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

Newtonian Gravity: curved time

  1. Jan 24, 2009 #1
    I'd appreciate any explanations and maybe an available source that discusses how and why time is relatively more curved than space. Is there is particular source for time curvature in the stress/energy/momentum tensor?? Thank you.
  2. jcsd
  3. Jan 24, 2009 #2


    User Avatar
    Science Advisor
    Gold Member

    This depends which units you use when putting space & time into one manifold. Using geometric units (1 second in time =~ 1 light second in space) where c = 1, it is not much curved.

    Pluging c=1 into the Schwarzschild metric:
    shows that at radial coordinate r both: the proper-time dimension and the space dimension are "stretched" by the same factor 1/SQRT(1-Rs/r)
    Last edited: Jan 24, 2009
  4. Jan 24, 2009 #3


    Staff: Mentor

    I don't think that it is that it is more curved in time, I think it is just that humans are much longer in the time dimension than any of the space dimensions (~75 light years long in time vs ~6 light nanoseconds long in height).

    That also explains why the curvature in the time dimension is more important for things like projectiles, a good baseball pitch goes about 60 feet in space but about 400 million feet in time. So curvature in the time dimension makes a bigger difference.
    Last edited: Jan 24, 2009
  5. Jan 24, 2009 #4
    To add to what Dale was saying, you might use cylindrical coordinates for planetary orbits, where the axial dimension is time. For one orbit of the Earth around the Sun, the equivalent 'distance' covered in time is one light year. It's a very stretched-out spiral.

    I'm not sure this really helps, but the Newtonian gravitational potential is expressed in terms of the mass density as
    [tex]\nabla^{2}\Phi = 4\pi G \rho[/tex]​
    [tex]\ \rho[/tex] is mass density.The mass density is the time-time component of the stress energy tensor [tex]\ T_{00}[/tex].
    [tex]\nabla^{2}\Phi = 4\pi G T_{00}[/tex]​
    Where we're only concerned about the gravitational field of stationary mass, the gravitational potential, in the weak field limit is related to the metric
    [tex]g_{00} = - \left(1 + 2\Phi \right)[/tex]​
    [tex]\ \Phi[/tex] is a small perturbation on the Minkowski metric.

    I should add that this is an approximation, not an exact solution, but where [tex]\ \Phi << 1[/tex]
    How this squares with what A.T. was saying, I don't know. This is how Sean Carroll presents it.
    Last edited: Jan 24, 2009
  6. Jan 24, 2009 #5
    This should suppliment what A.T. was saying. We can take the full Schwarzchild metric, throw out the parts in theta and phi that don't effect the result, and keep the temporal and radial components.

    [tex]c^2 d\tau^2 = \left( 1- \frac{r_s}{r}\right) c^2 dt^2 - \left( 1- \frac{r_s}{r}\right)^{-1} dr^2[/tex]

    Where the Schwarzchild radius is very small for objects like the Sun and Earth,

    [tex]r_s << r[/tex]

    so that

    [tex]\left( 1- \frac{r_s}{r}\right)^{-1} = \left( 1+ \frac{r_s}{r}\right)[/tex]

    The approximated metric is written as

    [tex]c^2 d\tau^2 = \left( 1- \frac{r_s}{r}\right) c^2 dt^2 - \left( 1+ \frac{r_s}{r}\right) dr^2[/tex]

    The metric can now be separated into two parts: a flat part, and a small perturbation.

    [tex]c^2 d\tau_{0}\; ^2 = c^2 dt^2 - dr^2[/tex]

    [tex]c^2 d\tau_{p}\; ^2 = - \frac{r_s}{r}c^2 dt^2 - \frac{r_s}{r}dr^2[/tex]

    The flat part is the Minkowski metric,


    in one spatial dimension.

