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Curvature of Space: What is it?

  1. Jan 15, 2014 #1
    Curvature of Space-time: What is it?

    General relativity talks about curvature of space-time due to mass. What does it actually mean by 'curve'. Is the space made of something that we can say is curving? If space is purely 'empty', then what is getting curve? Or is it that curving is just an analogy used?
    Last edited: Jan 15, 2014
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  3. Jan 15, 2014 #2


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    Both space and space-time are curved in general relativity. The easiest way to describe how to measure curvature involves being able to measure angles, but you can also infer and compute curvature by only being able to measure distances.

    For simplicity I'll talk about how to measure the curvature of space.

    The "easy way" involves selecting three nearby points, and drawing a triangle made out of geodesics (geodesics are the shortest lines connecting those points*). You then measure and sum the interior angles of the triangle.

    If the sum is not exactly 180 degrees, you have curvature. The amount of the curvature is the ratio of the angular excess divided by the area enclosed by the triangle.

    (For a proof on the sphere, see http://nrich.maths.org/1434)

    *This is actually true in the normal convex neighborhood of the points, if the points are too far apart, the line of shortest distance between two points will still be a geodesic, but the presence of multiple geodesics connecting the two points means that some of the geodesics might not minimize the distance.

    Because you can measure distances (and angles) even in empty space, you don't need a "thing" to curve. The ability to measure distances between points is necessary and sufficient to define the geometry, and its curvature.

    The technique above is one way of measuring what is called the "sectional curvature" of a surface, see for instance the wiki article http://en.wikipedia.org/wiki/Sectional_curvature.

    The descriptions of curvature used in GR are the Riemann curvature tensor and various contractions of it (such as the Ricci and Einstein tensors). It's possible in principle to determine the full Riemann curvature tensor from a knowledge of all the sectional curvatures for the various plane "slices" of the geometry, though I don't know the details offhand.
  4. Jan 15, 2014 #3


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    To add to what pervect posted, in GR, curvature of *spacetime* is the same as tidal gravity. We measure tidal gravity by looking at the motion of freely falling objects that are close together, to see if there is relative acceleration between them.

    For example, suppose we have two objects that start out momentarily at rest at slightly different altitudes above the Earth. As the objects free-fall towards the Earth, their radial separation will increase (because the one that is slightly closer to the Earth will accelerate downward, relative to an observer at rest, slightly more). This shows that tidal gravity is present.

    The reason tidal gravity, as just described, can be interpreted as spacetime curvature is that free-falling objects move on geodesics of spacetime--the spacetime analogues of curves like great circles on a sphere. So the two free-falling objects that I described above are moving on nearby geodesics which start out parallel--meaning parallel in spacetime, because they are both at rest at some instant of time, so if we drew a spacetime diagram with time, according to an observer at rest relative to the Earth, on one axis, and radial distance from Earth on the other axis, and drew the worldlines of both objects on the diagram, they would both cross the "t = 0" line at right angles. But these geodesics don't *stay* parallel--the objects don't stay at rest relative to each other, as I showed above. That means spacetime must be curved.

    Note that this is a different definition of "curvature" than pervect gave, but the two definitions are actually equivalent. For example, on a sphere, triangles drawn on the sphere's surface have angles summing to more than 180 degrees; but also, if you take two great circles (which are the geodesics on a sphere) that start out parallel--for example, two meridians of longitude both crossing the equator at right angles--they don't stay parallel; the meridians get closer together, eventually meeting at the poles. It can be shown mathematically that the two manifestations of curvature always occur together: any space or spacetime that shows one will also show the other. So you can use either one to tell whether curvature is present.
  5. Jan 15, 2014 #4


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    Ill add a bit too. "Parallel transport" of a vector along a closed curve is actually the way that most textbooks in GR discuss curvature, but I felt that the angle-summing method was more intuitive.

    Geodesic deviation is the technical name for the tendency of the distance between nearby parallel geodesics to accelerate relative to one another. This value of the accleration can be expressed by the Riemann curvature tensor, see for instance the wiki article http://en.wikipedia.org/wiki/Geodesic_deviation
    Last edited: Jan 15, 2014
  6. Jan 15, 2014 #5


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    Some pictures to explain intrinsic curvature:

    Positve curvature (circumference / diameter < PI) :


    Negative curvature (circumference / diameter > PI) :


    For triangles (three points connected geodesically) you have:

    Positve curvature : inner angle sum > PI
    Negative curvature : inner angle sum < PI
    Flat : inner angle sum = PI


    More info:

    However, as others said, this intrinsic curvature is related to tidal effects, or the non-uniformity of gravity. In a small area, where tidal effects are negligible, the Equivalence Principle applies, and gravity can be modeled without intrinsic curvature, just like in a rocket accelerating in flat space time. This is shown in the animation below. Note that the cone surface has no intrinsic curvature, because it can be rolled out flat. Unlike a sphere or a saddle surface.


