Thread: stars brighter and distances further due to gravitational lensing

1. stars brighter and distances further due to gravitational lensing

Okay, so now I want to know what the effect gravitational lensing has on the distance to luminosity ratio and the distance to redshift ratio of stars and galaxies. I am not considering gravitational redshift in this yet, though, just that amount light will bend as it is emitted tangent from the surface of a star along the edges of our line of sight. Since the lines of sight across the edges of a star are not a straight line, then, we will see a little more of the surface area than we otherwise would as the path of the light bends slightly around the edge of the star, making it a little brighter than it would otherwise be. Since most of the bending occurs in the immediate vicinity of the star, that is where most of the curvature would take place, so the line of sight itself would be almost precisely the same as that for the straight-line tangent points from the observer, but I will still figure in an approximation for the difference with this as well when d/r is large, which is usually the case anyway, to be as close to precisely accurate as possible.

For a path of the light forming a hyperbola with the vertices at (a,0) and the focus at (c,0), the center of mass of the star will be at the focus at (c,0). We will place an observer at (x,y). Since we have placed the origin of the hyperbola at (0,0) in this case, the formula for the hyperbola is just x^2/a^2-y^2/b^2=1. c^2=a^2+b^2, so it becomes x^2/a^2-y^2/(c^2-a^2)=1. Also, since the vertice lies at the edge of the star, the radius is r=c-a, so the formula in terms of this now becomes x^2/(c-r)^2-y^2/(2cr-r^2)=1.

The distance of the center of mass from the observer is d=sqrt[(x-c)^2+y^2]. The original angle of the path of light is zero along the y axis from the vertice. To find the angle of the path the light is travelling when it reaches the observer, which is the total amount of curvature the light has undergone, we need to find the slope between points on the hyperbola at (x,y) and some infinitesimal distance along that path. This is where the approximation takes place, however, since the total amount of curvature would be found first at points approaching an infinite distance, but since most of the curvature occurs near the surface of the star, then if d is very much greater than r, which it would be in almost every case for an observer, then the discrepency would be tremendously small, even many, many times smaller than the difference in the curvature to straight line paths to begin with.

For a given y, x would be x=sqrt[1+y^2/(2cr-r^2)](c-r). For another point at y'=y+dy, x' would be x'=sqrt[1+(y+dy)^2/(2cr-r^2)](c-r). The angle the path has curved, then, is L=tan-1 [(x'-x)/(y'-y)] = tan-1 [sqrt(1+(y+dy)^2/(2cr-r^2))-sqrt(1+y^2/(2cr-r^2))(c-r)/dy], which for large d, and therefore large x and y, reduces to a very good approximation of just L=tan-1 [(c-r)/sqrt(2cr-r^2)]. So with a known c, r, x, and y, we can find the curvature, which would then tell us the mass of the object. Normally, one would start with the mass and find the curvature to find c, though, but we can work back and forth with it if the mass is known, to determine what c must then be.

Now, from this we can tell the angle of the path of the light as it reaches the observer from a line parallel with the x axis to be just L1=90-L. We can also find the total angle from the x axis at the observer to the center of mass with L2=tan-1 [(y-c)/x]. The difference in the angle from this last angle to that of the straight line tangent points is L3=sin-1 [r/d], and to the end of the hyperbola near the observer is L4=L2 - L1.

So the actual radius of the star is r=sin(L3)*d, so the radius across our line of sight is rT=sin(L3)*cos(L3)*d=r*cos(L3), but the curve of the hyperbola due to the lensing will make it appear as rL=sin(L4)*cos(L4)*d. The ratio of the area of sight, then, will be rL^2/rT^2. Now we need to find out how much of the surface area of the star is actually seen. For this, we know that the way this is set up, we can actually see all of the way around to (a,0) on one side, where the original angle at that point is zero. For the straight line tangent point on this same side, however, we can only see around to an angle that is cut short of that by L5=90 - L2 + L3. The angle between the tangent points around the circumference, then, is LT=180-2*L3, while the angle around the circumference for the lensing is LL=2*(90 - L3 +L5)=180 - 2*L3 + 180 - 2*L2 + 2*L3=360-2*L2 (L2 is nearly 90 degrees greater than L3 the way I've done this, which is why there is a 180 degree difference in the formulas for the two).