    We can call the perturbations [tex]\ h_{\mu\nu}[/tex]

    [tex]h_{00} = - \frac{r_s}{r}[/tex]

    [tex]h_{rr} = - \frac{r_s}{r}[/tex]
  7. Jan 25, 2009 #6


    Staff: Mentor

    Sometimes I really dislike the font used for LaTeX. I have a really hard time distinguising between "tau" and "r". But thanks for posting that Phrak, is the perturbation that you mention in post number 4 the same as the perturbation in post number 5?
  8. Jan 25, 2009 #7
    after reading that "time is more curved than space" several times in different threads here I came to the conclusion maybe I missing something...looks like maybe not.
  9. Jan 25, 2009 #8


    User Avatar
    Science Advisor
    Gold Member

    DaleSpam put it quite nicely: Most things we observe advance mainly trough time, so time curvature affects them more. Light advances trough space with maximal velocity and is affected by space curvature and time curvature equally.
  10. Jan 25, 2009 #9
    Post #4 was a red herring, I'm afraid. Note the mass density is variable, where we want vacuum conditions. A.T. was on the right track with the Schwarzchild metric.
  11. Jan 25, 2009 #10
    I compared the LaTex in my posts side by side with the thread Introduction to Latex Typesetting, https://www.physicsforums.com/showthread.php?t=8997. I don't know why my LaTex became washed-out. But tau and r are some of the hardest characters to distinguish, as you say. My only use of tau was in the metric on the left hand side.

    I was scanning S.C.'s chapter on Schwarzchild and found this

    [tex]g_{00} = -(1+2\Phi)[/tex]

    [tex]g_{rr} = (1-2\Phi)[/tex]

    [tex]\Phi = -GM/r[/tex]

    Interestingly, this, and finding stress energy tensor, would be working somewhat backwards the easy way from Swarzchilds' work: solving for the metric from constraints on the stress energy tensor.
  12. Jan 25, 2009 #11
    It is not exactly right to say time is relatively more curved than space.
    Look at this metric equation, [tex]c^2 d\tau^2 = \left( 1- \frac{r_s}{r}\right) c^2 dt^2 - \left( 1- \frac{r_s}{r}\right)^{-1} dr^2[/tex]
    There is a [tex]c^2[/tex] in front of the [tex]dt^2[/tex] instead of the [tex]dr^2[/tex]. What that means is the change of time will be multiplied by "c" compared with just the change of distance.
    So that equation can be reduced to [tex]c^2 d\tau^2 = \left( 1- \frac{r_s}{r}\right) c^2 dt^2 - dr^2[/tex],
    It can be shown this equation is equivalent to Newtonian law of gravity
    Last edited: Jan 25, 2009
  13. Jan 25, 2009 #12
    I don't know what it means to say 'more curved' unless that comparison can be made in the same units. For relativity, that means setting c=1. But metric elements themselves are not the measure of curvature; the components of the Riemann and Ricci tensors are. Those are still waiting to be posted by someone.
  14. Jan 26, 2009 #13


    User Avatar
    Science Advisor

    Last edited: Jan 26, 2009
  15. Jan 27, 2009 #14
    My apologies, feynmann.

    I read Naty's question, about curvature and say 'Naturally it has to do with the Riemann curvature tensor, so we need to know what the Riemann curvature tensor looks like', but that's wrong if we want to know how things are in orbit around bodies with weak fields. It's not the curvature we want to look at--that will just lead to the stress-energy tensor, and tell us how vectors change direction transported around loops.

    We know what the trajectories of freely falling bodies look like where gravity is weak; circles and ellipses, or segments of either. So we want to know what the geodesic equations look like for these trajectories. This is where the time-like and space-like elements of the metric come in, so we can compare how each of the two contribute where the metric is tweaked from nice, flat Minkowski spacetime.

    For this, we don't need the Riemann tensor but we do need the Christoffel connection which contains first derivatives of the metric.
  16. Jan 29, 2009 #15
    63024 AU's = 1 ly

    In one orbit around the Sun, the Earth is moved 6.28 AU in space.

    The temperal path length is 10,000 times as long as it's spatical pathlength.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?