Robittybob1

Just clear up one thing please. Is the event horizon the same thing as the Schwarzschild radius?

Why would you want to do that?

It is pretty hard to tell what happens beyond the event horizon but for every particle that enters the BH the event horizon increases in area, that is weird.

In what way?

And prior to Buddhism there was?

Even the aquatic mammals e.g seals didn't become fully aquatic (get their oxygen from the water like fish do) either, they still need to come up for air now and then. So even they drown I believe. http://news.nationalgeographic.com/news/2012/01/120106-harp-seals-global-warming-sea-ice-science-environment/ - see there is at least one article confirming that.

OK I have set up the spreadsheet again and used very accurate values, and see that the equation for light deflection when the mass of the star (M) is contained within its Schwarzschild Radius (SR) the Newtonian formula would yield a value of 1 ie 2*M*G/(c^2*SR)=1 now at larger values of the stars radius ® in (non BH situations) that would produce a value of the angle of deflection in Radians =2*M*G/(c^2*r)=1 So does that 1 still represent Radians or 1 steradian of deflection? [they talk of solid angle which is measured in steradians, which I'd never heard of before today. But you can't convert steradians to arcseconds but only square arcseconds and no one is mentioning them ] In using Einsteins formulas the factor was 4 rather than 2 so that would result in an answer of 2 for the BH situation so that brings it up to 2 radians??? Does it? still not a full circle! There must be another factor somewhere?

Will that only happen with a naturally formed BH? Like if they somehow formed a miniscule BH in the LHC will it have enough mass to make light circle it? Or does this only happen when a star forms with at least 3 times the mass of the Sun? Then might had full circulation in the formula Einstein's bending of light calculation. 2GM/(c^2*R) yields a value in radians, so if M was 3 times the mass of the Sun and R would be possibly the Schwarzschild Radius (vary that and see what happens to the angle). http://www.scienceforums.net/topic/87209-is-this-for-real/page-5#entry846371and http://www.scienceforums.net/topic/87209-is-this-for-real/page-5#entry846509

http://www.scienceforums.net/topic/87209-is-this-for-real/page-5#entry846509 I probably used the wrong formula but I had found the right answers for the light deflection and then I compressed the Sun making the radius smaller and the deflection got greater but it never got full circle even at the Schwarzschild Radius for the Sun. Please have a look at what I did. The formulas are written in the style they go into Excel.

How did you convince yourself of that? Can light leave at a tangent to the Schwarzschild Radius (SR)? When calculated the bending of light from the Sun's SR it wasn't a full circle.

So do you agree light could possibly radiate from within the Schwarzschild radius but not to infinity, this light being so diffuse and scattered that it doesn't form an image? If "it is wrong to think that it is the escape velocity that prevents light escaping", what is right way to think of it?

If we think of the Earth with its man made satellites even though they have velocity they don't have escape velocity and they definitely leave the surface of the Earth, so could when they calculate that light can't escape a black hole, that still could mean that light can go some distance from the black hole before it is deflected (bent) back down again?

I think the moderators here are very fair.

Yes, you have to also consider the effective radius (the distance from the center of mass). So the density of the material is important.

The combination of 2GM/(c^2*R) has a part which is basically the Schwarzschild Radius i.e. the 2GM/(c^2). So I set up a spreadsheet and explored what effect varying the Sun's radius has on the answer that is coming out of the deflection formula. I then converted the answer from radians into proportion of a circle to see what happens to the deflection angle of light as the radius was brought closer to the Schwarzschild Radius of the Sun (about 2963 m from the previously listed estimates), and since the Relativistic deflection result being twice the Newtonian one I can follow that side by side. As I approached the Schwarzschild Radius of the Sun the angle of deflection increased but it never become anything like a full circle (there being 1,296,000 arcseconds in a circle). The highest deflection was at the Schwarzschild. Radius but maximized at 0.314 of a circle and half that for the Newtonian side. That was the first time I have worked with the Schwarzschild Radius so I was a little surprised to see that the deflection was not a full circle, for I thought at the Schwarzschild Radius light was bent to that extent???? Has anyone got an explanation for what I should have expected at the Schwarzschild Radius?

