What is the stress-energy-momentum tensor and its role in general relativity?

[Naty1]

kevin: You might find this discussion of interest: Spacetime Curvature Observer and/or Coordinate Dependent?

https://www.physicsforums.com/showthread.php?t=596224
April, 2012

...If by "gravity" you mean "particular effects of gravity", then yes. As you point out, particular effects of gravity on particular observers will always be dependent on the observer's 4-velocity…. this is a question of terminology, not physics. Whether or not "spacetime curvature" is observer-dependent depends on what you define "spacetime curvature" to mean.

edit: I just came across this John Baez/Ted Bunn discussion comment [and if I were the least bit organized, would have included it in my original post]:

"In general relativity, knowing all about the sources (the stress-energy tensor T) isn't enough to tell you all about the curvature." which complements the MTW quote.

 

[kevinferreira]

Not being an expert of GR or DiffGeo*, but I think a tangent space to a diff. manifold is not a topological space, merely an algebraic construction, a vector space (it has no neighbourhoods). And related to Lie groups, the exponential mapping takes indeed an infinite set of points from the Lie group, to be precise from an open neighbourhood of the origin (neutral element) to an infinity of vectors in the Lie algebra, in the general case and from one point from the Lie group to one vector in the Lie algebra.

Allow me to disagree, every vector space may be trivially defined as a topological space, and hence also the tangent space. You see, you can use the notion of norm on the vector space to have a notion of distance, and with the distance you can build a topology. From that you may trivially see it as a manifold. A vector space is the most nice manifold you may ask for.

 

[kevinferreira]

So additional relative velocity DOES cause physical effects,as your quote shows, but THAT curvature was not considered 'gravitational curvature'...That is, the 'amount of gravity' ...

I found this quite confusing. I think the disturbing cause is that we are considering only energy as the cause of gravity, and that is not true, energy is not a Lorentz scalar. Whenever you may want to consider velocity, you will see that you add a momentum density to the energy-momentum tensor, but you will also change the relativistic mass density, so that in the end (just like with the energy momentum 4 vector) nothing changes in general. I wonder if this is a correct view of it.

 

[dextercioby]

Allow me to disagree, every vector space may be trivially defined as a topological space, and hence also the tangent space. You see, you can use the notion of norm on the vector space to have a notion of distance, and with the distance you can build a topology [...]

http://en.wikipedia.org/wiki/Tangent_space Both definitions here (first and second) ascribe an algebraic (i.e. according to the axioms here http://en.wikipedia.org/wiki/Vector_space) character to the tangent space in a point x of a general manifold. Where does the norm (which would induce the topology on T_x (M)) come from ?

Your first sentence induces the set inclusion

{vector spaces} \subset {topological spaces}

which is not not correct (the axioms of a vector space don't mention a norm, so there wouldn't be any norm-induced topology
).

The right set connection is:

{topological vector spaces} = {vector spaces} \cap {topological spaces}

 

[Naty1]

nothing changes in general. I wonder if this is a correct view of it.

I found that confusing ! really...

Let's consider an object moving in free fall relative to an observer following a particular path. If an identical particle at a different velocity, or an identical particle with the same velocity but also angular momentum then passes the observer, the particles will in general follow three different paths.

That's the physics ,I think we can agree.

In the view I gave, the second particle with angular momentum changes gravitational spacetime curvature relative to the first; the third particle with only additional velocity has a different 'visible curvature'...but it is not part of 'the amount of gravity'.

That is just a convention, but one that seems to make sense to me regarding fast moving particles not becoming black holes.

 

[kevinferreira]

http://en.wikipedia.org/wiki/Tangent_space Both definitions here (first and second) ascribe an algebraic (i.e. according to the axioms here http://en.wikipedia.org/wiki/Vector_space) character to the tangent space in a point x of a general manifold. Where does the norm come from which would induce the topology on T_x (M) ?

Your first sentence induces the set inclusion

{vector spaces} \subset {topological spaces}

which is not not correct (the axioms of a vector space don't mention a norm, so there wouldn't be any norm-induced topology
).

The right set connection is:

{topological vector spaces} = {vector spaces} \cap {topological spaces}

Oh, yes, you're absolutely right, I was wrong in admitting a norm as an axiom (or a direct consequence of an axiom) of vector space definition. I guess it comes from my time spent studying functional analysis!
Anyway, thanks for clearing that up for me.
But that's troubling, as in the wikipedia article of the exponential map it is said explicitly

The radius of the largest ball about the origin in TpM that can be mapped diffeomorphically...

So a notion of distance is used... I wonder how.

 

[micromass]

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