From here we can find the surface area of the spherical cap on the surface of the star between each set of angles around the circumference. For the straight line tangent points, this is S=2pihr, where h=r-cos(90-L3)*r, so S=2pir^2*[1-cos(90-L3)], and for the curved path we get S'=2pih'r, where h'=r-cos(180-L2)]. The star is then brighter by the ratio of the areas for the line of sight and the ratio of the surface area seen than that of the straight line tangent. This gives us a total ratio between the two of

[rT^2/rL^2]*[S'/S]=

[sin(r/d) * cos(r/d)]^2 / [(sin[tan-1 ((y-c)/x) - 90 + tan-1 ((c-r)/sqrt(2cr-r^2))]) / (cos[tan-1 ((y-c)/x) - 90 + tan-1 ((c-r)/sqrt(2cr-r^2))])]^2 * [1-cos(180 - tan-1 ((y-c)/x))] / [1-cos(90 - sin-1 (r/d))]

This is the final formula, but before I continue on too much further with this, I'm going to try to plug in some quick values where d>>r to attempt to reduce this last formula as well in order to replace it with an extremely close approximation, so we'll have something a little easier to work with.
Last edited by grav; 2007-Jun-18 at 02:12 AM. Reason: correcting formulas

2. Originally Posted by grav
So with a known c, r, x, and y, we can find the curvature, which would then tell us the mass of the object. Normally, one would start with the mass and find the curvature to find c, though, but we can work back and forth with it if the mass is known, to determine what c must then be.
Okay, instead of working back and forth, if we know the mass and radius of a star, then we can use the total angle of the curvature of gravitational lensing to find c. Now, we had L=tan-1 [(c-r)/sqrt(2cr-r^2)], so tan(L)=(c-r)/sqrt(2cr-r^2), which gives us (2cr-r^2)*tan(L)^2=(c-r)^2. Expanding this out and applying the quadratic formula to find c, we get

c^2 - 2cr + r^2 = 2cr(tan L)^2 - r^2(tan L)^2
c^2 - 2rc*(1+(tan L)^2) + r^2*(1+(tan L)^2) = 0, so

If we make z=1+(tan L)^2, then

c^2 - 2rcz + r^2*z = 0 and then

c = [2rz +/- sqrt(4r^2*z^2 - 4r^2*z)] / 2
= rz +/- r*sqrt(z^2 - z)

I find we use the plus sign for this, and when z approaches one (tan L is very small), then

c approximates (1 + sqrt(z-1))*r and we finally get just

c = (1 + tan L)*r

3. I had a couple of errors in that first post. I fixed one of them, the second is that I should have used (x-c)/y, not (y-c)/x for a couple of the angles. I've also found a much simpler way of finding the final formula anyway, so I'll start over. Here are the initial angles.

L = 2GM/c^2/r * (360/2pi), where c is the speed of light here.

L2 = 180 - tan-1 (y/(c-x))

L3 = sin-1 (r/d)

L4 = L2 - L1

We can find the ratio of the cross section of the area seen to that of the straight line tangent for the same distance for the middle of each along the line of sight with just

(rT/rL)^2 = [ (tan L3) / (tan L4) ]^2

and we can find the ratio of the surface areas of the sphere that are viewed in each case with

(hL/hT) = (sin L4) / (sin L3)

The whole formula, then, is just

(rT/rL)^2 * (hL/hT) = [ (tan L3) / (tan L4) ]^2 * (sin L4) / (sin L3)

That is much better than the last one, for sure. As it turns out, though, I've screwed up the whole title of this thread. I thought for sure the cross sectional areas along the line of sight would remain about the same while the surface area we see increased, but it appears they are both about the same for one exponential factor, but inverse. So since the area for the line of sight squares while the other does not, the entire formula reduces to just one factor of

1 - 2GM/c^2/r

for "normal" conditions without extremes such as very small d and such. This means that a body is actually less as bright with a greater M/r ratio, and will appear to be further away than it should otherwise be. Now for a single solar mass and radius of that of the sun, this would only make a body appear a few parts in a million times further than it really is. For more massive and more compact stars, however, this may be more extreme.

The gravitational redshift doesn't really figure into all of this much over a large distance because it reaches a limit which is only subtracted, not divided into as part of the ratio of the distance or anything.

4. The formula I gave for L2 above only works for c>x, so I have redefined it as L2=90 + tan-1 ((c-x)/y).

The resulting formula for this of 1-2GM/c^2/r makes sense because as the M/r ratio grows higher, the body gets dimmer. When 2GM/c^2/r=1, therefore, the body will not be seen at all, and will become a black hole. According to this, then, it is at the Schwarzschild radius. The same thing will occur for light that is travelling radially and so is not gravitationally lensed, but redshifted to oblivion as it leaves the body just the same.

I have been exerimenting with the resulting formula for different distances. It is a very good approximation for even where d approaches r. I have noticed, however, that 1-2GM/c^2/r as the typical result for x<c does not appear to hold for x>>c, where d>>r. At very large distances, according to this formula, which may or may not be near precise enough, I can't tell at this point, the brightness seems to begin to drop with the cube of the distance. For a typical star, which defines the brightness of the galaxies as well, the brightness falls off dramatically at a distance equal to about that of the known universe, which may mean seeing to that distance and beyond would become exceedingly difficult, although the sizes of the galaxies at this distance would appear much, much larger than they really are. The resulting formula for the brightness deviates noticeably beyond a fraction of a light year from 1-2GM/r^2/c, and at very large distances, it becomes approximately (3.89842795*10^19 m)/d dimmer than it should otherwise be for a star like our own sun.

5. Originally Posted by grav
The resulting formula for the brightness deviates noticeably beyond a fraction of a light year from 1-2GM/r^2/c, and at very large distances, it becomes approximately (3.89842795*10^19 m)/d dimmer than it should otherwise be for a star like our own sun.
I've run different values for r and M for large distances as well, and the resulting formula for a large distance comes out to be a brightness that is (r/d)/(2GM/c^2/r)^2 (=(3.898*10^19 m)/d for our sun) dimmer than usual for the lensing as compared to a straight line tangent. Since the formula for a straight line tangent is generally just proportional to 1/d^2, with the square of the distance, that for lensing, then, is proportional to (r/d^3)/(2GM/c^2/r)^2, inversely proportional to the cube of the distance if I've done all of this correctly.

6. Originally Posted by grav
It is a very good approximation for even where d approaches r.
r is the radius of the star, and d is the distance from the observer to the star, right?

Why do you assume that the path of light is a hyperbola?

7. Originally Posted by hhEb09'1
r is the radius of the star, and d is the distance from the observer to the star, right?

Why do you assume that the path of light is a hyperbola?
Oops. Wouldn't it be? I thought something had to be wrong with what I came up with. It would just make too much difference for the distances we measure. It would explain dark matter right off the bat, though, despite the preciseness of the geometry, since it shows stars like our own sun being up to four times closer than we measure just within a distance of our own galaxy, which would then appear to be moving at speeds that are up to twice as great than they should be for the distance we would view them according to their brightness, but stars that are different than our own would be viewed at different distances, according to their mass and radius. It would also affect our measurements for dark energy tremendously, I'm sure. But if not a hyperbola, then what should I have used? Let me know and I'll begin again. I can't think of what else it could be, though. Isn't that the natural path of the effect of gravity for an open path as compared to the closed path of an ellipse? I'll look into that to make sure I'm thinking about it correctly.

8. Originally Posted by grav
Isn't that the natural path of the effect of gravity for an open path as compared to the closed path of an ellipse? I'll look into that to make sure I'm thinking about it correctly.
You're trying to use Newtonian mechanics and gravity, right? Hence, the ATM forum? It's going to be tough to check the gravitational lensing effects--the Newtonian value is smaller than the one for general relativity.

Sooner or later, you're going to have to move into the twentieth century.

9. Originally Posted by hhEb09'1
You're trying to use Newtonian mechanics and gravity, right? Hence, the ATM forum? It's going to be tough to check the gravitational lensing effects--the Newtonian value is smaller than the one for general relativity.

Sooner or later, you're going to have to move into the twentieth century.
No, I'm using Einstein's calculation for gravitational lensing, which comes out twice as great as that of Newton's for light. The total angle for the curvature for light coming from infinity, passing tangent to the surface at a distance of r from the center of mass, and continuing to infinity, then, is 4GM/c^2/r in radians. I am taking half of that because I am only considering half of the path for the light coming from the surface at a distance of r which is the radius of the star. For half the path, the light light will only curve half as much from the tangent point on, since I am using the tangent point as my reference point for the angles to begin with, so what the path before the tangent point would have been if it had started from infinity no longer applies.

10. I'm sorry, but I cannot follow your line of thought. Early in your thread you acknowledged that an ordinary star's effect on its own light will be very small. Later, after revising your algebraic setup and number crunching, you asserted that when viewed from many light years away, the apparent brightness of a star should vary inversely as the cube of its distance, not as its square. That would require that the gravitational refracting effect be large, and that the geometric properties be strongly dependent on the viewing distance.

I suspect that an unrecognized rogue term in your algebraic setup is blowing up on you. I have chosen not to try to find it because the algebra in this thread is too difficult to read and comprehend in light of the limitations of the plain text presentation and the lack of visual geometric assistance in following it.

I am analyzing the problem by simple geometric means. At viewing distances very large compared to the star's radius, we would see a portion of the surface that, in the absence of gravitational refraction, asymptotically approaches an exact hemisphere as the distance increases. At interstellar distances this would be virtually exact regardless of the distance, so the apparent brightness will vary inversely as the square of the distance.

To visualize the gravitational lensing wraparound of a star like our Sun, I simply sketched a hyperbola and its asymptotes from standard formulas, and sketched the star tangent to the hyperbola with its center at the focus. We can tell, from half of Einstein's prediction for the deflection of a background star, that the light from the tangent point would bend by less than an arcsecond and converge with the asymptote as it move out toward the observer. The asymptote leans away from the perpendicular axis by the same angle, and as such is barely separated from the limb of the star as seen by a distant observer. We should thus see a slightly enlarged disk with the added light wrapped around the limb. The proportion of this effect with the basic disk should be independent of the distance and thus the inverse square relation of the brightness should hold.

Note that I have not allowed for the gravitational redshift and weakening of the light from all parts of the star. Whatever they do, the proportions still should be independent of the distance.

Clear as mud? Draw it and see for yourself. A picture is worth a thousand words.

If anyone finds holes in my reasoning, please fire away. I am here to learn as well as to explain and sometimes refute.

11. 'rogue' algebra?

Grav, you said here in your above:
The resulting formula for this of 1-2GM/c^2/r makes sense because as the M/r ratio grows higher, the body gets dimmer. When 2GM/c^2/r=1, therefore, the body will not be seen at all, and will become a black hole. According to this, then, it is at the Schwarzschild radius. The same thing will occur for light that is travelling radially and so is not gravitationally lensed, but redshifted to oblivion as it leaves the body just the same. (bold mine)
But something doesn’t seem right. If you plug in known numbers for our Sun, you get a strange reading. Viz., G = 6.67E-11 m^3 kg^-1 s^-2, M = 2E+30 kg, c^2 = 9E+16 m^2 s^-2, and r = 6.95E+8 m, so when you plug them all into your equation above, the units work out fine, but the result is not = 1, but instead works out to = 4.303E-3 (dimensionless). Otherwise, pretty fancy footwork in the math.

Not sure what you’re trying to say here, but am all ears. [deleted]
Last edited by nutant gene 71; 2007-Jun-22 at 08:44 PM. Reason: deleted last line, not Sun's R_s = 8.85E-3m, it's Earth's (hypothetical)

12. Originally Posted by nutant gene 71
Grav, you said here in your above: But something doesn’t seem right. If you plug in known numbers for our Sun, you get a strange reading. Viz., G = 6.67E-11 m^3 kg^-1 s^-2, M = 2E+30 kg, c^2 = 9E+16 m^2 s^-2, and r = 6.95E+8 m, so when you plug them all into your equation above, the units work out fine, but the result is not = 1, but instead works out to = 4.303E-3 (dimensionless). Otherwise, pretty fancy footwork in the math.

Not sure what you’re trying to say here, but am all ears.
[deleted]
Ah, I was wondering where that came from: "deleted last line, not Sun's R_s = 8.85E-3m, it's Earth's (hypothetical)"

Grav's equation is the equation for the Schwarzchild radius.

13. Originally Posted by hhEb09'1
Ah, I was wondering where that came from: "deleted last line, not Sun's R_s = 8.85E-3m, it's Earth's (hypothetical)"

Grav's equation is the equation for the Schwarzchild radius.
Yep, I recognized it right off also, but my last line was silly. It said that his (dimensionless) result was about half of Sun's (hyptotheical) Schwarzchild radius, which is in fact wrong. The Sun's R_s is about 3 meters, per that equation. So I don't know what grav found here, which is why I'm all ears. But there seems to be a 'rogue' in the ointment.

(My silly again! Corrected by hhEb09'1 below, Sun's R_s is about 3 km.)
Last edited by nutant gene 71; 2007-Jun-22 at 10:30 PM.

14. Originally Posted by nutant gene 71
The Sun's R_s is about 3 meters, per that equation.
3 kilometers, not 3 meters

15. Originally Posted by nutant gene 71
Grav, you said here in your above:
The resulting formula for this of 1-2GM/c^2/r makes sense because as the M/r ratio grows higher, the body gets dimmer. When 2GM/c^2/r=1, therefore, the body will not be seen at all, and will become a black hole. According to this, then, it is at the Schwarzschild radius. The same thing will occur for light that is travelling radially and so is not gravitationally lensed, but redshifted to oblivion as it leaves the body just the same.
(bold mine)

But something doesn’t seem right. If you plug in known numbers for our Sun, you get a strange reading. Viz., G = 6.67E-11 m^3 kg^-1 s^-2, M = 2E+30 kg, c^2 = 9E+16 m^2 s^-2, and r = 6.95E+8 m, so when you plug them all into your equation above, the units work out fine, but the result is not = 1, but instead works out to = 4.303E-3 (dimensionless). Otherwise, pretty fancy footwork in the math.

Not sure what you’re trying to say here, but am all ears.
Sorry, I think you misunderstood that part. I said I found the ratio of the brightness (for d approaches r, so x<c, anyway) to be 1-2GM/c^2/r, so when 2GM/c^2/r=1, then one will observe zero brightness through gravitational lensing. That is to say, when M/r for a star is equal to or greater than c^2*r/2G, then light will curve completely around the star continually at its surface and not be seen at all, not that that is the case with our own sun or anything.

16. Originally Posted by Hornblower
I suspect that an unrecognized rogue term in your algebraic setup is blowing up on you. I have chosen not to try to find it because the algebra in this thread is too difficult to read and comprehend in light of the limitations of the plain text presentation and the lack of visual geometric assistance in following it.
Yes, thanks, I'm sure that would help tremendously. Sorry about that. I'll post an image of what I have done soon.

17. Okay, here's an image.

18. Curvature and corners

I am no expert on this so please forgive my ignorance.

Would it be easier to consider the curvature as a minor corner given the overall distance between point of emission and observed point.

The majority of the trip would be straight lines and the lens area fairly small so would it be significant?

Cheers

19. Originally Posted by Michael Noonan
I am no expert on this so please forgive my ignorance.

Would it be easier to consider the curvature as a minor corner given the overall distance between point of emission and observed point.

The majority of the trip would be straight lines and the lens area fairly small so would it be significant?

Cheers
Yes, the curvature converges to a straight line with a large distance as the total curvature approaches a limit which is very small for typical stars, although it can be very great as well for a large M/r ratio for the body, as with black holes, but that is usually not the case, at least for what we would use for distance indicators according to their brightness. The total curvature that has occurred remains the same as the path of the light straightens out, but the angles between the line of sight along this path and the center of mass, and that which is tangent to the star and the center of mass become smaller with greater distance. The question is, though, in what ratio to each other? It would seem at first glance that the difference should not be significant, however, which is why I've either got an error somewhere, or need to be more precise in some way, or else this might be very important after all when it comes to measuring distances.

20. Here is a more detailed image. 'c' used in the formula for the first angle is the speed of light, not to be confused with 'c' in every other case, which is the distance from the origin to the center of mass. The vertice of the hyperbola is at (a,0), the center of mass is at (c,0), and the observer is at (x,y).

And just in case one has trouble, if you set your pointer over the larger version that appears after clicking on the thumbnail and then wait a couple of seconds, an enlargement button will appear that will allow you to see the details clearly (it does for me, anyway, but that might just be my software, I'm not sure).
Last edited by grav; 2007-Jun-23 at 02:24 PM.

21. I do not understand what you are calculating and charting in that sketch. In the second paragraph of your OP, you describe the path of a grazing ray as a hyperbola with the center of the star at its focus, and you started with a valid equation for such a curve. All of that makes perfect sense to me. However, the sketch looks like nothing of the sort.

For a hyperbola there should be a pair of straight-line asymptotes coming from the origin of the x and y axes in the upper part of the chart. The angle between the asymptote and the x axis should have a cosine of plus or minus a/c. It would be somewhere near 45o for this one. The ray should be sharply curved at the grazing tangent point, but it should straighten out quickly and converge toward the asymptote. That is in keeping with having strong curvature near the star, with the curvature greatly diminishing as we move away from it. Your curve, however, looks more like a circular arc, and it is nowhere near where the asymptote should be for the relative values of the star's radius and its distance from the origin.

I stand my my model. As before, I get hopelessly confused trying to follow the development of your calculations through the course of the thread. Any further clarification would be most welcome.

Now let me proceed with my take on this exercise in the form of a thought experiment. Suppose we are observing a distant neutron star that is out of parallax range even for Hipparchos, and using a monster spectrograph that can detect and measure all of its radiation. It is part of an eclipsing spectroscopic binary, so we can get its mass and thus its radius. Our spectrograph can determine its temperature and thus its luminosity. We make the necessary corrections for the gravitational lensing and redshift (assuming the two effects do not cancel each other) and proceed to get a good estimate of its distance. Failure to make the gravitational corrections would throw our distance estimate off.

For a distant observer, I stand by my opinion that the ratio of the corrected luminosity to the uncorrected one will be independent of the distance, as I argued in a previous post. I still am confused by that distance cubed remark around the middle of the thread. Please explain.

I think we can agree that these gravitational effects on the star's own light will be negligible for ordinary stars.

22. Originally Posted by Hornblower
For a hyperbola there should be a pair of straight-line asymptotes coming from the origin of the x and y axes in the upper part of the chart. The angle between the asymptote and the x axis should have a cosine of plus or minus a/c. It would be somewhere near 45o for this one. The ray should be sharply curved at the grazing tangent point, but it should straighten out quickly and converge toward the asymptote. That is in keeping with having strong curvature near the star, with the curvature greatly diminishing as we move away from it. Your curve, however, looks more like a circular arc, and it is nowhere near where the asymptote should be for the relative values of the star's radius and its distance from the origin.
Yes, you're right about that. I wanted to place a quick image on here and didn't have a hyperbola in the set of shapes for the software I was using, so I used the side of an ellipse as the next best thing, and so it is exeragerated somewhat. I hope that didn't throw you off. The formulas, however, remain the same for that of a hyperbola. The asymptote runs from the origin to tangent to the hyperbola at an infinite distance.

I ran back through all of the angles again last night using a different method and used the precise formulas all the way through this time. Enough time has passed since I did it before that I could forget how I had found them before except in general. Everything worked out almost exactly the same as before with the formulas and angles for the hyperbola, coming out to very much the same as any approximations I had used, with barely a measurable difference (although the subtraction of the two angles for L4 may still make one. I haven't run it yet). But when I got to the ratio of the areas for the spherical caps, something seems different this time around. I haven't had enough time to finish yet, but I think this is where any error might lie. I'll keep working on it.

Also, thanks for your interest and help on this.

23. Okay, I found it. I was just using approximations at first because I didn't expect to get so deep into it in the first place and just wanted an idea of what the result might be. Although the approximations were off by no more than a few parts in a million at best, L1 and L2 become almost identical for large d for some reason, I suppose because their overall difference of L4 becomes very small, so when they are then subtracted to find L4 using those approximations for the initial values, the largest part of their values are cancelled out and all that is left is the small difference in the approximation, which causes that difference to become first order and very important beyond a certain distance where L4 becomes very small. I noticed this when I first ran the numbers, but that was just the preliminary, so I ran them anyway, and then forgot that that was where a large difference might take place until I ran them again last night. That was very irresponsible of me and I should have known better than to continue with that.

I have now performed it again in a different way but similarly. I found the exact formula for a line tangent to a given point on a hyperbola and used that to find the angle for the line of sight and so forth. I haven't run back through it for the spherical cap yet, but the angles work out now without anything too wierd resulting like I was getting before. I have found that I should be using [(sin L3) / (sin L4)]^2 for the ratio of the areas viewed along the line of sight for lensing compared to the line tangent to a spherical body, though, whereas the tangent of the angles is for a flat plane view only, although they still come out almost precisely the same for when d is even just slightly greater than r, but the final formula for the ratio of brightness all in all now just reduces to simply (sin L3) / (sin L4) , depending on what I find for the precise values for the spherical cap.

The precise calculations that I have found and the order in which they are to be performed are

Given: M, r, y

z = 2GM/cl^2/r
c = r*(z^2+1)*[1 + sqrt(1-1/(z^2+1))]
x = sqrt[1 + y^2/(2c-r)/r] * (c-r)
d = sqrt[(x-c)^2 + y^2]
L1 = tan-1 [(y + r*(2c-r)/y)/x]
L = 90 - L1
L2 = 90 + tan-1 [(c-x)/y]
L3 = sin-1 (r/d)
L4 = L2 - L1
ratio of brightness = (sin L3) / (sin L4)

One will notice that these will reduce to the same thing as found before as an approximation except for L4, which makes all the difference in the final formula. The resulting difference for the ratio of brightness is now very small at any distance, not varying too much from 1 for the M/r ratio of a star like our sun.
Last edited by grav; 2007-Jun-24 at 11:48 PM.

24. Well, I'm not even sure now what I did to find the ratio for the spherical caps before, probably took 'h' for the conic sections or something, but it wasn't right. This time I find the ratio to be [1 + (cos L2)] / [1 - (sin L3)], and I have an image for it below so one can check to see how I found it. I double-checked to make sure the angles work out whether positive or negative and that h' also works out the same way for the spherical cap when h' > r. The final formula now, then, is

[(sin L3) / (sin L4)]^2 * [1 + (cos L2)] / [1 - (sin L3)]

This gives a result where as the brightness dims with distance, it approaches a limit of (1 - 2GM/c^2/r) dimmer than the light would ordinarily be with the square of the distance for the tangent. It approaches this limit very quickly, however, similar to that of the gravitational redshift for the radial path of light. It would seem these two are very tightly related.

So now I have redone all of my angles and formulas as precisely and carefully as I can, and all of this seems very natural now, with the resulting formula quickly approaching the limit with greater distance as it does and all, so everything seems to be the way it should be, except for just one more little thing. I tried these new more precise formulas for a distance that is very close to r as well. According to this, as d is almost equal to r, where one would almost be standing upon the surface of a star, instead of approaching just 1 for the brightness on the surface, its light actually appears brighter due to the gravitational lensing. It says it is brighter by a factor of 1 + (2GM/c^2/r) / (1 + sqrt(2)). I'm not sure why exactly, except that we can see a little further around the horizon with lensing than we otherwise could. So now I have to think further about how we might define the original brightness to begin with. If I go by what we would normally observe with no lensing, defining that as 1, then these formulas stand. But if I define it by what we might normally observe on the surface including any lensing that takes place there, defining that as 1 instead, then the resulting formula must be divided further by 1 + (2GM/c^2/r) / (1 + sqrt(2)), making the resulting limit for the dimness for what we would view at a large distance equal to [1 - 2GM/c^2/r] / [1 + (2GM/c^2/r) / (1 + sqrt(2))], which is approximately 1 - 2GM/c^2/r * sqrt(2) for small M/r.
Last edited by grav; 2007-Jun-25 at 04:44 AM.

25. Originally Posted by hhEb09'1
Why do you assume that the path of light is a hyperbola?
You know... I've been thinking about this. The path is a hyperbola in classical physics. It's what we usually ascribe to the path of a body in a gravitational field, right? Well, with the advent of GR, has that changed? Is the path slightly different from this with the mathematics being slightly different from Newtonian, or is it still hyperbolic, but taking on a different route, like an extended or widened hyperbola or something?

26. I'm also wondering why the brightness at the surface should be greater by a factor of 1 + (2GM/c^2/r) / (1 + sqrt(2)). I've found and used all of the correct and exact formulas for observing gravitational lensing along the path of a hyperbola at this point, as far as I know, and it works out as it would seem it should for a large distance now, but I still thinking that it should be just 1 right at the surface, because there is no distance left for the light to curve around when d=r. I'm thinking that might tell us something about the actual path the light takes that is slightly different from that of a 'regular' hyperbola, where it would come out to just 1 at the surface.

27. Well, I posted this question about the precise shape of the path in Q&A and antisoneb reminded me that GR also includes precession, causing an ellipse to spiral around as a body orbits, so we cannot use just the usual formula for an ellipse with that. Now, I'm sure the same would apply in the case of an open orbit, as with a hyperbola. The thing is, though, that this would cause the curvature of the path to bend around further than it otherwise should when considering it in terms of classical physics only, but I think this is already included here anyway, in terms of time dilation in GR, and so at least in the case of light itself, would create an overall curvature that is twice as great. If so, then this is already accounted for in this thread, as the curvature is figured at twice as great as that of the Newtonian value, in accordance with GR.

But, the precession of an ellipse can be found by adding an extra amount of distance along the line of motion in the line of orbit regardless of distance from the center of mass but depending upon the angle to it. This might warp a hyperbola somewhat, even if the curvature is figured twice as great, I'm still not sure. If I can, I might try to work with the spiralling of a hyperbola in the same way as for that of an ellipse to account for precession and see if the shape still remains that of a hyperbola overall. The more I think about it, the less I see how it could.

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