19 Apr 16
Originally posted by twhiteheadDelta f is easy 1 r =~500 AU. each doubling of r results in f being 4X further out.
Version 2:
http://whereitsat.co.za/GraviationalLensing2.jpg
Challenge for the reader:
find delta f in terms of d, r, R and K
That is a nice drawing. I want to make a visualization of what the energy would look like where the beam would be like a long cloud you would see as some other color similar to the color of Sirius. Then another for Alpha Centauri, a star like our sun so the cloud would be somewhat yellowish and then the red dwarf's with a more reddish cloud and so forth. And on to all the 150 odd stars in that 20 ly radius, some 33,000 cublic light years of space.
And I want to do it full scale🙂
19 Apr 16
1 edit
Originally posted by sonhouseWhat do you mean by this?
1 r =~500 AU.
In my diagram, r is an arbitrary distance from the centre of the sun. There is certainly no special r at 500 AU. Perhaps you are talking about the point where alpha = beta? There is nothing really special about that.
20 Apr 16
Originally posted by sonhouseRight, ok, with you now. I was automatically including that without regarding it as an effect. The difficulty with "seeing" any kind of effect is that the line of focus is on the line from Sirius to the Sun as it extends past the Sun. That means that at, say, 1000 AUs what one would see is a slightly too bright and too blue Sun rather than an image of Sirius. Although light from Sirius gets there the light from the Sun would be swamping it.
The other effect is simply stars have a spherical wavefront.
20 Apr 16
Originally posted by DeepThoughtAt 1000 AU light from the sun is 1/1000^2 or about 1.4 MILLIwatts/meter squared. (~1400 watts/meter squared at 1 AU)
Right, ok, with you now. I was automatically including that without regarding it as an effect. The difficulty with "seeing" any kind of effect is that the line of focus is on the line from Sirius to the Sun as it extends past the Sun. That means that at, say, 1000 AUs what one would see is a slightly too bright and too blue Sun rather than an image of Sirius. Although light from Sirius gets there the light from the Sun would be swamping it.
Whereas at 1000 AU, the light focused there would come from an area roughly 1.5 r. maybe 1.2 r, something like that. If you look at a disc 1 meter square, each one of which would contain about 10 nanowatts of light, there are over 5 billion of those 1 square meter blobs and 10 nanowatts/m^2 times 5 billion = 51 watts/ meter squared.
That looks to me like it would overwhelm the light from the sun about 36,000:1
You do the arithmetic, see what you find.
For starters, the intensity of light from Sirius reaching us is given as 9.8 E-9 watts per meter squared. At 1.2r there are over 5 billion 1 meter squared areas.
Seems a simple multiply to me. If I am wrong, tell me what I did wrong.
BTW, I never said this focus would be like an image you would get from a real lens the size of the sun, I don't think there would be much in the way of image, just a bunch of those 1 meter area's coming together in one place.
20 Apr 16
Originally posted by sonhouseYou don't have a formula, and still don't understand the diagram.
Seems a simple multiply to me. If I am wrong, tell me what I did wrong.
What do you even mean by "an area roughly 1.5 r"? What is 'r'?
I suspect you are under the impression that once light is focused it then travels along the axis of focus. It doesn't. It goes right passed it. The energy arriving at a given point along the line comes from a thin ring around the sun.
I'll try and add a bit more to the diagram later today to help with energy calculations.
20 Apr 16
Originally posted by sonhouseThere is an image. its called an Einstein ring. For large lenses with interesting things behind, it can be quite detailed - multiple images of magnified whole galaxies. For stars it will just look like a ring of light when viewed from the exact line of focus, but makes interesting patterns when you are slightly off the line of focus.
BTW, I never said this focus would be like an image you would get from a real lens the size of the sun, I don't think there would be much in the way of image, just a bunch of those 1 meter area's coming together in one place.
20 Apr 16
2 edits
Originally posted by twhiteheadr is the radius of the sun. 1 r is right at the surface and using 1r I calculated you could put about 5 billion one meter square boxes around the surface of the sun and knowing the emission from Sirius clocking in at 9.8 nanowatts per meter squared you can get a rough idea of how much energy would be deposited at that 550 AU area, about 50 watts.
You don't have a formula, and still don't understand the diagram.
What do you even mean by "an area roughly 1.5 r"? What is 'r'?
I suspect you are under the impression that once light is focused it then travels along the axis of focus. It doesn't. It goes right passed it. The energy arriving at a given point along the line comes from a thin ring aro ...[text shortened]... sun.
I'll try and add a bit more to the diagram later today to help with energy calculations.
