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Some Q on GR (Novice+)

  1. Apr 28, 2014 #1
    I have been reading GR but I have these gaps in understanding. Can someone help clear them?

    1) SR has a transformation to new co ordinate system that is a linear one. When GR is described, the textbooks mainly discuss Geodesic and distance but I ave not seen the transformation from old to new coordinate system. I understand that it is non linear but where can I find the transformation from system such as (say) Cartesian? Is it even possible? As I see it, even if space is curved, Cartesian system should be defined. For example we can define 2D curved surface in 3D Cartesian frame, correct?

    2) In GR, I see Geodesic equation and Einstein's equation derived from it. The derivation looks logical until a point where suddenly Energy Momentum tensor is introduced in the equation from no where. Where did that come from? What is the basis to add it out of no where? That adds confusion. It looks as it there is conspiracy to make the equation work (sorry, J Kidding)

    3) Curvature is based on Riemann tensor which looks logical but then all of a sudden this tensor is shrunk (contracted) to Ricci tensor. I do not see any explanation why it is done and what is the logic behind it or how it affects 4 dimension surface in higher dimension calculation. If our intention was not to get it Riemann, why is it there in first place, can be get to Ricci directly from equation?

    Appreciate response.
     
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  3. Apr 28, 2014 #2

    Matterwave

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    1. The reason for this is that in general relativity, all possible coordinate transformations are "valid" so to speak. In SR, we like to stick with Lorentzian coordinates which preserve the form of the metric in all coordinate systems. However, in GR, we must allow arbitrary coordinate transformations, so it's no longer a matrix equation! Let's say that we have a manifold ##\mathcal{M}## of dimension n. On a patch (subset) of the manifold ##U\subset\mathcal{M}##, we may define a map ##\phi## from our manifold into a subset of ##\mathbb{R}^n##:

    $$\phi:U\rightarrow\mathbb{R}^n\quad\{\phi(P)=(x^1,x^2,...,x^n)|P\in U\}$$

    This map takes points on the manifold and assigns them n numbers, which we call the coordinates of P. In another patch ##V\subset\mathcal{M}## we may define a different map ##\psi##:

    $$\psi:V\rightarrow\mathbb{R}^n\quad\{\psi(Q)=(y^1,y^2,...,y^n)|Q\in V\}$$

    Both of these maps have inverses (they are 1 to 1, and actually should be differentiable) and so, in the areas where the patches overlap ##U\cap V##, then what we can do is use ##(\psi)^{-1}## to map a point in ##\mathbb{R}^n## to a point ##Q\in U\cap V## and then use ##\phi## to map this point Q into ##\mathbb{R}^n## to obtain my coordinate transformation:

    $$\phi\circ\psi^{-1}:\mathbb{R}^n\rightarrow\mathbb{R}^n\quad\{\phi\circ\psi^{-1}(Q)=x^i(\psi^{-1}(y^1,y^2,...,y^n))|Q\in U\cap V\}$$

    And...there you have it. This notation is usually abbreviated:

    $$x^i=x^i(y^1,y^2,...,y^n)\quad i=1,..,n$$

    There's your coordinate transformation, valid only on parts of the manifold where the coordinate patches overlap. This is probably why you don't see it very often if you don't study differential geometry. The form of the transformation, doesn't really enlighten us since all possible (differentiable anyways) transformations are allowed.

    The second part of your question, defining embedded curved surfaces would work well for us, if we always wanted to work in 2n dimensional space instead of n-dimensional space. We can embed our space-time manifold in a 2n dimensional Euclidean manifold, but, then we'd have to worry about working on this 2n dimensional Euclidean manifold. This embedding would seem pretty artificial. We want to work inside the space we are given.

    2. One might also ask the question "where did the m's come from in Newton's law of gravity? Why is F=Gmm/r^2?". This is a postulate of general relativity, and can be shown to work via experiment. We postulate that space-time is a 4-dimensional, curved, manifold, and that curvature tells matter how to move and matter tells space-time how to curve. These are postulates of the theory.

    Of course, just because it's a postulate, doesn't mean it came out of nowhere. It makes sense that if the universe's curvature is described by a curvature tensor, then the "thing" making the universe curve would be another tensor relating to the matter content of the universe. It just so happens that the stress-energy tensor is just such a tensor. It tells us the matter content of the universe.

    Alternatively, one can use the Hilbert action to "derive" the Einstein field equations from an action principle point of view. In this point of view, the matter Lagrangian's functional derivative with respect to the metric IS (defined as) the stress-energy tensor (at least, it's a non-symmetrized version of it). But this is no more fundamental than what Einstein did by postulating his field equations.

    3. For this, we go back to the stress-energy tensor. The stress-energy tensor tells us the matter content of the universe and it happens to be a symmetric rank 2 divergenceless tensor. What symmetric, rank 2, divergenceless tensor is there to describe the curvature? Well, the Einstein tensor! And there you go.
     
    Last edited: Apr 28, 2014
  4. Apr 28, 2014 #3

    pervect

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    There isn't any such thing as a Cartesian coordinate system in curved space-time.

