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B Resistance of Space-Time fabric

  1. Nov 20, 2016 #1
    According to relativity gravity is just an illusion caused by wrapping of space-time fabric by mass, right? So, what is the elasticity of of space-time? There must be some king of resistance for the wrapping, otherwise even a tiny mass would plunge in the space time for eternity.
     
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  3. Nov 20, 2016 #2

    Dale

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    Hi Ashiquee, welcome to PF!

    The "rubber sheet" analogy is pretty flawed, so you can easily get misled. The basic equation of general relativity looks something like G=kT. In that equation, T is called the stress-energy tensor and actually contains the stress tensor from Hookes law and G is the curvature which seems like a geometric distortion similar to strain. So you might be tempted to think of k as being the stiffness and G the strain, but the units don't work out. Strain is dimensionless, but curvature has units of 1/m^2, so they are different things. Similarly, in Hooke's law the stiffness is measured in Pascals, but here it is Pascal/m^2.
     
  4. Nov 20, 2016 #3

    Vanadium 50

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    It's just an analogy. One might as well ask what detergent is best for removing stains from the fabric of space-time.
     
  5. Nov 20, 2016 #4

    pervect

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    We can tell that space-time is wapred without knowing exactly how it deforms. In order to describe the warping in terms of elasticity, we would need a model of the deformation process, but all we have is a mode of the warped geometry.

    To give an example, let's talk about the warping not of space-time, but of a plane. A plane is flat, but if we consider the surface of a sphere, another 2-dimensional manifold, we can say that the surface of the sphere is curved.

    We can imagine a flat-lander, from Abbot's book "Flatland", a fictional 2d being, living on the surface of a plane, and another flat-lander living on the surface of the sphere. We'll call the second flat-lander a "sphere-lander' for reasons that I hope are obvious. How could the sphere-lander come to realize that he was living in a curved geometry, and how could the flat-lander come to realize he was living in a flat geometry?

    There's a simple procedure, which requires that the two entities be able to find the shortest path between two points. For instance they might have a stretchy string and they pull it taut. They call the path that the string follows a straightline, though from our perspective the sphere-lander's straighlines are great circles.

    There's a fairly simple test for flatness. You move a set distance in a straightline, might a right angle turn, move the same distance again, turn 90 degrees, move again, turn 90 degress again, and move again.

    If you're a flat-lander living on a flat plane, you'll always wind up right where you starte, after tracing out a square. If you are a sphere-lander, in general you will not. This is not the most sensitive test for curvature, but it's simpler to describe than other tests that are more sensitive. A specific example would be the case where the distance you move is 1/4 the circumference of the sphere, in that specific case you wind up back at your starting point after two right angle turns and 3 moves.
     
  6. Nov 22, 2016 #5
    Sir, what I'm asking is.. Mass wraps Space-Time. So, to how much extent it can be wrapped? or What is the min mass requires to wrap the space-time.
     
  7. Nov 22, 2016 #6

    Orodruin

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    See post #3.
     
  8. Nov 22, 2016 #7

    Ibix

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    Better not to describe it as warping since that does kind of imply a fabric. Gravity is spacetime curvature, meaning that the rules of geometry are not Euclidean near masses.

    There is no minimum mass for causing spacetime curvature so far as we are aware. Any mass will do it, no matter how small. It's seriously hard to spot non-Newtonian effects for something the size of Earth, though.

    It's a bit more difficult to say what is the maximum curvature because the full description of curvature at a point requires twenty numbers. You can summarise them in a number of physically meaningful ways, though, and many of those summaries can go to infinity. So the best answer is probably that there is no upper limit.

    Both of these facts may be revised by a quantum theory of gravity when we work it out. But we haven't got it yet, so who knows.
     
  9. Nov 22, 2016 #8

    pervect

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    The mathematical entity that describes the warping or curving of space-time is called the Riemann curvature tensor. It's not a single number, it requires at least 20 unique numbers to describe - in the form of 21 numbers and plus constraint equation. So this makes the "how much" question difficult to answer.

    The components of most interest in the 4-d space-time Riemann tensor are physically represented by tidal forces in Newtonian gravity, and are described by six of the twenty-one numbers.

    The same Riemann curvature tensor also describes curved surfaces in space. The simplest and most familiar examples are the curvature of two dimensional surfaces, where the Riemann tensor only has one unique component, and thus one can describe curvature by a single number. Thus I'l describe this simpler example to try to give some intuitive insight. Actually, I'll go one step further in the simplification process - I'll consider only one of the most simple curved 2-d surfaces, the surface of a sphere.

    Probably the simplest popularized description of the Riemann curvature tensor of a sphere would be to say that it describes the amount of angular excess in triangles. To take a specific example, consider the curved geometry of a sphere. This is described by spherical trigonometry, something you may not be familiar with but something that's a lot easier to learn than general relativity.

    In spherical trigonometry, the sum of the angles of a spherical triangle, made up of great circles (which play the same role as straight lines do in Euclidean geometry) is 180 degrees plus some constant times the area of the triangle. To give some more specifics, if you have a spherical triangle with three right angles (which is obviously impossible on a plane, but not on a sphere), it has a total area of 1/8 the surface area of the sphere.

    Modulo some constant factors, which I haven't worked out, the ratio of the angular excess to area is proportional to the value of the Riemann tensor. One can see then that for a sphere of diameter D, the value of the single number representing the Riemann curvature of the sphere is proportional to 1/D^2.

    So for 2-D surfaces, we can answer the "how much" question, though it doesn't really generalize well to the 4-d case.
     
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