Black holes

[Complexity]

I wish you guys would stop calling it FGR.

I think that 'Fake General Relativity' is a very apt name for what you keep proposing.

 

[Complexity]

I'm sure that Farsight will claim that light is virtually motionless at that point, so we'd never actually see the gamma-rays coming out of the black-holes.

I see what you did there.

Now I have horrible images in my head.

Try not to think of the Gilligan's Island theme song.

(payback)

 

[dafydd]

I think that 'Fake General Relativity' is a very apt name for what you keep proposing.

There is no better name.

 

[Brian-M]

I thought you weren't going to waste any more time on this Brian? LOL, good thread through isn't it? If you want me to go back to the posts of yours I skipped, let me know.

Not so much that I'm not going to waste any time in the thread, but more that I'm not going to waste time in areas of futile and pointless argument.

(Which is why I'm not going to reply to all your comments, some things will just take far too much time. At this point I'm just trying to make some of the points I've already made clearer.)

Huh? Both pairs of mirrors are just sitting there. This is no illusion.

How do you know they're just sitting there, and that you're not just moving along at a constant distance behind them?

That's what I'm talking about with the rotating space station scenario, you'd see exactly the same thing. Just because you see the two beams of light moving at different speeds (and you've put a lot of emphasis on the seeing part in earlier threads) doesn't actually mean they are moving at different speeds.

What? I'm just sitting there too.

It only looks like you're just sitting there too because you, your chair, desk, computer, room, and everything around you are all moving together.

Brian, either I've totally misunderstood you here or you're kidding yourself.

I think you've misunderstood me here. I was talking about the space station scenario, from the point of view of someone not moving with the space-station.

What rotating space station?

The rotating space station I've mentioned in posts 676, 805, 807 and of course 837 (which you're replying to).

We could do this parallel-mirror thing in a gravitational field regardless of rotation.

The point of the rotating space-station in post 676 was to point out that acceleration is acceleration, regardless of whether it comes from being in a gravitational field, a spaceship, or rotating ring, and that there's no reason to assume that the mechanism causing time dilation differs in any of these cases. That doing it on a rotating ring is exactly the same thing as doing it in a gravitational field.

The point of the rotating space-station in posts 805, 807 and 837 was to produce an easily understandable situation where you see the exact same phenomenon where the light in two light clocks at different elevations appears to be traveling at different velocities exactly as it as appears to do in a gravitational field, to demonstrate that the fact that you see the light traveling at different speeds doesn't necessarily mean the light actually is traveling at different speeds, which would then have been the basis for making a comparison to the same observed phenomenon in gravity.

The so-called paradox is that two twins with relative velocity each sees the other one's clocks going slower than his own. That's the symmetry. It's broken when one twin turns round and comes back.

Even in a purely Newtonian universe each twin would see the other's clock going slower than their own as the traveling twin is on his outward journey, and then they'd each see the other's clock going faster than their own during the return journey.

 

[ben m]

I wish you guys would stop calling it FGR.

How about FIROOPOE? ("Farsight's Idiosyncratic Reading Of One Paper Of Einstein's") Or FOMPOWHIGRI? ("Farsight's Odd Mental Picture Of What He Thinks GR Is")

The thing is, Farsight, everyone on Earth except you sees the words "General Relativity" and takes them to mean "The theory developed by Einstein, widely taught by MTW/Wheeler/Wald/Hartle/Schutz, tested experimentally, etc., in which the laws of physics are generally covariant."

I know that's not what you *want* them to think. You want future physics-language-speakers to use "GR" to mean "Farsight's reading of Einstein's 1916 paper", and to use some other word to mean "The predominant 1917-2012 reading of Einstein's 1916 paper and all subsequent work". You know what? Tough. If you want to change the language, you will have to get people to want "GR" to mean "Farsight's reading" first.

Why would you deliberately use language-no-one-understands? Actually, I know exactly why. Because you want to be able to yammer puffily about your knowledge of "GR", which is "experimentally verified, of course", and the "obvious" thought experiment that falsified "a surprisingly common, but ultimately overturned, mid-20th-century misinterpretation" since "everyone now knows they're gravastars." Which a lot of casual readers will mistake for an uncontroversial discussion of the same GR theory that they saw in a NASA press release, right.

