Question about energy at one wavelength:

Originally posted by DeepThought
No, the inner limit is not due to the shadow of this sun. Imagine building a gigantic glass lens where the profile of the lens is chosen to exactly mimic the gravitational effect of the sun on light passing through it. In this case there is no possibility of a shadow effect. Parallel rays of light impinging on a lens are focused to the point at the fo ...[text shortened]... al length of the lens (which is 8.0E7 clicks from the sun). I'm not sure what happens any more.

I don't know if you can call a light path that goes parallel to theta as being focused. Focus to me means concentration of energy. Besides, that is only for one critical angle of theta.

Above the theta angle the light is no longer even in parallel, just very slightly bent inwards. It turns out though there is a minimum focus at 8E7 km but also a maximum focus at whatever distance is between the distance of the two stars, for Sirius, 8 ly, for Alpha Centauri, 4 odd light years. And of course, the amount of actual energy in the focus line depends directly on the amount of energy emitted by the distant star.

My son in law has a Phd in statistical physics and teaches in Natal Brazil and we went over these numbers and he agrees with me. He at first wanted to do a formal study but his own teaching and research activity severely limited his available time so I am pretty much on my own.

The thing I see here is a different reading of how energy is distributed in the universe.

I think it is a glorious extension of what we see with telescopes.

Every star focuses light from every other star so there are porcupine like shafts of light coming off every star in the universe and I think it will eventually be shown that light has effects on objects in the Oort cloud and such, the added momentum could cause shifts in the orbital mechanics of objects in the cloud and could cause them to start spiraling in to the central solar system.

I at first thought we were dealing with megawatts of energy but I was wrong on that count, a 1 meter circular disc of that energy from Sirius amounts to about 50 watts but it increases as you go away from the sun because of the geometry of the larger discs simply contains more energy in a linear relation, 2X the distance in r, you get 2X the energy to be dumped out at 4X the distance of the first focus and so forth.

But that is not the end of the story, there will be more energy in the fringe of that beam also so if you have a surface area of say a million km square, or a disc about 1000 km across, roughly, then if it intercepted that beam there could be enough momentum imparted solar sail wise to get it to interrupt its normal stately orbit out say one light year or a half light year from the sun.

And of course those changes would be weak and it would not get down to the center of the solar system for maybe millions of years AND the probability for even an intersection like that to happen is low but there are millions of objects out in the Oort cloud so that ups the statistics of running into one of those beams.
Remember, EVERY star nearby is going to have it's own beam and THAT also ups the probability of it hitting some object in the Oort.

What I would love to have is help from a graphics guru to get all this false colored to show what every star inside of say 30 light years from us, what that set of beams would look like and if it could form the basis for an interstellar propulsion as a solar sail, the idea being, ride the beam to some destination but for that to work, even if the solar sail thing works, what destinations would be available, since the beams go out from the sun at odd 3 dimensional angles, not every beam would lead to a useful destination.

So the number of beam riders would be small at best but it would be an interesting study to see if there is enough energy in those beams to actually go from point A to point B.

It might be that energy could be used in a system that uses the energy to power some kind of ion drive, that would probably produce more long lasting thrust than a simple solar sail.

I want to build up a picture of where those beams go in a three dimensional analysis of the nearby stars but that is a mind boggling study. Imagine following a line from every star going who knows where in a map of the 3D volume around say the 30 ly radius I mentioned.

My guess is the beam could extend the range of some kind of nuclear/antimatter/fusion, whatever kind of drive you could safely stick on an interstellar craft, the Hawking postage stamp solution notwithstanding.

So I can see the beam, after much analysis, we find one beam that goes close to a desirable destination and at some point you have to hop off the beam and continue with your antimatter or fusion rocket changing directions to get to your destination. Kind of like taking a bus but it only goes within 40 miles of your destination and you have to continue on foot or bicycle the rest of the way at a different angle the bus dropped you off.

