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.(adsbygoogle = window.adsbygoogle || []).push({});

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)

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proper Time in the Rain Frame

Loading...

Similar Threads - Proper Rain Frame | Date |
---|---|

I Difference between proper time and coordinate time | Monday at 2:13 PM |

B Proper time of photon in Friedman metric | Feb 26, 2018 |

B Rest frame in GR | Feb 20, 2018 |

I Proper time to reach the singularity - rocket acceleration | Sep 27, 2017 |

A Maximizing survival time when falling into a black hole | Sep 20, 2017 |

**Physics Forums - The Fusion of Science and Community**