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

I Radial proper distance between objects in free fall

  1. Mar 17, 2017 #1

    timmdeeg

    User Avatar
    Gold Member

    If we consider 2 shells in Schwarzschild spacetime their radial proper distance ##d s^2 = \frac{1}{1-\frac{r_s}{r}}d r^2## follows from the metric with ##dt=0## and ##d\Phi=0##. Integrating yields ##\Delta s## as a function of the r-coordinates of the shells. A reference with the formula for this is available, if of interest.

    In contrast to the separation of the shells the radial proper distance between objects in free fall is increasing (accelerated, if I see it correctly). But how would we calculate ##\Delta s## in this case? Naively I was tempted to say the same way, but am not sure at all. So any help is appreciated.
     
  2. jcsd
  3. Mar 17, 2017 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes.

    You have to do the same calculation multiple times, each time corresponding to a different Schwarzschild coordinate time ##t##. For each calculation you use the ##r## coordinates of the two objects at that time ##t##.

    In other words, the calculation for the shells only has to be done once because the shells are static--they stay at the same ##r## for all times ##t##. But the calculation itself is done at a single time ##t##, so as long as you meet that condition you can do it for non-static objects as well.
     
  4. Mar 17, 2017 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    It is also worth noting, by the way, that doing the calculation as I described it in my last post--using the ##r## coordinates of the two objects at the same Schwarzschild coordinate time ##t##--is not the only possible way to do it. It just happens to be the easiest if you've already done the calculation for the static shells. But the surfaces of constant Schwarzschild time ##t## are not the ones that the two free-falling objects would treat as "surfaces of constant time" locally; if you want to use the latter surfaces as your definition of "at the same time" (so you can calculate the distance between the objects at a given "time" according to the objects themselves), the calculation gets more complicated and will give a different quantitative answer (although qualitatively it is still the same, the proper distance is still increasing with time).
     
  5. Mar 17, 2017 #4

    timmdeeg

    User Avatar
    Gold Member

    Here I have a problem. If the proper distance between the free-falling objects depends on my choice of the time, constant Schwarzschild time or sloppy the object time, then the definition of their proper distance seems to be not invariant or at least ambiguous. How would you define "a given "time" according to the objects themselves"?
     
  6. Mar 17, 2017 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    That's correct, it is. See below.

    For the case of the shells, since they are at rest relative to each other and to the hole, the proper distance as calculated in Schwarzschild coordinates turns out to be the right definition of this. But for the free-falling case there is no single "time" according to the objects themselves, since they are not at rest relative to each other.

    There is an invariant that captures the fact that the free-falling objects are diverging, but it's somewhat technical. It's called the expansion scalar. For the case of the shells, this scalar is zero (which it should be because the objects are at rest relative to each other). For the case of the free-falling objects, it's positive, which indicates that they are diverging. This scalar's physical interpretation is, heuristically, the rate of change with respect to the closest thing you can define to "time according to the objects themselves", of the closest thing you can define to "distance according to the objects themselves", when the objects aren't at rest relative to each other.
     
  7. Mar 18, 2017 #6

    timmdeeg

    User Avatar
    Gold Member

    Yes, in contrast to the FRW-model, where one can define a surface of simultaneity according to ##a(t)##. It seems, the decisive difference is that here comoving observers see themselves redshifted symmetrically.

    Thanks for this hint. Wikipedia https://en.wikipedia.org/wiki/Congruence_(general_relativity) says:
    "the expansion scalar represents the fractional rate at which the volume of a small initially spherical cloud of test particles changes with respect to proper time of the particle at the center of the cloud".

    Would this applied to Schwarzschild spacetime mean that the expansion scalar is represented by the relative acceleration of the free-fall objects with respect to the proper time of said particle (the volume makes no sense, because its constant here)?

    So in principle one could calculate the proper distance between the two objects in free-fall according to the two methods you explained assuming the same r-coordinates in both cases and would obtain different results. In both cases ##\Delta s## would be increasingly larger than ##\Delta r## closer and closer to the black hole. However approaching the event horizon ##ds/dr## goes to infinity in the shell scenario, where by I'm not sure about the ratio ##\Delta s/\Delta r## here. Does eventually the close-to-the horizon behaviour allow to prefer one of the two methods?
     
  8. Mar 18, 2017 #7

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The underlying difference is that the FRW model has uniform density everywhere and is homogeneous, whereas Schwarzschild spacetime describes a vacuum region which is not homogeneous (the radial direction is different from the tangential directions).

    Not quite. The Wikipedia page example uses a small ball of test particles, which includes particles separated tangentially as well as particles separated radially. See below.

    Yes, because the spacetime curvature in the Schwarzschild vacuum region is just Weyl curvature, and Weyl curvature is volume-preserving. So, as above, a small ball of test particles will stretch radially, but squeeze tangentially, and the two effects on the volume cancel out so the volume doesn't change, and the expansion scalar of the ball as a whole is zero.

    But if you just look at particles separated radially, the expansion scalar restricted to just those particles (i.e., to a radial "slice" through the small ball) is positive. (I might be making a minor abuse of terminology in calling this an "expansion scalar".) And if you just look at particles separated tangentially, then the expansion scalar restricted to just those particles is negative. Heuristically, the sum of the two is the expansion scalar of the whole ball of test particles, and is zero since the radial stretch and the tangential squeeze cancel.
     
  9. Mar 19, 2017 #8

    timmdeeg

    User Avatar
    Gold Member

    Thanks, your explanations have been very helpful.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Radial proper distance between objects in free fall
Loading...