What is the relationship between black holes and time dilation?

[pervect]

Hi Chris, you wrote:

The velocity of the Lemaitre observer at Schwarzschild radial coordinate r relative to a locally stationary inertial observer equals the negative of the radial escape velocity, Ve = sqrt(2M/r) in geometric units.

Yes, I've gotten that result in the past. Note that it corresponds to setting E=1 in the more general expression for the velocity (relative to a stationary observer) of a radially infalling observer

<br /> \frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}<br />

and that E=1 corresponds to a LeMaitre observer (someone who's velocity is zero at infinity).

Likewise, The Lemaitre observer will receive the transmited wavelength of the distant static observer as redshifted by a factor:
sqrt((1+Ve)/(1-Ve))*sqrt(1-2M/r) = 1+Ve, approaching the value 2 as Ve -> 1.

I've gotten this result before too, and two is the correct answer for the redshift of a radially infallling light beam as seen be a LeMaitre observer at the event horizon.

 

[Jorrie]

Yes, I've gotten that result in the past...

I've gotten this result before too, and two is the correct answer for the redshift of a radially infallling light beam as seen be a LeMaitre observer at the event horizon.

Thanks Pervect, it's comforting, but - I wonder what Chris would say...

Jorrie

 

[Chris Hillman]

Well done!

Hi again, Jorrie,

I have tried my hand for a Lemaitre observer. Despite expecting hand slaps for not using Painleve coordinates, I worked in a good old Schwarzschild
chart and hence had to stay outside of the event horizon.

Right, you used the method suggested by pervect, which combines the usual gravitational redshift (for a distant static observer and a static observer at some finite Schwarzschild radius) with the usual Doppler shift (valid for two local frames at event E), which is perfectly valid in the exterior region. And good, you used the correct velocity v = \sqrt{2m/r} (taking the positive sign since the Lemaitre observer is falling away from the distant static observer), and then as you say
\frac{ \lambda_{{\rm received}} }{ \lambda_{{\rm emitted}} } = \sqrt{1-2m/r} \, \frac{ \sqrt{1+\sqrt{2m/r}} }{ \sqrt{1-\sqrt{2m/r}} } = 1+\sqrt{2m/r}
and as you observed this approaches a wavelength ratio of two as the Lemaitre observer approaches the horizon. So if our Lemaitre observer has a rear pointing spectrograph, he knows he is near the event horizon when he sees a redshift in his rear view mirror of about two. But if he only performs experiments inside the cabin of his rocket ship, then he won't notice anything in particular as he passes the horizon!

Next, because the interior region belongs to an real analytic extension of the result you found, by a computation which is valid only the exterior region, and since the formula you found continues to make sense on 0&lt; r &lt; 2m, we might expect that it is also valid in the interior. To confirm that I suggested using a second method to directly compute the frequency ratio, which is valid on both regions. (More on that below.)

Also, it is worthwhile finding the radius as a function of the proper time of the Lemaitre observer remaining until impact at r=0, and then plugging into the expression you found above, you should find a certain power law expression for the observed redshift as a function of the proper time remaining until impact. So if our Lemaitre observer keeps an eye on his rearpointing spectrograph, he can estimate his remaining lifetime (assuming he doesn't turn on his rocket engine).

Likewise, going the other way, you found the correct redshift for signals sent radially from the Lemaitre observer back up to the distant static observer. It is interesting to express this in the proper time of the latter observer. His proper time agrees with Schwarzschild coordinate time (by definition of the exterior Schwarzschild chart!), so you need only compute r(t) for the Lemaitre observer. The coordinate speed is dt/dr = -1/(1-2m/r)/\sqrt{2m/r}, so integrating we find
t-t_0 = -\sqrt{\frac{2r^3}{9 m}} - \sqrt{8 m r} + 2 m \, \log \left( \frac{1+\sqrt{2m/r}}{1-\sqrt{2m/r}} \right)
This is hard to invert! But notice that the first term dominates when r \gg 2m, while the third term dominates when r \approx 2m. So you should be able to find two approximate expressions from which you can see that the redshift obeys a power law for most of the infall, but very near the horizon obeys an exponential law.

Yes, I've gotten that result in the past. Note that it corresponds to setting E=1 in the more general expression for the velocity (relative to a stationary observer) of a radially infalling observer
<br /> \frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}<br />
and that E=1 corresponds to a LeMaitre observer (someone who's velocity is zero at infinity).