Originally posted by sonhouse
I don't know if you can call a light path that goes parallel to theta as being focused. Focus to me means concentration of energy. Besides, that is only for one critical angle of theta.

Above the theta angle the light is no longer even in parallel, just very slightly bent inwards. It turns out though there is a minimum focus at 8E7 km but also a maximum ...[text shortened]... ind of ion drive, that would probably produce more long lasting thrust than a simple solar sail.

If you are just claiming there's a maximum of intensity there then that's different. Light must be focussed to points further out than that though. Given your formula:

D = K/r

where K = 4GM/c^2 (if I've remembered it correctly), then assuming that r is the distance of the light ray from the sun at the point of closest approach and the radius of the sun is R then r is in the range [R, infinity] and so D is in the range [0, K/R], so the line of foci should extend from R^2/K to infinity. I'll look at this again tomorrow and try to do the calculation properly as I've been using handwaving arguments up until now.

Originally posted by DeepThought
If you are just claiming there's a maximum of intensity there then that's different. Light must be focussed to points further out than that though. Given your formula:

D = K/r

where K = 4GM/c^2 (if I've remembered it correctly), then assuming that r is the distance of the light ray from the sun at the point of closest approach and the radius of t ...[text shortened]... row and try to do the calculation properly as I've been using handwaving arguments up until now.

This much is true: You will find an end to the focus. You might call light that runs parallel some kind of focus that goes on to infinity but I am talking about focus that actually focuses usable amounts of energy. Einstein did the inner focus point very early on, which for the sun is about 80 billion Km out.

There are other considerations too but I'll leave you with that one to work out.

BTW, K=5907.845 at least on my trusty rusty HP11C.

My HP48GX extends that to 5907.84590714

That number is done using the best numbers we have available now for G, mass of the sun and c^2.

G comes in at 6.67384 E -11, mass of sun at 1.989 E30 Kg and C^2 at 8.987551787 E16. so that 5907 is the result of 4GM/c^2. Probably 5907 is about as good as you can count on since the digits there extend to only 4 significant figures.

So 5907/ 6.963 E8 (radius of the sun in meters) gives that angle of 1.75 arc seconds but in radians, 8.48 E-6.

I go to an online app converting radians to arc seconds and it says 1.749 and change. close enough for government work🙂

If you extend r to 100X, that is 69 E6 km up from the sun. From the distance of Sirius, that roughly represents .5 arc second give or take.

The thing is, at 100r the angle of deflection is 17.5 milliarcseconds, a LOT smaller than that divergence angle of about 0.5 arc second so at 100r it can no longer focus.

The trick is to solve for that angle to be parallel with theta after the two effects are coupled together.

I forget the exact number for r right now, something like 34 r comes to mind for Sirius.

Have to go over my figures again. It would be nice to have a formula I can just crank this stuff out rather than having to solve it from multiple guesses.
dang. 34 r is too far out also. Deflection angle about .05 arc second and the divergent angle about 0.4 arc second way too big an angle for that .o5 to deflect even to parallel much less focus.

At 8 ly, one arc second = about 3.6E8 km above the sun.

So 20r looks like a deflection angle of about .09 arc second and the deflection angle of Sirius at that r # comes in about 0.01 so the solar deflection exceeds the Sirius deflection so light passing by that distance will be deflected by an angle of around 0.08 seconds of arc.That focuses at about 3.4 light years. Not close to the limit yet.

Originally posted by DeepThought
No, the inner limit is not due to the shadow of this sun. Imagine building a gigantic glass lens where the profile of the lens is chosen to exactly mimic the gravitational effect of the sun on light passing through it. In this case there is no possibility of a shadow effect. Parallel rays of light impinging on a lens are focused to the point at the focal length of the lens.

Except the lens in question does not have a focal length. It is not equivalent to a convex or concave lens. It is equivalent to a lens with a point in the middle. It does not have a point of focus but a line of focus.
Edit: Correction: it is equivalent to a lens with a dip at the centre. The closer the light is to the centre the more it is bent. Both concave and convex lenses are flat at the centre and bend light more the further out you go. A convex lens is required for a focal point.

