Conservation of energy in general relativity

[dx]

Why? We only need the "usual" Levi-Civita symbol which is "pseudo"-tensor (tensor density) by itself.
Dualization is given exactly by it and by "pseudo"-tensor reciprocal to it. Metric is an extra structure.

The hodge dual of a p-form β can be defined as

*β(v_p+1, ..., v_n) = β(v₁, ..., v_p)

where v₁, ..., v_n are oriented orthonormal vectors in T_xM.

 

[vanhees71]

How can be vectors orthonormal without a (pseudo-)metric?

 

[dx]

I was not suggesting that they can be. Just sharing a useful way of defining it.

 

[vanhees71]

But you claimed you can define hodge dualization in a manifold without a (pseudo-)metric. To define orthonormal bases you need a (pseudo-)metric. AFAIK you cannot define a hodge dualization without a (pseudo-)metric.

 

[robphy]

Possibly useful:
https://en.m.wikipedia.org/wiki/Hodge_star_operator

In mathematics, the Hodge star operator or Hodge star is a linear map defined on the exterior algebra of a finite-dimensional oriented vector space endowed with a nondegenerate symmetric bilinear form.

 

[Qwet]

I don't know, how you want to even formulate Gauss's theorem without Hodge duality and how to define the Levi-Civita tensor without a metric.

Well, I would do it like this.
Let's write 3-dimensional Gauss's theorem similarly to the form, which Ostrogradskiy got.

$\int_V [\frac{∂A^1}{∂x^1}+\frac{∂A^2}{∂x^2}+\frac{∂A^3}{∂x^3}]dx^1dx^2dx^3= \oint_{∂V} [A^1sgn(n_1)|dx^2dx^3|+A^2sgn(n_2)|dx^1dx^3| + A^3sgn(n_3)|dx^1dx^2|]$

If $A^i$ is vector density, then $\frac {∂A^i}{∂x^i}$ (summation over paired indices) appears to be scalar density, so the volume integral on the left side is a scalar, and the theorem gets covariant meaning, i.e independent of change of variables. At the same time, the surface integral on the right side appears to be a scalar, of course.

Where is any need for metric here?

 

[phinds]

Reference, please?

I do not understand how partial derivatives (from calculus of function of multiple variables) relate to "space", and Gauss's theorem to metric. In any book in a section about Gauss's theorem, there should not be any word about metric.

That's your idea of a reference, just making another statement?

 

[Qwet]

That's your idea of a reference, just making another statement?

No, I just wanted to say, that I don't have references on this point; I did't read any book, where there would be said, that Gauss's theorem cannot be used without metric. Of course, I did't read every book on Earth, so I said that I don't understand, how this two concepts are related.

 

[pervect]

That's your idea of a reference, just making another statement?

No, I just wanted to say, that I don't have references on this point; I did't read any book, where there would be said, that Gauss's theorem cannot be used without metric. Of course, I did't read every book on Earth, so I said that I don't understand, how this two concepts are related.

You (Qwet) are very confused, I'm afraid. If you start trying to actually find some references, you have at least a small chance of realizing how confused you are.

Even your example is confused. It's not possible to integrate a vector over a volume element in a curved space and/or space-time.

I'll provide a reference for this remark. See for example the preliminary section of "The Meaning of Einstein's Equations" by Baez et al.

An html link is https://math.ucr.edu/home/baez/einstein/, there's also a pdf version on arxiv.

Before stating Einstein's equation, we need a little preparation. We assume the reader is somewhat familiar with special relativity -- otherwise general relativity will be too hard. But there are some big differences between special and general relativity, which can cause immense confusion if neglected.

In special relativity, we cannot talk about absolute velocities, but only relative velocities. For example, we cannot sensibly ask if a particle is at rest, only whether it is at rest relative to another. The reason is that in this theory, velocities are described as vectors in 4-dimensional spacetime. Switching to a different inertial coordinate system can change which way these vectors point relative to our coordinate axes, but not whether two of them point the same way.

In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime -- that is, at the same place at the same instant. The reason is that in general relativity, we take very seriously the notion that a vector is a little arrow sitting at a particular point in spacetime. To compare vectors at different points of spacetime, we must carry one over to the other. The process of carrying a vector along a path without turning or stretching it is called `parallel transport'. When spacetime is curved, the result of parallel transport from one point to another depends on the path taken! In fact, this is the very definition of what it means for spacetime to be curved. Thus it is ambiguous to ask whether two particles have the same velocity vector unless they are at the same point of spacetime.