Exactly. This is a third method (valid only in the exterior region, where the standard analysis via effective potentials of the null geodesics is valid).

Before we move on to the second method, note that the first method can be applied to other radially infalling observers, such as the slowfall observers. In the exterior Schwarzschild chart, the slowfall observer has tangent vector
\frac{1-m/r}{1-2m/r} \, \partial_t - \frac{m}{r} \, \partial_r
That is, in the exterior he has a world line of form
t-t_0 = -\frac{r^2}{2 m} - r - 2m \log ( r-2m)
Using the second method, can you find the wavelength ratio for signals sent from this observer back up to the distant static observer? And vice versa!

Here, recall that the slowfall observers, by definition, accelerate radially outward with just the right magnitude of acceleration (as a function of position) which would suffice to maintain their position according to Newton's theory of gravitation, namely m/r^2. But in gtr, the energy of the gravitational field itself gravitates, so the gravitational attraction is a bit stronger (roughly speaking), so these observers slowly fall radially inwards. How does this agree with the result you found for signals sent from the distant static observer to a slowfall observer? Can you find the wavelength ratio measured by the slowfall observer as a function of his proper time, and (suitable approximationate expressions for) the wavelength ratio measured by the distant static observer is his proper time?

Earlier I remarked that for signals sent radially from a distant static observer down to a radially falling observer, we should expect the usual gravitational blue shift to oppose the Doppler red shift. We could have guessed which would win out in the case of Lemaitre observers from your answer to the question about slowfall observers. This should help to explain why we found a redshift for both ingoing and outgoing signals in the case of a Lemaitre observer and a distant static observer.

OK, on to method two:

The ingoing Eddington chart is defined so that the ingoing radial null geodesics appears as horizontal coordinate lines. We can ensure this by setting du = dt + dr/(1-2m/r), which integrates to u = t + r + 2m \, \log(r-2m). The line element in the new chart is
ds^2 = -(1-2m/r) \, du^2 + 2 \, du \, dr + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
-\infty &lt; u &lt; \infty, \; \; 0 &lt; r &lt; \infty, \; \; 0 &lt; \theta &lt; \pi, \; \; -\pi &lt; \phi &lt; \pi
This chart is well defined on the right exterior and future interior regions, the same places where congruences of infalling observers are defined. It is perfectly suited for analyzing wavelength ratios for pairs of observers in which a more distant observer is sending signals radially downward to a closer observer. Can you figure out how to draw the light cones in this chart? The frame fields for static, Lemaitre, and slowfall observers? (Hint: if you know how to transform a vector field into a new chart, that's all you need since the frame fields are made up of unit vector fields.) Can you draw a diagram from which, using similar triangles, you can confirm our guess above about the general expressions for signals sent from the distant static observer down to the Lemaitre or slowfall observers? (The article "Frame fields in general relativity" in the version listed at http://en.wikipedia.org/wiki/User:Hillman/Archive should help if any of this seems confusing.)

The outgoing Eddington chart is defined so that the outgoing null geodesics appear as straight lines, by an expression very similar to the above. This chart is perfectly suited for analyzing wavelength ratios for pairs of observers in which a closer observer is sending signals radially outward to a more distant observer. Can you use the second method with this chart to confirm the expressions we found earlier? Note: the Lemaitre and slowfall frame fields are defined on the right exterior region and future interior region, whereas the outgoing Eddington chart is defined on the past interior and right exterior region (referring to the usual block diagram exhibiting the global causal structure of the maximal analytic extension of the exterior Schwarzschild vacuum). So, our results here only make sense for the exterior region. Of course, that is just what we expect since infalling observers cannot send signals outside the horizon if they have fallen into the future interior region!

You may be familiar with the (past) interior Schwarzschild chart in which the line element becomes
ds^2 = \frac{-1}{2m/t-1} \, dt^2 + \left( 2m/t-1 \right) \, dz^2 + t^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
0 &lt; t &lt; 2m, \; \; -\infty &lt; z &lt; \infty, \; \; 0 &lt; \theta &lt; \pi, \; \; -\pi &lt; \phi &lt; \pi
Observers who maintain constant spatial coordinates in this chart are called Frolov observers and they are geodesic observers who never emerge into either exterior region, although the Frolov congruence is also defined in the future interior region (it has a "caustic" at "the center of the X" in the usual block diagram, i.e. some pairs of world lines of Frolov observers intersect at that two-sphere). If you have seen the fine embedding diagrams of various spatial hyperslices inside the Schwarzschild vacuum as discussed in MTW, Gravitation, then you might recognize the spatial hyperslices above as the expanding (shrinking) cylinders {\mathbold R} \times S^2; in the Kruskal-Szekeres chart they appear inside the past (future) interior regions as hyperbolic arcs nested between the horizon and the past (future) curvature singularities.

