- 09 Nov '08 03:31The equation is the deflection angle of light skimming the sun, Einstein's equation is Angle=4GM/C^2*R

G I have two values for, 6.67428E-11 and 6.67259E-11. Don't know which one is more the accepted value but it shouldn't make much difference in the answer which one you use.

M I have as 1.98892E30 Kg * 4G as 5.3098E20 (rounding out a bit)

and C^2 as 8.98755178737E16 * R, which I assume to be the radius of the sun in meters, or 6.96E8 which comes out to about 6.25533E25.

So 5.3098E20/6.25533E25=8.4884E-06.

Is that answer supposed to be degrees? Radians? Not sure of the units in that answer.

The G value is in M^3 Kg^-1 S-2 units and C^2 in m^/S^2? M/S squared, gives you M squared/seconds squared?

And mass in Kg so what does the answer end up being, what units? And see if you get 8.488E-6 and change just to check my work. Thanks. Don. - 09 Nov '08 05:19 / 1 edit

G is in m^3/kg s^2*Originally posted by sonhouse***The equation is the deflection angle of light skimming the sun, Einstein's equation is Angle=4GM/C^2*R**

G I have two values for, 6.67428E-11 and 6.67259E-11. Don't know which one is more the accepted value but it shouldn't make much difference in the answer which one you use.

M I have as 1.98892E30 Kg * 4G as 5.3098E20 (rounding out a bit)

and C^2 as 8.98 ...[text shortened]... being, what units? And see if you get 8.488E-6 and change just to check my work. Thanks. Don.

M is kg probably.

The numerator is therefore m^3/s^2

This cancels nicely with the demoniator, leaving us an answer that should be in radians. Radians, of course, being a ratio of a length to a length and therefore unitless! - 09 Nov '08 14:06 / 4 edits

Did anyone do the math, did anyone get 8.48E-6? So radians it is, wasn't sure what that answer meant since everyone puts the number at 1.7 Seconds of arc. MLPrior, this is the equation and the proof from Eddington that put Einstein on the map where he measured the change in the position of a star during a total eclipse of the sun, which is a pretty dam tricky measurement especially for the time frame, nearly 100 years ago.*Originally posted by KazetNagorra***If you use SI units in your equation, the answer will always be in radians.**

I was just trying to redo the math to see if I came out with the same thing.

Hey Thou, thanks for that, never thought about those units, didn't realize radians are dimensionless units. So degrees are units with dimensions and radians are not? That is a revelation, always wondered why radians were the measurement of choice.

So to turn radians into degrees, don't you just multiply by 57 and change? The answer is yes, with all the digits of the HP48, it comes out to 8.48850254198E-6 and rounding out, comes out to 1.7509 arc seconds (the good old HP48 has HMS conversions which makes that easier).

I wonder how you go from there to proving the Schwartzchild radius for the sun? For instance, if you divide R by ten, the angle goes up by ten to 17.5 arc seconds and so you should reach a point where the angle = some number that represents the SC radius but not sure what that would be. Would it be where that angle = 180 degrees? Not sure.

I am assuming you are taking the size of the sun progressively smaller and smaller with the same mass so R would also get progressively smaller. - 09 Nov '08 15:24 / 1 editA bit of google sniffing shows the Schwarzie radius equation to be

2GM/C^2 which turns out to be about 2950 meters or a diameter of about 5908 meters. Am trying to see how that relates to the angle, what is the equivalent angle of the Schwartz. May the Schwartz be with you

If I have it correct then, it should be dividing the two radii, 696,000,000/2950 = 235932 which seems to be the multiplier of the angle, IE, 234932 * 1.75 seconds of arc= 412,881 seconds of arc which if you divide by 1296000 = 0.318581293 ths of a circle so * 360= 114.68 degrees which seems to be 2 radians for some reason. That is just weird.

Anyone want to confirm all this? - 09 Nov '08 16:42

Degrees are weird. They aren't defined with respect to other measurements, which is why mathematicians prefer radians I believe.*Originally posted by sonhouse***Did anyone do the math, did anyone get 8.48E-6? So radians it is, wasn't sure what that answer meant since everyone puts the number at 1.7 Seconds of arc. MLPrior, this is the equation and the proof from Eddington that put Einstein on the map where he measured the change in the position of a star during a total eclipse of the sun, which is a pretty dam tric ...[text shortened]... ogressively smaller and smaller with the same mass so R would also get progressively smaller.**

In fact one degree is properly defined as pi/180 radians, giving about 57.2958 degrees per radian. It's also dimensionless, but when you get a dimensionless number from an equation that is an angle it's always in radians I think. - 09 Nov '08 16:52 / 1 edit

