Question about relativistic acceleration:

Originally posted by sonhouse
If you look at our theoretical laser just skimming the surface and we have eliminated absorption from corona and other stuff, the famous number Einstein came up with, 1.75 arc seconds is what the laser beam would be deflected going by the sun when the beam just skims the surface.
This is all mass VS diameter thing. Nothing to do with heliopause. Actually, ...[text shortened]... 000 Km (radius of sun), comes out to about 81 billion kilometers or 54 billion miles or 580 AU.

It's a good idea to convert to radians for this small angle stuff. In radians there are 2*pi radians to a full circle = 360 degrees. The formulae are then:

sin(x) ~ x - x^3/6 + O(x^5)
cos(x) ~ 1 - x^2/2 + O(x^4)

So:
tan(pi/2 - x) = sin(pi/2-x)/cos(pi/2-x) = cos(x)/sin(x) ~ (1 - x^2/2)/x = 1/x - x/2

So for 1.75 arcseconds we convert to radians by 2*pi*1.75/(3600*360) = 8.482 * 10^-6
tan(pi/2 - x) = 117896 - O(10^-6) = 117,896

So I agree with your maths, but from an engineering point of view the idea is unfeasible. To get something at the focal point you need to get it to 580 AU's which is just too far away, Voyager hasn't got that far yet.

Originally posted by sonhouse
If you look at our theoretical laser just skimming the surface and we have eliminated absorption from corona and other stuff,

I must be missing something. What is the point of all this? So you get two laser beams that meet up at some distant point. How is this different from merely pointing them directly at the spacecraft?
They do not get focused by passing near the sun, in fact its the opposite, each beam gets spread out.

Originally posted by twhitehead
I must be missing something. What is the point of all this? So you get two laser beams that meet up at some distant point. How is this different from merely pointing them directly at the spacecraft?
They do not get focused by passing near the sun, in fact its the opposite, each beam gets spread out.

That spread out thing is what I considered at first but if you look at any physics 101 book you can see them chart it out where a beam gets deflected inwards not outwards and that deflection is at the 1.75 arc seconds angle.
I was just using the laser idea as an illustration of the effect. There would be not much actual use for those beams.

So a beam on the left side of the sun and a beam on the right side of the sun all aimed skimming the surface will converge at 580 AU. Remember, the largest deflection angle is at the surface. It turns out the deflection angle does not follow the inverse square law. It is just a linear function, where the deflection angle at 2 r, 700,000 km above the surface is not 1/4th of 1.75 but 1/2 of 1.75 or 0.875 arc second. So that means the focal point where the two beams converge if they are 700,000 km above the surface but before the encounter with the sun the beams are in parallel, they now focus at 4 times the 580 AU or 2320 AU.

So that means that the further a beam gets from the sun, the less deflection it has so the main deflection is already done by just skimming the surface or if the beams are 2 R up the residual deflection is that much less.

So the actual convergence might not be exactly at 580 AU, probably a bit less because of the residual effects of the deflection still going on but I don't have a good enough math background to figure out what the calculus of all that would be, it would be the sum of all the distances the beams go by. Maybe the actual focus would be half the 580 au but it would for sure be somewhere between 290 and 580 AU.

Anyone able to do the math on that? I just simplified it down to where there was no further deflection after the beams skim the surface. But the beams do go through a large but reducing gravitational gradient so I imagine the true focus is not exactly at 580 but at that distance, it still would be on the focal line so it might be light that would pass say at 1.5r above the sun.

Anyone have enough math to figure that one out?

Originally posted by sonhouse
That spread out thing is what I considered at first but if you look at any physics 101 book you can see them chart it out where a beam gets deflected inwards not outwards and that deflection is at the 1.75 arc seconds angle.

I am not saying they would bend away from the sun. I am saying that a beam of light being bent by the suns gravity gets spread out, not focused. If you shine a laser beam that is 1 cm diameter passed the sun, when it get bent, the side of the beam closest to the sun gets bent more than the side further away.
So, if you sent two beams either side of the sun, the 'focal point' or the place where the two beams cross, will be a long stretched out line somewhere on the far side of the sun because the two beams will have been spread out by the process.
If you instead just aimed the beams at a single point, you would get much higher focus.

