Since I am trying to come up with a simpler method than what I had in the other thread, that is precisely the dilemma I face as well. At the least, the rays would just curve a little further around the sun and there would be no overall magnification, just a slight distortion for the extra amount of surface area that is observed. At the most, the rays could indeed follow some apparently arbitrarily increased angle away from the tangent, just as long as the overall change between the angle of emission of the light and the angle of reception is equal to half the angle for gravitational lensing. It appears so far that only by knowing the shape of the path the light takes can the correct magnification be found. Even with some question as to the shape of the path in the other thread, a hyperbola should be correct to at least first order.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

I suspect that the magnitude of the deflection is the same as the
magnitude of the Shapiro delay. Hence my earlier reference to the
quote of Einstein in the Wikipedia article.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

Here is my stab at estimating how far beyond the classical limb of the Sun we could see, and how much the apparent disk is enlarged.

Suppose the gravitationally bent ray is a hyperbola, which would be the Newtonian solution for a particle moving fast enough. That may not be completely accurate for this relativistic exercise, but I think it will be close enough for a "ball park" figure.

A ray grazing the photosphere will bend about 0.9 arcsecond as it proceeds toward us from that point, and will approach an asymptote that is separated from the photosphere by a small distance. That asymptote will be our line of sight to the apparent position of the tangent point. This point will be about 2 miles beyond the classical geometric limb, and my plotting of the asymptote using standard calculations for a hyperbola places it approximately the same distance above the photosphere. Thus the apparent radius is increased by about 1 part in 220,000, probably unobservable by any practical means. A two-mile elevation at the Sun's distance subtends only about 4 milliarcseconds.

As I said earlier, the numbers from a rigorous relativistic calculation may be significantly different, but I really think we are within an order of magnitude here.

Yes, that's true. That is exactly the magnitude of the deflection of the ray. However, as chornedsnorkack also mentioned, the effect would produce a combination of magnification plus extra surface area observed, so if the ray just bends around the mass an extra .875 arcseconds from the straight line tangent point, then there is no magnification but an extra 4.24*10^(-6) of the surface will be observed. On the other hand, if we figure the ray is deflected from the same tangent point, so that we just observe an extra .875 arcseconds of sight for the same distance, then we would always observe the sun to take up at least 1.75 arcseconds of sky regardless of the distance, which is of course way too large of an effect to be seen at that end of the scale, so we'll need something a little more precise to find for each.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Okay, I think I might be getting somewhere finally. I was finding ways to cancel out the unknowns when dealing with the hyperbola directly, but they kept leading to cubic equations which I had no desire to try to work through. So I decided to concentrate more on the line that runs tangent to the hyperbola as a limit to see where that might lead, and found a really easy way of dealing with the unknown values, such as c and a in the image below.

The formula for the hyperbola is x^2 / a^2 - y^2 / (c^2 - a^2) = 1. But if we run the tangent line toward infinity, where x and y become very large while c and a stay the same, then the 1 on the right side of the formula becomes insignificant and the formula reduces to x^2 / a^2 = y^2 / (c^2 - a^2), providing the formula for the line itself of x = y [a / sqrt(c^2 - a^2)], with a slope of sqrt(c^2 - a^2) / a.

We can see from the image that the overall curvature of the hyperbola runs from vertical at the point tangent to the sphere of the sun and approaches the limit for the slope of the line tangent to the hyperbola at infinity, with an overall deflected angle of z = 2GM / rc^2. The definition for the line in this case is x = (tan z) y, so 1 / (tan z) is the slope when doing it that way.

So now we have (tan z) = a / sqrt(c^2 - a^2). We can also see from the image that c = a + r, where r is the radius of the sun, so we get (tan z)^2 = a^2 / (c^2 - a^2) = a^2 / [(a + r)^2 - a^2] = a^2 / (2ar + r^2). Solving for a, we find a = (tan z) r [(tan z) +/- sqrt((tan z)^2 + 1)]. Then, taking the positive result since the negative would give a negative value for a and extending (tan z) to a couple of terms with only one necessary under the square root, we get a = (z + z^3 / 3) r [(z + z^3 / 3) + sqrt(z^2 + 1)]. Extending the square root gives a = (z + z^3 / 3) r [(z + z^3 /3) + (1 + z^2 / 2)], then working through the terms and grouping them together gives us the result of a = z r (1 + z + 5 z^2 / 6 + 2 z^3 / 3) to four terms of z. Then, finding for c with c = a + r provides us with c = r (1 + z + z^2 + 5 z^3 / 6 + 2 z^4 / 3) to five terms.

