a_g_marut

[Edit: no. The λ in question is the wavelength of the light before it reaches the pupil because that wavelength is what determines how many more wavelengths light has to travel from the source to reach one side of the pupil than the other. The lens and vitreous humour focus the light onto the retina by ensuring that the light from a point source travels the same amount of time to reach the corresponding point on the retina, regardless of which point on the pupil it passes through. Because all the light travels the same amount of time from the source to the corresponding point on the retina, the light waves’ maxima all arrive at the same time, so the light waves interfere constructively at that point and produce a bright spot. Near that point, the light travels almost, but not quite, the same amount of time, so the point source illuminates a region around the point slightly less than it illuminates the point where it is focused. When light comes from two sources close to each other, the difference between the amount of time the light from one source takes to reach one side of the pupil and the amount of time the light from the same source takes to reach the other side of the pupil is close to the difference between the amount of time the light from the second source takes to reach one side of the pupil and the amount of time the light from the second source takes to reach the other side of the pupil. And there is nothing that the lens or vitreous humour can do about that.]

In The Two Towers, the elf Legolas, at a distance of five leagues, observed once, “there are one hundred and five [riders on horses]. Yellow is their hair, and bright are their spears. Their leader is very tall.” In 2014, a viral video made the claim that this was impossible, based on the equation θ≈1.22λ/d, where θ is the angular size of the Airy disk produced by a point source of light, λ is wavelength, and d is the diameter of the pupil. My idea is that, in a material with a high refractive index, λ would be proportionally less than it is in air, resulting in a smaller θ, and with it an image with better resolution.

(This post’s image and alt text are not my work; Wikipedia user Inductiveload released them into the public domain.)

I've read that a distant observer watching matter fall into a black hole will see the matter get closer and closer to the horizon but never reach it. At the same time, the black hole is producing Hawking radiation and will vanish in a finite amount of time. Therefore, the observer should see the black hole vanish before they see the infalling matter cross the horizon. If the observer, the infalling matter, the velocity of the infalling matter, and the center of the black hole are collinear, then the final flash of radiation that the observer sees will pass right through the infalling matter. This implies that, before (from the perspective of the infalling matter) the infalling matter reaches the horizon, the mass will disappear, the horizon will disappear, and the infalling matter will end up in flat spacetime rather than cross any horizon.

The Wikipedia article https://en.wikipedia.org/wiki/Vaidya_metric gives an equation relating coordinates to distances (equation 6 on the current version of that page) when radiation with a non-negligible amount of energy is escaping the black hole:

ds² = -(1 - 2M(u)/r) du² - 2 du dr + r² (dθ² + sin² θ dφ²)

(I assume c = G = 1)

For a distant observer at rest with respect to the black hole, this becomes

ds² = -(1 - 2M(u)/r) du²

with the coefficient of du² approaching -1 as r goes to infinity. From this I infer that the time that an observer at infinity measures between observations of two events is the difference between the u-coordinates of those events.

From https://en.wikipedia.org/wiki/Hawking_radiation#Black_hole_evaporation the time (as measured by an observer at infinity) for a black hole to evaporate is

tₑᵥ = 5120πG²M³/ℏc⁴

= 5120πM³/ℏ

therefore, M = ∛(ℏtₑᵥ/5120π)

Let u=0 be the u-coordinate at which the black hole vanishes. Therefore, to an observer at infinity, for a given value of u less than zero, the time until evaporation tₑᵥ = 0 - u = -u. Therefore,

M(u) = ∛((ℏ/5120π)(-u)) when u < 0, or 0 otherwise

To shorten the metric equation, let

r′ = 2r

dΩ² = dθ² + sin² θ dφ²

ℏ = 80π

Then, for u < 0,

2M(u) = 2∛((1/64)(-u)) = -(∛u)/2

Substituting into the metric equation,

ds² = -(1 - (-∛u/2)/(r′/2)) du² - 2 du d(r′/2) + (r′/2)² dΩ²

= -(1 + ∛u/r′) du² - du dr′ + r′² dΩ²/4

At the horizon, the equation r = 2M becomes

r′/2 = -∛u/2

r′ = -∛u

∛u/r′ = -1

Therefore, at the horizon,

ds² = -(1 - 1) du² - du dr′ + r′² dΩ²/4

ds² = -du dr′ + r′² dΩ²/4

For a timelike interval,

0 > ds² = -du dr′ + r′² dΩ²/4

du dr′ > (r′ dΩ/2)²

The right-hand side of this inequality is the square of a real number and therefore must be nonnegative, so the left-hand side must be positive. This implies that, at the horizon, if du is positive (that is, going forward in time), then dr′ must be positive (that is, away from the center of the black hole). So, not only can matter not enter the black hole through the horizon, but any matter passing through the horizon is doing so from the inside of the black hole to the outside. Therefore, all infalling matter must stay outside the horizon until the black hole vanishes.

Anyway, my country has just become a fascist dictatorship, so I'm just putting this out there so I, A. G. Marut, can be remembered for something (if only my proof because the conclusion was already known) after they come for me in the Niemöllerian sense.