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tmdarkmatter

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Posts posted by tmdarkmatter

  1. 22 hours ago, KJW said:

    Calculating in radians allows you to calculate without the explicit use of trigonometric functions.

    Thank you for your assistance! I can see the mistake I made above. But now that we have this angle in radians, how can I convert the observer who is just a dot into a surface or at least a two-dimensional line? Because I am not looking for the object needed at a distance of the diameter of Earth that would be big enough to block Proxima Centauri for one point, I want to calculate the new angle that is created having in mind that the "observer" is for example an object of a diameter of 1 meter and how big the object creating the shadow on the other side of Earth should be. The main issue is to convert the observer from a point into a surface. You might say that the size of the observer can be disregarded, but it is clearly not the same to block the light of Proxima Centauri for just one point than to block it for a certain surface. And it does not seem that it is just an addition of the surface of the observer to the object calculated with this amount of radians, because the angle of the light coming from Proxima Centauri in total changes. We would see the star bigger if our eyes would be more separated.

    I have been analyzing this and it is as if the point of the observer would be closer to the object creating the shadow, so an inverted shadow is arriving at the surface. It´s like moving the apex towards the opposite side and creating a cross with a second side where the observer is located, converting the observer into a surface.

    This would mean that the bigger the observer becomes, the more the cross would displace towards the opposite side and the bigger the angle would become.

  2. Ok, I have done these simple calculations and the results are as follows:

    If the observer is 1 km over the surface of earth, according to the tangent ecuation, the object on the surface of earth must have the size of 4,8481 mm (the size of a very small coin) to block Proxima Centauri. So if you are staring at a rooftop that is 1 km away and Proxima Centuari is just showing up behind the rooftop, you might estimate that a tiny coin would be enough to block the light coming from Proxima Centauri. But what is interesting is that the blocking object on the other side of earth at a distance of 12,742 km would also increase in size by 4,8481 mm, so the equation is totally linear. At a distance of 100 km above earth, the object on the surface would have a size of 0,48481 meters to block the light coming from Proxima Centauri. But what is interesting is that if I insert the total distance to Proxima Centauri in that equation, the result is an object 907 times the size of Proxima Centauri, meaning that the light we see coming from Proxima Centauri is 907 times bigger than the star itself, because we do not only see the star but also its bright surroundings. (Correct me if I am wrong)

    But the ratio between blocking object on the surface and blocking object on the opposite side of earth would begin at zero when the observer is on the surface to a value of almost 1 when the observer is light years away.

    And the size of the blocking object would in both cases become bigger than earth itself.

    So if we are standing on earth and would fit in this area of 0,1844 square meters, the object to block the light from Alpha Centauri from us on the opposite of earth (maybe I can call it antishadow) would have to be 16.630,78 times bigger than ourselves. This seems to be a constant for all objects of all sizes on the surface in the case of the angle of Alpha Centauri.

  3. 8 hours ago, swansont said:

    s = r*theta

    Yes, this is getting closer to what I need. We can also write it as theta = s/r indicating that if r increases with an increasing distance, s also has to increase for theta to be the same.

    Later today I will do some calculations moving the observer away from the surface of earth. This would provide me a certain surface on earth as a blocker which can be compared to the blocker surface on the opposite side of earth. This way I can compare them and figure out how both values increase if the observer is moving away from the surface.

  4. 1 minute ago, swansont said:

    The beam illuminates the whole earth, i.e. you could observe the star from any point on the surface unless earth itself is in the way.

    Yes, of course, this is obvious. But if I am standing on earth, I only see a dot in the sky. But this dot means a certain circle on the opposite side of earth, a circle of a certain size that is necessary to completely block that beam. Imagine if I am standing on a mountain and asking somebody to climb on another mountain about 100 kilometres away, I can see at the horizon and that Alpha Centauri will be visible just a few meters above that mountain from my perspective. So I am asking him to block Alpha Centauri for me. What is he going to use? A coin? A big sign? A house?

    But even worse, I want him to create a shadow to Alpha Centauri that has the size of 1 square meter in my position. How do I calculate that?

