BTW, I always thought GR _is_ about exotic solutions (just look at the arXiv statistics), just like science fiction (speaking of which, your ‘Permutation City’ was - at last - published here in Poland four days ago )!

In Schwarzschild geometry, you suffer tidal stretching in the direction you’re falling, and (half as much) tidal squeezing in each direction perpendicular to it. The red shift you see aligns with the stretch, the blue shift with the squeeze.

In BKL geometry, the same arrangement just oscillates: each of the 3 directions takes turns being the stretch while the other two are the squeeze.

As you integrate down to the singularity, I think you’d just be averaging (and possibly diluting) the same net effect as you’d get from the Schwarzschild geometry. Basically, if you integrate an oscillating function (of constant peak-to-peak size) multiplied by some original function, you’re not going to get an infinite result when the integral of the original function was finite.

My other argument is this: suppose Bob falls into the hole, and he gets to see every wavefront ever emitted by some radiation source. I think that would mean that the family of null geodesics describing those wavefronts would have to converge on a particular null curve (you could think of this limit curve as “the light from the infinitely far future”, though of course there is no actual light taking this path). That null curve would intersect the singularity at some distinguished event E. But the BKL geometry, although it’s not exactly time-invariant like the Schwarzschild geometry, is always the same on average; it’s just oscillating, it’s not evolving. So there is no “distinguished event E” coming from the geometry.

Putting this another way, suppose the time for Bob when he observes wavefront N, tau(N), converges, as N goes to infinity, on the time when he hits the singularity. If Alice fell into the hole before Bob, I think it’s inevitable that she’d fail to see the entire history of the radiation. And if Carol fell into the hole after Bob, I think she’d see the entire history completed some finite time before she struck the singularity. But what’s so special about Bob? There’s nothing about his relationship to the black hole or the radiation source that can make him special like this.

I haven’t read the paper you cited on Malament-Hogarth spacetimes, but it looks like fun. In Scott’s review article, I think the only real objection he raised to Malament-Hogarth spacetimes was quantum effects. FWIW I’m generally sceptical about these kinds of exotic solutions of classical GR (or at least I am when I’m not writing science fiction).

As for the link between conservation of energy, red and blue shift, and the time you see pass in the outside universe, the reasoning goes like this: starting with the Schwarzschild geometry, you can make use of the fact that it has a “time translation symmetry” to get something like a conservation law along the world lines of the incoming photons. (In general, conservation of energy is a more complicated notion in curved spacetime than it is in special relativity or Newtonian physics.) That’s how you derive the red and blue shifts.

Once you know those shifts, it’s just a simple counting argument to deduce the amount of time you see passing. If a one-hertz wave is emitted from a distant star, and what you observe due to blue shift is a ten-hertz wave, then for every second that passes for you, you must see 10 seconds pass for that star.

For an observer who free-falls into a black hole, she sees light that falls directly from the zenith to be red-shifted, but as she looks towards the edge of her view, the red-shift changes to a blue shift.

Where she sees a red shift, she is seeing fewer periods of the light per unit of her proper time than if she stayed outside; where she sees a blue shift, she is seeing more periods of the light.

The exact formula for the red/blue shift is a bit complicated, but I’ve worked this out in detail here.

As the observer approaches the singularity at r=0, the blue shift at the edge of her view goes to infinity like 1/r. However, dr/dtau for her (where tau is her proper time), also goes to (minus) infinity, like -1/sqrt[r]. So if I’m analysing this correctly, the number of periods of blue-shifted radiation will be the integral of some constant times 1/sqrt[r], which is a finite integral. So even for the most blue-shifted radiation she sees, she only sees a finite time period of the history of the radiation source.