That's not a mistake that these readers would make if you didn't call it GR. If you called it "FGR" off the bat, you wouldn't get the benefit of those ten posts worth of nonphysicist's-heads-nodding-in-agreement before the physicists notice you and disagree on substance. As soon as you get physicists disagreeing on substance, you're forced straight into "I don't use math" and "I reject your conclusions because my mental picture doesn't include them" and so on, which even nonexperts recognize as crackpottery.

In other words, the words "General Relativity" have built up a lot of respect over the past century. You want to steal those words, and momentarily dress yourself in the words' respect, even though the things that merited that respect (general covariance, experimental tests, deep mathematical rigor) have nothing to do with what you're talking about.

Famous con-man Frank Abagnale, on stealing a pilot's uniform, hoped to garner a bunch of respectful smiles and salutes---from people who, of course, thought they were saluting the daring and romance associated with the actual flying of airplanes. Now imagine if Abagnale had stuck around (rather than hightailing it) after people called him out for not knowing an aileron from an airsick bag. "I wish you guys would stop calling it pilot impersonation," he might say. "I am wearing an authentic pilot uniform, and I insist on calling myself a pilot. You will have to invent another word for the airplane-flying subtype of pilot, some of whose uniforms I can see are less authentic than my own."

 

[DeiRenDopa]

Ya know, sometimes Farsight is right.

I really should try harder, to think for myself ...

(snip)
C

Because the universe is expanding, and the solar system isn't getting smaller. And yes they are.

No problem with that. There's empirical evidence that the speed of light varies with gravitational potential, which means the maths of Kruskal-Szekeres coordinates is flawed. It isn't wrong as such, it's flawed in that it presents you with a description that does not match the behaviour of the universe.

Yes. I've given it repeatedly. The speed of light varies with gravitational potential just like Einstein said. But people who are convinced that it's absolutely constant absolutely refuse to accept it. If I arranged two parallel-mirror light clocks at different elevations, you know that they won't stay synchronised. You also know that there's no literal time flowing between the mirrors, just light moving. If we used say dust in space, you'd be able to see the light beams moving like this:

|-----------------|
|-----------------|

Think of the light beams as racehorses. When one gets ahead of the other you say it's moving faster than the other. If somebody tried to tell you they were going at the same speed, you'd laugh at them. When they then tried to tell you that space was falling down, you'd tell them to stop wasting your time with fairytales.

(snip)

Farsight does not explain how this happens, but let's invent some magical dust which has the following properties²:

-> it is massless

-> it instantly emits light isotropically when a light pulse hits it

-> its refractive index is the same as the vacuum through which the light pulses are travelling

-> it does not change the trajectory or speed of the light pulses in any way.

One, very important, additional property is this:

-> the dust emits light only when the 'clock' light pulse hits it; light which comes from other dust (particles) does not cause it to emit light (i.e. there will be no light echoes on the film).

But where is this camera located? Well, Farsight doesn't say. In fact, in one of his descriptions (B) he refers to two cameras.

So I've taken the liberty of setting up three cameras: one, called U, is in the same horizontal plane as the upper parallel-mirror light-clock; one, called L, in the same horizontal plane as the lower parallel-mirror light-clock; and one, called M, mid-way between U and L. U, M, and L are in the same vertical plane, and the same vertical plane as the two right-hand mirrors. This plane is orthogonal to the (vertical) plane containing the two parallel-mirror light-clocks.

Farsight doesn't say how he does it, but let's suppose the emission of light pulses from the two clocks (i.e. from the left-hand "mirror") is synchronized.

Clearly, the only possible location for any camera is in the horizontal plane "mid-way" between the horizontal plane of the upper clock and that of the lower clock.

Look, again, at Farsight's signature diagram, and read the text again.

Here's what the film from any camera observing the two light beams must record, at the instant the beams were emitted:

|-----------------|
|-----------------|

Now no camera located in the horizontal plane of the upper clock (i.e. all U clocks) can possibly produce a film any of whose frames look like this¹! The same goes for all possible L clocks.

In fact, what we've discovered is something Farsight never bothered to tell us: an objective, independently verifiable way to synchronize the clocks²! All we do is move our equipment around until we find a place where the first frame of the film from the camera looks like the above.

So let's change our definition of the M clock to be one in any horizontal, "mid-way" location, so that the first frame from its film looks something like the above.

