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Proper Time in the Rain Frame

  1. Aug 6, 2015 #1
    So, I've been working through "Exploring Black Holes: Introduction to General Relativity" by Taylor and Wheeler, and I'm somewhat puzzled by a term in the "Rain Frame." The "Rain Frame" is meant to be a frame of reference of an object initially released from rest at infinity as it free-falls into a black hole as described by the Schwarzchild Solution.

    To derive this frame, on page B-13, there's a box that starts with "shell coordinates." These express dr and dt for a frame of reference held at a constant distance from the black hole in the Schwarzchild Solution, as though on a "shell" over it (ie, us sitting on the surface of the Earth is a shell frame in these terms).

    [itex]dr_{shell} = \frac{dr}{\sqrt{1 - \frac{2M}{r}}}[/itex]
    [itex]dt_{shell} = \sqrt{1 - \frac{2M}{r}} dt[/itex]

    (Where dt and dr without subscripts are the terms from the Schwarzchild metric)

    We then uses special relativistic transformations to switch to a frame that is freely falling into the black hole from rest at infinity. An object's velocity as a function of radial distance (for the case of being dropped from rest at an infinite distance) was originally derived from the Schwarzchild Metric, then run through the shell transformations for:

    [itex]\frac{dr_{shell}}{dt_{shell}} = -\sqrt{\frac{2M}{r}}[/itex]

    So, using special relativistic transformations for differentials from the shell frame to this passing "rain" frame:

    [itex]dt_{rain} = -v_{rel}γdr_{shell} + γdt_{shell}[/itex]

    Substituting [itex]dr_{shell}[/itex] and [itex]dt_{shell}[/itex] with [itex]dr[/itex] and [itex]dt[/itex] from the Schwarzchild metric, then solving for [itex]dt[/itex]:

    [itex]dt = \frac{dt_{rain}}{ γ \sqrt{1-2M/r } } + \frac{ v_{rel} dr }{ (1-2M/r) } [/itex]

    As is shown elsewhere in the book, in the case of a free-falling object released from rest at infinity,

    [itex]γ ≡ (1-v_{rel}^2)^{-1/2} = (1-\frac{2M}{r})^{-1/2}[/itex]

    (Using units where c = 1)

    And substituting this [itex]dt[/itex] back into the Schwarzchild metric:

    (For describing motions on a plane)

    [itex](dτ)^{2} = (1-\frac{2M}{r})(dt)^{2} - \frac{ (dr)^{2} }{1-2M/r} - r^{2}(d \phi )^{2}[/itex]

    And we finally get:

    [itex](dτ)^{2} = (1-\frac{2M}{r})(dt_{rain})^{2} - 2\sqrt{ \frac{2M}{r} } dt_{rain}dr - (dr)^{2} - r^{2}(d \phi )^{2}[/itex]

    So, I've gone through the process of deriving it in this much detail, because I'm confused, now...

    [itex]dt_{rain}[/itex] is the proper time for the observer, the rate of passage of time the observer sees on their wristwatch as they fall through the event horizon. If you integrated it, you'd get the amount of time as recorded by the plunging observer in-between two events. [itex]dt[/itex] would be outside coordinate time - the time passage as measured by someone far away from the black hole. In the Schwarzchild Metric, as I understand it, [itex]dτ[/itex] is, as it usually is, used to denote the "wristwatch"/proper time of the observer, as well, as opposed to the time of some other observer (in this case, [itex]dt[/itex], coordinate time).

    But if we're using [itex]dt_{rain}[/itex] for our "wristwatch time" of the plunging observer, then what in the world is [itex]dτ[/itex] referring to in this final metric? Who's passage of time is that? Is that the passage of time of someone moving relative to this frame?

    As I'm writing this, though, I think the answer has dawned on me - so this is an entirely new metric. Derived from the Schwarzchild, of course, so closely related, but it's a different metric, and so the [itex]dt_{rain}[/itex] here has taken the role of [itex]dt[/itex], and so instead of comparing the proper time of some observer in the vicinity of a black hole to a distant coordinate time, this is comparing the proper time of some observer in the vicinity of a black hole to the time passage of this freely-falling observer. So the dτ is some third, new observer, that is neither the coordinate nor freely-falling one, but one who is in the immediate vicinity of the free-fall observer(? - if not, then whose r-value is being used? The free-fall/rain observer's?) and whose clock ticks at a different rate. Is this correct?

    (Despite thinking I've figured it out, still asking since I've already written this, others might have a similar question and this would help them, and to verify what I think the answer is)
  2. jcsd
  3. Aug 7, 2015 #2


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    Gold Member

    You may find this helpful. The key is the transformations that take the metric to the new form ( which shows explicitly the flat spatial slices). I'm glossing over the fact that I don't know whose time ##\tau## is either.

