Johnny5

Suppose that you are located at 29 degrees 58 minues 24 seconds North latitude, and 31 degrees 7 minutes 45 seconds East longitude And you have a compass in your hand, and its pointing north. If you look directly at the North star from that position, at 8:00 PM in the evening, what angle will you have to tilt your head back to see it?

I'd prefer to learn it from you, but if you're too busy don't worry about it. Riemann tensor... here is a wikipedia link to Reimann tensor. I can't make any sense out of that.

Yes, I think that's what they are. I have a specific star chart that I am trying to understand. SC2 constellation chart Northern Circumpolar region- Epoch 1925 From 30 degrees north latitude to 90 N At the center of the star chart is polaris. Then there is a series of concentric circles. The first circle, the one with the smallest radius, is labeled 80 degrees. Within that circle are two stars which are part of ursa minor. Polaris is labeled [math] \alpha [/math]. Then the closest star to polaris is labeled [math] \delta [/math].

Can you do an example for me Matt? I give you the following two numbers: 468, 950 Demonstrate the algorithm please.

Does anyone know what "right ascension means" and "declination"? Thank you

That's right Matt. I will go back and change it right now.

I need to review this myself Sarah, so first let me get this question down. I don't see a part c, so I'm just going to do them all. [math] \mathbf{u} \equiv \left [ \begin{array}{c} 3\\ 2\\ -4\\ \end{array} \right ] [/math] [math] \mathbf{v} \equiv \left [ \begin{array}{c} -6\\ 1\\ -7\\ \end{array} \right ] [/math] [math] \mathbf{w} \equiv \left [ \begin{array}{c} 0\\ -5\\ 2\\ \end{array} \right ] [/math] [math] \mathbf{z} \equiv \left [ \begin{array}{c} 3\\ 7\\ -5\\ \end{array} \right ] [/math] Question 1: Are the sets {u,v}, {u,w}, {u,z}, {v,w}, {v,z}, {w,z} linearly dependent? Why, or why not? In order to answer this question, we must know the definition of linear dependence. Let us just cover the case where the number of vectors is finite. A set of vectors [math] \mathcal{f} v_1,v_2,v_3,...,v_n \mathcal{g} [/math] is said to be linearly dependent if and only if we can find scalars {r1,r2, r3...rn} such that: [math] \sum_{i=1}^{i=n} r_i \cdot \mathbf{v}_i = \vec {\mathbf{0}} [/math] The symbol on the RHS denotes the zero vector of an arbitrary vector space. This symbol shows up in the axioms of a vector space. Now, there are different 'fields' which the scalars can come from, so we should take this into account. For example, in a given problem, the scalars might come from the real numbers, while in another, the scalars might come from the complext numbers. So we should expect this to vary, from one problem to another. But in all cases, the scalars must come from a field, and therefore obey the field axioms, which means that the scalar 0, and the scalar 1, must be elements of the field. So let us incorporate this into the previous definition of linear dependence. Definition: A set of vectors [math] \mathcal{f} v_1,v_2,v_3,...,v_n \mathcal{g} [/math] chosen from some given vector space, is said to be linearly dependent over some chosen field [math] \mathbb{F} [/math] of scalars, if and only if we can find scalars {r1,r2, r3...rn}, not all zero, ( such that: [math] \sum_{i=1}^{i=n} r_n \cdot \mathbf{v}_i = \vec {\mathbf{0}} [/math] Online definition Part 1: Are the vectors u,v linearly dependent? u = <3,2,-4> v = <-6,1,-7> Choose two scalars at random, r1, r2. r1u = r1<3,2,-4> = <3r1,2r1,-4r1> r2v = r2<6,1,-7> = <-6r2,r2,-7r2> r1u + r2v = <3r1-6r2,2r1+r2,-4r1-7r2> These two vectors will be linearly dependent if there are r1,r2, not both zero, and: 3r1-6r2 = 0 2r1+r2 = 0 -4r1-7r2 = 0 So now, we have to find two different scalars, r1,r2, which makes the three equations above true simultaneously. The first equation gives us a constraint, which is this: 3r1=6r2 or equivalently r1=2r2 Now, putting this into the second and third equations leads to: 2(2r2)+r2 = 0 -4(2r2)-7r2 = 0 4(r2)+r2 = 0 -8(r2)-7r2 = 0 5r2 = 0 -15r2 = 0 The only way for either mathematical statement above to be true, is if r2=0, and using the constraint, it follows that r1 must also be zero. Thus, it is not the case that there are least two scalars r1,r2, not both zero, for which: r1u + r2v= 0 Hence, by definition, it is not the case that u,v are linearly dependent.

Yes' date=' I know, that's straight out of classical Maxwellian electrodynamics. An electric current generates a magnetic field. Can this be related to the energy-momentum tensor of Professor Einstein's theory of General Relativity?