    [add]
    For a simpler more intuitive example why there isn't any such thing - try to find a "Cartesian coordinate system" on the 2 dimensional curved manifold of the surface of a sphere, for instance a globe of the Earth. You can map the surface of the globe to a plane, but not in a manner that generally preserves distances. You can make the flat map work near a point, but as you get further and further away, the effects of curvature become too large to ignore.

    But you might be interested in Fermi normal coordinates, which are a reasonable approximation of a Cartesian coordinate near a particular world-line. See for instance http://arxiv.org/abs/gr-qc/9402010 "On the physical meaning of Fermi coordinates." Note in advance that in spite of their appealing qualities near a reference worldline, that Fermi normal coordinates aren't necessarily defined everywhere. Also note that it's rather hard to find closed form solutions for them.

    [add]
    I've been thinking for some time that mapping the surface of the globe to a plane using Fermi normal coordinates on the 2d manifold of a surface of a sphere (ignoring time, one might argue that they are Riemann normal coordinates instead) would be instructive - at the moment I don't have any such visual aids.

    I've seen the Geodesic equation derived from Einstein's equation, but not vica-versa.

    The energy-momentum tensor is actually needed in special relativity to describe, in a covariant manner, the distribution of matter. Probably the easiest way to motivate the energy-momentum tensor is via a swarm of particles. You'll find you need a rank 2 tensor to describe how the energy-momentum of a swarm of particles transforms in a covariant manner in SR, this can be thought of as the tensor product of the particle-flux 4-vector, (if it's not familiar, it's worth reading about, see wiki and http://web.mit.edu/edbert/GR/gr2b.pdf ) and the energy momentum associated with each particle in the swarm.

    Not too many SR books talk about the issue, Rindler has something about it I think (the notation isn't very modern). If you can find my blog, I have some (not very finished) discussion of this issue too, it also has a link to the MIT paper I gave above.

    This was the heart of the theory. Matter is naturally described by the rank 2 energy-momentum tensor, and the simplest possible theory would suggest that some reduction of the curvature tensor from rank 4 to rank 2 which is then set proportional to the stress energy tensor.

    An additional refinement though is that the divergence of the stress energy tensor is zero. As a consequence, the left hand side of Einstein's equation is not the Ricci, but the Einstein tensor, a modification of the Ricci which makes the divergence zero.
     
    Last edited: Apr 28, 2014
  5. May 20, 2014 #4
    Thanks for help. I get the explanation for 1 and 2 but not well for 3 as yet. I understand that Einstein Equation has 2 ranks so tensor with 4 ranks wont work however when we started from basics and said that space is curved when the parallel transport returns back in different direction OR Curvature tensor is non zero, we talked about rank 4, not 2. Then we say that that does not fit the equation so let us contract it. It looks too artificial. So the Q is what is physical interpretation of Ricci Tensor/scalar. We are still talking about 4 Dimensions but with 2 ranks instead of 4? What exactly does not mean?
    If they are Zero, can we say for sure the space is flat. (I think not since Ricci tensor = 0 does not necessarily mean Curvature Tensor is zero) So I have not been able to grasp that.
     
  6. May 20, 2014 #5

    PAllen

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    That would be fun. I'd love to read it if you ever write it up. Intuitive analysis suggests it would show a number of relevant properties of FN coords in GR, including that it can only cover relatively small patches of the sphere, compared to lat/long which cover all except poles.

    [Edit: analogs of both RN and FN coords should be possible. The former, I think, would cover a much larger patch on a 2-sphere, but have a near flat metric only very close to the origin; the latter (FN) would have near flat metric all along (and near) a chosen defining axis, but cover a much smaller patch.]
     
    Last edited: May 20, 2014
  7. May 20, 2014 #6

    Matterwave

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    At some point in any physical theory, we must basically postulate a dynamic equation. This is true for e.g. the Dirac equation, the laws of Newton, the Maxwell's equations, etc. These equations should be based off of experiment, but otherwise, they cannot be derived from more basic principles (sure you can "derive" them from an action principle, but then you are just postulating that action). Einstein basically postulated his field equations. They are not, strictly speaking, the only ones which might work (see f(R) theories, etc.), but they are the simplest, most elegant ones.

    All we can tell you about how to arrive at these field equations are motivations, not derivations.

    Some motivations: 1) Space should tell matter how to move and matter should tell space how to bend. 2) The matter content of the universe is described in a rank 2 stress-energy tensor which is divergence free and symmetrical.

    From this, Einstein basically postulated that the field equation would be G=8πT, because the Einstein tensor G has the property that it's divergence free (due to the Bianchi identities) and symmetric (by construction), which it has to be if it's to be made equal to the stress-energy tensor.

    There's no way to know, a priori, that the Riemann tensor is NOT the tensor you want to use in your field equation (except perhaps you don't want to over-specify the curvature, since you want there to be the possibility of curvature in a vacuum). But it turns out that Einstein's equation is based on the Einstein tensor, not the Riemann tensor.

    And indeed, if R=0, it does not mean that the manifold is without curvature. R=0 is true for the vacuum, but space-time can have curvature in a vacuum due to sources (boundary conditions) far away. This is actually a good thing, because you don't want gravitational interactions to be impossible through a vacuum.
     
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