Does it matter where any M clock is located? It does if we want to have frames from the clock which look like the relevant Farsight diagram. However, for simplicity, let's consider only cameras located in the same vertical plane as the clocks. True, the film frames won't look like the above, but we can apply image processing software to them to make them look like the above, without messing up any simultaneity information. Let's start by putting M between the right-hand mirrors, but a teensy weensy bit to the left. Like this (M is the location of the camera; I've omitted any and all signals from light beams in the two clocks):

|-----------------|
|----------------M|
|-----------------|

Let's find the frame in the film which shows light from dust a teensy weensy bit to the left of upper clock's right-hand mirror. Where will light from dust in the lower clock appear, on this frame? Per Farsight, it will be a tad further to the left; something like this:

|-----------------|
|-----------------|

But what do we actually see is the objective, independently verifiable evidence (i.e. a hard copy of the film frame)?

Stay tuned, for the next exciting episode of "FGR, breakthrough of the century, or not!"

¹ except, of course, if the clocks were not synchronized; however, that's an explicit condition ...
² Important caveat: this is just one (possible) way to achieve synchronization. There may well be others, and it's quite possible this method is incompatible with FGR (but we have no way of knowing that; Farsight has never said how this kind of clock synchronization is to be done, in FGR).

 

[sol invictus]

In other words, the words "General Relativity" have built up a lot of respect over the past century. You want to steal those words, and momentarily dress yourself in the words' respect, even though the things that merited that respect (general covariance, experimental tests, deep mathematical rigor) have nothing to do with what you're talking about.

Famous con-man Frank Abagnale, on stealing a pilot's uniform, hoped to garner a bunch of respectful smiles and salutes---from people who, of course, thought they were saluting the daring and romance associated with the actual flying of airplanes. Now imagine if Abagnale had stuck around (rather than hightailing it) after people called him out for not knowing an aileron from an airsick bag. "I wish you guys would stop calling it pilot impersonation," he might say. "I am wearing an authentic pilot uniform, and I insist on calling myself a pilot. You will have to invent another word for the airplane-flying subtype of pilot, some of whose uniforms I can see are less authentic than my own."

Exactly right.

(ben, I often read your posts with admiration and envy of your ability to cut down quacks, but you've outdone yourself with that one.)

 

[TubbaBlubba]

I had a little "incident" with black holes the other day in (High school) physics class. Another student whined that it was demeaning of the textbook to say that a black hole is a collapsed star, because that's "obvious."

I felt like being a bit cheeky, so I said that it's not at all obvious - in fact, I could prove that a black hole was not a collapsed star with a thought experiment. I argued something like this: As the mass of the star approaches the Schwarzschild radius (that is, as the Schwarzschild radius of the center becomes greater than that of the center of the star's mass - when the first atoms collapse, if you will) it could not grow any more from our point of view. Because, as the rest of the star approaches it, time slows down from our point of view, and as such it never passes the event horizon. As such, we could never observe a collapsed star with a Schwarzschild radius any more than some infinitesimal amount, because the central singularity could obviously not grow in size.

Now, of course, there's something seriously wrong with the above argumentation. The other student couldn't figure it out (of course) which was what I had intended. Any way, I realized that I'm not quite sure what's wrong with it myself. I understand, mostly from reading these threads, that as something approaches the event horizon:
1. It sends fewer signals
2. The signals it sends are increasingly redshifted.
At some point, it sends its last signal. After that signal, can we say that it has crossed the EH? Is this a meaningful thing to say from our point of view? What about space-time distortion? Something at the event horizon should have a gravitational field indistinguishable from the same mass inside the singularity according to Newton's laws - I'm guessing the same applies to GR.

I think the answer lies somewhere around the lines that the Schwarzschild radius doesn't act as an event horizon prior to compression. But does this mean that the radius can never increase from our point of view, no matter how much junk we toss at the black hole? Or can we measure an increase of it after lobbing some massive object into it?

Yeah, I hope my questions made sense.

 

[sol invictus]

I had a little "incident" with black holes the other day in (High school) physics class. Another student whined that it was demeaning of the textbook to say that a black hole is a collapsed star, because that's "obvious."

I felt like being a bit cheeky, so I said that it's not at all obvious - in fact, I could prove that a black hole was not a collapsed star with a thought experiment. I argued something like this: As the mass of the star approaches the Schwarzschild radius (that is, as the Schwarzschild radius of the center becomes greater than that of the center of the star's mass - when the first atoms collapse, if you will) it could not grow any more from our point of view. Because, as the rest of the star approaches it, time slows down from our point of view, and as such it never passes the event horizon. As such, we could never observe a collapsed star with a Schwarzschild radius any more than some infinitesimal amount, because the central singularity could obviously not grow in size.