I am interested in learning how to read star charts, and in familiarizing myself with at least some of the various constellations. Does anyone here know how I can use the internet properly, in order to accomplish this? Thank you

[math] \text{Integration By Parts Lecture} [/math] Suppose you are given two functions at random: 1. f(x) 2. g(x) Now, take their product: [math] \text{f(x)} \cdot \text{g(x)} [/math] Which can also be written without the dot, multiplication implicit instead of explicit, as follows: [math] \text{f(x) g(x)} [/math] Now, take the derivative with respect to x, of their product: [math] \text{f(x) g(x)} [/math] [math] \frac{d}{dx} [\text{f(x) g(x)} ] = \frac{df}{dx} g(x) + f(x) \frac{dg}{dx} [/math] Where I have made use of the product rule of the differential calculus, to be proven at the end of the lecture. Now multiply both sides of the equation above by dx, to obtain: [math] d[\text{f(x) g(x)} ] = df g(x) + f(x) dg [/math] Now, integrate both sides of the formula above: [math] \int d[\text{f(x) g(x)} ] = \int[df g(x) + f(x) dg] [/math] Now, the integral of a sum is the sum of the integrals, that is: [math] \int[df g(x) + f(x) dg] = \int df g(x) + \int f(x) dg [/math] Therefore, using the transitive property of equality: [math] \int d[\text{f(x) g(x)} ] = \int df g(x) + \int f(x) dg [/math] Integral d(anything) = anything + C Ignoring the arbitrary constant of integration we have:

I don't get this in the least. I was thinking of a stream. Imagining earth to be traveling through a swarm of magnetic particles, some passing through the earth, others interacting with the earth, to give it its elliptical path around the sun. Suppose that there is something in the way of the path of the earth. The earth has some momentum, its barelling through space... following the law of inertia. Particles that get in the way, earth just pushes aside. But in the process, during the magnetic/graviational interaction, the center of mass of earth changes position slightly, in the solar system rest frame. The net effect must lead to an elliptical orbit, with the center of mass of the solar system at one foci. The formula to be obtained is: Link to Wolfram on the ellipse [math] \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 [/math] The equation above was the one I was thinking of. Notice, that in the discussion of the ellipse at Wolfram, the center of the ellipse is defined to be the point midway between the two foci of the ellipse. Wolfram also has a picture of how to construct an ellipse, using two tacks, and a string, and your pencil. Professor Richard Feynman discussed this method for constructing an ellipse in his now famous Feynman lectures. In the lecture I am thinking of, which took place in California, I think Caltech, professor Feynman was talking about gravity. The whole lecture was about Kepler's laws. What Professor Feynman was attempting to do, was use primitive geometry to derive Kepler's laws of planetary motion. The equation above is not true for the center of an ellipse at some arbitrary point in a frame, the center of the ellipse has to be located at the origin of the frame in order for that definition to be true. But given that it is, you can interpret a,b just as they do at wolfram. The key thing to remember about the ellipse is that the sum of the distances from foci to point which is in orbit, is a constant. You can see this fact from the Wolfram diagram. A special case of an ellipse is a circle. For a circle a=b=R, where R is the radius of the circle. And in this case, the two foci are located at the origin. Now, if you look at the orbit of the point in the Wolfram diagram, you should notice that there is an angle defined right where the two lines meet. As a later excercise we are going to compute that angle. For right now, it suffices to just learn what a,b denote, and the answer is in the figure at Wolfram.

I don't recall seeing that formula at wikipedia before. If I were you, I would attempt to derive it. Why are you trying to compute the hall potential difference? Exactly what is the purpose? At any rate, here is the definition of Magnetic flux density: [math] \Phi_m = \oint \mathbf{B} \bullet d\vec a [/math] The direction of the differential area vector with magnitude da, is normal to the surface you are integrating over.

What is the EM lift/drag ratio?

I see that they use Z, I also see a whole lot more.

You're welcome. Regards

Well I know EM theory, but it sounds like you have your hands full. But I am enjoying learning the general theory of relativity from you, so any way that you want to teach it, is ok by me. What is it i should study between now and tomorrow... the Riemann tensor?

Riemann tensor... ok I still haven't seen that yet. My linear algebra review is going well by the way. Also, do the formulas for the general theory of relativity, connect to the concept of path of least resistance? Like suppose that the sun is a giant magnetic field source. Sending out magnetic monopoles that interact with the earth's Van Allen Belt. Then the earth is moving through a swarm of particles. Can the formulas be extended to this idea? "Path of least resistance?"

One definition of 'exist' is this... To exist, is to be in the current moment in time. By this definition, if time exists, then it inside of itself, in German... ding an sich. So you should ask yourself what you mean by the word 'exist' in your sentence. For that matter, you should also ask yourself what you mean by 'time'. You cannot taste,touch,smell,hear, or see time. I would ask you only this, is time something that you can sense? Regards