Can you find the Frolov frame (in the future interior region) in terms of the ingoing Eddington chart? Can you then extend your computations above to obtain the wavelength ratio observed by a Frolov observer in the future interior region, for signals sent radially inward by a distant static observer? Can you re-express your answer as a function of the proper time of our Frolov observer? You should a blue shift evolving into a red shift! (Where does the transition occur?) Can you find the world lines of the Frolov observers in terms of the ingoing Eddington and Painleve charts? If you plot the world lines Lemaitre, slowfall and Frolov observers in the ingoing Eddington chart, can you "see" why we obtain redshift or blueshift?

 

[pervect]

I'd like to make a few comments about conserved quantities.

An easy way to solve for the equations of radial motion is to take advantage of the existence of conserved quantities.

This general approach is not restricted to the exterior region - one simply has to use well behaved coordinates such as the Painleve coordinates currently being discussed. While the effective potential was limited to the exterior region, this is a consequence of using the Schwarzschild coordinates to derive the conserved quantities.

I've talked about this before, briefly, BTW
https://www.physicsforums.com/showthread.php?t=126307

Consider the Painleve metric

<br /> -{d{{T}}}^{2}+ \left( d{{r}}+\sqrt {{\frac {2M}{r}}}d{{T}}<br /> \right) ^{2}+{r}^{2} \left( {d{{\theta}}}^{2}+ \sin ^2<br /> \theta d {{\phi}}}^{2} \right)<br />

Question: there appears to be a discrepancy between this and CH's version in
https://www.physicsforums.com/showpost.php?p=1205700&postcount=48?

Continuing:

This metric is not a function of T. Therefore it has "time-translation symmetry" over T. Therefore \xi^a = (1,0,0,0) is a time-like Killing vector field.

It is an important theorem that if \xi^a is a Killing vector field, and that \gamma is a geodesic with a tangent vector (4-velocity) u^a, that \xi_a u^a is a constant everywhere on the curve \gamma. See for instance Wald, pg 442.

Considering the above expression, we find that \xi_a = g_{ab} \xi^b, and because the only non-zero component of \xi^b is \xi^0 which has a value of 1, we can say that

\xi_a = g_{a0}

Thus
\xi_0 = -1 + \frac{2M}{r}
\xi_1 = \sqrt{\frac{2M}{r}}

and the other components are zero

This in turn yields the expression

<br /> \left(-1 + \frac{2M}{r} \right) \frac{dT}{d\tau} + \sqrt{\frac{2M}{r}} \frac{dr}{d\tau} = \mathrm{constant}

For a radially infalling observer, the existence of this conserved quantity, allows us to solve a number of problems, such as finding the 4-velocity of a radially infalling observer. Note that we get a second equation from the fact that the magnitude of a 4-velocity is always equal to one (or minus one, depending on the sign convention being used). Thus the magnitude of the conserved quantity above (which can be considered to be a conserved energy since it is due to a time translation symmetry) is all we need to find both components dT/d\tau and dr/d\tau of a radially infalling observer.

 

[Chris Hillman]

I don't see any disagreement!

Hi, pervect,

This general approach is not restricted to the exterior region - one simply has to use well behaved coordinates such as the Painleve coordinates currently being discussed. While the effective potential was limited to the exterior region, this is a consequence of using the Schwarzschild coordinates to derive the conserved quantities.

Agreed, if you can define conserved quantities in a chart valid on a larger region, you can use them on that larger region.

<br /> -{d{{T}}}^{2}+ \left( d{{r}}+\sqrt {{\frac {2M}{r}}}d{{T}}<br /> \right) ^{2}+{r}^{2} \left( {d{{\theta}}}^{2}+ \sin ^2<br /> \theta d {{\phi}}}^{2} \right)<br />
Question: there appears to be a discrepancy between this and CH's version in
https://www.physicsforums.com/showpost.php?p=1205700&postcount=48?

