it is derived in a proof. As in "constant scalar curvature" giving [math]R=4\Lambda[/math].

As an example, the weak field approximation for Newtonian gravity is normally derived from [math]g_{\mu\nu}=\eta_{\mu\nu}-\epsilon_{\mu\nu}[/math] where [math]\eta_{\mu\nu}=-1[/math] and [math]-\epsilon_{\mu\nu}[/math] is some tiny perturbation. Keeping normal GR notation, the way I am proposing would start out with flat space time as [math]-\Lambda^1+\Lambda^2=0[/math] instead of just a -1. A small perturbation of [math]\Lambda^2[/math] would be used to generate the field. A bit less exciting than deSitter space, but lots to go over in the future.

[JMessenger]

I had someone in another forum (where speculations aren't allowed so have to be careful how one phrases a question) pose the following, which is something I had not considered before:

 

 

What do you mean by "no proof"? The EFE including cosmological constant is actually easier, in a sense, to "prove" than the standard EFE, at least if you derive it from a Lagrangian. The Einstein-Hilbert Lagrangian, from which the standard EFE is derived by variation with respect to the metric (after integrating over all spacetime), is

1 16π R−g − − − √

Hilbert and Einstein claimed that this was the right Lagrangian because it was the most general one using no greater than second derivatives of the metric. But they hadn't realized that that was, strictly speaking, false: the most general such Lagrangian also includes a constant term:

1 16π (R+2Λ)−g − − − √

Varying this with respect to the metric gives the EFE with a cosmological constant, as I wrote it down. So there is certainly "proof" of that version. So no, I'm not bothered

 

I was aware of Lagrangian mechanics, but not aware of this derivation. The two formulas read

[math]\frac{1}{16\pi} R\sqrt{-g}[/math]

and

[math]\frac{1}{16\pi}(R+2\Lambda)\sqrt{-g}[/math]

 

which is interesting in that the theory is founded upon [math]L=T-V[/math], where T is the kinetic and V is the potential of the system. In a system where the baseline is represented by [math]\Lambda^V-\Lambda^T[/math] this would seem to mean that with no curvature there is no Lagrangian. It would require a change in the magnitudes of the scalar values of the [math]\Lambda^T[/math] to create it (which would change R above).

 

In other words

[math]\frac{1}{16\pi} R\sqrt{-g}[/math] with [math]R\rightarrow 0[/math]

.
[math]\frac{1}{16\pi}(R+2\Lambda)\sqrt{-g}[/math] with [math]R\rightarrow -2\Lambda[/math]

[ajb]

As an example, the weak field approximation for Newtonian gravity is normally derived from [math]g_{\mu\nu}=\eta_{\mu\nu}-\epsilon_{\mu\nu}[/math] where [math]\eta_{\mu\nu}=-1[/math] and [math]-\epsilon_{\mu\nu}[/math] is some tiny perturbation.

 

 

You linearise the field equations by considering small fluctuations about the Minkowski metric, or you could consider other metrics. These small functuation are gravitational waves or the "graviton field".

 

Keeping normal GR notation, the way I am proposing would start out with flat space time as [math]-\Lambda^1+\Lambda^2=0[/math] instead of just a -1. A small perturbation of [math]\Lambda^2[/math] would be used to generate the field. A bit less exciting than deSitter space, but lots to go over in the future.

 

Would that not just have the effect of having a small cosmological constant? Based on what we know, the cosmological constant is small and positive.

 

Anyway, Perturbations are very important in cosmology, however one does not think of perturbations of the cosmological constant. It is not dynamical.

[JMessenger]

 

Would that not just have the effect of having a small cosmological constant?

 

 

It does seem that way to me.

 

Anyway, Perturbations are very important in cosmology, however one does not think of perturbations of the cosmological constant. It is not dynamical.

 

I was using [math]\Lambda[/math] in the sense of the scalar value, and meant [math]\Lambda^1[/math] as representing the magnitude of the cosmological constant (which as you say is not dynamical) but was considering [math]\Lambda^2[/math] as a rank 0 tensor which, when perturbed, becomes the dynamical higher rank 2 Einstein tensor.