    Over a larger area you need intrinsic curvature to make those local cone-like patches fit together. This looks something like this:


    It is "just an analogy" in the sense that it is an abstract mathematical model, like anything in physics. The mathematical concept of curvature doesn't require space to be filled with something, in order to be curved. Curvature is a certain distortion of distances between coordinates.
    Last edited: Jan 15, 2014
  7. Jan 15, 2014 #6
  8. Jan 16, 2014 #7

    Jonathan Scott

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    The curvature mentioned so far here is intrinsic curvature, which is like the curvature of part of a ball and causes local geometry to diverge from flat space. This is effectively present wherever mass density is present, but not outside the mass in empty space.

    When free fall motion through empty space under gravity is described as following the curvature of space-time, this is a bit different. This is mathematically more like the curvature of a cone of paper, which makes lines appear curved from some points of view (when watching from a distance) but to someone moving around the cone, the local geometry is like ordinary flat paper. This is related to the way in which a free-falling object appears to be accelerating and/or following a curved path to a distant observer, but an observer with the object would not feel any acceleration.

    Note that free fall motion under gravity is normally dominated by curvature of space with respect to time. That is, as a particle travels around an orbit or otherwise follows a free fall path, a line showing its motion in space plotted against time curves towards the gravitational source. Gravity also curves space by the same curvature (in units where c=1) but this only has a significant effect on motion at relativistic speed, including light, which is deflected equally by the curvature with respect to time and space, so it ends up being deflected by twice the amount that Newtonian gravity theory would predict (as that only includes the time part).

    Anyway, the important point to note is that there's more than one sort of curvature involved, and when empty space is described as curved, that's a different sort of curvature from the curvature at a location with a non-zero mass density.
  9. Jan 16, 2014 #8


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    This is not correct. Spacetime curvature in GR is tidal gravity, and tidal gravity is present even in empty space. There is a sense in which the curvature in the presence of mass density (or more generally stress-energy) is different from the curvature present in empty space, but it's not a matter of "intrinsic" curvature vs. some other kind. See further comments below.

    I don't think this is a very good analogy. The cone of paper does have extrinsic curvature, but the only reason we know this is that we can perceive and measure the higher-dimensional space that the cone is embedded in. We can't perceive or measure any higher-dimensional manifold that spacetime is embedded in, so the only curvature of spacetime we can measure is intrinsic curvature. The intrinsic geometry of the cone is flat globally (except at the apex of the cone), not just locally; but the intrinsic geometry of spacetime is only flat locally, not globally.

    That's because free-falling objects follow geodesics. But spacetime curvature--meaning intrinsic curvature, which, as above, is the only kind we can measure in spacetime--is not a matter of whether or not objects are following geodesics. It's a matter of whether or not tidal gravity is present. As I described in my earlier post, you can directly measure tidal gravity using only free-falling objects, following geodesics; each individual geodesic is "straight", but the relationship between them shows the presence of spacetime curvature.

    As I noted above, this isn't true; spacetime curvature in general is just tidal gravity, and is present everywhere.

    However, relativists often divide up spacetime curvature into two types: Weyl curvature and Ricci curvature. The kind of curvature I described in my earlier post, which is what we usually think of as "tidal gravity", is Weyl curvature; it can be present in empty space. Ricci curvature, however, can only be present when nonzero stress-energy is present; that's because Ricci curvature is the kind that appears in the Einstein Field Equation, which relates curvature to stress-energy. In the idealized case of a perfectly spherically symmetric gravitating body, Ricci curvature is only present inside the body, and Weyl curvature is only present in the empty space surrounding the body (at least, I think the latter is true, but I'd have to check to be sure). But both of them are "intrinsic" curvature.
  10. Jan 16, 2014 #9


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    Indeed as Peter noted there is obviously tidal gravity in vacuum space-times (excluding trivial solutions) and this is due to the non-vanishing Riemann curvature; the Ricci curvature does not determine the Riemann curvature in 4 dimensions so just because the Ricci curvature vanishes in vacuum space-times doesn't mean the Riemann curvature will also vanish. Indeed the Weyl curvature (which comes from the Weyl tensor i.e. the conformal tensor) will still be non-vanishing.