Now, of course, there's something seriously wrong with the above argumentation. The other student couldn't figure it out (of course) which was what I had intended. Any way, I realized that I'm not quite sure what's wrong with it myself. I understand, mostly from reading these threads, that as something approaches the event horizon:
1. It sends fewer signals
2. The signals it sends are increasingly redshifted.
At some point, it sends its last signal. After that signal, can we say that it has crossed the EH? Is this a meaningful thing to say from our point of view? What about space-time distortion? Something at the event horizon should have a gravitational field indistinguishable from the same mass inside the singularity according to Newton's laws - I'm guessing the same applies to GR.

I think the answer lies somewhere around the lines that the Schwarzschild radius doesn't act as an event horizon prior to compression. But does this mean that the radius can never increase from our point of view, no matter how much junk we toss at the black hole? Or can we measure an increase of it after lobbing some massive object into it?

Yeah, I hope my questions made sense.

What you have to remember is that the event horizon isn't a static spherical surface in any kind of ordinary sense, and its location is defined by future events as well as present and past.

Consider a collapsing star. At some time before the star collapses, a spherical event horizon appears at its center and proceeds to expand outwards at the speed of light. The horizon appears before the collapse because even though the star hasn't yet started to collapse, a light signal emitted from its center might not have time to escape before it does start to collapse.

Further, the entire star cannot be inside the Schwarzschild radius defined by its total mass, else it is already a black hole. In fact, no part of the star can be smaller than the Schwarzschild radius defined by that part of the star, else that part is already a black hole.

Does that help?

 

[TubbaBlubba]

What you have to remember is that the event horizon isn't a static spherical surface in any kind of ordinary sense, and its location is defined by future events as well as present and past.

Consider a collapsing star. At some time before the star collapses, a spherical event horizon appears at its center and proceeds to expand outwards at the speed of light. The horizon appears before the collapse because even though the star hasn't yet become a black hole, a light signal emitted from its center might not have time to escape before it does collapse.

Further, the entire star cannot be inside the Schwarzschild radius defined by its total mass, else it is already a black hole. In fact, no part of the star can be smaller than the Schwarzschild radius defined by that part of the star, else that part is already a black hole.

Does that help?

Okay, I think I'm getting it. But could say, the core of a star become a black hole first, and then swallow the outer parts of the star? Or must (by virtue of the math involved, or some property of stars, or something) the entire star collapse behind its Schwarzschild radius at once?

If the earlier could happen, that is, the core becomes a black hole, and swallows the outer part... Would you say that the Schwarzschild radius has increased after the black hole swallows the rest of the star?

Furthermore, let's say we have a hypothetical binary star system. One of the stars goes black hole. It then "swallows" the other star. Can we meaningfully said that the black hole has a Schwarzschild radius defined by both the mass of the original star and the star the subsequent black hole consumed?

... I shouldn't be doing this, I can't even handle relativistic electrodynamics.

 

[sol invictus]

Okay, I think I'm getting it. But could say, the core of a star become a black hole first, and then swallow the outer parts of the star?

Yes - you could say that. In any kind of spherically symmetric collapse it will always be the case that the event horizon appears at the center and then expands - but it might not be the case that the star has started collapsing yet during at least part of that growth. But it should be noted that in realistic situations the maximum possible time difference between "horizon first appears" and "star begins to collapse" is very short.

If the earlier could happen, that is, the core becomes a black hole, and swallows the outer part... Would you say that the Schwarzschild radius has increased after the black hole swallows the rest of the star?

Well, the Schwarzschild radius of the star is something that just depends on the total mass, which doesn't change when it collapses (at least neglecting mass or energy that gets blasted away - but that reduces the mass and the final Schwarzschild radius) - so, no.

Furthermore, let's say we have a hypothetical binary star system. One of the stars goes black hole. It then "swallows" the other star. Can we meaningfully said that the black hole has a Schwarzschild radius defined by both the mass of the original star and the star the subsequent black hole consumed?

Sure - after the capture, the mass of the resulting hole will be something a bit less than the sum of the masses of the two stars (less because some mass/energy gets radiated away during the swallowing).