Well, the Lemaitre coframe field can be written in the Painleve chart
\sigma^0 = -dT, \; \; \sigma^1 = dr + \sqrt{2m/r} \, dT, \; \; \sigma^2 = r \, d\theta, \; \; \sigma^3 = r \, \sin(\theta) \, d\phi
Then
ds^2 = -\sigma^0 \otimes \sigma^0 + \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 + \sigma^3 \otimes \sigma^3
which gives
ds^2 = -(1-2m/r) \, dT^2 + 2 \, \sqrt{2m/r} \, dT \, dr + dr^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right)
Hmm... looks I inadvertently ommitted a two in one of the equations of one of my posts in the thread you cited... is that the "discrepancy" you meant?

If anyone is worried that we might be writing the metric incorrectly in the Painleve chart, see Frolov and Novikov, Black Hole Physics, or http://www.arxiv.org/abs/gr-qc/0001069, where the Painleve chart is discussed extensively.

This metric is not a function of T. Therforre it has "time-translation symmetry" over T. Therfore \xi^a = (1,0,0,0) is a time-like Killing vector field.

That's really awful notation (unfortunately very common) which is used only in physics; the entire rest of the mathematical world would write this (timelike) Killing vector as \vec{\xi} = \partial_T. See for example Stephani et al., Exact Solutions of Einstein's Field Equations, Cambridge University Press, 2nd edition, 2001, which has some fine introductory chapters surveying basic mathematical techniques useful for gtr.

For a radially infalling observer, the existence of this conserved quantity, allows us to solve a number of problems, such as finding the 4-velocity of a radially infalling obsever.

With respect to... what? My point was the static frame is only defined in the exterior.

Thus the magnitude of the conserved quantity above (which can be considered to be a conserved energy since it is due to a time translation symmetry) is all we need to find both components dT/d\tau and dr/d\tau of a radially infalling observer.

Just to be clear (I hope) to all: here, pervect is referring to a quantity which is invariant along (in particular) the proper time parametrized world line of a Lemaitre observer in the Painleve chart. As a parameterized curve, this contains more information than giving T(r), say. Also, this is not the same thing at all as computing the physical velocity of this observer at some event wrt to another observer whose world line passes through that event. The latter amounts to expressing one frame field (anholonomic basis for the tangent space at each event on some world line) in terms of another.

 

[pervect]

Hi, pervect,

Hmm... looks I inadvertently ommitted a two in one of the equations of one of my posts in the thread you cited... is that the "discrepancy" you meant?

Yes -exactly.

If anyone is worried that we might be writing the metric incorrectly in the Painleve chart, see Frolov and Novikov, Black Hole Physics, or http://www.arxiv.org/abs/gr-qc/0001069, where the Painleve chart is discussed extensively.

I was worried about that very thing - when I had worked on this (quite a while ago), I had scrounged my metric from a homework problem someone had posted. When I saw that you had a different expression for the metric, I wasn't sure whether you had made a simple typo, or I'd somehow scrounged the wrong metric.

With respect to... what? My point was the static frame is only defined in the exterior.

Not really with respect to anyone in particular - as you point out, there isn't any static observer in the interior. However, the 4-velocity still exists as a mathematical entity.

The main use is that knowing the 4-velocity, i.e. dT/dtau and dr/dtau allows one to integrate to find T(tau) and r(tau), i.e. to solve for the geodesic curve corresponding to some particular energy. The LeMaitre observer will be a particular case having a specific value for the energy parameter. I think I did more on this topic in the other thread I mentioned.

 

[Chris Hillman]

The perils of transcribing from notes in a different format

Yes -exactly.

OK--- I apologize to all for my unfortunate transcription error (from my notes on computations related to the Painleve chart), which caused the confusion. Alas, it is all too easy to make an goof of this kind...

 

[Chris Hillman]

A small but important correction

Unfortunately, I can't edit the original post, but let me note a correct here. Thanks to pervect for noticing a typo in the following:

In the Painleve chart (1921), the metric tensor can be written
ds^2 = -dT^2 + \left( dr + \sqrt{m/r} \, dT \right)^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right)
= -(1-2m/r) \, dT^2 + 2 \, \sqrt{2m/r} \, dT \, dr + dr^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
-\infty &lt; T &lt; \infty, \; 0 &lt; r &lt; \infty, \; 0 &lt; \theta &lt; \pi, \; -\pi &lt; \phi &lt; \pi

If you multiply out the first equation, it doesn't match the second, because I accidently omitted a factor of two, writing \sqrt{m/r} when I meant to write \sqrt{2m/r}. Sorry for any confusion!

