Johnny5

how can the poles reverse every 11 years, and us not notice any kind of effect? What is the strength of the sun's magnetic field, at the location of earth? In say units of Gauss. Thank you

I was picturing the magnetic field of the sun as toroidal in shape. Giving the sun two poles, a north pole, and south pole. Then imagining these lines extending outwards into space, and passing through the ecliptic plane, practically perpendicular to it. So I meant north/south pole using this model. So if it's wrong I need to make it right. Thank you

Thank you Dr Swanson

I realized that too Jacques, but I still don't get the significance. Let me sort of start off something here, and maybe someone can help out. I have a very important goal currently, and that is to find a reference frame in which to mathematically analyze the motion of the earth around the sun. But it seems that has already been done by Kepler (yes i knew that ). Ok so anyways... I have to build the whole thing up one step at a time. The first thing is to find the origin of the frame. The best place would be the center of mass of the solar system, but I have no idea how to compute it yet. However, to a good approximation, the center of mass of the sun is the center of mass of the solar system, since most of the mass in the solar system comes from the sun. I of course know how to compute it, but in order to do that I would need the exact location of all the planets right now, relative to the sun, and their distances away, and also their masses. Now, with the knowledge that the earth orbits the sun in an ellipse, we can carry out the analysis with the XY axes in the plane of earth's orbit. It just so happens that most of the other planets also have their orbits in this plane. This plane is called the ecliptic plane. The Z axis of the frame is perpendicular to the ecliptic plane. Now, the earth is spinning on an axis, but the axis isn't perpendicular to the ecliptic plane. If it were perpendicular to the ecliptic plane, then the equator of earth would always lie in the ecliptic plane. But in reality, the earth is spinning around an axis which points towards the north star. Now, the earth has regular seasons, marked by the spring equinox, autumn equinox, and summer and winter solstices. First let's start with the summer solstice. The date of the summer solstice is June 21st. Use the following picture: Notice that all points above arctic circle recieve sunlight 24 hours/day on June 21st In the above picture, start with the summer solstice, which occurs on June 21st. Focus on the vector which starts at the center of inertia of the earth, and passes through the north pole, and which points at the north star. Because of the 23 degree angle this vector makes with the ecliptic plane, the arctic circle is tilted towards the sun, and the antarctic circle is tilted away from the sun. Thus, for a whole 24 hours, all places on earth above the arctic circle recieve sunlight all day long, while all places on earth below the antarctic circle are in darkness all day long. Then sixth months later, the opposite is true. The vector still points to the north star, but now the earth is on the opposite side of the frame. So on December 21st (The winter solstice), it is now all points below the antarctic circle which receive sunlight 24 hours a day, while all points above the arctic circle are in darkness 24 hours a day. And so clearly, this tilt is what is responsible for the seasons. Oh I think I get it now, I think I just answered my own question. On June 21st, the earth is where it is in the picture. sunlight is approaching from right to left, and strikes half the globe at any moment, while the other half of the globe is in darkness (nightime on the other side of the world). But the earth is tilted and spinning, so imagine a conic section, a plane which cuts the earth right through that Arctic circle. Here is a definition of the arctic circle Now focus on the curved arc, which runs from the highest point on earth receiving sun [math] \alpha [/math], to the north pole. In one 24 hour period, the point alpha traces out the arctic circle. That is a way to understand, and remember what the arctic circle is. Of course the earth will have moved further along in its elliptical path, but the point is clear. If you stopped time on june 21st, and spun the earth around its spin axis one time, the point alpha, would trace out the arctic circle in the rest frame of the sun. And the antarctic circle can be similarly understood, and remembered. So the location 66.5 isn't arbitrary at all. It is 90 degrees minus the earth's exact tilt. Thank you Jacques, I thought so, but I subtracted 23 from 90, and got 67, not 66.5, i should have just rounded. Regards

Can the sun change its polarity?

Well which is it? i know the earth is tilted at 23 degrees relative to the ecliptic plane, but is this why the arctic and antarctic circles are at 66.5?