Originally posted by twhitehead
I must be missing something. What is the point of all this? So you get two laser beams that meet up at some distant point. How is this different from merely pointing them directly at the spacecraft?
They do not get focused by passing near the sun, in fact its the opposite, each beam gets spread out.

Gravitational lensing of neutrino flux might be some use to increase detection rate. But I have to agree with you, there's not much point trying to use it to focus lasers for pretty much any purpose other than an unbelievably expensive way of demonstrating Einstein's theory of gravity.

Originally posted by twhitehead
I am not saying they would bend away from the sun. I am saying that a beam of light being bent by the suns gravity gets spread out, not focused. If you shine a laser beam that is 1 cm diameter passed the sun, when it get bent, the side of the beam closest to the sun gets bent more than the side further away.
So, if you sent two beams either side of the s ...[text shortened]... rocess.
If you instead just aimed the beams at a single point, you would get much higher focus.

The beams would already be being spread out by not ever able to be exactly parallel but the gravitational effect would not add much to that spreading, the beam would still be pretty much the same leaving the sun as entering.

It was just an illustration of the focus effect.

The real effect would be a bit more complicated to explain. It involves wavefronts of incoming light from a distant star and I have trouble just using words to describe what I see, you need drawings to visualize what I am talking about.

If you visualize the wavefront of light coming off a star, say Alpha Centauri, if you imagine the light starting from some kind of explosion, say a small nova, the wavefront when it reaches the vicinity of the sun would be a sphere 8.8 light years in diameter.

If you look at the small slice of that wavefront coming into the solar system, it is nearly a straight line but not quite.

If you had a wavefront coming from a literal infinite distance, the line would be indistinguishable from a total vertical line of energy impacting the solar system.

But closer in sources like local stars, the wavefront is almost but not quite vertical so a close look at the angles involved shows at some point the angles are too far from that 1.75 arc seconds to be able to be involved in the focal line effect so it turns out the line of energy focused from Alpha Centauri is just about the same as the distance between the two stars and therefore there is a spike of energy leaving the solar system in the direction opposite from AC, a line of focus 4 light years long but no more.

And the same with Sirius, it has a line of focus 8 light years long but no more.

Originally posted by sonhouse
The beams would already be being spread out by not ever able to be exactly parallel but the gravitational effect would not add much to that spreading, the beam would still be pretty much the same leaving the sun as entering.

It was just an illustration of the focus effect.

The real effect would be a bit more complicated to explain. It involves wavefro ...[text shortened]... but no more.

And the same with Sirius, it has a line of focus 8 light years long but no more.

Some vocabulary to help the discussion:

A plane wave - this is the idealised case of a wave from a source at infinity, each wavefront forms a perfect infinite plane.
A spherical wave - This is the wave that comes from an instantaneous point like source.
A Green's function - This is a spherical wave coming from a point-like source which only emits for an instant (I think this is what you mean with your Nova).

But closer in sources like local stars, the wavefront is almost but not quite vertical so a close look at the angles involved shows at some point the angles are too far from that 1.75 arc seconds to be able to be involved in the focal line effect so it turns out the line of energy focused from Alpha Centauri is just about the same as the distance between the two stars and therefore there is a spike of energy leaving the solar system in the direction opposite from AC, a line of focus 4 light years long but no more.

I don't know what you mean by "focal line effect". Can you try to rephrase what you are saying.

Originally posted by DeepThought
Some vocabulary to help the discussion:

A [b]plane wave - this is the idealised case of a wave from a source at infinity, each wavefront forms a perfect infinite plane.
A spherical wave - This is the wave that comes from an instantaneous point like source.
A Green's function - This is a spherical wave coming from a point-like sourc ...[text shortened]... I don't know what you mean by "focal line effect". Can you try to rephrase what you are saying.[/b]

I saw that if you had a set of lasers skimming the surface (just a thought experiment, not looking for anything real) where one laser was on far right side of the sun and the other on the other side, if the laser beams were parallel, they would converge at 580 AU.

So that represents the first focal "point".