 

 

I think some of the confusion of what I am stating stems from there being no hard definition of what it means when a tensor "vanishes".

 

 

 

Take for instance equation 1.2 from http://eprints.ma.man.ac.uk/894/02/0-19-852868-X.pdf

 

[math]\alpha u + \beta \frac{\partial u}{\partial n}=g[/math]

 

If we set [math]\alpha u=0[/math], which was a rank 0 tensor, and then [math]\mid\beta \frac{\partial u}{\partial n}\mid\rightarrow 0[/math], which is a higher rank tensor, would we define that as the higher rank tensor "vanishing"?

 

If [math]\alpha u \neq 0[/math], but still [math]\mid\beta \frac{\partial u}{\partial n}\mid\rightarrow 0[/math], do we still consider the higher rank tensor to have "vanished" if the rank 0 tensor remains?

 

 

So my main problem is the a priori assumption that when the Einstein tensor "vanishes" there is no rank 0 tensor remaining. I don't think the equation (or physical evidence) supports this.

 

So for my post #3 where MTW "redefine" the Einstein tensor. [math]"G_{\alpha\beta}"=G_{\alpha\beta}+\Lambda g_{\alpha\beta}[/math]. It seems to me that it was redefined in order to ignore that there is a rank 0 tensor remaining after [math]G_{\alpha\beta}[/math] "vanishes".

Edited December 7, 2012 by JMessenger

[ajb]

 

I was aware of Lagrangian mechanics, but not aware of this derivation. The two formulas read

[math]\frac{1}{16\pi} R\sqrt{-g}[/math]

and

[math]\frac{1}{16\pi}(R+2\Lambda)\sqrt{-g}[/math]

 

 

This is absolutly standard.

 

I was using [math]\Lambda[/math] in the sense of the scalar value, and meant [math]\Lambda^1[/math] as representing the magnitude of the cosmological constant (which as you say is not dynamical) but was considering [math]\Lambda^2[/math] as a rank 0 tensor which, when perturbed, becomes the dynamical higher rank 2 Einstein tensor.

 

 

I think some of the confusion of what I am stating stems from there being no hard definition of what it means when a tensor "vanishes".

 

 

This this an additional sclar field in your theory. People do think about how you should couple scalar fields to gravity; minimally or not.

 

Anyway, to me the theory still looks rather undynamical, if you simply "split the cosmological constant in two". I imagine you need to add somthing to the action.

[JMessenger]

 

This is absolutly standard.

 

 

This this an additional sclar field in your theory. People do think about how you should couple scalar fields to gravity; minimally or not.

 

Anyway, to me the theory still looks rather undynamical, if you simply "split the cosmological constant in two". I imagine you need to add somthing to the action.

I did't mean to imply I am splitting the cosmological constant in two, just that the dynamical portion (Einstein tensor) becomes a rank 0 tensor with scalar values equal to the cosmological constant for flat space-time.

 

Here is where I think the difficulty arose concerning scalar fields in GR:

 

Normally for [math]\alpha u +\beta\frac{\partial u}{\partial n}=g[/math] we see that [math]\beta\frac{\partial u}{\partial n}[/math] is some derivative of [math]u[/math] with respect to something else. Suppose we now have an equation of two of these scalar fields, which is [math]\gamma u +\eta\frac{\partial u}{\partial n}+ \alpha u +\beta\frac{\partial u}{\partial n}=g[/math] where for what want to use the equation for it appears physically that [math]\eta\frac{\partial u}{\partial n}=0[/math] and that [math]\gamma u[/math] is sooooo tiny that it has no bearing in our solar system. If we have no need for [math]\gamma u[/math] then we might be tempted to ignore the required [math]\alpha u[/math]. The difficulty with this though, is if we ignore [math]\alpha u[/math], what exactly are we finding the partial derivative of in [math]\beta\frac{\partial u}{\partial n}[/math] and in reference to what? The easy way out conceptually (but not formulaically) is to simply take partial derivatives of one dimension with respect to the others. If there is no partial derivative of one dimension with another (they aren't functions of each other), you simply have a higher rank tensor that now appears as a Euclidean rank 0 tensor with no scalar values.