I am concerned that one - or any number of, for that matter - references won't get through to you, but I can at least make the attempt, and hope for the best.

The key concept here is "parallel transport". I assume you're probably unfamiliar with the concept, and you may be unfamiliar with th math needed to talk about the concept too. I don't have any easy solution for this, unfortunately, except to suggest that you read more. I can also add that parallel transport is vital to understanding general relativity and curvature, and is discussed in the field of differential geometry. Unfortunately, vector calculus on flat manifolds is not going to be sufficient for what you want to understand.

On a general level, you have been assuming that you can just integrate the energy-momentum 4-vector over a volume element to "add together the energies". But Baez's remark illustrates why this can't work in a curved space / space-time.

 

[PeterDonis]

If $A^i$ is vector density, then $\frac {∂A^i}{∂x^i}$ (summation over paired indices) appears to be scalar density

Not in a curved spacetime, it isn't. Not even in curvilinear coordinates in flat spacetime, for that matter.

 

[Qwet]

You (Qwet) are very confused, I'm afraid.

Yes, actually. That is the reason for my question.

The key concept here is "parallel transport". I assume you're probably unfamiliar with the concept, and you may be unfamiliar with th math needed to talk about the concept too.

I know, what parallel transport is. I understand that we cannot get the law of energy conservation due to ambiguity of the ways to transport vectors in the space:

As time is inhmogeneous, we don't have energy-momentum 4-vector which would be preserved during system's dynamical change. It is only possible to define 4-vector locally. And next, the problem regarding how to sum this vectors arises.

On a general level, you have been assuming that you can just integrate the energy-momentum 4-vector over a volume element to "add together the energies".

Yes, I understand, that we cannot integrate 4-vector over a volume element. However we can integrate 4 scalar densities 4 times over a volume element.

Not in a curved spacetime, it isn't. Not even in curvilinear coordinates in flat spacetime, for that matter.

I don't understand, why. Because straight calculations show, it is right for any space, even non-metric one, that if $A^i$ is vector density, then $\frac {∂A^i} {∂x^i}$ is scalar density.

We can integrate scalar density over a volume element even in non-metric space. The point is that a volume element $dV$ is scalar anti-density.

The product of scalar density and scalar anti-density $dV$ is a scalar. As the scalar doesn't change, when transported (independent of the way of transport), we can sum up scalars defined in different points of space. This makes corresponding volume integral be invariant over any coordinate transformations.

Similarly, in the surface integral

$\oint_{∂V}A^idS_i$

$dS_i$ surface element is covector anti-density.

The scalar product of the vector density $A^i$ and the covector anti-density $dS_i$ is a scalar, and it makes this surface integral be invariant over any coordinate transformations.

All this operations are independent of metric.

Duality lays in the fact, that density of antisymmetric covariant tensor anti-density $e_{ijk}$, for which $e_{123}=\frac 1 {3!}$, absolutely corresponds to antisymmetric contravariant tensor density $e^{ijk}$, for which $e^{123}=1$. Easy to notice, that $e^{ijk}e_{ijk} = 1$. Thus, for example, every antisymmetric covariant tensor $T_{ij}$ has corresponding vector density $B^i = e^{ijk}T_{jk}$. Reverse method to get antisymmetric covariant tensor is $T_{ij} = e_{ijk}B^{k}$

These operations are independent of metric, as well (i.e defined in non-metric spaces).

 

[PeterDonis]

Yes, actually.

No, actually, you aren't acting like you know you are confused. You keep making confident statements that are wrong, and you continue to repeat them even after you have been told they are wrong.

That is the reason for my question.

Then you should listen to the answers you have gotten to your questions, instead of continuing to make confident statements that contradict those answers.

it is right for any space, even non-metric one, that if $A^i$ is vector density, then $\frac {∂A^i} {∂x^i}$ is scalar density.

No. It is wrong. You have already been told this.

All this operations are independent of metric.

No. They are not. You have already been told this.

 

[PeterDonis]

The OP question has been answered, and further discussion does not seem like it will be productive. Thread closed.