Incidently, I just noticed a new paper http://www.arxiv.org/abs/gr-qc/0701061 coauthored by J. B. Griffiths, author of an important monograph on CPW solutions, which mentions the Ferrari-Ibanez CPW solution. I can't resist mentioning that in one of my questions to Jorrie above (where does the transition occur?), I was actually looking ahead to the time when, perhaps, we can discuss this important exact solution!

I suppose I really should add that this solution models two linearly polarized gravitational plane waves, having aligned polarizations and with specially chosen "amplitude profiles", which collide head on; this CPW model has the remarkable property that the "nonlinear interaction zone" which models the "aftermath" of the collision is locally isometric to one of the regions in the future interior of the Schwarzschild solution! Isn't that nifty?

The C-vacuum is related to one of those Weyl vacuums I just mentioned in another thread as a counterexample to the idea that gtr might not be quite as tricky as I sometimes claim!

 

[Jorrie]

Thanks, but Wow!

Hi again Chris.

Right, you used the method suggested by pervect, which combines the usual gravitational redshift (for a distant static observer and a static observer at some finite Schwarzschild radius) with the usual Doppler shift (valid for two local frames at event E), which is perfectly valid in the exterior region. ...

Thanks, I'm glad to be on the right track and appreciate your very detailed reply. But wow, it is a BIG mouthful for me.

As engineer, I was very chaffed when I got to grips with the Schwarzschild metric and chart. I more or less stopped there and only lately started to look a bit deeper (to the 'inside', pun intended). The guidance from yourself and Pervect is invaluable - I will surely work on the problems that you posted!

Regards, Jorrie

 

[George Jones]

Yes, I've gotten that result in the past. Note that it corresponds to setting E=1 in the more general expression for the velocity (relative to a stationary observer) of a radially infalling observer

<br /> \frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}<br />

and that E=1 corresponds to a LeMaitre observer (someone who's velocity is zero at infinity).

Here's another way to look at it.

Suppose the radially infalling observer starts from rest (with respect to a static observer at r = r_0 &gt; 2M) at r = r_0. Then, because the above speed is zero at r_0,

<br /> E = \sqrt{1- \frac{2M}{r_0}},<br />

so the above speed becomes

<br /> \sqrt{1 - \frac{1- \frac{2M}{r}}{1- \frac{2M}{r_0}}}.<br />

 

[pervect]

Nice reformulation! - it's less abstract and more physical.

I'll take this opportunity to provide what I think is a simpler derivation of the original formula.

First, a very quick review of 4-velocities (I know George and Chris know all this, but I'm not sure of how much our other readers know).

If we have coordinates T, r, \theta, and \phi, we can write out the components of the 4-velocity as

<br /> u^a = \left[ \frac{dT}{d\tau}, \frac{dr}{d\tau}, \frac{d\theta}{d\tau}, \frac{d\phi}{d\tau} \right]

using geometric units where c=1. \tau[/itex] should be interpreted as the proper time of the observer, i.e. we can parameterize the wordline of the observer by four functions T(\tau),r(\tau),\theta(\tau),\phi(\tau) where \tau represents the proper time or &#039;age&#039; of the observer, and the above set of derivatives of the coordinates with respect to the proper time is the 4-velocity.<br /> <br /> It should also be apparent that<br /> g_{ab} u^a u^b has a value of plus or minus 1, +1 if we have a +--- metric signature, -1 if we have a -+++ metric signature.<br /> <br /> Now let&#039;s consider how to find the magnitude of the relative 3-velocity of two observers if one observer has a 4-velocity of u^a and the other has a 4-velocity of v^b.<br /> <br /> The product \vec{u} \cdot \vec{v} = g_{ab} u^a v^b is going to be independent of the choice of frame or coordinate system. It will determine the angle between the two 4-vectors on a space-time diagram, and hence it will determine the magnitude of the 3-velocity.<br /> <br /> By considering the case in flat space-time where u = (1,0,0,0) and v = ( 1/\sqrt{1-\beta^2},\beta/\sqrt{1-\beta^2},0,0), we can see that the magnitude of the 3-velocity is just<br /> <br /> \beta = \sqrt{1 - \left(\frac{1}{\vec{u}\cdot\vec{v}}\right)^2}}<br /> <br /> To apply this formula to the original problem, we need only the 4-velocity of a stationary observer at radius r, and the 4-velocity of an infalling observer with energy E at radius r.<br /> <br /> In Schwarzschild coordinates, using a +--- metric, the 4-velocity of a stationary observer will just be<br /> <br /> \left[ \frac{1}{\sqrt{1-2M/r}},0,0,0 \right]<br /> <br /> This results from the normalization requirement that g_{00} u^a u^b = 1<br /> <br /> We only need the first component, v^0 of the 4-velocity of an infalling observer to compute the dot product. <br /> <br /> <a href="http://www.fourmilab.ch/gravitation/orbits/" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.fourmilab.ch/gravitation/orbits/</a>, or the argument I presented about finding conserved quantites from Killing vectors, gives the value of this quantity.<br /> <br /> v^0 = \frac{E}{1-2M/r}<br /> <br /> Therfore the dot product \vec{u} \cdot \vec{v} is just<br /> <br /> g_{00} \left( \frac{1}{\sqrt{1-2M/r}} \right) \left( \frac{E}{1-2M/r} \right) = \frac{E}{\sqrt{1-2M/r}}<br /> <br /> since g_{00} = 1-2M/r using the +--- sign convention.<br /> <br /> So the relative velocity is just<br /> <br /> \beta = \sqrt{1 - \frac{1-2M/r}{E^2}}<br /> <br /> We can also re-write this in the form suggested by George by solving for the energy E in terms of the maximum height.

