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What info do I get from these tensors?

  1. Jul 4, 2015 #1
    I have calculated the metric tensor, inverse metric tensor, Christoffel symbols, Ricci tensor, curvature scalar and the Einstein tensor for the Robertson Walker Metric:

    ds2= (cdt)2 - R2(t)[dr2/(1- kr2) + r2(dθ2 + sin2(θ)dΦ2)]

    Here is the metric tensor:

    g00 = 1
    g11 = - R2(t) / (1- kr2)
    g22 = - R2(t)r2
    g33 = - R2(t)r2sin2(θ)

    Every other element is 0.

    Now from this tensor, I know that I can calculate the proper times and proper distances between events in this space time using the formula ds2=gμνdxμdxν. I can also gain info on whether two events in this metric are space-like separated, time-like separated, or light-like separated.

    That is the info that I can gain about the space time using the metric tensor. (Feel free to let me know if I am missing anything).

    Now let us skip to the Ricci tensor:

    R00 = - 3R''(t) / R(t) where R''(t) is the 2nd derivative of R(t) with respect to t.

    R11 = [2(R'(t))2 + R(t)R''(t) + k] / (1- kr2)

    R22 = 2r2(R'(t))2 + r2R(t)R''(t) + 2kr2

    (Note: The term r and the function R(t) are not the same. I am just saying this to avoid confusion between the r and R)

    R33 = 2r2(R'(t))2sin2(θ) + r2R(t)R''(t)sin2(θ) + 2kr2sin2(θ)

    Every other element is 0.

    Now, what information do I actually get from the Ricci tensor (in general, but you can apply this question to the Ricci tensor that I just derived as an example)? Here is what I mean when I ask this: You know how I was able to derive proper times and proper distances using the metric tensor? Well what information about events in a metric (or the structure of the space time itself) can I derive from the Ricci tensor? Please be specific and technical. I know it tells you something about the curvature of space time, but what is that something? As things stand right now, I only know that a non-zero Ricci tensor implies that the metric has curvature (and even then it can't even be said that a 0 Ricci tensor implies no curvature because the Ricci tensor is just a contraction of the Riemann tensor which may have some non-zero elements that simply aren't covered in the Ricci tensor). I know that with the Riemann tensor, you can find the change in a vector's orientation if you parallel transport it around a manifold, but I don't think you can do that with the Ricci tensor.

    Now let us head to the curvature scalar:

    The curvature scalar = [ - 6R''(t)R(t) - 6(R'(t))2 - 5k] / R2(t)

    The same question that I had for the Ricci tensor applies to the curvature scalar. What info do I get from it? All I know about it for right now, is that a non-zero curvature scalar means that the metric has curvature, and that a 0 curvature scalar means that the metric is flat space.

    Now for the Einstein tensor:

    G00 = [6(R'(t))2 + 5k] / (2R2(t))
    G11= [ - 4R''(t)R(t) - 2(R'(t))2 - 3k] / (2 - 2kr2)
    G22 = [ - 4r2R''(t)R(t) - 2r2(R'(t))2 - kr2] / 2

    G33= [ - 4r2R''(t)R(t)sin2(θ) - 2r2(R'(t))2sin2(θ) - kr2sin2(θ)] / 2

    Every other element was 0.

    As you may expect at this point, my question remains the same. What information can I extrapolate with this tensor? Is there any formula containing the Einstein tensor that I can use to derive some info about events and curvature in space-time like there is with the metric tensor?

    Now here is a big one that I have been trying to grasp:

    I haven't yet derived the stress energy momentum tensor for this particular metric because I am still trying to find out whether or not this metric has a non-zero cosmological constant, and if so what that constant is (if anybody can tell me this then that would be nice).

    However, with other stress energy momentum tensors that I have derived, I have noticed that even when I derive the tensor itself, I can never seem to get any info from it.

    I know the meanings of the various elements of the tensor. I know that T00
    is energy density, the other elements in the temporal row and the temporal column are momentum density, and the rest of the elements are momentum flux.

    However, it is the energy density of what? The momentum flux of what? The matter/energy in the metric? Is it the energy density of the matter in the metric at any given event in the space time, or is it the energy density of matter required for the creation of such a region of space time? How exactly do I get useful information from the stress energy momentum tensor? I've asked in threads before if quantities such as the T00 in a stress energy momentum tensor dictate the energy density that it takes to create such a space time. Every time I have asked if this was the correct interpretation, I have been told that this was not how to interpret it.

    To make matters worse, whenever I look at papers on that particular metric, they don't even seem to focus on these tensors that much, as they more so seem to focus on vectors and co-vectors that aren't even in the Einstein field equations and seem to come out of nowhere.

    For example, when I did the Godel metric, the paper talked more about the rotation vector ωμ and the fact that Tμν = ρuμuν where ρ is the density of the matter in the metric and uμ is the co-vector that corresponds to the 4-velocity of the matter in the metric. It didn't even tell me how ρ was derived or where it came from. How would I know what the density of the matter in the metric is when the sites that I go to in order to find the metrics that I study usually only give me the line element? They don't even give me the cosmological constant. Not to mention the fact that according to the wiki, the formula Tμν = ρuμuν only works with dust solutions to the EFE's. Other media goes on to mention killing vectors and other differential geometry topics that don't appear in the Einstein field equations.

    It almost feels as though the tensors in the EFE's are useless because I never seem to get any info out of them except for what I get from the metric tensor.

    This is why I ask, what information about space- time curvature and events do the tensors in the Einstein field equations actually give you? Please be specific. You can use the metric shown in this thread as an example to explain your points if you want.

    Thank you
     
  2. jcsd
  3. Jul 4, 2015 #2

    PeterDonis

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    Staff: Mentor

    space-time, as a general note, you might find this article by John Baez, on the meaning of the Einstein Field Equation, helpful:

    http://math.ucr.edu/home/baez/einstein/

    A fair bit of what I'll say below is discussed in more detail there.

    ##R_{00}## tells you, roughly speaking, about the "acceleration due to gravity". In your example, you have

    $$
    R_{00} = - \frac{3 \ddot{R}}{R}
    $$

    where I have used the double dot to denote the second time derivative, which is standard notation. Suppose we have two objects that are both at rest in the FRW coordinate chart, at slightly different values of the ##r## coordinate (but with the same ##\theta## and ##\phi##, so we can leave those out). These objects are standardly referred to as neighboring "comoving" objects. The proper distance between the objects, at some instant of time, is ##R(t) dr##, where ##dr## is the separation in the ##r## coordinate; the rate at which this distance is changing with time is ##\dot{R} dr##; and the "acceleration" of this distance (the rate at which the rate of change of distance is changing) is ##\ddot{R} dr##. We will assume that ##\dot{R}## is positive, i.e., that the universe is expanding, since that's what we actually observe.

    Now suppose that ##R_{00}## is positive. (We'll see below how we can figure out whether it is or not.) Then we have that ##ddot{R}## is negative, which means that the rate of change of the distance between the objects is decreasing; i.e., their relative motion is decelerating. This can be thought of as "attractive gravity": the matter in the universe is exerting gravitational pull which is slowing down its expansion.

    OTOH, if ##R_{00}## is negative, then we have that ##\ddot{R}## is positive, which means that the relative motion of comoving objects is accelerating. This can be thought of as "repulsive gravity", and it is what we actually observe comoving objects in our universe to be doing at present. Ordinary matter can't cause this (we'll see why below); the only known thing that can is dark energy, which is why the accelerating expansion of the universe is taken as evidence for the existence of dark energy.

    The other components of the Ricci tensor actually don't give you any additional information; the Baez article I linked to discusses why.

    The curvature scalar just gives you a portion of the information in the Ricci tensor. If you have the Ricci tensor, the curvature scalar adds no information.

    The Einstein tensor is what links the dynamics of the metric (in your example, the behavior of the scale factor ##R(t)## as a function of time) to the presence of stress-energy. The Einstein Field Equation just equates the Einstein tensor with the stress-energy tensor; that gives you, in the general case, ten partial differential equations relating the metric and its derivatives to the components of the stress-energy tensor. In your example, there are actually only two equations with non-trivial content, again because the spacetime is so highly symmetric. These equations are called the Friedmann equations, and they give the dynamics of ##R(t)##.

    You can't tell from the metric alone, in the form you've written it, whether there is a nonzero cosmological constant or not. Nor can you tell what the stress-energy tensor is. To know those things, you would need to know the exact dependence of the scale factor ##R(t)## on the time ##t##.

    In other words, the FRW metric is not just one metric; it's a whole family of metrics, which can describe a variety of possible universes, depending on the specific form that ##R(t)## takes. The way you figure out a specific form of ##R(t)## is to make some assumptions about what kind of stress-energy is present, and whether or not there is a cosmological constant. Our best current model of the universe assumes a nonzero cosmological constant, and a stress-energy tensor that is dominated by cold ordinary matter, so that in the standard FRW coordinate chart, the only nonzero component is ##T_{00}##, the energy density.

    Of the matter and energy in the universe.

    It's not "in the metric", it's in the universe. The metric describes the spacetime geometry; the matter and energy (and cosmological constant, if present) determines the metric via the Einstein Field Equation. As I noted above, in your specific example, the EFE reduces to the two Friedmann equations; those relate all of the key quantities involved: the scale factor ##R(t)##, the energy density ##\rho## of matter, the pressure ##p## of matter (assumed to be zero in our model of the current universe, since all the matter is cold, but in the far past it wasn't zero), the cosmological constant, and the curvature constant ##k## (which is zero in our best current model).
     
  4. Jul 7, 2015 #3



    Now here is what I understand from this post thus far:

    First you assumed that R(t) and R'(t) were positive. You subsequently were able to derive that the proper distance between the two objects was R(t)dr because you set the angles θ and Φ equal to 0. You also set k equal to 0 since that is what we currently observe in our universe. Now here is my first question:

    Did you also set t equal to 0 since the objects were both in whatever locations they were in at the same time? This of course would get rid of the (cdt) term. However, if this is the case, then shouldn't the proper distance either be R(0)dr or otherwise shouldn't ds2 = the square root of the quantity [(cdt)2 - R2(t)dr2] ?

    Back to the stuff that I understand:

    If R(t)dr is the proper distance between the two objects, then it acts as a position function of time in a sense. Therefore R'(t)dr is the rate at which this distance is changing (the velocity so to speak), and the rate at which this rate is changing is R''(t)dr (the acceleration so to speak). A positive R''(t) means that the objects are accelerating away from each other (thus implying an expanding universe or a repulsive force) while a negative R''(t) means that the objects are decelerating and attracting towards each other. It is basically just like the basic kinematics you learn in first year calculus (about how the first time derivative of a position function is velocity and how the 2nd time derivative is acceleration and how one direction is a positive derivative while moving in the opposite direction indicates a negative derivative).

    Now, here is where a couple of more questions come about:

    The site where I actually found out about the FRW metric said that as of today, the R(t) that we observe is 1. Under this circumstance, both R'(t) and R''(t) = 0. Wouldn't this imply that the universe is neither expanding nor contracting, which would contradict what we have observed of our universe? Also, for metrics such as the Morris and Thorne wormhole metric (shown here: http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html ) which do not contain any functions of time aside from the (cdt) term, then assuming that you set t = 0 (like what I think you did in your example), differentiating the proper distance between events with respect to time would just yield 0. Wouldn't this also imply a stagnant universe?
    Back to what I understand:

    You went on to say that R00 was basically acceleration due to gravity. Based on the context, I am guessing that R''(t)dr would be the acceleration of the two objects away from each other in the specific situation that you gave, while R00 is the acceleration of the expansion of the universe due to gravity in general (as opposed to just the acceleration of those two objects away from each other). Am I correct in saying this?

    Also, you said that you are assuming that R'(t) is positive, but are you also assuming that R(t) is positive as well?

    Additionally, I get what you are saying about the whole "negative R00 means that the universe is expanding while the positive means that it is contracting. However, what does the specific number or value of R00 tell you? In other words, if R(t) = 5t2 then R00 = - 6 / t2

    Now aside from indicating an expanding universe, what would that - 6 / t2 actually mean? - 6 / t2 of what? Is it saying that the universe is expanding at - 6 / t2 meters per second per second (or some other unit of acceleration)? Also, if that site that I got the FRW metric from was right in saying that R(t)= 1 as of today, then R00 = 0 . Doesn't that contradict the observation that the universe is expanding?

    Additionally, it seems to me that R00 is just acceleration due to gravity only for the FRW metric specifically, but it doesn't seem to be the general case for all metrics. For example, the Godel metric (found here: https://en.wikipedia.org/wiki/Gödel_metric) has the Ricci tensor element R00 = 1 . It's Ricci tensor doesn't even have any functions of time in it, and this metric deals more with swirling dust particles and closed time like curves rather than the expansion of the universe. Would R00 still indicate acceleration or expansion/contraction for a metric that doesn't even deal with the expansion of the universe, but instead deals with things such as black holes, worm holes, closed time like curves, etc... ? If not, then what would R00 (and the rest of the Ricci tensor for that matter) indicate for a solution such as the Godel solution?

    Also, some solutions don't even have a non zero R00 element for their Ricci tensor. For example, the Morris-Thorne worm hole metric (http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html ) only has R11 = (-2b02) / (b02 + l2 )2

    That is actually the only non-zero Ricci tensor element for that particular metric. What info would a Ricci tensor like that give me?

    Finally, what is the point of deriving the entire Ricci tensor if only R00 gives me information? Is it just solely for the purpose of ultimately getting to the Einstein tensor?

    Thank you.
     
  5. Jul 7, 2015 #4

    PeterDonis

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    It's not necessary to set ##t = 0##. All you need is to realize that "distance" is defined as being between two events that are at the same time, i.e., have the same ##t##. That means ##dt = 0## in the line element, so, since we already know that ##d\theta = 0## and ##d\phi = 0##, the only nonzero coordinate differential is ##dr##. But ##t## can still change, so the proper distance ##R(t) dr## is still a function of ##t##.

