Question about energy at one wavelength:

Originally posted by DeepThought
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.

And at that point you get double the radiation collected on your disk.

You appear to have managed to do the calculation without ever working out dx. How did that happen? What would dx be in your scenario? How did you work out actual radiation per unit area without it?

Originally posted by sonhouse
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?

I think the formula for deflection of light we've been using is in the weak field approximation. That weak field approximation won't work near a black hole.

Originally posted by twhitehead
And at that point you get double the radiation collected on your disk.

You appear to have managed to do the calculation without ever working out dx. How did that happen? What would dx be in your scenario? How did you work out actual radiation per unit area without it?

In the calculation I had two expressions for dx which let me get rid of it. My objective was to get the width of the annulus through which light from Sirius hits the collector disk. One of the two formulas is:

h = rdx/x

Where h is the radius of the collector disk, r the radius of the Sun and x the distance to the first focus. So

dx = hx/r

x is 542 Astronomical units, Wikipedia helpfully gives the radius of the Sun in AUs and it is 0.00465047 AUs, the radius of my collector dish is 1 metre, so dx is 116 km.

https://en.wikipedia.org/wiki/Solar_radius

Originally posted by DeepThought
I think the formula for deflection of light we've been using is in the weak field approximation. That weak field approximation won't work near a black hole.

It won't work for the sun either for small r as it clearly tends to infinity as r tends to zero.

Originally posted by DeepThought
I think the formula for deflection of light we've been using is in the weak field approximation. That weak field approximation won't work near a black hole.

Well, if it is a weak field approximation, then going out from the center, bigger r #'s would be an even weaker field so with that logic, the linear r part works.

Originally posted by DeepThought
x is 542 Astronomical units, Wikipedia helpfully gives the radius of the Sun in AUs and it is 0.00465047 AUs, the radius of my collector dish is 1 metre, so dx is 116 km.

You are clearly completely wrong.
The width of the annulus is necessarily smaller than the radius of your collector dish as is evident in the diagram.
As I keep on pointing out, light diverges after it passes the sun because the angle of deflection is larger for smaller r.

Originally posted by twhitehead
You are clearly completely wrong.
The width of the annulus is necessarily smaller than the radius of your collector dish as is evident in the diagram.
As I keep on pointing out, light diverges after it passes the sun because the angle of deflection is larger for smaller r.

I think it would be a variable, and of course only working in a real sense for neutrino's able to penetrate into the sun but it seemed to me the angle would reach an internal max at around 0.7 r then get much lower till a neutrino passing through the center of the sun would have zero deflection.

Other relativistic effects notwithstanding. I based that assumption on the idea that as a neutrino passes deeper and deeper into the sun the gravity of the sun above the neutrino would tend to cancel the effects of the total mass of the sun.

I of course retain the right to be totally wrong🙂 Just trying to visualize a pretend neutrino laser beam say one meter across and aiming two such beams from each side of the sun into the sun moving them in parallel and seeing what happens.

I expect the outer shell gravity to partially cancel the inner shell gravity till the center when deflection is zero since all the gravity is canceled in the center.

Originally posted by twhitehead
It won't work for the sun either for small r as it clearly tends to infinity as r tends to zero.

Are you saying you think the deflection goes to infinity at the center of the sun?

Originally posted by sonhouse
Are you saying you think the deflection goes to infinity at the center of the sun?

No, I am saying the equation clearly says that and in reality it doesn't, therefore the equation doesn't hold for very small r. If it was a black hole, then it would go to infinity and thus be rational although possibly still not the correct equation.

See updated diagram for shape of gravity wells, plus some known figures for future calculations.

http://whereitsat.co.za/GraviationalLensing4.jpg

Originally posted by sonhouse
I think it would be a variable, and of course only working in a real sense for neutrino's able to penetrate into the sun but it seemed to me the angle would reach an internal max at around 0.7 r then get much lower till a neutrino passing through the center of the sun would have zero deflection.

Yes, I agree. There must be an equation that describes it, but it is clearly not the one we have available to us.
My point being that any calculation you did with the equation available to us would not apply to neutrinos passing through the sun as they are clearly governed by a different equation. The overall effect is that you actually get double the radiation (dual focus) along the focus line, although we would need the equations to work out exactly where the overall effect is strongest.

Originally posted by twhitehead
You are clearly completely wrong.
The width of the annulus is necessarily smaller than the radius of your collector dish as is evident in the diagram.
As I keep on pointing out, light diverges after it passes the sun because the angle of deflection is larger for smaller r.

That is dr in my notation and comes out at half the radius of the disk. dx, in my notation is the distance of the focus of light just skimming the collector disk from the point of first focus. The point of my calculation was to get dr. You might want to read it properly before claiming that it's wrong.

Originally posted by twhitehead
It won't work for the sun either for small r as it clearly tends to infinity as r tends to zero.

It won't work for the Sun for small r because the material of the Sun gets in the way.

Originally posted by twhitehead
Yes, I agree. There must be an equation that describes it, but it is clearly not the one we have available to us.
My point being that any calculation you did with the equation available to us would not apply to neutrinos passing through the sun as they are clearly governed by a different equation. The overall effect is that you actually get double the ra ...[text shortened]... although we would need the equations to work out exactly where the overall effect is strongest.

Can't you do it by simply integrating small slices of outer to inner core kind of thing, where the mass of one slice (spherical slices) reduces the gravitational effects of the core beneath that one?

I am assuming if you had say two worlds almost touching but still in orbit around each other, say you had a km clearance between the two worlds and we ignore such pesky effects like the Roche limit, and there you are with your nice spacesuit.

You jump into the lack of air and the gravity at the center of those two exactly matching masses, that gravity field goes to zero between them, wouldn't that effect be the same in a spherical body considering concentric shells and the mass of the outer shell partially offsetting the gravity of the inner shells till you get to dead center and if you for instance, had a chamber carved out you could float around in the center totally weightless. Isn't that right?

I think two facts are true: One, at the surface you get the famous 1.75 arc second, and two, dead center, zero arc seconds of deflection.

I think that is a reasonable starting point.

The question would be what would be the actual distribution of variability of the deflection going deeper and deeper into the sun.

Originally posted by sonhouse
Can't you do it by simply integrating small slices of outer to inner core kind of thing, where the mass of one slice (spherical slices) reduces the gravitational effects of the core beneath that one?

I am assuming if you had say two worlds almost touching but still in orbit around each other, say you had a km clearance between the two worlds and we igno ...[text shortened]... a chamber carved out you could float around in the center totally weightless. Isn't that right?

A hollow, but massive, shell has no internal gravitational field - at least in the Newtonian limit - for exactly the same reason that that is the case in electrostatics.

Originally posted by twhitehead
Yes, I agree. There must be an equation that describes it, but it is clearly not the one we have available to us.
My point being that any calculation you did with the equation available to us would not apply to neutrinos passing through the sun as they are clearly governed by a different equation. The overall effect is that you actually get double the ra ...[text shortened]... although we would need the equations to work out exactly where the overall effect is strongest.

My point being that any calculation you did with the equation available to us would not apply to neutrinos passing through the sun as they are clearly governed by a different equation.
Why do you think this?

Edit: I see you want something that will penetrate the Sun's material. The formula:

Deflection angle = K/r

has K = 4GM/c^2, but M goes as r^3. So if we introduce a new constant k where K = kR^3 for R = radius of Sun, we have:

Deflection angle = kr^2. For r < R.

This is the formula we need, we don't need to use the full technology of General Relativity as the weak field approximation is fine inside the Sun. The force of Gravity at the Suns centre is zero.