Well, now that we have very precise values for c and a, it shouldn't be too much more difficult to find for the magnification and extended surface area observed from here.

Attached Images

lensing02.GIF (2.8 KB, 6 views)

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Okay, one can see from the image below that the angle of deflection will be the same for the angle between the x axis and a line perpendicular to the tangent line that runs through the center of the sphere of the sun. Normally the tangent line would just run tangent to the sphere, so the length of q would be r, and so the size of the sun would be proportional to r as it actually is. In this case, however, the length of q would depend upon the line of sight which approaches the line tangent to the hyperbola for an observer at a very large distance d as compared to r, or d >> r. This would give a maximum magnification at a very large distance where the line of sight tangent to the sun would become the line tangent to the hyperbola, whereas q = (cos z) c.

The ratio of magnification, then would just be the ratio of (cos z) c to r, or (cos z) c / r. Expanding that for (cos z) and c, we get a maximum magnification at a very large distance of mag = (cos z) c / r = (1 - z^2 /2 + z^4 / 24) (1 + z + z^2 + 5 z^3 / 6 + 2 z^4 / 3) = 1 + z + z^2 / 2+ z^3 / 3- z^4 / 8, which approximates 1 - log(1 - z) to the first few terms. Spherical massive bodies viewed from a large distance will appear this much bigger than they really are.

Attached Images

lense05.GIF (3.4 KB, 6 views)

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

From the image in the last post, the lines tangent to two sets of hyperbolas running across each side of the sun would become more and more parallel with greater distance, so that the surface area normally observed would verge upon 2 pi r^2, but as the image shows, the rays of light curve an extra z radians further around, so that the maximum surface area observed becomes 2 pi r^2 + [2 pi r^2 - 2 pi r (r - (sin z) r)] as found from the area of spherical caps, so that the maximum observed area is increased by a factor of [2 pi r^2 + 2 pi r^2 (sin z)] / (2 pi r^2) = 1 + (sin z) = 1 + z - z^3 / 6 + z^5 / 120 - ...

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

That's interesting grav.


Can you get a figure?

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Yes. For a distance from the sun that is many times its radius, the sun's apparent diameter and the surface area that can be observed are both increased by 4.24 parts in a million over what they would be with no gravitational lensing.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

That's about what I guessed as base upon the 2 parts per million time dilation leading to a 2 parts per million increase in width..

I think...

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Yes. The deflection of the ray along the path from another source is 4GM/rc^2, so z = 2GM/rc^2 for only half the path from the same source. The Shapiro delay, time dilation, and gravitational redshift are all sqrt(1 - 2GM/rc^2), which approximates just 1 - GM/rc^2, for half the difference that one would find for the increase in width.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

it seems to show, to me, that light continues to appear to travel at c, even in a strong gravitational field.

even though there is time dilation, the light appears to be traveling further, so the speed stays the same.

the Shapiro delay, I think, only accounts for the greater distance of travel.

__________________

Here is an interesting paper I found a while back. Section 3a refers particularly to what you are describing.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

so if one imagines a star, that is too massive to form a neutron star, collapsing what is the formula for the apparent size of the collapsing mass?

My thoughts are that maybe the event horizon theory of such collapsing objects is false, and that, like the Sun appearing bigger at a distance, than it is up close, that a collapsing mass looks a lot bigger too. And what you have is a quite small collapsing mass(maybe a tennis ball size, I don't know) that looks several miles across, with a radius approaching the Schwarzschild radius.
The small collapsing mass experiencing, from the outside, huge time dilation, but to its self collapsing in seconds, while its mass is radiated away, over the millions of outside years, as Hawking radiation.

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The formula for a black hole mass would be different. I have only given the magnification for light grazing a star that is extremely far from the Schwarzchild radius, producing an open hyperbola to first order. The path of light still far from but approaching the event horizon of a black hole would form that of a closed ellipse instead. Even closer, it would orbit partially as an ellipse, fall inward, and circle a few times before climbing back out, as Grant Hutchison mentioned in another thread.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I'm certain that isn't right. Light would never follow an elliptical path
around a non-rotating black hole. Almost certainly not around a
rotating black hole, either, but rotating black holes are so pooky
that I'm never sure of anything about them.