  5. 2 minutes ago, Bufofrog said:

    Assuming you are serious, here is a link from NASA on how eclipses work.

    I think you misunderstood my questions. I am not questioning how eclipses work. I am just trying to find a solution to my problem with my calculations. But as we can see in your images, the sun is much bigger than the moon and the moon is much bigger than its shadow on earth. Therefore, the size of an object at a distance of the diameter of earth should be much bigger than our thumb to cover Proxima Centauri, but how big should it be?

  6. In that case how do you explain the existence of solar eclipses if the diameter of the moon is 3,474.8 kilometres and the diameter of the sun 1,392,700 kilometres? And even worse, how do you explain that the solar eclipe is only visible on a line of just a few hundred kilometres of width on Earth considering the size of the moon?

  7. 2 hours ago, swansont said:

    This makes no sense. An angle, by definition, assumes a vertex, which is a point.

    That´s the problem. I need to go from a point to an area or distance.

    2 hours ago, swansont said:

    The light from Proxima Centauri hitting the earth can be treated as parallel over such a short distance as the diameter of the earth. 

    The angular size of the star is not the same as the divergence of light from it.

    Maybe I can also ask you the question: How big must an object be so at a distance of the diameter of earth it is able to block the light coming from Proxima Centauri so it creates a shadow that is just big enough to cover an object of the size of 1 metre, 2 metres, 10 metres?

    If you consider the trajectory of the light from Proxima Centauri as parallel, that would mean that a coin would block the light and that is not the case.

  8. This time I am contacting you because I need the help of talented and experienced people for a math problem related to astronomy. This will not be a discussion and my purpose is not to criticize or question your skills and knowledge. I really need your help and all kind of help is appreciated. And if you are a member of this website, you already qualify to help me and will earn the corresponding respect. It is also allowed to make mistakes at first and correct them later to obtain an even better solution (we are all humans after all). Now this is the problem:

    Imagine we are standing on Earth and staring at the star Proxima Centauri which has a size of about 1.02 ± 0.08 milliarcsec in the sky. This is about 0,00027778 degrees (correct me if I am wrong).

    Now, imagine if the earth rotates and Proxima Centauri is situated just below us behind Earth so we can no longer see it. What I wanted to calculate is the area Proxima Centauri would be shining on on the opposite site of earth corresponding to the point we would see if earth was not in the way. So I took the tangent equation and using the angle alpha of 0,000277778 degrees and the diameter of earth which is about 12.742 kilometres, I obtained the solution that the diameter of the area beeing hit by radiation coming from Proxima Centauri which would become a dot after crossing the diameter of earth and hit us is of about 61,77 meters (once again correct me if I am wrong).

    The problem is that this diameter was calculated having in mind that the observer standing on earth is just a dot, so the angle is solely based on the size of Proxima Centauri. But what I now need to know is how this angle would change if the observer was for example an object of the size of 1 meter or 2 meters or 10 meters. And how would increasing the size of the observer increase the diameter of the area on the opposite side of Earth that is beeing hit by radiation coming from Proxima Centauri? Of course, I am only interested in the area of Earth beeing hit that is clearly in the way between Proxima Centauri and the observer. I am aware that Proxima Centauri is always hitting about half the surface of Earth with its radiation, if there are no other objects in the way, but only a small portion of this area would indeed be in the way between Proxima Centauri and the observer.

    Please, to make the calculations easier and to first obtain a certain notion of what I am dealing with, consider the surface on the other side as if it was "flat", because this calculation would become incredibly complex if we would incorporate the curvature of Earth, and this might not be important as long as the observer and the area beeing hit by the radiation stay small. Of course, once we are talking about kilometres, Earth´s curvature will start playing an important role and if the observer has the size of Earth him/herself, the total area would be about half the area of Earth.

    Any formulas are welcome, but please try to explain things so easy that even I can understand them.

    And if you have any questions or do not understand a part of the text, just let me know.

    Thank you in advance for your help!

