xyzt Posted January 7, 2015 Share Posted January 7, 2015 (edited) 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. The trajectory of the lead ball depends on its initial velocity, so, if it thrown parallel to the ground, it will not fall vertically but it will follow a parabola. Light will also follow a (different) parabola. Instead of posting rubbish questions, as you have been doing forever on this forum (and others), why don't you check into getting into a class, as you promised earlier? Edited January 7, 2015 by xyzt Link to comment Share on other sites More sharing options...
Robittybob1 Posted January 7, 2015 Share Posted January 7, 2015 I don't understand the point of this question. Unless you have a specific problem to solve, what do you expect to gain from this piece of information? It won't help you understand anything more general. 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. The original disagreement stems from Robbitybob's post #27 where he asks about a free falling light clock, a beam of light bouncing back and forth, from side to side, say in a free-falling elevator. He wants to know whether the light beam will show a deflection from the horizontal or not . To which xyzt replied that since they follow different trajectories, yes, there will be a deflection from the horizontal. That, is what I had a problem with. And I would imagine its now time to move on. 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. Link to comment Share on other sites More sharing options...
xyzt Posted January 7, 2015 Share Posted January 7, 2015 (edited) 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). I would recommend a chapter of GR ("Photon Trajectories" or "Light Bending by Gravitating Bodies") that deals with these issues but I know by now that it is futile. You have no intention to study anything, you are just spraying the forums with questions, only to ignore the answers. 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. You will never do any calculation. 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. You keep trying to compare accelerated frames results with inertial frame results. Physics doesn't work this way. This is the fourth time I pointed this fallacy to you in this thread. Edited January 7, 2015 by xyzt Link to comment Share on other sites More sharing options...
elfmotat Posted January 7, 2015 Share Posted January 7, 2015 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. If you drop an elevator into free-fall, as long as the elevator is small compared to the gravitating planet/star/etc., any light inside will behave completely normally as if you were in an inertial reference frame. So as long as the elevator isn't a significant fraction of the size of the gravitating body, you won't be able to tell that you're falling by measuring light deflection. If the elevator was very very big, then you would be able see deviations in the path of the light. Sorry it took so long for you to receive an answer. You keep trying to compare accelerated frames results with inertial frame results. Physics doesn't work this way. This is the fourth time I pointed this fallacy to you in this thread. No, he's comparing free-fall frames with inertial frames, which is a perfectly valid thing to do locally. Link to comment Share on other sites More sharing options...
xyzt Posted January 7, 2015 Share Posted January 7, 2015 (edited) No, he's comparing free-fall frames with inertial frames, which is a perfectly valid thing to do locally. He's comparing free-falling frames with the Earth-bound frame. You seem to have a serious reading comprehension. Edited January 7, 2015 by xyzt Link to comment Share on other sites More sharing options...
elfmotat Posted January 7, 2015 Share Posted January 7, 2015 He's comparing free-falling frames with the Earth-bound frame. You seem to have a serious reading comprehension. Read MigL's post. Then read the "this is right" reply he received. Do you disagree with anything I said in my previous post, or do you just enjoy bickering? Link to comment Share on other sites More sharing options...
xyzt Posted January 7, 2015 Share Posted January 7, 2015 Read MigL's post. Then read the "this is right" reply he received. Do you disagree with anything I said in my previous post, or do you just enjoy bickering? No, I am just pointing out that you are trolling while posting mistakes. You have been doing it since I started posting in this thread. Link to comment Share on other sites More sharing options...
elfmotat Posted January 7, 2015 Share Posted January 7, 2015 No, I am just pointing out that you are trolling while posting mistakes. You have been doing it since I started posting in this thread. Please quote any and all mistakes I made, and explain why they are mistakes. I have already given you the same courtesy, as can be found here. Although you chose to ignore your mistake and pretend it didn't happen, I promise I won't do the same. Link to comment Share on other sites More sharing options...
Robittybob1 Posted January 7, 2015 Share Posted January 7, 2015 I would recommend a chapter of GR ("Photon Trajectories" or "Light Bending by Gravitating Bodies") that deals with these issues but I know by now that it is futile. You have no intention to study anything, you are just spraying the forums with questions, only to ignore the answers. You will never do any calculation. You keep trying to compare accelerated frames results with inertial frame results. Physics doesn't work this way. This is the fourth time I pointed this fallacy to you in this thread. 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). He's comparing free-falling frames with the Earth-bound frame. You seem to have a serious reading comprehension. I'll try to calculate them separately in the end. I just need a hand to get going. Link to comment Share on other sites More sharing options...
Robittybob1 Posted January 8, 2015 Share Posted January 8, 2015 (edited) The trajectory of the lead ball depends on its initial velocity, so, if it thrown parallel to the ground, it will not fall vertically but it will follow a parabola. Light will also follow a (different) parabola. Instead of posting rubbish questions, as you have been doing forever on this forum (and others), why don't you check into getting into a class, as you promised earlier? 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 Edited January 8, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted January 8, 2015 Share Posted January 8, 2015 ! Moderator Note Unlike Data, I do not enjoy petty bickering. If you can't discuss things in a civil fashion, consider that non-participation is an option. Do not respond to this Link to comment Share on other sites More sharing options...
Robittybob1 Posted January 8, 2015 Share Posted January 8, 2015 (edited) 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? Edited January 8, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
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