[JMessenger]

On the subject of minimally coupled scalar fields, take the Wikipedia example: http://en.wikipedia.org/wiki/Scalar_field_solution under Einstein tensor

 

[math]G^{\alpha\beta}=8\pi\sigma diag[-1,1,1,1][/math]

 

It states that [math] \sigma [/math] is the energy density. That is the major difference between this theory (and I believe what is the only thing that can be derived from the Riemann) is that [math] \sigma [/math] isn't the energy density of the field. You can call it the density of the quintessence if it makes you feel better but the energy density we are familiar with has to be the difference between the cosmological constant and [math] \sigma [/math]. No difference=no energy=uniform Euclidean scalar field=uniform density quintessence field=flat space-time. If you want energy, then there must be a decrease in the scalar value of [math]\sigma[/math] away from the scalar value of the cosmological constant.

 

Edit: Just to point it out, the tensor above is a rank 0 tensor with constant scalar values. This is showing that in a scalar interpretation, the Einstein tensor can turn into a rank 0 tensor with non-zero components.. The metric signature of [math]diag[-1,1,1,1][/math] and the definition of "energy" in General Relativity are mutually exclusive so the [math]\sigma[/math] cannot be the energy density with which we are familiar with.

Edited December 9, 2012 by JMessenger

[JMessenger]

This may seem like rambling, but practicing how to relate this to those not familiar with tensors and for benefit of the forum to bring up failures in my logic.

 

Taking into consideration the redefinition of the Einstein tensor that Misner, Thorne and Wheeler spoke about of [math]"G_{\alpha\beta}"=G_{\alpha\beta}+\Lambda g_{\mu\nu}[/math], and that there is no mathematical definition of "vanish", which is what I technically view as required (but lacking) for the equation [math]R_{\mu\nu}+\frac{1}{2}R g_{\mu\nu}=0[/math], I propose breaking it down to the most simple of mathematical arguments:

 

Tensors in GR are represented through the use of 4x4 arrays or matrices. In order to sum two matrices, each component is summed with the corresponding component of another matrix. I propose the true definition for "vanishing" of the sum of two matrices is that ALL components must be exactly zero. For the special case of a matrix that can be written as

 

[math]\left(\begin{array}{cccc} X & 0 & 0 & 0 \\ 0 & X & 0 & 0 \\ 0 & 0 & X & 0 \\ 0 & 0 & 0 & X\end{array}\right)[/math]

 

we can sum this with a similar matrix of

 

[math]\left(\begin{array}{cccc} Y & 0 & 0 & 0 \\ 0 & Y & 0 & 0 \\ 0 & 0 & Y & 0 \\ 0 & 0 & 0 & Y\end{array}\right)[/math]

 

by individually adding each component to the corresponding component such that

 

[math]X_{11}+Y_{11}=(X+Y)_{11}[/math] etc.. meaning

 

[math]\left(\begin{array}{cccc} X+Y & 0 & 0 & 0 \\ 0 & X+Y & 0 & 0 \\ 0 & 0 & X+Y & 0 \\ 0 & 0 & 0 & X+Y\end{array}\right)[/math].

 

These matrices can also be written

 

[math]X\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)[/math]

 

such that the sum is

 

[math](X+Y)\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)[/math].

 

For the sum of these two matrices to "vanish" absolutely, then

[math]X=0,Y=0[/math]

 

or

 

[math]Y=-X[/math].

 

For the sum of two matrices (tensors) to allow [math]R_{\mu\nu}+\frac{1}{2}R g_{\mu\nu}[/math] to vanish exactly, then the form of the two tensors must be either [math]X=0,Y=0[/math]

 

or

 

[math](X-Y)\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)[/math].

[ajb]

By non-vanishing, I would take it to mean that at every point the tensor is not zero. This is independent of the coordinate used.