 

[thothicabob]

I've been following this thread with interest. Tho the math is beyond me for the most part, a lot of questions have been cleared - or so I think. If it's not considered impolite, I'd like to drop a couple of my own to check my understanding, albeit well after this thread apparently ends, as it seems the discussion is the 'right type' for them. I'm hoping Chris H. or pervect will have time to answer them:

The questions are "Observer" type questions...

1. To an observer looking "out" from within the black hole (inside the event horizon), how would the outisde universe appear to him? I am thinking one possibility is that it would appear "smeared" over the boundary of the event horizon, but that could be due to a faulty understanding on my part. Further, would 'time' outide appear to have been 'infinitely' accelerated, which would (could) add to that "smearing" effect?.

2. As a black hole gains mass, as I understand it, the diameter of the sphere bounded by its event horizon increases, at least from the point of view of an outside Observer. How would this appear to an Observer within the event horizon? Would his "universe" (for lack of a better term) appear to be expanding? Would he notice it at all?

Thanks in advance - I'll go look for my Hefty(tm) brand protective garment in the meantime... )

 

[pervect]

I haven't worked out in detail how the outside universe appears to someone falling into a black hole. One thing you'll have to specify is the velocity of the infalling observer - he won't be stationary, no stationary observers exist inside the event horizon. Equivalently, you could specify his energy or the height from which he fell.

I have worked out the redshift such an observer sees for radially infalling rays, and while it depends on the energy parameter of the observer, it is finite.

While I haven't worked this out personally, there's a reliable website that discusses this issue, which includes some movies of obsevers falling into a Schwarzschild black hole as well.

See http://casa.colorado.edu/~ajsh/schw.shtml

I believe he also discusses the issue of redshift vs angle.

Sort of as an aside, a real physical black hole probably won't have an interior Schwarzschild geometry, especially if the black hole is rotating as is very likely.

 

[thothicabob]

Thanks, Pervect. Your answer and that page you suggested were helpful, and they also allow me to better formulate my questions in other ways.

Unfortunately, it also lends fuel to one idea I've wanted to be able to 'debunk or confirm' (for myself) that I'll need to work out somehow (to wit: that for all practical purposes, the Universe itself can be thought of as a black hole, and we exist inside it). I know it's been treated here before, so I'll not ask you to do so again - I'll just re-read the stuff, and also do some more pokin' around and try to find some quality grokkage time. The thing is, it seems that, even after all this time and all we've learned, there's still no conclusive answer that's not dependent on "how you look at it" or what initial assumptions you accept, and I've not yet been able to resolve some "logical" questions that I had hoped would at least be satisfactory enough for "me".

On the one hand, it would be a neat and tidy package if it were the case, in some respects. On the other, it opens other cans of worms that serve mainly ony to push some basic questions backwards, away, or calls them "silly" (or meaningless, as you prefer), and not really answer them.

Thanks again!


And yes, I know I really should get my butt back to school and stuff some "real math" into my head, but in a life-context, I'm a afraid I've already passed my own personal event-horizon for that sort of thing, so the best I can do is to try to reason things best I can, and - when necessary and possible - ask people who: a) know better, and; 2) are willing to take the time to answer questions seriously. In that sense, I appreciate the time you (and others here like you) take out for people like me (and others here like me). I doubt I'm alone in that, either.