Not quite. It's the ratio of a distance to a distance; arclength over radius.*Originally posted by Eladar***In other words, radians is a distance that can be measured with linear units, while degrees can't.**

Degrees are the ratio of arclength to radius multiplied by 180 and divided by pi. It has that extra factor to make everything more confusing. - 09 Nov '08 16:54

No, radians are not a distance, radians are dimensionless (you cannot take a sine/cosine/log/exp of a non-dimensionless variable). Degrees are dimensionless as well, there is just a linear conversion factor between them.*Originally posted by Eladar***In other words, radians is a distance that can be measured with linear units, while degrees can't.** - 09 Nov '08 17:10 / 1 edit

That is interesting in itself. So you cannot take a sin of a unit with dimensions. So the sin of 6 pounds will give you a math error*Originally posted by KazetNagorra***No, radians are not a distance, radians are dimensionless (you cannot take a sine/cosine/log/exp of a non-dimensionless variable). Degrees are dimensionless as well, there is just a linear conversion factor between them.**

So does anyone want to figure out if I am right about the deflection angle of the schwartzchild radius, it seems to be 2 radians. My next question would be why? I thought for some odd reason the SC radius would be where light goes into a perpetual orbit but if the 2 radian # is correct, light still escapes, which is true, light only gets captured by a black hole. So I guess the next step in this is to figure the deflection angle of a black hole.

I found a nice intro to black hole theory that derives the radius, and it turns out to be the same as the Schwartie. So something must be wrong with my math. It seems not to add up. The site, here BTW:

http://design.lbl.gov/education/blackholes/index.html

seems to indicate the SC radius as the point of no return but if my numbers are correct (don't see how they could be) there is some disconnect to the deflection angle. - 09 Nov '08 17:28 / 1 edit

Indeed. You can understand why if you Taylor expand the sine; for example taking the sine of 1m:*Originally posted by sonhouse***That is interesting in itself. So you cannot take a sin of a unit with dimensions. So the sin of 6 pounds will give you a math error**

sin(1m) = 1m - 1/3! m^3 + 1/5! m^5 - ... ???

As for the SC radius: maybe there is a factor pi missing somewhere. - 09 Nov '08 19:34

non-dimensionless is a rather odd term. Shouldn't that be with dimension? In other words, shouldn't you be saying, you can only take a sine/cosine/log/exp of a variable with dimension? But I think you meant to say that you can only take sine/cosine/log/exp of non-dimensional variables.*Originally posted by KazetNagorra***No, radians are not a distance, radians are dimensionless (you cannot take a sine/cosine/log/exp of a non-dimensionless variable). Degrees are dimensionless as well, there is just a linear conversion factor between them.**

I see what you are saying. You are correct. The difference between degrees and radians is that radians have a more direct conversion to linear measurement. - 09 Nov '08 21:30

I think maybe it might just be accidental, coincidence. I am going to try other masses and see if it comes out the same. I think I did the math right, but I will double check.*Originally posted by KazetNagorra***Indeed. You can understand why if you Taylor expand the sine; for example taking the sine of 1m:**

sin(1m) = 1m - 1/3! m^3 + 1/5! m^5 - ... ???

As for the SC radius: maybe there is a factor pi missing somewhere.

Anyone else want to tackle this? - 10 Nov '08 01:52 / 5 edits

A = 4GM/c^2R*Originally posted by sonhouse***The equation is the deflection angle of light skimming the sun, Einstein's equation is Angle=4GM/C^2*R**

G I have two values for, 6.67428E-11 and 6.67259E-11. Don't know which one is more the accepted value but it shouldn't make much difference in the answer which one you use.

M I have as 1.98892E30 Kg * 4G as 5.3098E20 (rounding out a bit)

and C^2 as 8.98 ...[text shortened]... being, what units? And see if you get 8.488E-6 and change just to check my work. Thanks. Don.

G = 6.67428 x 10^-11 m^3/kg*s^2

M = 1.9891 ×10^30 kg

c = 299,792,458 m/s

R = 1.392×10^9/2 m = 6.96x10^8 m

4GM = 5.31032 x 10^20 m^3/s^2

c^2 = 8.9876 x 10^16 m^2/s^2

4GM/c^2 = 5.9085 x 10^3 m

4GM/c^2R = 8.489 x 10^-6 radians

I was careless with rounding; I rounded as I felt like it, not according to any kind of strict rule like significant digits. The small difference between your answer and mine should be insignificant.