Question about energy at one wavelength:

Originally posted by twhitehead
I'll double check all that later, but I suspect you are getting it wrong.
It would help if you stuck with the symbols used in the diagram:
http://whereitsat.co.za/GraviationalLensing3.jpg

I'll try to also add more to the diagram such as known figures.

The problem is you have d as one of the lengths and I wanted d to mean delta. To show a small change in d I'd have needed dd, which isn't the most helpful notation...

My calculation is correct, the only potential sources of error are that I've assumed that the light from Sirius is parallel with the line between Sirius and the Sun, and the various small angle approximations. I really do think these corrections are tiny, they'll be of the order of one part in a million so I think the calculations correct.

Originally posted by DeepThought
The problem is you have d as one of the lengths and I wanted d to mean delta. To show a small change in d I'd have needed dd, which isn't the most helpful notation...

d doesn't bother me, but I also have other letters in use. You can of course use 'delta' or Δ (but apparently a lower case delta confuses RHP)

My calculation is correct, the only potential sources of error are that I've assumed that the light from Sirius is parallel with the line between Sirius and the Sun,
Parallel before or after it passes the sun? What happens to it after it passes the sun?

It is the x in my latest diagram that is critical.

Originally posted by twhitehead
d doesn't bother me, but I also have other letters in use. You can of course use 'delta' or Δ (but apparently a lower case delta confuses RHP)

[b]My calculation is correct, the only potential sources of error are that I've assumed that the light from Sirius is parallel with the line between Sirius and the Sun,
Parallel before or after it passes th ...[text shortened]... hat happens to it after it passes the sun?

It is the x in my latest diagram that is critical.[/b]

Clearly, that light goes on forever or at least till it hits some kind of nebula cloud or some such. One thing it never will be, is focused.

Originally posted by sonhouse
Clearly, that light goes on forever or at least till it hits some kind of nebula cloud or some such. One thing it never will be, is focused.

It appears you are not understanding what I asked. DeepThought assumed that angle alpha is so small that he could ignore it. I am happy with that approximation. However, after light passes the sun, two rays that are initially parallel but are at slightly different distances from the sun, actually diverge from each other. I am sure you don't get this, and suspect DeepThought hasn't quite got it.

The light is focused, but the light that converges is from a ring around the sun. Consecutive rings focus at different points, and not in an discrete manner as you seem to assume.

Originally posted by twhitehead
It appears you are not understanding what I asked. DeepThought assumed that angle alpha is so small that he could ignore it. I am happy with that approximation. However, after light passes the sun, two rays that are initially parallel but are at slightly different distances from the sun, actually diverge from each other. I am sure you don't get this, and ...[text shortened]... onsecutive rings focus at different points, and not in an discrete manner as you seem to assume.

You mean one beam say one second behind the other but at whatever, 60 r where I would expect two parallel beams to be, why would they be at different distances if they came from the same source? Suppose the source was pulsing at a billion times per second, are you saying the relative phase of the pulses would somehow be different after they leave the gravity field of our sun? That would be pulses about 30 cm long.

Originally posted by sonhouse
You mean one beam say one second behind the other ....

No, I am saying that in my diagram, light that passes through r and light that passes through R will diverge.
If r and R are 1 metre away from each other as you have used in your calculations, the light diverges to a considerable size x at the line of focus. You have made the erroneous assumption that they remain 1 metre apart.
If they diverge to 2 metres apart (ie x is 1) then the area of the collector is four times larger than you calculated so your energy per square metre falls by a factor of four. If they diverge by kilometres, then you are off by a factor of millions.
If you don't know x, then you don't know anything - not even ballpark figures.

Originally posted by twhitehead
No, I am saying that in my diagram, light that passes through r and light that passes through R will diverge.
If r and R are 1 metre away from each other as you have used in your calculations, the light diverges to a considerable size x at the line of focus. You have made the erroneous assumption that they remain 1 metre apart.
If they diverge to 2 metr ...[text shortened]... tor of millions.
If you don't know x, then you don't know anything - not even ballpark figures.

