Strees-energy-momentum tensor

kevinferreira

Yes, I see that, all objects used in Einstein equations (and the associated mathematical treatment) have an 'independet reality', just like a vector in euclidean space have a unique defining direction and magnitude no matter how we wish to represent it via our coordinates. So, I just have to think that the Riemann curvature tensor, the Ricci tensor, the metric tensor, etc. have a mathematically objective reality on spacetime to get to the conclusion with the help of Einstein's to see that the energy-momentum tensor shares the same properties.

But....
If I'm not mistaken, all these objects are defined locally, i.e. through their action on elements of the tangent space to a point (I take here the definition of tensor as a function acting on any number of cartesian products between the tangent and the cotangent spaces).
Therefore, even if we may define a tensorfield in every point of our spacetime as the collection of these definitions at each tangent space for each point, how can GR tell us something about large scale phenomena if all objects in our treatment are defined on tangent spaces, which are nothing else but 'nice' and local aproximations of our manifold?
Also, another thing that bothers me is that since locally the laws of special relativity hold, and we can 'transform away' gravity by just considering an accelerated frame (just as Einstein presented it in his 1916 paper), how is it that this locality and the locality used to define our tensors may in the end tell us something about the manifold itself?
This bothers me alot, we use the word 'locally' everywhere, but in the end we can actually tell something about spacetime itself.
Is it implicit in our definitions and mathematical treatment of the theory that what we actually do is the same as for example the way we would study the curvature of a graph, just observe the tangent lines at each point and then compare them in order to know how 'non-flat' is our graph? I think this might be somehow true, as the Riemann and Ricci tensors' components are calculated in some particular coordinates with the derivatives of the metric components...
And I also remember having done an exercise where we show that locally the metric is the Minkowski flat metric up to a second order term, related with the curvature tensors. But again, this 'locally' is not of the same kind as the previous ones, or is it?

I'm sorry I'm getting out of the topic, but I think all this has to do with the energy-momentum tensor, it's the other side of the equation!

atyy

The first "locally" just means that the objects in the theory are tensor fields - they exist at every point on the manifold, and act on tangent vectors. (Actually, I've never seen this called "locally" until now.)

The second "locally the metric is the Minkowski flat metric up to a second order term" shows that the EP is true only at a point, and only up to first derivatives of the metric. It's the second derivatives that are crucial in distinguishing curved and flat spacetime. This is why in writing the matter action that Pervect and I talked about in posts 14 and 15, the matter fields must be coupled to the metric and not its derivatives. This prescription from the equivalence principle is also called "minimal coupling".

When you differentiate the full GR action L_EH+L_matter wrt the various fields, you then get the Einstein field equation with the correct definition of the stress tensor, and the equations of motion for the various matter fields. Because the equations came from an action in which matter was minimally coupled to the metric, the stress tensor derived in this way obeys the EP in the sense that in coordinates in which the metric (but not its second derivatives) have the same form as a Lorentz inertial frame, the components of the stress-energy tensor then have their SR meaning.

Altabeh

Good question. Well, the general approach is based on a covariant formalism, providing one with global conservation laws in asymptotically flat spacetimes. For example, it can be used to give the mass of an isolated black-hole (which is asymptotically flat). A more recent and general approach can be found in arXiv:1202.1905.

LL pseudo-potential is related to curvature via Einstein field equations with some modifications. More technically, it is related with Ricci tensor density and scalar curvature density through an identity derived according to Noether's prescription for finding the conserved quantities of a Lagrangian which is a scalar density.

I hope this is enough. If you need more details, we can discuss it.

P

kevinferreira

Well, the tangent space is defined pointwise on the manifold, so that it describes linearly and 'locally' the manifold... This was the sense I was trying to give to it. What bothers me is that this pointwise definition is implicit on tensors, and then in the end they can give us some information about non-pointwise properties of the manifold... Or I'm confused and maybe they don't, after all when we say that test particles follow geodesics, we are just saying they follow a particular set of points on spacetime, and maybe these points form a geodesic by some combination of pointwise properties. Hmm...

Aah, I see here the whole beauty of starting from the action with this minimal coupling which lead us to the EP... Hmm, what do you mean more precisely by the 'SR meaning' of the components of the stress-energy tensor?

kevinferreira

Ok, I see, I'll check on that. Thanks a lot for your help!

atyy

Well, just think of something simple like position as a function of time in classical Newtonian mechanics. If the function is analytic, from its derivatives at a point, through Taylor series, you can get the function away from that point. In this sense, higher and higher derivatives are more and more "nonlocal" (although strictly speaking, the derivatives are all well defined at that point, which is the point of calculus and taking limits). The rough idea is that the first derivative contains the difference of two points, so it is more "nonlocal" that the value of the function itself at that point. And of course, once you have specified the function at every point, you have also specified all the derivatives.