Then if you move the lasers so the beams are now 1R above the surface of the sun with the same angles, going by tangentially but parallel now the focus point should be 4 times more distant at 2320 AU, way far away for sure, so the line I speak of is drawing a line from that first focus to the second focus point, that line represents light from a real spherical wavefront coming in where light impinging on the surface of the sun does its 1.75 arc second deflection thing and focuses at 580 AU. But the wave front is spherical. So that means that light coming in at 1R above the surface will be slightly bent away from the center line of the sun, the tangential line that crosses through the center of the sun. It's really hard to put this in words, a drawing would clear it up instantly. But at 1R above the sun that distance is now 2320 AU but slightly modified because although the deflection angle at that altitude above the surface of the sun is 0.875 arc seconds now, the actual flight path of the spherical wave front is pointed slightly away from the tangential line going through the center of the sun.

So the focal point for 1R will be a fit further than the 2320 I calculated.

And it gets worse the further above the surface that spherical wave front impinges the tangent line, the line that would be perpendicular to a perfectly flat wave front (don't know if I worded that right).
Anyway, there will be some point where the smaller and smaller deflection angle will meet up with the greater and greater angle of the spherical wave front to that perpendicular line that the spherical wave front will no longer be focused by the sun.

It turns out that line of focus extends only as far as the distance between the two objects, the distant star and the sun.

At some distance above the surface of the sun the angle of deflection will be so small that the waves coming in from the spherical wave front will only be able to change the rf flight to parallel from its previous course, not converging in the distance.

Even further up the spherical wavefront will take over again and it will be back to the same spherical wave front as if the sun was not there.

Which is why the 'spike' of focused radiation only exists for the distance between the objects.

I discovered this effect after very carefully doing the trig for smaller and smaller angles and you are right, att I used radians.

I was really surprised I could use a common Casio calculator to make those calculations.

Originally posted by sonhouse
It turns out that line of focus extends only as far as the distance between the two objects, the distant star and the sun.

That doesn't add up. I claim that if we assume the sun is a point, then for any distance beyond the sun, we can always find a ring around the sun such that light is deflected from the other star to that point in question. The fact that the sun as a fairly large radius means there is in practice a minimum focus distance corresponding to the surface of the sun, but there is no maximum.

Originally posted by sonhouse
I saw that if you had a set of lasers skimming the surface (just a thought experiment, not looking for anything real) where one laser was on far right side of the sun and the other on the other side, if the laser beams were parallel, they would converge at 580 AU.

So that represents the first focal "point".

Then if you move the lasers so the beams are ...[text shortened]... ans.

I was really surprised I could use a common Casio calculator to make those calculations.

Ok., now I just need to find out what you mean by 1R. The limiting distance where a point source's emissions can be focussed to a point is 580AU, imagining your parallel lasers working in reverse gives us that result. So a spherical wave originating from a point nearer to the sun than that will diverge rather than converge. Given the focal length is 580AU which is of the order of 3 light days, Alpha Centauri is 4 light years or about 495 times the focal length away. Using a hack's guess based on the lens makers equation, the focus point will move by O(1/495) focal lengths or about 1 A.U..

Originally posted by DeepThought
Ok., now I just need to find out what you mean by 1R. The limiting distance where a point source's emissions can be focussed to a point is 580AU, imagining your parallel lasers working in reverse gives us that result. So a spherical wave originating from a point nearer to the sun than that will diverge rather than converge. Given the focal length is 5 ...[text shortened]... n the lens makers equation, the focus point will move by O(1/495) focal lengths or about 1 A.U..

My assumption reading sonhouse's post was that "1R above the surface of the sun"
corresponded to 2*[solar radius].

Ie: 1 solar radius above the solar surface... what I would normally designate 2*r[sol]

This matches with the bending of light being 1/4 of the strength of light passing just
over the surface as we have doubled the distance and thus quartered the gravitational
field strength.

Originally posted by googlefudge
My assumption reading sonhouse's post was that "1R above the surface of the sun"
corresponded to 2*[solar radius].

Ie: 1 solar radius above the solar surface... what I would normally designate 2*r[sol]

This matches with the bending of light being 1/4 of the strength of light passing just
over the surface as we have doubled the distance and thus quartered the gravitational
field strength.

Right, I thought it was something like that - the rays of light won't converge in that case.

Originally posted by DeepThought
Right, I thought it was something like that - the rays of light won't converge in that case.