    This is a definition, not an observation. It is not the only possible definition, but it's a useful one.

    Why? R(t) being 1 (or any value) at some instant of time does not mean it's constant, not changing with time.

    Note first that the key is ##dt = 0## (i.e., we are evaluating the distance between two events that are at the same time), not ##t = 0##, as above. That said, the key thing that makes the time derivative of the proper distance zero is that none of the metric coefficients are functions of time. Obviously that's not true of the FRW metric; all of the spatial metric coefficients are functions of time.

    (Btw, the cdt term, in itself, is not a function of time; it's the difference in time between two events.)

    Yes; that just corresponds to assuming that the universe has some nonzero size. Note that R(t) can't be negative because all of the terms in the metric in which it appears must be positive, since any interval with ##dt = 0## must be a spacelike interval and so must have a positive ##ds^2##.

    Roughly speaking, the magnitude of the acceleration or deceleration of nearby "comoving" objects relative to each other. (The Baez article I linked to gives a more detailed discussion, which is applicable to any spacetime.) But the intuitive interpretation of this can vary from spacetime to spacetime. See below.

    Where is this coming from?

    No, it isn't. See above.

    Remember that, as I said above, ##R_{00}## tells you about the relative acceleration or deceleration of "comoving" objects. That's not the same as acceleration or deceleration of the expansion of the universe. The Gödel spacetime is not expanding (or contracting); heuristically, it is "rotating". But, as you note, it still has a positive ##R_{00}##, indicating (again heuristically) that nearby objects that are "rotating with the universe" will decelerate (if I have my signs right) relative to each other.

    Note that this is in a particular coordinate chart. In other coordinates, ##R_{00}## will be nonzero.

    Because, as your example of the wormhole metric shows, the "obvious" coordinates for expressing the solution are not always the ones that give you a simple nonzero value for ##R_{00}##. Strictly speaking, a direct physical interpretation of ##R_{00}##, like the one Baez describes in his article, requires a local inertial frame. But to transform your solution into a local inertial frame, you need all the components of the Ricci tensor (or any tensor) in the coordinates you start from.
     
  6. Jul 9, 2015 #5
    I was studying the Baez article that you linked me to, and I came across what you are talking about (about how the Ricci tensor tells about how the volume of matter in free fall changes). One key thing however that I came across was information about the Weyl tensor. Now I've got the formula for the Weyl tensor, but I just want to make sure of something.

    You said that if a Ricci tensor such as the one for the wormhole metric doesn't have a non-zero R00 element, then that could simply mean that you have to convert the tensor to a different coordinate system where there is a non-zero R00. Now, the formula for the Weyl tensor involves the Ricci tensor. If I want to derive the Weyl tensor for a metric, but the R00 for the metric is 0, then should I first convert the tensor to different coordinates before deriving the Weyl tensor? Would it make it so that I get no information from the Weyl tensor if I derived the Weyl tensor using a Ricci tensor whose R00 is 0?

    If I should convert the Ricci tensor to different coordinates, how do I do that without having to derive the metric tensors and Christoffel symbols all over again?
     
  7. Jul 9, 2015 #6

    PeterDonis

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    Are you sure? Standard usage denotes both the Ricci tensor and the Riemann tensor by the same symbol, ##R##. You have to look at the number of indexes to tell which is which (Ricci has two, Riemann has four). The Weyl tensor is defined in terms of the Riemann tensor.

    You don't. Changing coordinates changes the forms of all those things.
     
  8. Jul 9, 2015 #7
    This is the formula that a website gave me for the Weyl tensor:

    Cabcd = Rabcd + (1/2) [- Racgbd + Radgbc + Rbcgad - Rbdgac + (1/3) (gacgbd - gadgbc)R]

    In this tensor I see all 3, the Riemann tensor, the Ricci tensor and the curvature scalar.

    Is this incorrect?
     
  9. Jul 9, 2015 #8

    PeterDonis

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    Ah, that's right--the Weyl tensor can be thought of as the Riemann tensor with the Ricci tensor portion "subtracted off" (this is obviously just heuristic). So the Ricci tensor does appear in its definition. But thinking of the Weyl tensor as "depending on" the Ricci tensor is kind of misleading; the point of defining the Weyl tensor is to pick out the part of the Riemann tensor that does not depend on what matter and energy is present locally (the part that does is the Ricci tensor, via the Einstein Field Equation). So physically, the Weyl tensor and the Ricci tensor are best viewed as two separate "pieces" of the Riemann tensor.
     
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