 

[TubbaBlubba]

Sure - after the capture, the mass of the resulting hole will be something a bit less than the sum of the masses of the two stars (less because some mass/energy gets radiated away during the swallowing).

All right. So, this means we can meaningfully speak of something crossing the event horizon from our point of view (it would have to, otherwise it would not be part of the hole) even though we can't OBSERVE it by means of signals emitted from the object?

 

[Mr.D]

I need to get into non-accepted electron models to justify this spectacular, and since it's off-topic and I don't have patent evidence and Einstein to back me up, I'll leave it there.

We should move on from this discussion and talk about them. Or electromagnetism. It's amazing how people nowadays just don't understand it. It's like nobody has read the original Maxwell.

GR, Quantum Mechanics, E&M. Crikey; is there anything you think that modern physics gets right?


ETA: And really, your "non-accepted" electron model isn't really off topic - at some point you're going to have to show us how to reconcile your notions of fermions with structure/spin as something rotating with SR.

 

[Roboramma]

Take care. Susskind will end up telling you about an elephant that's in two places at once. Take care with light cones too. Like reference frames and coordinate systems, they're abstract things. You can't point up to the sky and say "Look, there's a light cone. Your retina isn't actually "on the surface of many light cones". It's on the inside surface of your eye, your eye is in space, and light moves through space and terminates on your retina. It's important to stay very grounded with all this.

You can't actually look anywhere and experience reality directly, so I don't really see your point here. If I look at an apple, I may be seeing a real apple, or I may be seeing a very well done plastic apple and not be able to tell the difference. The apple may be red, or it may be that the lighting in the room creates the illusion that the apple is red.

Given that vision itself is an impossible problem that requires the brain to make assumptions about the world, your "if you can't see it, it's not real" stance makes little sense.

We have models that make accurate predictions. When we make different kinds of observations (like going and picking up the apple and finding it's solid, tasting it and finding it sweet, etc.) those different lines of evidence give more and more support for the underlying models.

This isn't just how physics works: It's the only way that any accurate understanding of the universe (in any part) can work.

 

[W.D.Clinger]

At what point does his velocity begin reducing to zero?

I'm sorry, I don't know. It's quite close in. Maybe Clinger might like to go with the flow and earn his keep by working that out for you. Or maybe somebody can point to something already online.

So far as I know, FGR (where the F stands for Farsight and/or fake) doesn't answer dlorde's question. FGR definitely forbids the coordinate transformation I'll use below to calculate the answer in standard GR. So far as I know, the only way to answer this question in FGR is to use Farsight as an oracle, but he's already said he doesn't know the answer.

In standard GR, I'd solve the problem as follows. As the problem was originally stated by Farsight, the object is "dropped from infinity". That suggests we use Painlevé-Gullstrand coordinates t_f, r, θ, ϕ. Letting β be the square root of 2m/r, the infalling object's coordinate velocity at r will be

[latex]
\[
\frac{d r}{d t_f} = - \beta = - \sqrt{\frac{2m}{r}}
\]
[/latex]​

Integrating dt_f gives the infalling object's proper time.

That's high school algebra: Start with the Painlevé-Gullstrand metric form and plug in dr = - β dt_f and dθ=dϕ=0.​

(That's why Painlevé-Gullstrand coordinates are convenient for this problem.)

Because the problem involves the infalling object's velocity as observed by an observer at infinity (or by an observer whose position at some fixed r is maintained by a radial acceleration away from the black hole), we need to transform the Painlevé-Gullstrand coordinate velocity to the Schwarzschild coordinate velocity.

That coordinate transformation is forbidden by FGR, but it's easy in Einstein's standard GR (and it's especially easy after we've done exercise 24):

[latex]
\[
\begin{align*}
dt &= dt_f - \frac{\beta}{1 - \beta^2} dr \\
&= \left( - \frac{1}{\beta} - \frac{\beta}{1 - \beta^2} \right) dr \\
&= - \left( \frac{1}{\beta (1 - \beta^2)} \right) dr
\end{align*}
\]
[/latex]​

so the Schwarzschild coordinate velocity is

[latex]
\[
\frac{dr}{dt} = - \beta (1 - \beta^2)
\]
[/latex]​

Differentiating that velocity with respect to t yields a formula for the acceleration (as measured in Schwarzschild coordinates). The most extreme (negative) coordinate velocity of the infalling object is attained when that coordinate acceleration is zero.