You'd need the incoming light to be already converging (before being influenced by the lens) for it to reach a point of focus closer than the focal point.
As above, there is no such thing as a focal point. Further in real life the lens is not a flat plane but could be considered spherical although in reality is it infinite. The light starts converging the moment it leaves Sirius, so in an ideal universe with only two stars the focal line goes between the two stars as well as out both ends - yes the light is focused on the far side of Sirius too!

As theta increases to the critical value, when the rays are left travelling parallel to the imaginary line joining Sirius and our lens, in other words the light is focussed on the point at infinity, by continuity the point of focus must pass through each point between the focal point and the point at infinity.
That I agree with and is where I disagree with sonhouse who has an unexplained discontinuity.

Originally posted by sonhouse
It turns out though there is a minimum focus at 8E7 km but also a maximum focus at whatever distance is between the distance of the two stars, for Sirius, 8 ly, for Alpha Centauri, 4 odd light years.

And you have given no reasoning to back that up.

And of course, the amount of actual energy in the focus line depends directly on the amount of energy emitted by the distant star.
No, the energy is spread out along the line and varies depending on where along the line you are.

The thing I see here is a different reading of how energy is distributed in the universe.

I think it is a glorious extension of what we see with telescopes.
What you appear to not realise is that we have telescopes and have observed the effect with telescopes directly so in reality you have discovered nothing new.

Every star focuses light from every other star so there are porcupine like shafts of light coming off every star in the universe
I think you are seriously mistaken about how it works. The light along the focal line is not travelling parallel to the line. It is not a single shaft of bright light.

and I think it will eventually be shown that light has effects on objects in the Oort cloud and such, the added momentum could cause shifts in the orbital mechanics of objects in the cloud and could cause them to start spiraling in to the central solar system.
You are wrong. The star is significantly brighter when observed along the focal line but only in the order of a couple of orders of magnitude. In addition the objects in question are not points and so in reality the energy is spread out over quite a wide area. Overall, you are suggesting that a star like Sirius could affect the Oort cloud without any lensing given that it is orders of magnitude brighter than most other stars.

I at first thought we were dealing with megawatts of energy but I was wrong on that count, a 1 meter circular disc of that energy from Sirius amounts to about 50 watts but it increases as you go away from the sun because of the geometry of the larger discs simply contains more energy in a linear relation, 2X the distance in r, you get 2X the energy to be dumped out at 4X the distance of the first focus and so forth.
But that is countered by the greater spread along the focal line.

I repeat - gravitational lensing has been observed! And no, it didn't burn out the telescopes.

A couple of references:
http://spiff.rit.edu/classes/phys240/lectures/lens_results/lens_results.html
The reference discusses a number of uses of gravitation lensing including actual data of actual observed lensing that contradicts sonhouse's claim of a finite limit to the lensing effect. I assume that sonhouse believes the limit would be a lot smaller for planets rather than stars, yet we observe lensing events for planet sized objects.

The second reference clearly states that sonhouse is wrong.
http://www.centauri-dreams.org/?p=785

The distance is 550 AU as a minimum, with continuing use of the instrument after that. There is no upper limit on the distance for FOCAL because, unlike the case with optical lenses, the gravity-focused radiation stays on the focal axis after 550 AU. In other words, the focal line extends to infinity.

If you want highly technical analysis then try this:
http://arxiv.org/pdf/1105.3544v1.pdf

Originally posted by twhitehead
A couple of references:
http://spiff.rit.edu/classes/phys240/lectures/lens_results/lens_results.html
The reference discusses a number of uses of gravitation lensing including actual data of actual observed lensing that contradicts sonhouse's claim of a finite limit to the lensing effect. I assume that sonhouse believes the limit would be a lot smaller f ...[text shortened]... ays on the focal axis after 550 AU. In other words, the focal line extends to infinity. [/quote]

That is clearly wrong. It is not a focal 'length', it is a series of focal points in a line. And there IS an end to the focus. I of course know all about Einstein rings. Did you think I didn't know about all that? What I am focusing on is local effects and that has not been pursued in any significant way.

