22 Apr 16
Originally posted by DeepThoughtYour formula would not be completely accurate as the neutrino experiences gravity from well outside the sun and is deflected by that for some of its path and by your formula while it is inside the sun and then again back to the old formula once it leaves the inside of the sun.
We don't need a more sophisticated calculation,
In addition, the original formula is for a constant G whereas in yours the G varies throughout its path through the sun, so some integration would be necessary.
22 Apr 16
1 edit
Originally posted by twhiteheadG as far as we know is a constant of nature and doesn't vary. You mean g, for gravitational acceleration. I'd have to have a look at how the formula is derived to see how much it matters, bear in mind that for most of its path the neutrino is almost directly approaching or departing from the Sun - it's only close to where the direction of force is at right angles to the line of flight. The real problem is that I've assumed that the Sun has a constant density which it doesn't and I've assumed it's a sphere, which it isn't.
Your formula would not be completely accurate as the neutrino experiences gravity from well outside the sun and is deflected by that for some of its path and by your formula while it is inside the sun and then again back to the old formula once it leaves the inside of the sun.
In addition, the original formula is for a constant G whereas in yours the G varies throughout its path through the sun, so some integration would be necessary.
In the meantime it's occurred to me one could use this effect to measure the radial variation of density of the Sun by seeing how the neutrino flux from Sirius changes as one moves away from the Sun along the line connecting the Sun and Sirius.
22 Apr 16
Originally posted by DeepThoughtI think I meant M.
G as far as we know is a constant of nature and doesn't vary. You mean g, for gravitational acceleration.
In the meantime it's occurred to me one could use this effect to measure the radial variation of density of the Sun by seeing how the neutrino flux from Sirius changes as one moves away from the Sun along the line connecting the Sun and Sirius.
Sounds great except that we cant detect neutrinos very easily, and the neutrinos form the sun would completely overwhelm those from Sirius.
22 Apr 16
1 edit
Originally posted by twhiteheadK=5907 or thereabouts. divided by the radius in meters, which is 696300000 = 8.48E-6 radians which is 1.75 seconds of arc. Invert the radian number, you get 117800 or so and times the radius, now in km, 696300 = 8.2E10 km (82 billion km or about 51 billion miles)
It is my belief at this point that the first focal point for light is at 0.54 AU. and not 540 AU. I believe sonhouse forgot to convert from kilometres to metres.
I would like confirmation from someone before I proceed that I have got it right so far.
Did you get 5907 for K?
22 Apr 16
Originally posted by twhiteheadhttps://en.wikipedia.org/wiki/Gravitational_lens#Solar_gravitational_lens
It is my belief at this point that the first focal point for light is at 0.54 AU. and not 540 AU. I believe sonhouse forgot to convert from kilometres to metres.
I would like confirmation from someone before I proceed that I have got it right so far.
22 Apr 16
Originally posted by twhitehead1 AU is 1.496 E11 m not 1.49 E14 m which accounts for the discrepancy.
OK, please tell me what I am doing wrong.
See updated diagram with new formulas and figures.
http://whereitsat.co.za/GraviationalLensing4.jpg
I have:
y = r / tan ( (K / r) - alpha)
In reality alpha can be ignored.
And it gives me the first focus at 8.2e13 metres or 0.54 AU.
This clearly disagrees with your figure of 540 AU.
At first I thought I might be mixing up radians and degrees, but I don't think so.
https://en.wikipedia.org/wiki/Astronomical_unit
22 Apr 16
Originally posted by sonhousePlease check my diagram and all figures:
K=5907 or thereabouts. divided by the radius in meters, which is 696300000 = 8.48E-6 radians which is 1.75 seconds of arc. Invert the radian number, you get 117800 or so and times the radius, now in km, 696300 = 8.2E10 km (82 billion km or about 51 billion miles)
Did you get 5907 for K?
http://whereitsat.co.za/GraviationalLensing5.jpg
K=5908
22 Apr 16
1 edit
Originally posted by twhiteheadAs a minor point, in your last jpeg you have the units for the speed of light as ms^-2, that should be -1. How do you get your figure? I get it from similar triangles, using your notation, but with d for delta rather than the distance to Sirius (can you replace that with D, so we don't get this problem).
If I use an annulus of 0.5 m, I get a delta f (dx, I believe for Deepthought) of 6.944e+9 m
This would seem to be different from Deepthoughts figure.
To be specific,
focus for r=696300000 is 8.205617E+13
focus for r=696300000.5 is 8.205617E+13
difference =6.944428E+09
x/df = R/y
x = 1m so:
df = y/R = 540/0.00465047 = 116,129 m = 116 km.
Let's work out the angle. we have beta ~ tan(beta) = x/df = 1/(116129) = 8E-6 radians, which converted to degrees is 1.77 arc seconds, which rings a bell.
Your df = 6.944E9 would give 2.97E-005 arc seconds which is too small.
Edit: this might be a numerical stability problem.
22 Apr 16
Originally posted by DeepThoughty = r / tan ( (K / r) - alpha)
How do you get your figure?
y: distance to focus
r: radius at sun
K: 5908
Alpha: angle at Sirius. It can be safely ignored I believe.
I get it from similar triangles, using your notation, but with d for delta rather than the distance to Sirius (can you replace that with D, so we don't get this problem).
Will do.
22 Apr 16
Originally posted by twhiteheadUsing a spreadsheet I first checked on alpha, it comes to 0.0018 arc seconds so it's a thousand times smaller. Then I substituted in the numbers and got:
y = r / tan ( (K / r) - alpha)
y: distance to focus
r: radius at sun
K: 5908
Alpha: angle at Sirius. It can be safely ignored I believe.
[b]I get it from similar triangles, using your notation, but with d for delta rather than the distance to Sirius (can you replace that with D, so we don't get this problem).
Will do.[/b]
R = 6.96E+008 metres (radius of Sun)
K = 5908 metres (Those being the units K is measured in, note that K is twice the Schwartzschild radius)
y1 = R/tan(K/R) = 81922560932357
y2 = (R + 0.5)/tan(K/(R + 0.5)) = 81922561050112.6
dy = y2 - y1 = 117755.59375 metres
Which is the number I got earlier. If you were using the Windows calculator it may be you got some sort of rounding error from the precision the calculator works with.
Looking at your earlier post you quote:
y1 = 8.205617E+13
y2 = 8.205617E+13 ( = y1 at the precision you're quoting to)
but y1 + 6.944428E+09 = 8.205617E+13 + 6.944428E+09 = 82063114428000
(adding 117000 to your y1 gives your y2 at the precision you quoted). To hazard a guess you used the calculator, wrote down y1 and y2 and missed a digit entering the numbers (although I can't reproduce the error by leaving out the first 6 in y1).