If I did that calculation correctly, the Schwarzschild coordinate acceleration goes to zero at r=6m, where the Schwarzschild coordinate velocity will be c times negative 2/9 times the square root of 3, which is approximately -.385c.

Let me emphasize once again that my calculation above depends upon a coordinate transformation Farsight has rejected. So far as I know, the answer to dlorde's question cannot be calculated in FGR.

 

[Vorpal]

Clinger, I disagree, conditionally on that I don't really understand FGR besides its "Schwarzschild-Koordinaten sind uber alles." But from the fact ∂_t is a Killing field (or integrating the t-component geodesic equation directly),
[latex]\[ \frac{dt}{d\tau} = \frac{e}{1-2M/r}, \][/latex]
where e is the specific energy at infinity. Then we can just use the normalization of timelike four-velocities and the Schwarzschild metric directly to find
[latex]\[ \frac{1}{2}(e^2-1) = \frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 - \frac{GM}{r} \][/latex]
and therefore:
[latex]\[
\frac{1}{2}\left(\frac{dr}{dt}\right)^2 = \frac{1}{2}\left(\frac{dr}{d\tau}\right)^2\left[\frac{1-2M/r}{e}\right]^2 = \left(\frac{e^2-1}{2}+\frac{M}{r}\right)\left(\frac{1-2M/r}{e}\right)^2 \][/latex]
Differentiate with respect to t and divide through by dr/dt:
[latex]\[
\frac{d^2r}{dt^2} = \frac{M}{e^2r^3}\left(1-\frac{2M}{r}\right)\left(6M-(3-2e^2)r\right)
\][/latex]
This has a root at r = 6M/(3-2e²). For freefall from rest at infinity, e = 1, and we have your r = 6M result.

 

[Complexity]

Clinger, I disagree, conditionally on that I don't really understand FGR besides its "Schwarzschild-Koordinaten sind uber alles." But from the fact ∂_t is a Killing field (or integrating the t-component geodesic equation directly),

where e is the specific energy at infinity. Then we can just use the normalization of timelike four-velocities and the Schwarzschild metric directly to find

and therefore:

Differentiate with respect to t and divide through by dr/dt:

This has a root at r = 6M/(3-2e²). For freefall from rest at infinity, e = 1, and we have your r = 6M result.

We've seen a great many Walls of Words here, but far too few Walls of Math.

Makes me happy.

(I'm a push-over for Greek letters)

 

[W.D.Clinger]

Clinger, I disagree, conditionally on that I don't really understand FGR besides its "Schwarzschild-Koordinaten sind uber alles." But from the fact ∂_t is a Killing field (or integrating the t-component geodesic equation directly),

I didn't mean to suggest that people who know what they're doing have to use a coordinate transformation to answer dlorde's question. I meant only to say I answered dlorde's question by employing a coordinate transformation that is definitely not allowed by FGR.

So far as I can tell, FGR involves no math at all. Farsight has rejected or ridiculed all mathematical discussion of general relativity, in several different threads, so I have concluded that FGR is a math-free simplification of GR.

If Farsight wishes to bless the math you posted, however, then I'll have to admit it's part of FGR. So far as I can determine, the only way to tell what is or is not part of FGR is to wait for its prophet, Farsight, to bless us with one of his pronouncements.

I do know that Farsight is dead set against coordinate transformations, so none of Einstein's covariant/contravariant stuff is part of FGR, so none of Einstein's sections 8 5 through 22 (in his most important paper on general relativity) are part of FGR.

This has a root at r = 6M/(3-2e²). For freefall from rest at infinity, e = 1, and we have your r = 6M result.

One of the nice things about math is that people who know what they're doing tend to come up with the same answers.

 

[DeiRenDopa]

Revolutionary FGR breakthrough: ansibles possible!

Like Brian-M a bit earlier, I had an epiphany!

By thinking for myself (as oracle Farsight commands) and doing my own research (ditto), I have discovered the truly revolutionary insight at the heart of FGR: the ansible^WP!

I have a truly marvelous demonstration of this proposition which, unfortunately, this margin post is too small to contain (more News at 10).

 

[Mashuna]

I need to get into non-accepted electron models to justify this spectacular, and since it's off-topic and I don't have patent evidence and Einstein to back me up, I'll leave it there.

You've not had either of those things in any of your previous arguments either - why stop now?