True if that was the case, but in this experiment the ball is dropped from absolutely stationary so to an observer next to it it would appear to fall "straight down". Same with Galileo's balls, Newton's apple and Einstein's free-fall elevator, falling to the COM of the massive attractor. The curvature of the Earth could be problem and so could the rotation of the Earth on its axis. From a Newtonian perspective I can see there is a problem in defining where "dead level" is for the surface of Earth is curved, and even if I took a line at right angles to the line that the ball dropped down (presumably dead straight up and down, vertical) this would still be a tangent and any light particle following this tangent will be in effect climbing out of the gravitational well with the pull of gravity coming from the COM and will find this force getting further and further behind it as the distance the photon travels increases. http://arxiv.org/pdf/physics/0508030v4.pdf I found this study calculation the Newtonian and Relativity based deflection and the formulas show one exactly twice the value of the other but when I put numbers into the equation I get an answer but it does NOT relate to the amount in arc seconds. What are the units of 2GM/(c^2*R) G= gravitational constant M mass of the Sun c = speed of light R = radius of the Sun. 1.97465E-08 is the value I got from putting the best values into the equation. How do I convert the output of that equation to arcseconds? Please anyone. I must have used some incorrect values for I have done it again using these estimates c = 300000000 Mo = 2E+30 Ro = 700000000 G = 6.66667E-11 and got an output of 4.2328E-06 which I multipled by a factor to convert radians to arcseconds 206264.806 and the answer came out to 0.87... arcseconds Horray! With the help of this paper. http://home.fnal.gov/~syphers/Education/Notes/lightbend.pdf

Just keep my name out of your bickering for I'm only posting relevant information. I'm not stopping you doing anything. Keep going.

I will do the calculations alright, as soon as I can find a formula that seems relevant. Even if I have to work up to something more complicated. But remember I'm pretty bad with algebra and calculus etc so it has to be more than explanatory. I read yesterday that the free-fall frame is equivalent to an Inertial frame, so when you say I can't "compare accelerated frames results with inertial frame results", which is which? When does a free fall frame become accelerated? Standing on a surface in a gravitational field was equivalent to an accelerated frame. Is that right? So the comparison of a lead ball to a photon was done in an accelerated frame (but am I allowed to ignore the Earth's rotation and orbital motion). I'll try to calculate them separately in the end. I just need a hand to get going.

There were 5 possible results: 1. A photon could move higher. 2. The photon could move in a straight line and not be affected by gravity 3. The photon maybe curve less than the space-time curvature. 4. The photon could exactly follow the space-time curvature. 5. or the Einstein concept of the photon falling additional to the space-time curvature (nearly twice the Newtonian value). Not only that but it will be the first time I actually plug values into these space-time curvature equations, so I'm hoping to begin to understand them. This is right I was trying to understand whether the light will behave in the same way as it did before the elevator began the free-fall.

5.45755E-17 m this is how far a lead ball will fall in the time light travels 1 meter. The geodesic for the lead ball is vertical as it has no lateral motion. If the light follows the space-time curvature of the Earth how much curvature would we see over 1 meter horizontal to the Earth surface?

We are not dealing with free falling frames now but the surface of the Earth, with gravity at g and light traveling at c.

I fully concur as Acme is always OK.

How would you compare how far light will fall due to gravity over 1 meter parallel to the surface of the Earth compared to how far a lead ball will fall in the same amount of time? This is nearly the same question on the net. https://van.physics.illinois.edu/qa/listing.php?id=2010 From this site http://www.einstein-online.info/spotlights/light_deflection this quote: What does that mean? Is light falling twice as fast as lead balls?

I have feeling you are still disagreeing that photons fall as fast as lead balls. One is a mass-less particle and the other massive. I appreciate they will take different trajectories but will the downward component due to gravity be the same in the same period of time. OK the time period has to be infinitesimal for the photon moves off at the speed of light?

Don't answer the question and I'll look at starting a thread for the first time on this forum.