[JMessenger]

By non-vanishing, I would take it to mean that at every point the tensor is not zero. This is independent of the coordinate used.

 

But can a point in a region of zero curvature still have a non-vanishing tensor?

 

 

ajb,

 

 

 

You got me thinking about how textbooks on GR define certain things, and thumbing through a copy of The Large Scale Structure of Space-Time by Hawking and Ellis, and I ran across their definition of a vector:

 

 

 

..., that is [math](\partial f/\partial t)_{\lambda}[/math] is the derivative of f in the direction of [math]\lambda (t)[/math] with respect to the parameter t. Explicitly,

[math]\left(\frac{\partial f}{\partial t}\right)_{\lambda}|_{t}=\lim_{s \to 0} \frac{1}{s} \{f(\lambda(t+s))-f(\lambda(t))\} [/math].

 

Technically, one could argue that this definition is incomplete, as we could equally write

 

[math]\left(\frac{\partial f}{\partial t}\right)_{\lambda}|_{t}=\lim_{s \to 0} \frac{1}{s} \{f(\lambda(t+s))-K-f(\lambda(t))+K\} [/math]

 

or

 

[math]\left(\frac{\partial f}{\partial t}\right)_{\lambda}|_{t}=\lim_{s \to 0} \frac{1}{s} \{K-f(\lambda(t+s))-(K-f(\lambda(t)))\} [/math]

 

where K is some arbitrary constant. In other words, their definition of a vector excludes the equally valid definition of a vector as shown in the below graph:

Edited December 10, 2012 by JMessenger

[ajb]

But can a point in a region of zero curvature still have a non-vanishing tensor?

A tensor will usually depend on the point at which it is "over", the exact value of the components will depend on the coordinate system used. However, due to the transformation law of tensors, if the components of a tensor are zero in some coordinate system (in the neighbourhood of some point) then they are zero in all coordinate systems.

 

It is quite possible to have pieces of a manifold that are flat while other pieces are not. You have to be a little careful about how you want to state the curvature. Anyway, a specified tensor can be zero at certain points; such a thing is "not no-where vanishing".

 

A no-where vanishing tensor is a tensor that is non-zero at all points on the manifold. We can't say that the components have specific values at a given point in any given coordinate system, only that it is not zero.

You got me thinking about how textbooks on GR define certain things, and thumbing through a copy of The Large Scale Structure of Space-Time by Hawking and Ellis, and I ran across their definition of a vector:

This is the standard way of thinking of vectors, in terms of equivalence classes of velocities of curves. There are equivalent ways of thinking of them, I tend to think of vector fields as the derivations of the sections of the structure sheaf. You build the tangent sheaf from this.

 

This is just a fancy way of saying that "vector fields are partial derivatives".

 

 

[JMessenger]

A tensor will usually depend on the point at which it is "over", the exact value of the components will depend on the coordinate system used. However, due to the transformation law of tensors, if the components of a tensor are zero in some coordinate system (in the neighbourhood of some point) then they are zero in all coordinate systems.

 

It is quite possible to have pieces of a manifold that are flat while other pieces are not. You have to be a little careful about how you want to state the curvature. Anyway, a specified tensor can be zero at certain points; such a thing is "not no-where vanishing".

 

A no-where vanishing tensor is a tensor that is non-zero at all points on the manifold. We can't say that the components have specific values at a given point in any given coordinate system, only that it is not zero.

 

Interesting, had not run across the phrase "not no-where vanishing".

 

I can understand the part of having a manifold that is flat "in pieces", and that a no-where vanishing tensor that is non-zero at all points on a manifold, but can a manifold be completely flat, at all points and pieces, with a no-where vanishing tensor at all points and pieces?

[ajb]

Interesting, had not run across the phrase "not no-where vanishing".

 

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"!

 

 

I can understand the part of having a manifold that is flat "in pieces", and that a no-where vanishing tensor that is non-zero at all points on a manifold, but can a manifold be completely flat, at all points and pieces, with a no-where vanishing tensor at all points and pieces?

 

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.