I had actually thought about it but not sure, the energy would be out of time but when a collector gathers the light, a bit of displacement wouldn't mean much since 3 nanoseconds they would all be at the collector.

I was wondering if that effect would change the frequency of the photons, say a frequency of 300 mhz, which has a wavelength of 1 meter, would it still be 300 mhz after it leave the diffraction area.

Originally posted by twhitehead
It appears you are not understanding what I asked. DeepThought assumed that angle alpha is so small that he could ignore it. I am happy with that approximation. However, after light passes the sun, two rays that are initially parallel but are at slightly different distances from the sun, actually diverge from each other. I am sure you don't get this, and ...[text shortened]... onsecutive rings focus at different points, and not in an discrete manner as you seem to assume.

I took a look at your diagram. The thing with my calculation is that it is for the first focus, so I don't have light going under the centre line the way you do. Your diagram has light that was focused to what you are calling y1, but for my calculation that isn't there because the body of the Sun is in the way. This means my r and your r are different.

Originally posted by DeepThought
I took a look at your diagram. The thing with my calculation is that it is for the first focus, so I don't have light going under the centre line the way you do. Your diagram has light that was focused to what you are calling y1, but for my calculation that isn't there because the body of the Sun is in the way. This means my r and your r are different.

My r, R, y1, y2, x Δr, Δf are all variables. As such they apply to any point for which you are doing calculations. They are to be used to get the general formula, but once you have that formula they would work for the first focus just as easily.

You can't have been calculating just the first focus, because you had energy calculations, and you can't do energy calculations without area. To get area you need a ring around the sun of a given width (Δr). When you do that, the light going through the ring does not all focus at the first focus, but along the focal line for a length Δf. To calculate the energy received at a given point along the focal line you need to calculate the area of the required collector to gather all the energy, for which you need x. (the collector must be a circle of radius x.). Once you have x, you divide your total energy by the collectors area pi x^2 to get the energy per unit area. Because of this a large x will make a very significant difference.

Originally posted by DeepThought
I took a look at your diagram. The thing with my calculation is that it is for the first focus, so I don't have light going under the centre line the way you do. Your diagram has light that was focused to what you are calling y1, but for my calculation that isn't there because the body of the Sun is in the way. This means my r and your r are different.

But there should be a calculation like that anyway when you consider neutrino's which pass right through the sun unimpeded. That means there would probably be a focus line starting somewhat sooner than that for EM radiation since the mass of the sun gets smaller as r gets smaller but the bending effect should max out for neutrinos at something like 0.7 r and further in, the deflection would get greater again till at dead center, neutrinos passing the center of the sun would just go on its merry way unaltered, at least in its direction of travel. I think that would be awesome to be able to accurately track the path of neutrinos through the sun coming from interstellar distances, even if the sun produces a huge number. Neutrino's still follow the inverse square law so the intensity would fall of with distance also and therefore the 'beam' of focus should also be stronger than the emission from the sun at the focus line. So it looks like neutrinos would start coming together at around 380 AU not 542 as is the case with EM.

Originally posted by sonhouse
That means there would probably be a focus line starting somewhat sooner than that for EM radiation...

As I have noted before, the focus line actually extends all the way to infinity in both directions, including on the far side of Sirius!
You won't see it in my current diagram because the lens is an idealized vertical plane. I will try and add another diagram showing the actual gravity fields.

Light from Sirius experiences the sun's gravity before it reaches the sun. Therefore there is some light that is bent towards, and crosses the focus line before reaching the sun.

The maximum effect is for light passing through the sun but we haven't yet worked out where the maximum intensity will be. It will be interesting to find that out.

Originally posted by twhitehead
As I have noted before, the focus line actually extends all the way to infinity in both directions, including on the far side of Sirius!
You won't see it in my current diagram because the lens is an idealized vertical plane. I will try and add another diagram showing the actual gravity fields.

Light from Sirius experiences the sun's gravity before it ...[text shortened]... n't yet worked out where the maximum intensity will be. It will be interesting to find that out.