I am looking for help with the graphics of what I think is going on. The calculation I made showing that 3 light year figure was made by comparing the divergence of light from Sirius at that 20r volume of space above the sun which is about 0.01 arc seconds divergence from theta, the name you gave to the line going between the center of Sirius to the center of the sun. If you divide the 1.75 number by 20 you get a convergence angle of 0.08 arc seconds which overwhelms the 0.01 arc second angle by 8 to 1. Therefore at that r level, it is easy to figure the focal point for that r. Which turns out to be about 3 light years away from the sun.

That is simple trig. Is there something in that not correct? Do you disagree with the divergence angle of light coming from Sirius at 20r?

Do you disagree the angle of convergence is about 0.08 arc second at 20r?

Why don't you go with those numbers and see where the ring of light at 20r around the sun converges.

This isn't rocket science, simple trig.

One parsec is that distance that = 1 arc second of divergence at 1 AU.

That is 3.26 ly. The distance to Sirius=8 light years.which is 2.4 parsec. That means 1 arc second of divergence is 2.4 AU which is about 150 E 6 km above the sun.

So you can cut that in half to see 0.5 arc second= 75 million km above the sun.

Divide by 10 and you get 0.05 arc second of divergence at 7.5 million km above the sun.

Now at 7.5 million km above the sun is 10.7 r. Divide 1.75 by 10.7 gives you a convergence angle of 0.16 seconds of arc. 0.16 minus 0.05 and the light is converged at an angle of 0.11 arc second. Convert that to radians and it is 5.33E-7 radians. Invert that to 1.6E6 and multiply by 10.7r or 696300 and you get 1.1E12. One light year is 9.461 E12 Km. 1.1E12/9.461E12 = 0.11 light year focus point for light passing above the sun at 10.7r.

Not as much as I calculated before, but I crossed all the t's and dotted all the I's this time.

Originally posted by sonhouse
That is clearly wrong.

No, it is correct. If it was 'clearly' wrong you would be able to explain why. You haven't. You just keep insisting that you are correct with no actual explanation.

It is not a focal 'length', it is a series of focal points in a line.
And the piece I quoted called it a focal axis.

And there IS an end to the focus.
No there isn't, and all my sources which include observational evidence clearly contradict that claim.

I of course know all about Einstein rings. Did you think I didn't know about all that?
No, I of course know that you know about them. They are, after all, what we are discussing.

What I am focusing on is local effects and that has not been pursued in any significant way.
And what do you mean by 'local effects'? If you are still talking about a laser-like high powered focus point that can push asteroids, you are mistaken.

I am looking for help with the graphics of what I think is going on.
Draw it on a piece of paper by hand.

The calculation I made showing that 3 light year figure was made by comparing the divergence of light from Sirius at that 20r volume of space above the sun which is about 0.01 arc seconds divergence from theta, the name you gave to the line going between the center of Sirius to the center of the sun. If you divide the 1.75 number by 20 you get a convergence angle of 0.08 arc seconds which overwhelms the 0.01 arc second angle by 8 to 1. Therefore at that r level, it is easy to figure the focal point for that r. Which turns out to be about 3 light years away from the sun.

That is simple trig. Is there something in that not correct? Do you disagree with the divergence angle of light coming from Sirius at 20r?

Do you disagree the angle of convergence is about 0.08 arc second at 20r?

Why don't you go with those numbers and see where the ring of light at 20r around the sun converges.

This isn't rocket science, simple trig.
Yes, its simple trig but its based on flawed reasoning. Why did you arbitrarily pick a 20r volume of space?
Don't try to do the math until you understand the problem.

Originally posted by twhitehead
No, it is correct. If it was 'clearly' wrong you would be able to explain why. You haven't. You just keep insisting that you are correct with no actual explanation.