It all depends on the distant to the external light source. If you looked at the spherical wave front of something like a point source at the distance of Earth, there also wouldn't be any focusing because the alleged focal point would be way before a distant focus happens, that is to say not much will focus before 580 AU, maybe neutrino's because they can pass through a portion of the sun and the angle of deviation would change some. So the minimum distance there can be any kind of focus you either have to use parallel beams which could be ON the surface of the sun I think, or back a few million miles, those beams at 1 R, sorry if I got that wrong, 1R in this case being the surface, I should have said 2R in the last post. Anyway that focus will be at 580 AU. If you go up to 2R, 700,000 km above the surface, it will focus something like 2320 AU.

But spherical wave fronts are not like perfectly parallel beams, that would only happen from a point source at an infinite distance.

So at closer interstellar distances, since coming from a point source, the light at the source going straight at the sun would pass by the surface and get focused but as you go away from that angle I would call zero angle, you can see if you are at that point source, 90 degrees away, you don't get any of that energy, only at zero angle plus and minus some very small angle,
But that angle would be pointing away from the sun and therefore there would be a maximum angle pointing away from the sun that will still be deflected to a focal point somewhere down the line. As the angle goes past that, the most that can happen is the pointing angle of the light goes parallel to the line between the point source and the sun so that angle and greater ones can never be involved in the focus of the sun's gravitation.

Does that make sense? It would be a LOT easier if I had a blackboard🙂

It's a matter of the competition between the increasing angle of light away from the sun from the point source vs the continuously reducing angle of the gravitational bending, which like I said before, to my surprise, is linear, not inverse square like the actual gravity field. So at the surface, 1.75 arc seconds, at 700,000 km high, its 1/2 of 1.75 or 0.875 and at 1.4 million km it's half of that and so forth. So as that angle gets smaller the angle from the point source gets proportionally larger so there has to come an altitude above the sun VS the distance to the point source where focusing ceases, and what I determined is a source 4 light years away gives a spike of concentrated light 4 light years long, like a cosmic flashlight! And one 8 light years away gives an 8 light year long cosmic flashlight. And so forth for all the sources around the sun. So looking at the big picture, if you had a spacecraft able to accurately go round the sun at say 5 light years out you would find a spike from Sirius but if you went to the right angle for AC there would be no spike only whatever the point source would show from whatever distance you are away from AC. I think at Earth, AC comes in at 9.8 nanowatts per square meter, if I did the arithmetic right🙂 per square meter or per square foot, been a while since I did those numbers.

All you have to do to find out is know the total radiation coming from AC, how big it is and divide by how big the surface area of a sphere 4.4 light years in radius is, that gives you what we would receive here on Earth and you could go from there to find out exactly how much is actually focused in my focus line. I have not been able to figure out exactly how much energy is involved but I think its in the order of megawatts or gigawatts in those spikes and if so, it can be a place where if you put a solar sail, you could be getting a more or less a constant thrust in that direction. Maybe🙂

Originally posted by sonhouse
It all depends on the distant to the external light source. If you looked at the spherical wave front of something like a point source at the distance of Earth, there also wouldn't be any focusing because the alleged focal point would be way before a distant focus happens, that is to say not much will focus before 580 AU, maybe neutrino's because they can p ...[text shortened]... the sun's gravitation.

Does that make sense? It would be a LOT easier if I had a blackboard🙂

If you have a point source of light closer to the sun than 580 A.U.s then gravitational lensing will not focus the rays of light to a point. Also the light from your point source would be swamped by the light from the sun.

It's dawned on me what you are talking about with a focal line. It's easier to talk about this in terms of rays rather than waves. We can consider parallel rays a distance r from the centre of the sun. Rays at r = R (1 radius) are affected more than rays at r = 2R, so the rays passing the sun at a distance of 2R are focused to a point further out. Your line of focii would be starting at 580A.U. and carry on to infinity.

Originally posted by sonhouse
...and what I determined is a source 4 light years away gives a spike of concentrated light 4 light years long, like a cosmic flashlight!

You got it wrong. It will give a continuous spike from the focal point of a ray skimming the surface of the sun, and going off to infinity. In theory, if wavelengths were infinitely small, the point at infinity would be brightest because it corresponds to the largest circle. However, because wavelengths are finite and there is some spread going on, the reverse should be true, ie the point closest to the sun should be brightest, or possibly somewhere beyond that depending on the wavelength.