Max intensity is further out away from Sirius simply because the rings of energy from higher r numbers contains more total energy than lower r numbers. 2r, twice the energy in that ring than 1r ring. 10r, ten times the total energy, its a simple area thing. If 1r contains say 50 watts which is roughly what I calculated than at 10r the total there would be 500 watts/meter squared. Seems simple enough to me.

So it seems to me the 30 r distance I calculated as being about the max concentration would be about also 30 times that of the inner most ring or 1500 watts/meter squared, about the same as the sun here in our neck of the woods.

I also thought about the fact there would be bending of photons well before the light passes close to the sun but came to the conclusion it gets bent as a summation or integral of all the bending effects leading up to the closest passage and I think that is what added up to the figure big Al threw out, 1.75 arc seconds so it looks to me like it is safe to say we can ignore these second order effects.

I visualize the effect as kind of looking at a powered boat going round a whirlpool in a body of water, the outer fringe of the whirlpool will deflect the boat to max out when it passes as close as it can while avoiding catastrophe then out of the whirlpool to the new angle of travel which also includes changes of direction on the way out also. It is the sum of all those changes that is important, not worrying over whether distance A plus distance B plus distance C has some effect.

Originally posted by sonhouse
Max intensity is further out away from Sirius simply because the rings of energy from higher r numbers contains more total energy than lower r numbers.

So you keep claiming, but given that you have not got an actual formula, and still don't understand the scenario, you are almost certainly wrong.

2r, twice the energy in that ring than 1r ring. 10r, ten times the total energy, its a simple area thing.
And this is countered by delta f and x growing and the area of the collector growing as the square of x. Its a simple area thing.

Seems simple enough to me.
Because you continue to ignore the diagram.

I also thought about the fact there would be bending of photons well before the light passes close to the sun but came to the conclusion it gets bent as a summation or integral of all the bending effects leading up to the closest passage and I think that is what added up to the figure big Al threw out, 1.75 arc seconds so it looks to me like it is safe to say we can ignore these second order effects.
They are not 'second order effects' They are part of the overall effect. It can be safely ignored for light passing by the sun, but it doesn't mean that there isn't a focal axis prior to the sun - there is.

I visualize the effect as kind of looking at a powered boat going round a whirlpool in a body of water, the outer fringe of the whirlpool will deflect the boat to max out when it passes as close as it can while avoiding catastrophe....
No, the maximum deflection goes to the boat that fails to avoid catastrophe.

It is the sum of all those changes that is important, not worrying over whether distance A plus distance B plus distance C has some effect.
For light that passes the sun, that is largely (but not entirely) true.
The formula you gave for deflection will not hold for very small angles of alpha. The formula is an approximation of the true equation which would be more complex because the sun is not a black hole.

Originally posted by twhitehead
My r, R, y1, y2, x Δr, Δf are all variables. As such they apply to any point for which you are doing calculations. They are to be used to get the general formula, but once you have that formula they would work for the first focus just as easily.

You can't have been calculating just the first focus, because you had energy calculations, and you can't do ...[text shortened]... get the energy per unit area. Because of this a large x will make a very significant difference.

I have a collector disk of unit radius at the first focus - x in my notation. Rays of light that just skim the surface of the Sun hit the centre of the collector disk. Rays of light that pass at a distance dr above the surface of the sun skim the edge of my collector disk and would be focused at the point (x + dx).

Let's move my disk outwards so that the light that just skims the sun just skims the edge of the disk. That is the first point where your arrangement is valid.

Originally posted by twhitehead
So you keep claiming, but given that you have not got an actual formula, and still don't understand the scenario, you are almost certainly wrong.

[b] 2r, twice the energy in that ring than 1r ring. 10r, ten times the total energy, its a simple area thing.
And this is countered by delta f and x growing and the area of the collector growing as the ...[text shortened]... roximation of the true equation which would be more complex because the sun is not a black hole.[/b]

The bit about a black hole, what is the full story then? Is the formula given by Einstein not complete? It seems clear if that is complete, the angle of deflection is a simple linear function of r, 1 r =X, 2r =0.5x, 3r= 1/3 x. Is that wrong?