1.5 r, radius about 1 million km so about 6,2 billion boxes one meter size would fit there so maybe 60 watts would be deposited somewhere around 1000 odd AU. The ring of energy coming in at the 2r region would deposit 4 times further out, not a square thing, just the base is twice as large and the angle of deflection is half as much so the two effect together causes the best focus to come in at 4 times the first focus. So going out to 4 r, it would come in at 4 times the last focus, so 550 AU for 1 r, 2200 AU for 2r, 8800 AU for 3 r, 35000 AU for 4 r and so forth.
The last ring would clock in at about 500,000 AU or 8 ly. After that it would be parallel at best except for your infinitesimal stuff and then not even parallel.
20 Apr 16
Originally posted by sonhouseIt would appear you are using a delta r of 1m and r (from my diagram) at 1.5 radii of the sun. So you are calculating the amount of energy passing through a ring of one metre width. You are then making the erroneous assumption that all that energy gets focused to a point. It doesn't.
1.5 r, radius about 1 million km so about 6,2 billion boxes one meter size would fit there
You need to calculate delta f first.
There is a reason I asked you to find that formula first 🙂
20 Apr 16
1 edit
Originally posted by twhiteheadI never said it would focus into a point. All I said was the energy would come together in a volume, I never said anything about a point. That would be asking too much from such a flawed lens. In fact I rather think of it as the resultant being skewed and stretched along that line if anything, but some kind of collector, a solar mirror, telescope, whatever, would get most of that energy even if it wasn't nice and phased together. I expect no such thing. Just usable energy and with spectroscopes the ability to see the lines and show it comes from Sirius and so forth. And not a huge amount of energy either, 50 watts per meter squared is not going to give you a sunburn for sure.
It would appear you are using a delta r of 1m and r (from my diagram) at 1.5 radii of the sun. So you are calculating the amount of energy passing through a ring of one metre width. You are then making the erroneous assumption that all that energy gets focused to a point. It doesn't.
You need to calculate delta f first.
There is a reason I asked you to find that formula first 🙂
But a large enough bucket would be able to collect and use that energy and around the fringes of that one meter square area I expect there would be more energy to collect, how much I am not sure of right now.
One effect I was wondering about, would there be a frequency shift in the process or would say, a 1000 micron wave still be 1000 microns where they come together.
You can see for instance, my thought experiment of the two space craft say a light year apart on either side of the sun shooting progressively longer and longer wavelengths, there would come a maximum size wavelength that would no longer be focused simply because it would represent a wave say 10 million km across which would be a frequency of 0.03 hertz. As it passes the sun the gravity would smear it around but I think it would be fair to say there would be no amplification of that frequency simply because it is too big. So a chart of those frequencies would reveal information about the gravity and mass and so forth. And of course I know there are other methods, I just was pointing out there are other ways to scan a kit.
20 Apr 16
Originally posted by sonhouseAnd once you have found delta f, the next problem will be to find the size of the required collector. I'll update the diagram to show the necessary lines for the calculation when I get a chance.
In fact I rather think of it as the resultant being skewed and stretched along that line if anything, but some kind of collector, a solar mirror, telescope, whatever, would get most of that energy even if it wasn't nice and phased together.
But a large enough bucket would be able to collect and use that energy and around the fringes of that one meter square area...
It would appear you erroneously believe that the energy would be focused to within a metre square collector if not a point.
20 Apr 16
1 edit
Originally posted by twhiteheadIt would appear not. I was just using that as a thought experiment reference point. Till we have some kind of light bucket out at 1000 AU everything else is simply guesswork.
And once you have found delta f, the next problem will be to find the size of the required collector. I'll update the diagram to show the necessary lines for the calculation when I get a chance.
[b]But a large enough bucket would be able to collect and use that energy and around the fringes of that one meter square area...
It would appear you err ...[text shortened]... usly believe that the energy would be focused to within a metre square collector if not a point.[/b]
I chose a 1 meter box because the power reading from Sirius is given in watts per square meter.
I used to try the same thing in microns but I got all bulloxed up doing that🙂
But it does seem clear there would be more energy outside that 1 meter square. How much, I would love to find out the hard way🙂
20 Apr 16
Originally posted by sonhouseWhat you are doing appears to be mostly guesswork and very wild calculations. What an astronomer can do is highly accurate calculations that can give a very good estimate of what we would see and such calculations can be verified by observations of other gravitational lenses. 'Simply guesswork' is not a good characterization.
It would appear not. I was just using that as a thought experiment reference point. Till we have some kind of light bucket out at 1000 AU everything else is simply guesswork.
I chose a 1 meter box because the power reading from Sirius is given in watts per square meter.
But that is a 1 meter box at the sun, not a 1 meter box at the focus.
There is a reason I asked you to work out the relationship between delta r and delta f. I want you to think about it.