Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metric of a static, spherically symmetric spacetime

  1. Jul 10, 2012 #1
    The (0,0) and (r,r) components are: [itex]g_{00}= -e^{2\phi}[/itex],[itex]g_{rr}=e^{2\Lambda}[/itex]. From the first component, combined with the fact that the dot product of the four velocity vector with itself is -1, one can find in the MCRF, [itex]U^0=e^{-\phi}[/itex]. What does this mean? In the MCRF, the rate of the two clocks is the same, hence in Minkowski spacetime, U0=1. Here, in this case, it isn't so.
    From the (r,r) component, one can deduce that if dt=dθ=d[itex]\phi[/itex] =0, i.e the proper radial distance is dl=[itex]e^\Lambda[/itex]dr. Again, what does this mean? In Minkowski space, the proper distance betwenn r2 and r1 is r2-r1.
     
  2. jcsd
  3. Jul 10, 2012 #2

    Dale

    Staff: Mentor

    It means that Schwarzschild coorinate time is not equal to proper time and that Schwarzschild r coordinate is not equal to proper distance.
     
  4. Jul 10, 2012 #3
    So the coordinate time is the time measured at infinity and time measured at radius r is [itex]e^\phi t[/itex] which is always less than t. Is that correct? Don't understand the significance of [itex]\Lambda[/itex] though. Why would the radial coordinate be anything other than the proper radial distance?
     
  5. Jul 10, 2012 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    The Schwarzschild r coordinate is defined as circumference of a circle centered on the star divided by 2π. Because space is not Euclidean, this is not the same as the radius of the circle (proper radial distance).
     
  6. Jul 11, 2012 #5
    I understand why a circle drawn on a non Euclidean surface like a sphere has a circumference lesser than 2πr but a circle centered on the centre of the star lies in a Euclidean plane.
     
  7. Jul 11, 2012 #6

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    The space enclosed by the circle is non Euclidean. This is quantified by the difference between the proper radial distance and the radial Schwarzschild coordinate. Look at:
    http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid

    Here the exterior and interior spatial Schwarzschild geometry combined:

    Schwarzschild_interior.jpg

    From: http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Vollst.C3.A4ndige_Schwarzschild-L.C3.B6sung
     
  8. Jul 13, 2012 #7
    I see how the surface θ=π/2, t=constant should have been a flat plane in flat spacetime but it's a paraboloid in Schwarzschild spacetime. What happens at r=rs? Flamm's paraboloid only depicts the non-Euclidean equatorial plane. Is there nothing that can be done to show the curvature of all such planes(entire space)?
     
    Last edited: Jul 13, 2012
  9. Jul 13, 2012 #8

    Dale

    Staff: Mentor

    It is hard to draw a 4D curved Lorentzian manifold on a 2D Euclidean flat display device.
     
  10. Jul 13, 2012 #9

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    I'm not sure whether you have grasped the point that the diagrams in post #6 represent the geometry of a 2D slice of 3D space around a black hole (Flamm's paraboloid) or a star/planet (the yellow diagram). If you were near the planet you would perceive the "slice" as a flat plane, but if you measured the geometry of the plane using rulers and protractors, you find the geometry was as if you were on the illustrated curved surface embedded in Euclidean 3D space. In these diagrams, the vertical dimension doesn't represent any direction in 3D space: it's an extra "fictitious" dimension we add to the diagram to illustrate the curvature within the 2D surface.

    As DaleSpam says, you'd need (at least) a 4D diagram to show the curvature of 3D space.

    The other point is that you can't extend the paraboloid inside the event horizon because we are looking at space from the point of view of an observer who is at rest relative to the black hole. No such observers can exist on or inside the horizon.
     
  11. Jul 14, 2012 #10
    If the curvature can be represented only with the aid of an additional fictitious dimension, how could it be measured using rulers and protractors?

    Thanks. I had been thinking of it as a real dimension.
    Why does the paraboloid shown on wiki have an open end? It should be closed like the one in post #6.
     
  12. Jul 14, 2012 #11

    Dale

    Staff: Mentor

    I don't understand the relationship you think exists between the dimensionality of a quantity and the devices used to measure it. Suppose you had a quantity which required 6 rulers and 4 protractors to measure at each point in space. How would you plot that quantity?

    The one on wiki shows only the exterior Schwarzschild curvature, and the diagram of the exterior solution falls apart at a certain radius. The one here shows the exterior and the interior solution.
     
  13. Jul 14, 2012 #12

    Nugatory

    User Avatar

    Staff: Mentor

    One example: Pick three points (not all lying on the same line). Because three points determines a plane, these three points will necessarily all lie in the same plane. Connect the three points with straight lines to form a triangle. Now use your protractor to measure the three internal angles of the triangle. Do they add up to 180 degrees? If not, you've just measured curvature, even though there's no way of representing (that is, drawing a picture of) the curvature without adding an additional fictitious dimension. But that representation is just an aid to visualizing the curvature; it's not needed to measure it, nor to work with it mathematically.

    Another example appears further up in this thread: You have a circle, you measure its circumference with your ruler, then you measure its diameter. If the diameter isn't equal to the circumference divided by ∏... you've measured curvature.
     
    Last edited: Jul 14, 2012
  14. Jul 14, 2012 #13

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    I thought I had answered that:

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Metric of a static, spherically symmetric spacetime
Loading...