Thanks Ed Regards

I am trying to understand exactly how the earth moves around the sun. Specifically, I need to be aware of any anomalies in the earth's orbit over the past 6000 years or so. I want to get an intuitive feel for just how much the positions of the other planets actually influence earths orbit. Copernicus supposedly went against Ptolemy's teachings in the 1500s by speculating that the earth, and other planets went around the sun. I don't really trust history, but at least you get a general feel for things. Copernicus` book was umm, On the revolutions of the celestial spheres... I think, and Ptolemy's book was Amalgest. But, and someone please correct me if I'm wrong, Copernicus still had the planets going around the sun in circles, it wasn't until Kepler came along, with his three laws (which I need to now derive), that the model of a planets path around the sun was changed to an ellipse. In fact, I think that in between them, Tycho Brahe actually went back to an earth centered system. But anyway, I was thinking of doing the analysis in the rest frame of the sun. I am going to ignore the motion of all the planets except earth, and model the earth's orbit as an ellipse, just as Kepler did. (A small modification can be made later, by assuming that the barycenter of the earth/moon is what travels in an elliptical path, not the earth only, but for now we can ignore the moon). And I am going to derive Kepler's three laws from Newton's formula for gravity, and basic assumptions about orbital motion. Now, I realize that Newton was fixated on the path of the earth's center of inertia around the sun, and ignored the motion of the whole rigid body of earth. But later I plan to take this into account, since the earth is precessing. Now, in order to develop an understanding of how the earth moves around the sun, in the rest frame of the sun, one has to wonder where the earth's north poles and south poles are, in the frame. One might guess that the line through the magnetic poles of earth, is perpendicular to earth's orbital plane, but that isn't the case. Instead, the earth's axis of spin is tilted relative to the plane at 23 degrees 26 minutes 22 seconds, and points towards the north star, polaris, in the constellation ursa minor. So, the analysis of motion is going to be carried out in a frame where the center of inertia of the sun is permanetly at rest. Then, it will be assumed that the earth traverses an elliptical path around the center of mass of the sun, and the plane this ellipse lies in, is called the ecliptic plane. Here are some links to diagrams with the ecliptic plane in them, and discussion of the "celestial sphere". Hyperphysics: Ecliptic plane In the previous link, take specific notice of 'declination' and 'right ascension.' Knowledge of these terms is necessary, in order to understand how to locate a given star in the "celestial sphere". In the next picture, you can actually see the positions of a few planets, simultaneously, in one image, and get a feel for the fact that there really is a "ecliptic plane." Clementine spacecraft picture of Saturn, Mars, & Mercury simultaneously You can call this frame S. The X,Y axes of frame S can lie in the plane, and the z axis runs through the center of the sun. Now, i think that the earth is closest to the sun in the summertime, around June 21st, which is the summer solstice. On this day, if you are located anywhere on the tropic of cancer, which is at 23 degrees 26 minutes 22 seconds north latitude, and you look straight up into the air at noon, you will be looking right at the sun. It may help you to take time to visualize this clearly. Suppose that wherever you are on the surface of the earth, that you are standing on a tangent plane, and that there is an axis perpendicular to this plane which runs through the center of inertia of the earth, and your feet are standing on the tangent plane. Thus, this tangent plane touches the earth at only one point. So at noon on this day, if you tilt your head back 90 degrees, at 12 noon, so that you are looking stright up into the atmosphere, you will be looking right at the sun (but you have to be at 23 degrees north. And that is June 21st. And I think this is when the earth is farthest from the sun, but If I'm wrong someone please correct me. So now we come to the first definition. Apehelion: Point lying on the elliptical path of earth, which is farthest away from the center of inertia of the sun. Opposite of... Perihelion: Point lying on the elliptical path of earth, which is closest to the center of inertia of the sun. Here are some good pictures (the first one especially): Solstices, and Equinoxes 23 degree tilted axis, in an elliptical orbit in the ecliptic plane Ecliptic Plane: Mercury, Venus, Earth, Mars

Does anyone know why the antarctic circle and the arctic circle are at 66.5S and 66.5 N respectively? Thank you

At n=0' date=' you have: [math']\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{1}\sqrt{k}}{0}[/math] [math]\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}(1)}{0}[/math] [math] log_b(x) = \frac{lnx}{lnb} [/math] [math] log_2(1) = \frac{ln 1}{ln2} [/math] The natural log of 1 is zero, and the natural log of 2 is around .69, so: [math]\displaystyle \lim_{n\rightarrow 0}\frac{0}{0}[/math] What does n range over?