[b]It is not a focal 'length', it is a series of focal points in a line.
And the piece I quoted called it a focal axis.

And there IS an end to the focus.
No there isn't, an ...[text shortened]... itrarily pick a 20r volume of space?
Don't try to do the math until you understand the problem.[/b]

We are talking at the same time. Read the full post I just made.I am trying to find the r number that makes the angle of convergence equal to the angle of divergence. Not there yet. At 21r, the divergence of Sirius is 0.1 arc seconds and the convergence is .08 so that will not focus.

Now I read 20r has divergence and convergence almost equal so that must be close to the angle that leaves the convergence parallel to theta.

At 15r the divergence from Sirius is 0.07 arc second and the convergence of the sun is 0.11 arc second that leaves 0.04 effective convergence.

.04 arc seconds is 1.9E-7 radians, inverted is 5.2E6 times 696300 is 3.66E12 km focus. divide by 9.4E12 and that focus comes out around 0.4 light year.

I am getting different numbers, not leading where I once thought.

It's a bit trickier than I first thought.

Originally posted by sonhouse
We are talking at the same time. Read the full post I just made.

Lots of rambling calculations with not much explanation as to why. As I said, don't do the math first, do the logic first. What are you trying to calculate and why?

It's a bit trickier than I first thought.
Clearly. And I gave you a reference with formulas in so that you don't need to be guessing in the dark, but you dismissed it offhand despite it being by an astronomer with the exactly appropriate speciality.

I think I get what you are doing. You are guessing a number of radii out from the sun, then checking whether or not the light converges. You get what you think is the largest integer radii that converges, then work out the focus point for that radii.
Your error is to restrict yourself to integers. What about 20.1562346778r ? You are forgetting that it is continuous and not discrete. You are also making the mistake of trying to do it all by hand instead of simply solving the equation.

Originally posted by twhitehead
I think I get what you are doing. You are guessing a number of radii out from the sun, then checking whether or not the light converges. You get what you think is the largest integer radii that converges, then work out the focus point for that radii.
Your error is to restrict yourself to integers. What about 20.1562346778r ? You are forgetting that it is ...[text shortened]... e also making the mistake of trying to do it all by hand instead of simply solving the equation.

If I knew what that equation was, I would use it. All I have now is the deflection angle equation, 4GM/c^2*r

So I made a chart of like you said, integers from 1 to 35 of the solutions to that equation. At 30 r the angle is .0586 arc seconds.

Using parsecs, 1 arc second of divergence from theta amounts to 2.45 AU which is 364.5 e9 meters. Using that, the angle of divergence at 30r is 0.053 arc seconds, very close to the above 0.058 arc seconds. Close enough to call it a draw, where the two effects cancel out and the distance for that is over 8 light years, like I said, the distance between Sirius and the sun and extending for that distance past the sun and pooping out as far as concentrated energy goes, at around 8 light years.

I did it more spread sheet fashion this time.

Just look at the angle above theta vs the deflection angle of the sun at 30 r.

They are very close to the same number. You have to subtract the angle above theta from the deflection angle to get the real left over deflection, which both numbers are in the vicinity of 0.058 arc seconds.

Originally posted by sonhouse
If I knew what that equation was, I would use it. All I have now is the deflection angle equation, 4GM/c^2*r

So combine that with the angle of incidence equation:
arctan ( r / d ) where d is the distance between the two stars (Sirius and the sun).

To find where the light will end up parallel you need to solve:
arctan ( r / d ) = GM / ( r* c ^ 2 )
Not trivial algebraically but with a computer you solve it with ease.

...like I said, the distance between Sirius and the sun and extending for that distance past the sun and pooping out as far as concentrated energy goes, at around 8 light years.
And that is where you are completely wrong, because you seem to be under the false impression that light travels in integer intervals. It doesn't.

As I keep saying, stop banging your head on the math, and get the logic right first.