I agree with what Grant said, as clarified by me: Light on a path
that comes close to the photon sphere at 1.5 Schwarzschild radii
spirals inward more and more gradually until it reaches periastron
somewhere not far above the photon sphere, then spirals out again,
gradually at first, then faster and faster. To be clear: When I say
"faster and faster" I don't mean that the speed of the photon is
changing (although that actually is possible this close to a black
hole, especially if it is rotating), but that the spiral is straightening
out faster and faster, so that the radial component of the speed
of the photon is increasing.

The fact that the radial component of the speed of the photon
increases as it gets farther from the black hole can be taken as
an explanation of why elliptical photon orbits are impossible.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

Thanks, Jeff. I guess I should have just said that the path of light that passes close to the photon sphere of a black hole will vary greatly from the first order calculations I made.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

part of the problem maybe that light isn't really a particle, it is a wave, or not really a wave either, but it is the way point A communicates with point B.

Is there a limit to how much light can blue shift?

If not, what is the problem with light heading straight through a collapsing star the size of a tennis ball, and then passing straight out the other side, if you ignore the interaction with the matter?

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Not that I know of. The formulas say it can blueshift to infinity, but that may be one one of those first order things as well when calculating in terms of extreme cases. If light is emitted at the event horizon of a black hole, it should be redshifted to oblivion, especially since the light never reaches us. The opposite may be true for light falling in.

It's the time dilation thing again. As far as what you said earlier about the speed of light never changing, as Grant Hutchison also mentions in the other thread I started about the path of light as an ellipse or open hyperbola, one can think of the speed of light as remaining constant at c in a gravitational field, but the time dilation increases as sqrt(1 - 2GM/rc^2) with proximity to the mass. Local observers in the field will continue to measure the speed of light at c, because they are time dilated in the same way as the light. The path of light still wouldn't be longer due to the deflection of light in the case of the Shapiro delay, however, as it is measured by a distant observer, but it may be possible to use some type of Lorentz transforms in conjunction with the time dilation whereas local observers might measure the distance travelled as being longer or something like that, I'm not sure, but for the distance a distant observer actually measures, they would have to figure in the time dilation of the light to find the delay.

As far as light falling into a black hole and out the other side, due to the time dilation, one will never see light or matter actually pass the event horizon. To a distant observer, it will appear to merely "freeze" as it approaches the horizon. It can be speculated what will be observed by an observer that falls in within their own time, but a distant observer will never see this within the lifetime of the universe. The irony is that singularities are also speculated to exist within a black hole, but the time dilation actually forbids that from ever being observed by a distant observer. Even the matter that forms the black hole itself will never be seen to fully cross the event horizon, only approach it, so a true event horizon, much less a true singularity, could never be observed from a distance, so light would never be observed falling in and coming out the other side due to the time dilation either, but as I said, with such extremes, it is always possible that the equations are only correct to first or second order, so could vary slightly from the ones we work with in such a way that even the slightest variations might make a great difference in extreme cases, so noone can say for sure without firsthand observations and measurements.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

grav, Frog March,

While it is true that a distant observer would never see anything cross
the event horizon, he would see a moderately sharp boundary where
everything vanishes. Above the event horizon, matter is falling away
from him, redshifting, and fading rapidly. It fades out completely at the
event horizon, and there is only blackness below. If all the matter in a
star collapsed into a black hole while your back was turned, and you
saw the light suddenly go out, by the time you turned around to see
what happened, the black hole would be completely black. All the
redshifted light that was emitted just above the event horizon would
already have gone past you. Where a moment ago you had a big, bright
sun shining on your back, now all you see is a black circle the diameter
of the photon sphere, surrounded by a ring of starlight. Even if the
quantum nature of photons didn't restrict how long they would still be
hanging around, the limited ability to detect long wavelengths (kilometre
waves) would make the remaining light undetectable in less than a
second. So I think Wheeler's term "black hole" is more apt than "frozen
star", though the latter is more socially acceptable in some places.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

I though the idea was that you would still be able to see objects as they fell towards the event horizon, if you forget about the red shift, it's just that they would get slower and slower, in a way that you would never see them cross over, or disappear.

__________________

Yes. The time required for the light to reach a distant observer
depends on where in the gravity well it is emitted. From the event
horizon, it takes forever.

But the redshift stretches visible light from an infalling body to
invisibility within a second. The fact that light is composed of a
finite number of photons means there will be a last photon of any
given wavelength, and the last visible photon reaches the distant
observer in less than a second after the body is seen to get near
the black hole.

There could still be some very long-wavelength photons reaching
the distant observer a minute or an hour later, but they would be
so long and so few in number that they would be undetectible.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves