19 Oct 17
1 edit
Originally posted by @sonhouseI'll have to read that a couple of more times I think. If I get the general jist, do you usually calculate R's starting at one, then going up by a fixed amount over and over again until you reach a certain R value?
It's basically a square function, 3XR, F=9times that of 1 R, 6R, 36X 1R and so forth. 6R is 4.17 million Km above the 'surface' of the sun and the focal point is 36 times 80 billion Km or 2.88E12 Km out.
That reaches a limit but that is another story.
The thing about a ring that high up is, visualize an ice cream cone with a small ball in the end. ...[text shortened]... into space a line of concentrated energy will be deposited and after that, not much of anything.
19 Oct 17
Originally posted by @eladarI suppose you would not solve R's but plug in R's.
I'll have to read that a couple of more times I think. If I get the general jist, do you usually calculate R's starting at one, then going up by a fixed amount over and over again until you reach a certain R value?
19 Oct 17
1 edit
Originally posted by @eladarIt's funny trying to show people when it is so obvious to me, but I studied this for a long time. There are two competing effects. One is the divergence angle of light where first it starts out where the ice cream cone is just touching the surface, that is where Einstein's angle of bending comes in, 1.75 ish arc seconds or about 8E-6 radians. I figured this whole business out using radians. Arc seconds are not much use past showing the angle.
Once the light goes past the focal point the rays would then appear upside down ? So you would get a opposing cones?
If you invert the radian number you come up with a multiplier where the first focus is that multiplier times the radius, which puts first focus at about 80 billion Km away from the sun on the opposite side following a line from the distant sun going through the center of our sun and continuing on, and 80 billion miles out (about 550 A.U.)
But as you analyse light not just skimming but higher up, the angle is slightly greater, so at 2R the bending is only half what it was at the surface, or about 7/8ths of an arc second so the distance is 4 times out, about 2000 AU.
So light from that source from higher and higher off the surface of the sun is bent more away from the bending that the mass of the sun provides and at some point the only thing that can happen is the light in that ring will be bent parallel to the line connecting the two stars.
Get that part? Increasing angle and decreasing bending, two opposing effects and at some point there can be no more focus.
And it turned out the line of focus is roughly = to the distance between the stars so Alpha Centauri cast's a focal line 4 ly long, Sirius, a line of focused energy 8 ly long and so forth. Does that make sense now?
R is a variable, whatever you want to make it.
My next goal is to somehow make a single formula that will tell the beginning AND end of the focal line between stars. But for now, I am satisfied to have shown the line ~= the distance between the stars.
Do you see how that happens? A more distance source has lines that are smaller in angle so the place where the equal bending vs angle takes place puts the focal point for that particular R number further down the line.
The cone of light converges like a smoke ring that gets closer and closer together as it goes forward, don't even know if that can happen with real smoke rings but that is the best analogy I can come up with and at some point the smoke ring will no longer be a ring but all the bits of the ring will be together in one place. That is the focus.
19 Oct 17
As I said originally, this is a rather specialized field. When talking to others in your field assumptions about the meaning of R is taken for granted. Then when you talk to people outside your field you try to explain things without realizing all the simple background isn't there so they don't understand what seems so basic for you.
I get to deal with this quite often as a teacher, but I experience the explaining to limited background people quite often. In a class of 30, there will be a few who have been socially promoted or studied to pass tests without understanding or simply cheated.
I am doing some reading on line.
Just to be clear, this topic only has to do with light that we see, not the description of light patterns in general. This focal line made up of many focal points is light that we see. Is that correct?
Shorter back and forths work better since longer ones lead to more places of misunderstanding.
19 Oct 17
Originally posted by @eladarThe bending of light around a mass is a fundamental aspect of spacetime. So EVERYTHING is bent, RF, light, cosmic rays, IR (all of that is EM radiation) but things like neutrinos also. Neutrinos are interesting, they focus closer than the 80 billion Km, first focus, that is, because they penetrate layers of the sun so if you imagine a flashlight of parallel neutrinos hitting the sun, the ones that penetrate about .7 R are bent closer than the ones at 1R and such. The interesting thing there is if you have a neutrino detector at that focus point, you could use it to learn a lot more about the cosmos than otherwise would be capable of on Earth since there would be an ampified beam of neutrinos and there is zero other way to do that except by the fundamental bending of spacetime around a mass.
As I said originally, this is a rather specialized field. When talking to others in your field assumptions about the meaning of R is taken for granted. Then when you talk to people outside your field you try to explain things without realizing all the simple background isn't there so they don't understand what seems so basic for you.
I get to deal with th ...[text shortened]...
Shorter back and forths work better since longer ones lead to more places of misunderstanding.
Of course the hard part is getting such a detector, whether neutrino, light, UV, IR, X ray, cosmic ray, whatever, just getting that stuff to 500+AU is a daunting project, considering the deepest we have gone into space with ANYTHING is the original Pioneer Voyagers which are out something like 40 AU away from the sun now and they have been at it for forty years. And that is just a small probe. We would have to launch really big stuff needed at that distance. A neutrino detector is pretty big at least here on Earth, tons of highly purified water and such lined with sensitive photodetectors registering a small invisible to human eyes flash of light when a neutrino slams head on I mean dead nuts head on into some atom, it can cause a chain reaction releasing a photon.
Out at the focus the amplification could be as much as 100 DB so the detector can be a lot smaller but still, getting ANYTHING to 500 AU is scifi for now. Maybe in a hundred years but not THIS decade for sure.
That was one of the talking points about the focus line, the ability to see things amplified in a way that would use the sun as a giant million mile wide lens. That will have to wait for a huge increase in propulsion capabilities to do that job though, chemical rockets would never be able to do it in any kind of reasonable time frame, just look at voyager, 30 or 40 AU out and they have been going for 40 years. At that rate, 400 years to get to 500 AU
and the problem there is actually trying to use the focus as part of a super telescope, the 500 AU place is going to get interference from the sun's corona so you may have to go out to the 2 R place or 3 R, 3R would focus around 4500 AU or something like 175 billion miles from Earth. Big problem taking advantage of all that.
19 Oct 17
1 edit
So what you are talking about is using something like the sun to magnify distant objects.
As far as I know, the only way we can detect planets in other solar systems is through Jupiter size plants in Mercury type orbits creating warbles in their sun which then can be detected by a Doppler effect. My cousin Paul's team passed the light through an iodine cell to measure the effect. He said you had to calculate out the earth's movement to then see the movement from the sun/star.
It would be cool to be able to use the sun to see far off galaxies.
19 Oct 17
Originally posted by @sonhouseSo the best way to measure is through neutrinos since you don't have to be as far away from the sun to use them. Could you use them to 'see' planets?
The bending of light around a mass is a fundamental aspect of spacetime. So EVERYTHING is bent, RF, light, cosmic rays, IR (all of that is EM radiation) but things like neutrinos also. Neutrinos are interesting, they focus closer than the 80 billion Km, first focus, that is, because they penetrate layers of the sun so if you imagine a flashlight of paralle ...[text shortened]... 500 AU or something like 175 billion miles from Earth. Big problem taking advantage of all that.
19 Oct 17
Originally posted by @eladarThat is the R in the equation. That is the variable that tells the bending angle. 1 R for the sun, the bending angle is 1.75 arc seconds or 8E-6 radians, take your pick. I prefer radians because inverting radians gives the multiplier you need to calculate the focal point for a given R number. 8E-6 inverted gives about 117,000 which that times R, in this case 1R so the focal point for light skimming the surface of the sun is 117,000 times R.
As for your calculations, how do you take into account the changing bend the further you get away from the surface?
So in miles, R is about 440,000 miles so focal point is about 51 billion miles out. Or 80 billion Km, chose your poison.
For 2R, the angle is about 2E-6 which inverted is 468,000 so 468000 times 880,000 clocks in about 411 billion miles out.
19 Oct 17
1 edit
Originally posted by @sonhouseYou said R was the ratio of the distance from the center of the star to the readius of the sun.
That is the R in the equation. That is the variable that tells the bending angle. 1 R for the sun, the bending angle is 1.75 arc seconds or 8E-6 radians, take your pick. I prefer radians because inverting radians gives the multiplier you need to calculate the focal point for a given R number. 8E-6 inverted gives about 117,000 which that times R, in this c ...[text shortened]... ut 2E-6 which inverted is 468,000 so 468000 times 880,000 clocks in about 411 billion miles out.
True, that distance is related to the amount of the bend, but I did not think it was the bend.
Although you did say the rate of change in the bend is linear.
So what is the bend at R = 1?
What is the bend at 2 and 3?
Inverse relationship the number of rings to amount of the bend.
The angle of the bend is super small. I didn't know the seconds you were talking about was 3600th of a degree.
19 Oct 17
3 edits
Originally posted by @eladarFunny, now my calc spits out 9.9 E11
8/n^2 E-6 Where n is equal to the radius of the ring divided by the radius of the sun.
3R would have a bend of 8/9 E-6
If you invert that you'd get 1125000.
Times 880000 gives 3.6 E17
Went up to check the last answer. It ANS the wrong number, way off.
E11 makes much more sense.
20 Oct 17
3 edits
Originally posted by @eladarThe bend at R=1 is what Einstein calculated, 1.75 arc seconds or about 8.5E-6 radians. I was just rounding out. R=2 is 1/2th of that, about 4.2E-6 radians or 0.875 arc seconds.
You said R was the ratio of the distance from the center of the star to the readius of the sun.
True, that distance is related to the amount of the bend, but I did not think it was the bend.
Although you did say the rate of change in the bend is linear.
So what is the bend at R = 1?
What is the bend at 2 and 3?
Inverse relationship the number ...[text shortened]... he bend is super small. I didn't know the seconds you were talking about was 3600th of a degree.
There are 1,296,000 arc seconds in a complete circle.
So if you have a telescope with resolution of 1 arc second it can split the circle into that many parts, 1.2 million. So if you extend that circle to say, the moon, at 240,000 miles away, that is radius x2 X pi is a circle about 1.5 million miles and divide that by 1296000 shows that telescope to be able to see objects 1.1 miles wide. The Hubble does about 20 times better so can resolve stuff on the moon around 300 feet wide.
What I noticed in Einsteins original formula, 4GM/C^2R was it was linear, the bend at 1 R is 1.75 and for a long time till I looked closely at that original from Einstein, I thought it would be like the inverse square law but it is just linear, so at 3 R it is 1.75/3 or about 0.58 arc seconds or 2.83E-6 radians. Invert the radian number and you get 352,000 and change so the focus there would be 352,000 times 440,000X3 = about 465 billion miles out. But my formula does the same thing, R^2/ZM but the units have to be in meters and Kg. R=3^2 so 9 * 7.07E8 (meters at R=1) =4.5E18/3E-27(Z) *3E30 (Mass of sun in Kg)
3E-27*2E30=6000 so it is 4.5E18/6000 = 7.5E14 meters or 7.5E11 Km or 750 E9 km or about 4.68E11 miles or 468 billion miles out. Differences due to rounding out.
That just shows two methods give the same answer and the second one is my own. F=R^2/ZM where F is in meters, R is in meters, M is in Kg. Z is a constant, 3E-27. The sun clocks in at about 2E30 Kg so you can use that figure to see any focal point up to the distance between the stars.
That also works for planets too. like Earth, which rounded out is 6E24 Kg and radius in meters is ~6.3E6 meters so for light skimming surface, R=1, R^2 is about 4 E13 / 3E-27* 6E24 is 4E13/.018 = 2.2E15 meters first focal point of Earth. 2.2 E12 meters or about 1.3 billion miles away from the source star.
My formula is the general solution to the focal point problem
I would like to extend that to include the end point too but all I have so far is I know the focus poops out at the distance to the star so Alpha Centauri makes a streak 4 ly long and Sirius makes a streak 8 ly long. Would like to include that in my formula but that seems to be above my pay grade🙂
Does that make sense?
The thing about the first focal point of Earth is the area around Earth, say a mile wide going all around Earth would contain much less energy from some distant star than the same 1 mile wide circle around the sun so from Sirius there may only be 5 watts or so focused at that 1.2 billion mile distant focal point. The sun would have a LOT more at it's focal point of 550 AU. (1 AU = the distance between Earth and sun, or about 93 million miles) AU stands for Astronomical Units.