I think of minimal coupling as another name for the EP. The SR (special relativity) meaning is just that in flat spacetime, and especially in coordinates where the flat spacetime metric is diag(-1,1,1,1), we understand the stress tensor components to have meanings such as T₀₀ is the "energy density". When we go to arbitrary coordinates in curved spacetime, it isn't clear why T₀₀ would still be the "energy density". So to get the meaning of the stress tensor components, we take advantage of the fact that locally you can have "flat spacetime" as long as you don't look at higher derivatives, and it is in those coordinates that the stress tensor components have their traditional SR meaning.

WannabeNewton

Not sure what you mean by the tangent space describes the manifold M locally at some p in M. Tp(M) is a subset of the tangent bundle of M; it is not a neighborhood of p. Local properties are expressed in terms of neighborhoods of points and / or bases for topologies.

kevinferreira

That makes sense, it is true that one says 'the derivative of f at x' even though in the definition of that object one has to evaluate f at a neighbour point of x (with the subsequent limit). It seems more clear now!

Yes, that is what I was thinking about, just wanted to be sure of it completely. It's amazing how we could think that Einstein's equations are plane and simple when looking at them, and think that we can forget about all the rest, but in fact all fundamental principles and axioms of GR are somehow encoded on those equations. I feel amazed! Thanks for your help!

kevinferreira

Indeed, Tp(M) is not a neighbourhood of p, but you can find a local homeomorphism (that is even a local diffeomorphism, I think...) between a neighbourhood of p in M and a neighbourhood of the origin on the tangent space. It is called the exponential map and is very useful in Lie group theory. You may want to check http://en.wikipedia.org/wiki/Exponential_map, the part about Riemannian geometry. This has also something to do with Riemann normal coordinates, although I don't want to go on that as I haven't revisited that part of my lectures yet.
You can see it says it kinda helps to understand how the tangent space may be seen as a 'linearisation' of the manifold. This makes sense if you think about a sphere and the tangent plane at the north pole. You can imagine that even if the only point of contact between the sphere and the plane is the north pole, you may describe very well the sphere around the pole by considering points on the tangent plane. This is my way of thinking about it.

WannabeNewton

Without getting side - tracked from the thread, I think the issue here is just what you mean by "local". The linearity part is fine but there are many "local" properties of and on manifolds that can't just be described by a tangent space at a point (for example how the co-variant derivative of a vector field along a direction depends only on the element of the tangent space at the given point corresponding to that direction whereas the lie derivative of the vector field along another requires knowledge of the other vector field in a neighborhood of the point).

dextercioby

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.

*My exposure to both GR and DiffGeo was minimal a while ago (and has not improved significantly in the mean time), the course was based on the 75 page brochure by Dirac! :D

dextercioby

I wouldn't say that, nothing really comes from the sky, not even the action (once you have the action, you definitely have a method to get an energy-momentum 4 vector/tensor). Every action integral has at least 3,4 justifying assumptions as to why that Lagrangian and not another one. Nothing is random in physics.

Naty1

pervect: Your post #17 was helpful.

In general I depend on you experts for current usage of terminology as well as physical interpretations of detailed mathematics. [I've been out of school way too long to try and catch up!]

from your post:


That term comes from PeterDonis and several old threads where you and DocAl
and probably others were kind enough to help me understand spacetime 'gravitational curvature'. I came away with the understanding ['real'] gravitational observations relating to a rapidly moving massive body can be answered as if the body is stationary so that relative velocities are NOT considered as part of gravitational curvature...

One aspect of THAT logic is the one I posted already: fast moving particles don't become black holes. A related perspective would be that for a single electron, as an example, the rest energy density of the electron is the only thing that causes spacetime gravitational curvature. The kinetic energy is frame-dependent, just as the velocity is....and does not contribute.

So 'velocity' does not yet enter into my understanding of 'the amount of gravity'.


Here is how docAl explained it from an old thread [ edited by me for brevity]:

If a flat sheet of graph paper represents two dimensional space without gravity, with the introduction of gravitation the paper itself becomes curved. [Curvature that cannot be "flattened" without distortion.] Gravitational "spacetime curvature" refers to this curvature of the graph paper, regardless of observer, whereas visible/perceived curvature in space is related to distorted, non-square grid lines drawn on the curved graph paper, and depends on the frame choice of the observer...."

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' .....

So while I believe the above is self consistent, I really do not know if such terminology is
generally understood that way in these forums. And when I first posted in this thread I was unsure if the OP, got the answer he thought he did.

PS: best thread on gravity in a while!!

Naty1

kevin: You might find this discussion of interest:


Spacetime Curvature Observer and/or Coordinate Dependent?

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


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

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

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

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} [itex] \subset [/itex] {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} [itex] \cap [/itex] {topological spaces}