    Given a space-time ##(M,g_{\mu\nu})## and a geodesic congruence ##\xi^{\mu}##, the expansion scalar ##\theta := \nabla_{\mu}\xi^{\mu}## will measure the convergence (resp. divergence) of neighboring geodesic integral curves of ##\xi^{\mu}## towards (resp. from) a reference geodesic from the same congruence. For example the congruence of geodesics in Scwarzschild space-time defined by test particles freely falling from rest at infinity has ##\theta = -\frac{3}{2}(\frac{2M}{r^3})^{1/2}## for ingoing geodesics as we would expect.

    Tidal gravity is second order in ##\nabla_{\mu}## since it relates to relative accelerations between integral curves of ##\xi^{\mu}##. Tidal gravity governs the proper time evolution of ##\theta## along the reference geodesic as well as the proper time evolution of the vorticity and shear of the congruence. Imagine for example a vector ##\eta^{\mu}## that extends from the reference geodesic to a neighboring geodesic in the congruence such that ##\eta_{\mu}\xi^{\mu} = 0## everywhere along the reference geodesic and such that ##\eta^{\mu}## remains locked to the neighboring geodesic (we can arrange for both of these by Lie transporting ##\eta^{\mu}## along ##\xi^{\mu}##: ##\mathcal{L}_{\xi}\eta^{\mu} = 0##). Then the acceleration of the neighboring geodesic relative to the reference geodesic is simply given by the equation of geodesic deviation: ##a^{\mu} = \xi^{\gamma}\nabla_{\gamma}(\xi^{\nu}\nabla_{\nu}\eta^{\mu}) = -R_{\gamma \delta \nu}{}{}^{\mu}\eta^{\delta}\xi^{\gamma}\xi^{\nu}##.

    Clearly this tidal gravity (relative acceleration) will manifest itself in non-trivial vacuum space-times such as Schwarzschild space-time and it will fully determine the Riemann curvature, which is a purely intrinsic measure of curvature. GR does not deal with extrinsic curvature for obvious reasons.

    Free fall motion is characterized by the geometric statement that ##\xi^{\nu}\nabla_{\nu}\xi^{\mu} = 0##; this is an invariant statement about the vanishing path curvature of geodesics. You're confusing this with the shape of the resulting integral curves of ##\xi^{\mu}## as represented in different coordinate systems. This has nothing to do with extrinsic curvature it is simply a result of using coordinate systems corresponding to different physical observers: the integral curves of ##\xi^{\mu}## are straight lines in locally inertial frames but, for example, will appear curved in the proper reference frame of a static observer in Schwarzschild space-time.

    EDIT: I would recommend reading section 13.6 of MTW and working through exercise 13.14.
    Last edited: Jan 16, 2014
  11. Jan 16, 2014 #10
    A complemenatry description...In another discussion I had posted a view from Roger Penrose,

    PeterDonis posted there:
    This also appeared in that discussion but I did not record the source:

  12. Jan 17, 2014 #11

    Jonathan Scott

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    I have no major disagreement with the technical details in this thread, but the original post seemed to be at a very basic level and I thought it could be useful to illustrate the ideas informally at a lower level.

    I hoped it would be clear that I was of course using the "ball" curvature analogy (intrinsic curvature in 3D) to relate to Ricci, and I was using the "cone" analogy for linear curvature in the sense of the shape of a ruler or light beam or the path of a falling object mapped relative to time using a typical coordinate system, as this is the common usage outside GR.

    Of course in a real gravitational situation, there is tidal curvature everywhere, but I consider that a less important and higher order effect.
  13. Jan 17, 2014 #12


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    I'm not sure I understand this. "Tidal curvature" is the only kind of spacetime curvature there is. It's only "higher order" in the sense that it depends on second derivatives of the metric; but the first derivatives of the metric (the connection coefficients) aren't properly described by the term "curvature", and they don't form a tensor anyway.
  14. Jan 17, 2014 #13
    Back to some real basics for rushikesh if he is still reading:

    from his original post

    yes. Gravitational spacetime curvature is also caused by other forms of energy. You can see a visualization of different sources of gravity here:


    In GR spacetime does curve, but there are multiple measures, multiple components, of such curvature. Tidal gravity is a physical way to think about 'curvature' in GR.