Does anyone know if there a way to find out the current distance between the earth, and the sun in real time, using an online resource? I need to know the distance between the earth and sun now. Regards

I am trying to explain gravity as magnetism. The only problem I am having, is that magnetism can be repulsive or attractive, and gravity is strictly attractive. The mathematics is a necessary review. Kind regards

That's what I'll do then. Regards

well i happen to have a book here, umm hold on... This was written by Jullius Adams Stratton, and it says... As you can see, the concept of magnetic monopoles was actually worked on. I'm not quite sure where he was going with that, but what is a magnetic field anyways? The tiniest bit of research shows that the sun is the source of a huge magnetic field B, so what is a B field composed of?

Don't do it twice.

Does he mean a speeded up version of Euclid's algorithm?

Well here's what bothers me... Consider this please: [math] \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+...+\frac{9}{10^n}+... [/math] Factor out 9/10' date=' using the distributive law of algebra, so that you have: [math'] \frac{9}{10} \sum_{n=0}^{n=\infty} x^n = \frac{9}{10}(1+x+x^2+x^3+...) [/math] Where x = 1/10. So here we have a case where |X|<1, and this is a geometric series, so that we can replace the infinite series by S, like so: [math] \frac{9}{10} \sum_{n=0}^{n=\infty} x^n = \frac{9}{10}S[/math] Where [math] S= \frac{1}{1-x} = \frac{1}{1-1/10} = \frac{1}{10/10-1/10}= \frac{1}{9/10} = \frac{10}{9} [/math] Therefore: [math] \frac{9}{10} \sum_{n=0}^{n=\infty} (\frac{1}{10})^n = \frac{9}{10} \cdot \frac{10}{9} =1 [/math] When we (erroneously conclude that) 1=.999999999999999... I feel there is an error in there. what do you think?

Is the following derivation correct? We want so compute the sum of the following infinite series: [math] 1+x+x^2+x^3+x^4+....x^n+... [/math] If x is greater than one, the series clearly diverges, but the interval where x is greater than or equal to zero is not so clear. Assume that the series is convergent, and the sum is S. Therefore: [math] S = 1+x+x^2+x^3+x^4+....x^n+... [/math] Therefore: [math] xS = x+x^2+x^3+x^4+....x^n+... [/math] Therefore: [math] S-Sx=1 [/math] By the distributive axiom of algebra: [math] S-Sx=S(1-x) [/math] By the transitive property of equality, it now follows that: [math] 1=S(1-x) [/math] Now, as long as [math] (1-x) \neq 0 [/math], we can divide both sides of the equation above by (1-x). Assume that not [(1-x)=0]. So... [math] S = \frac{1}{1-x} [/math] The formula on the right, is the famous formula for the sum of the terms of a geometric series, as seen here: Geometric Series So my first question, is whether or not the derivation is correct. Thank you

Do you think I can handle your discussion?

Starting with this (Planck time) write this: [math] 1.35 \times 10^{-43} \frac{\text{seconds}}{\text{state change}} [/math] 1) If the previous statement is true, then how many times does the universe change state per second? 2) What must be true for the above statement to be false? The answer to question one is trivial, the answer to question 2 is not. [math] \frac{1}{1.35 \times 10^{-43}} \frac{\text{state changes}}{\text{second}} = 7.4 \times 10^{42} \frac{\text{state changes}}{\text{second}} [/math] The previous number is enormous. Suppose that you are in a reference frame, in which a photon has the following speed: [math] c = 299792458 \frac{meters}{second} [/math] How many meters does the photon jump, in one state change? Answer: [math] (1.35 \times 10^{-43}) (299792458) = 4.05 \times 10^{-35} \frac{\text{meters}}{\text{state change}} [/math] With a magnifying glass, the motion of the center of mass of the photon would look like this: ............................................... The distance between any two points would be on the order of 10^-35th of a meter, and assuming the photon moves at a constant speed for the entire second, the distance between any two adjacent points would equal the distance between any other two adjacent points. To another person, moving at a constant speed relative to the photon, the amount of distance the photon jumps would be different, but, in one second, it would have been at 7.4 x 10^42 different places. So in different frames, the distance the photon jumps is different, but given that time flows equitably in all frames, the number of different positions occupied per second, would be the same in all frames, except ones in which the photon is at rest. However, according to the theory of SR, time does not flow equitably in all frames. The answer to question two is difficult to understand.