     

  9. So, if we play god and we have a star and Mercury in front of us and now take the sun and move it very close to their light, what actually happens is that the original light ray is deviated and absorbed by the surface of the sun. But why do we still see the star or Mercury? It is because another light ray that was not heading into our direction is now beeing deviated by the sun and reaches the observer. So if I take the highway, we are having a highway that is not heading into the direction of the observer but at the sun it is deviated towards the position of the observer, so when we watch from the observers position where the car is getting onto the highway at the next village (mercury) and compare it with the position when the car far away is getting onto the highway (the star), we would see the next town at an angle far closer to the position of the ray arriving at the observer than the city far away. Therefore the difference.

    But if we now draw lines to try to figure out the maximum angle, we can see that the angle always increases the further away the star is, but this increase of the angle becomes smaller and smaller and the distances to increase the angle by the same amount becomes longer and longer. So how do they figure out what the maximum angle should be for an infinite distance? I guess there is just a simple logarithmic equation for that. But do they really take the maximum distance of the observable universe to calculate the maximum?

    It is interesting to think that a straight line at an angle is actually not straight for the observer once this line reaches very long distances.

    And the same happens if we increase the distance of the observer. So the maximum we should consider would be the deviation of light coming from the furthest galaxy passing next to the sun and the observer standing at the other limit of the visible universe.

  10. 7 hours ago, swansont said:

    And, of course, it's shown visually in fig 4 - the lateral displacement is not the same for the planet P as the star, even though the light path is identical.

    So, I can imagine that if I was a photon, I would be like driving on a highway and once the highway gets closer to the sun it gets bent and I am forced to travel on this curve, obviously according to the gravitational force of the sun and the distance of the highway from the sun, but after the curve, the highway would just continue completely straight. This is what Einstein is saying. But why would it matter if I have been driving on the highway for a long time on a straight line or if I went on the highway just ahead of this curve? Why would that matter? Shouldn´t we rather think that the curve starts since the beginning at a very low intensity and the only part of really strong curvature is close to the sun? Otherwise, this seems like a refutation of GR, because it cannot explain why the angle is greater in the case of far distances. This space-time curvature is just not flexible enough to explain what we observe and after all we might yet be observing a force pulling the light to the sun and no bending. We can discuss the theory, but we cannot question what we are observing (unless there is something wrong with the measuring devices) and this is not according to what they are saying that the curvature is only significant close to the sun. In the case of Mercury, about 70% of the expected curvature is still missing, because this curvature should be created while light is travelling from infinite to the curve. Therefore, we cannot ignore the curvature created far away from the sun. Instead of having a few very high values of curvature (in a few light seconds), we are having millions of tiny values (during years of travelling) that should be added and altogether sum up for the maximum curvature. In order to observe this maximum curvature, we do not only need to observe stars at very long distances, the observer him/herself has to be very far away too. But at what distance would it still make sense to try to observe an eclipse to investigate this curvature? Did they ever observe an eclipse from Mars, from Jupiter or further away? I think instead of trying to always look ahead, they should first look back at our solar system from these distances to learn more about us. There are only very few images looking back.

    Now imagine if we were lucky enough to observe one star passing by in front of another star. What if the light from the star is bent so much that a star on the left switches to the right or a star above switches to below? This should make sense, even if the angles are small.

  11. No there is no explanation of that. The angle should be the same if the object at the sun´s limb is venus or if it is a star several light years away.

    On the other hand, they are saying that the angle can be ignored for beeing very small. What happens if we move away 1 light year from the sun. Would the angle change? Because if not and the sun and the star behind it are just two little dots, it might be possible that we would see the star behind the sun on the opposite site of the sun because of this little displacement, so we would be getting a totall wrong image. If that´s the case, all the galaxies we are seeing might be like upside down, because stars we would consider up or down or left or right would actually be on the opposite side, it´s only that the gravitational force of the center of the galaxy distorted the trajectory of the light of its own stars.

  12. 1 hour ago, swansont said:

    Different distances from the sun.

    Wait, the text says that these are the displacements of planets once we see them passing close to the sun´s limb from our perspective, right? So why does their displacement depend on the distance to the sun, if the "non negligible" bending only takes place less than 5 solar diameters away from the sun? It would not matter if the planet is 20 solar diameters or 1,000,000 solar diameters away from the sun, because there would be no important effect on the light travelling at this distance from the sun.