    Another way to think about it was explained to me some years ago in these forums: SR is modeled on flat graph paper; You might have 'curves' on that graph paper illustrating space and time but those are simply coordinate type curves: expressing a relationship in curved coordinates [like spherical coordinates] just changes the picture, not the physics. When the graph paper itself is curved [in such a way it cannot be flattened without tearing] THAT is instrinsic spacetime gravitational curvature.

    Dynamic illustration of curvature:

    It's spacetime that is curving not just space. But nobody really agrees on what spacetime 'is' anymore than they can explain what an electron 'is'. The best we do is to predict actions and confirm those predictions in experiments [make observations].

    Here are a few attempts to describe aspects of space and time which I like:

    John Archibald Wheeler,

    Even a 'vacuum' is not necessarily 'empty', depending on what you mean. Spacetime has vacuum energy, a cornerstone of quantum mechanics. And in GR there is an ever present sort of 'negative' pressure', very tiny, but very important, also called the cosmological constant and called by some 'dark energy'.

    Curving of space time is the geometrical interpretation originated by Einstein. Other physical physical forces are modeled via field theory. Einstein's view of gravity is unique and distinct. It works everywhere we can so far imagine except at the big bang and the center of black holes.
    If there is a better 'analogy' it may come from our next theory....quantum gravity.
  15. Jan 17, 2014 #14
    Please note that although general relativity talks about curvature of space-time , but actually there is another representation of geometry of space-time called Weitzenbock geometry , which the curvature of space-time is exactly zero, but information of gravity is hidden in a non-symmetric connection. This interpretation of gravity is also called Teleparallel gravity , the space time doesn't have curvature but instead the torsion(anti symmetric part of connection) is non-zero. So we encounter a convention, which do you like? the gravity as curvature of space-time or the torsion of it?
  16. Jan 28, 2014 #15
    The youtube video posted above was great for visualizing why the apple falls from the tree's point of view - thanks! In fact all those links to visualizations were really helpful.

    Got some questions ...

    When spacetime is 'bent' by mass is it being bent through a higher dimension than the 3+1 of spacetime itself? Or are the 3+1 dimensions all you need in this interpretation of gravity?

    On the Einstein vs Newton video it looks like the time axis was being bent into the space axis from the external point of view (and vice versa): is that true? Is that what's happening?

    Can I think of time like a spacial dimension at right angles (in a direction I can't perceive or point in) to the 3 of space? Or is that wrong?

    If responding to me please use simple words and pictures as much as is possible ;-) More links to brilliant visual descriptions like those one above would be perfect!
  17. Jan 28, 2014 #16
  18. Jan 28, 2014 #17
    The intuitive examples of curvature made here are related to curvature of space, but I didnt notice anyone related to the curvature of spacetime. I mean, the idea of curvature of space is mathematically given by connections [itex]\Gamma[/itex]ijk that mean something like when you go around the jk plane how an "i" arrow changes (Leonard Susskind explains it perfectly, I have to check again his course). My doubt is, what does the connection intuitively mean when i, j or k is the time direction?
  19. Jan 28, 2014 #18


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  20. Jan 28, 2014 #19


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    Remember that Post #2 explained that if the angles of a triangle do not add up to 180, then you are in curved space. There is no need, or reason to believe, that there are more dimensions that the space is "curved in". The diagrams of triangles in Post #5 are shown on a surface a higher dimensional space, but that can be misleading. If you make a triangle with "straight" lines and the angles do not add up to 180, then you know that "straight" lines in your world are really curved. Imagine a bug on the surfaces on Post #5. He thinks those grid lines are all straight lines and he knows nothing about higher dimensions. But a smart bug would still know he is on a curved surface.
  21. Jan 29, 2014 #20
    I feel like I'm still missing something. I think you're right - perhaps it's the diagrams that are confusing me because they all seem to be showing spacetime (represented by a 2 dimensional surface) being curved through a higher dimension to produce the geometrical effects mentioned above ... or are those diagrams representing 2 spacial dimensions and the 'higher' dimension that they're curving through is time? Is that it?

    If not, what other ways are there to explain why, when there are large masses about, the angles of triangles don't add up to 180 degrees and parallel lines meet other than with a higher dimension that we can't perceive to curve spacetime through?
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