    The River Model of Black Holes

    Andrew J. S. Hamilton, Jason P. Lisle

    Last edited: Aug 7, 2015
  4. Aug 7, 2015 #3
    It seems I am missing a crucial piece here, but I don't know what it is - why is it not correct to simply interpret

    [tex]\displaystyle{L=\int_{S}d\tau }[/tex]

    as the length of the geodesic segment S ? Would this not just mean that it is the amount of time physically accumulated on a free-fall rain frame clock along the segment S, and hence that τ is the proper time of a rain frame observer ? If not, why not ?
  5. Aug 7, 2015 #4


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    Gold Member

    Maybe ##\tau## is the proper time of a 'comoving' or fiducial observer ##u^\mu=\partial_t/\sqrt{-g_{00}}## and ##t_{rain}## is the proper time of the boosted fiducial frame which comoves with the river. Beats me.
    Last edited: Aug 7, 2015
  6. Aug 7, 2015 #5
    This seems reasonable. The radial velocity in this coordinate system is


    so the velocity becomes 1 at the event horizon, and then smoothly increases without bound for [tex]r\rightarrow \infty [/tex]. At the same time, we know that the total in-fall time from horizon to centre is finite and well defined - taken together, this only really makes sense if the rain frame time coordinate is physically the same as proper time along the geodesic. Or am I seeing this wrong ? The Wiki article on Gullstrand-Painleve coordinates seems to indicate a similar conclusion, though I am not sure I am interpreting correctly what is said there.
  7. Aug 7, 2015 #6


    Staff: Mentor

    This is correct as you state it, but it's also leaving out some crucial information. See below.

    ##d \tau## is the proper time experienced by an observer who is following a worldline with coordinate differentials dt, dr, etc.. In other words, to know the physical meaning of ##d\tau##, you have to know what kind of worldline you are trying to describe.

    For example, in the quote I gave above, where you say that ##dt## (I'll drop the "rain" subscript as all the coordinates I'll be referring to are rain frame coordinates) is the proper time for the observer free-falling from rest at infinity, what you really mean is this: we are describing a worldline in which ##dr / dt = - \sqrt{2M / r}##, since that condition describes the worldline of an observer free-falling from rest at infinity. This condition means that ##dr = - \sqrt{2M / r} dt##. If you use the line element you wrote down to calculate ##d \tau## for that case, you get:

    d\tau^2 = \left( 1 - \frac{2M}{r} \right) dt^2 - 2 \sqrt{\frac{2M}{r}} dt \left( - \sqrt{\frac{2M}{r}} dt \right) - \frac{2M}{r} dt^2 = \left( 1 - \frac{2M}{r} + 2 \frac{2M}{r} - \frac{2M}{r} \right) dt^2 = dt^2

    In other words, this worldline just happens to be one for which ##d\tau = dt##.

    But if we pick some other worldline, we will get a different value for ##d\tau##. For example, suppose we pick the worldline of an observer "hovering" at a constant ##r## (which must be outside the horizon). For this observer, the only nonzero coordinate differential is ##dt##, and we have

    d\tau = \sqrt{1 - \frac{2M}{r}} dt

    No, it isn't. It's the same spacetime geometry, just described in different coordinates. All these coordinates really do is reparameterize the time coordinate so that events at and inside the horizon have finite time coordinates, instead of the time coordinate going to infinity at the horizon, as it does in Schwarzschild coordinates.
  8. Aug 7, 2015 #7
    So this

    [itex](dτ)^{2} = (1-\frac{2M}{r})(dt_{rain})^{2} - 2\sqrt{ \frac{2M}{r} } dt_{rain}dr - (dr)^{2} - r^{2}(d \phi )^{2}[/itex]

    Is really just this?

    [itex](dt_{rain})^{2} = (1-\frac{2M}{r})(dt_{rain})^{2} - 2\sqrt{ \frac{2M}{r} } dt_{rain}dr - (dr)^{2} - r^{2}(d \phi )^{2}[/itex]

    So it's the same spacetime geometry, so it's the same metrics. I was just slipping up in my vocabulary, there; it's different coordinates is what I meant, thanks for catching my vocabulary error, I didn't realize that by saying "different metrics" I was saying "different spacetime geometry," but since metrics are used to measure spacetime geometry, I see now that it makes sense that different metrics would mean different spacetime geometries.
  9. Aug 7, 2015 #8


    Staff: Mentor

    No. ##dt_{rain}## is a coordinate differential, not a proper time differential. ##d\tau## is a proper time differential, i.e., it is the differential of proper time along a segment of a worldline whose coordinate differentials are ##dt_{rain}##, ##dr##, ##d\phi## (and ##d\theta##, which you left out--but we've been assuming purely radial motion, so ##d\theta## and ##d\phi## will be zero and we can leave those terms out). ##d\tau = dt## is the result of applying the first equation above (with ##d\tau^2## on the LHS) to a worldline for which ##dr = - \sqrt{2M / r} dt##; it is not a general expression that's always true.

    The second equation doesn't make sense anyway; for almost all worldlines it would be saying that ##dt_{rain}## is not equal to itself, which is absurd.
  10. Aug 7, 2015 #9
    Okay. Well, that actually was what I originally meant, heh, I just misunderstood what you were implying in the previous post. I understand this, though, that makes sense. Oh, and all of these equations have been for motion on a plane, hence leaving out any angular references other than [itex]d \phi[/itex] .
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