    What I mean is that according this text the light leaves Jupiter, travels in a straight line towards the sun and is only deviated by the sun´s gravity once it is less than 5 solar diameters away from the sun. And when this light is moving from the sun to earth it is again moving in a straight line. So why does it matter if the straight lines are 1 light second, 1 light minute or 1 light year long? The angle won´t change.

  13. 20 hours ago, swansont said:

    It is interesting that on the one hand beyond the distance of 5 solar radii the light ray should be pradically straight and on the other hand from earth we should consider a displacement of 0.49 for Mercury and a displacement of 1.69 for Neptune, while for the observer on Earth it should be considered that the distance to the sun is similar to "infinite". All this does not make sense. If there is only bending very close to the sun, why should there be a huge difference of the displacement between Mercury and Neptune? And what would be the displacement for all of these planets if the observer was further away than Pluto?

  14. 1 hour ago, swansont said:

    The mass of the Milky Way is about 10^12 solar masses. The central bulge is about 5000 light years. or ~10^11 bigger than the sun closest approach. The ratio, which tells us the deflection, is 10x bigger than for the sun.

    It is interesting how simplified this approach is. Instead of light having to pass by individual stars to reach us, the light just needs to pass by a "central point" of the galaxy where you just sum up all the mass of the stars and position that mass on a single point. So when the light passes by stars very closely, according to your calculations these stars would have no mass at all, because you had already summed up their mass in the center of the galaxy. Imagine light having to pass by thousands of stars, each of them deviating by 1.75 arc-sec or even much less. I think this approach is too simple but maybe I am wrong.

  15. 12 minutes ago, swansont said:

    Why don’t we see deflection for light that isn’t passing very close to the sun? 

    Did you check the images of the stars during the Eclipse that made Einstein famous? Not only the closest ones were displaced. Of course the further away from the sun, the lower is the displacement, according to the 1/r^2 equation, but this equation also never reaches an absolute zero. Only the stars that are completely opposite to the sun should maintain their original direction within the solarsystem, if not displaced by planets.

    30 minutes ago, Bufofrog said:

    That's not a very good conspiracy theory

    I am not interested in conspiracy theories. I am interested in understanding the universe a little better.

  16. 11 hours ago, swansont said:

    Light is bent according to known conditions in GR, i.e. by close proximity to sources of gravity. So saying “we don’t know” is not true. Do you have evidence of light being bent under other conditions, or sources of gravity in interstellar space?

    I think it is interesting that we can clearly observe and measure the famous Einstein rings, but currently we consider that the remaining light passing by galaxies, if not beeing part of an Einstein ring, just travels totally straight, and that´s a big mistake. If gravity has an infinite range, our entire image of the universe is completely distorted. Even our own sun is distorting the image of the universe before it reaches us. When it was discovered that the sun bends light, they were happy because they could give Einstein a Nobel prize, but they did not further investigate what this bending would actually mean to us. What I mean with "If 7 minutes of sun gravity can bend the light by 1.75 arc-sec" it´s just an example of what we should expect to happen when light passes by a galaxy. The deviation does not only depend on the force gravity exerts on it, but also on the time of exposure. I know that GR would just say that space-time is bent to make it easier, but a Jet also has to spend more fuel the longer it travels around earth. I think when we observe Einstein rings, we are just observing the most distorted light surrounded by other distorted light where we do not really know the position of its sources, so the entire image is distorted, not only the rings. So we are considering the light right next to the Einstein ring as travelling straight and compare the Einstein ring with that (normal, not distorted) light to calculate the mass of the object involved, when the light next to the ring is also distorted. Therefore, the mass can be different too.

     

     

    6 hours ago, MigL said:

    I apologize for not reading the previous 6 pages but I'm recuperating from eye surgery

    Don´t worry, you are not "forced" to do anything here. But I hope everything went well with your surgery and thank you for the video. I will watch it today when I have more time.

    11 hours ago, iNow said:

    Lots and lots of evidence, and a willingness to update our models when the old ones prove to be faulty. 

    A great evidence that would make me shut up would be images of galaxies becoming smaller and smaller due to expansion, but we would have to wait for millions of years.

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