Johnny5

Well, let me see what I have thus far... [math] \vec W + \vec F_e = \frac{dP}{dt} = 0 [/math] Now, you are right Fe is a roughly constant in this problem. According to the formulas, if the persons hand holds the pebble at a constant height h above the surface of the earth, the force of electric repulsion between the electrons in the persons hand, and the electrons at the surface of the pebble, helps keep the pebble from moving up and down. Also, I have neglected the use of centrifugal force, as JC pointed out, so where does that fit in swansont? I can replace Fe by C to denote a constant vector. Then I could do this: [math] - \frac{GMm}{r^2} \hat r = - \vec C [/math] Is that what you recommend, write Fe as a vector constant ?

Sarah, do you mean gravitational boost?

The original poster wanted a complete explanation that's a very tall order. The three dimensional form should be used, and the spherical harmonics for a hydrogen atom should be derived. Can anyone here do that for him? To Muralisankar: Do you know spherical coordinates?

I see now that you were referring to GR. So let me ask you this, what does [math] T_{\mu \nu} [/math] have to do with "shortest path" Thank you

It is obvious that S` is not an inertial reference frame' date=' since Newton's third law is untrue in frame S`. I never claimed that S` was an inertial reference frame. I know that it isn't so you misunderstood something i said. Moving on... S` is the rest frame of the spaceship. The origin of S` is the center of inertia of the ship. The center of inertia of the ship is permanently at rest in reference frame S`, so that it MUST be permanently attached to the ship. This was stipulated to be true. Moving on to your next objection... First of all, force and accleration are not equivalent; force and acceleration are not the same thing. The man in the ship feels force, not acceleration. It is incorrect to say that he is the one who can tell that he accelerated. Observers in both reference frames, see the other as accelerated. Instead, it is correct to say that he can tell that his spaceship is being subjected to an external force. One way for him to infer that he is in a non-inertial reference frame, is to see that his ships engines are on, and from that he can infer that Newton's third law is untrue in frame S`. Therefore, he can infer (from inside the ship) that his ship's rest frame is a non-inertial frame. So if he looks outside the window and sees a moon in free space, accelerating without an action/reaction pair, he will hardly be surprised. (Where I have assumed his ship is using a conventional propulsion system of course). This is a very complicated problem you know. Suppose that S` is an inertial reference frame. Then, it would be the case that the center of inertia of his ship is at rest in an inertial reference frame. Therefore, when he turns on the laser, we should expect the light beam to move in a straight line, away from the emitter. And by experiment, we know the speed of this is going to be 299792458 meters per second. Suppose there is a moon at rest in free space, relative to a point a distance D away from it. Now, let R denote the vector from the center of inertia of the moon, to this point. Let the direction of travel of the spaceship be in a straight line which is perpendicular to R. So this is a very simple situation to visualize. As the ship's center of inertia moves along this straight line, the center of inertia is accelerating in the rest frame of this moon. (There are no external forces acting upon the moon, therefore the rest frame of the moon is an inertial reference frame). Now, keep in mind that S` is really a non-inertial reference frame. Now, at the moment that the rocket is located at the tip of R, let the laser be turned on. The goal is to now find the path of the photon in reference frame S, as well as reference frame S`. Now, focus on the plane which contains the straight line which the rocket is moving along, and the center of inertia of the moon. Keep in mind that any straight line, and any point not on that line, lie in one and only one plane. Furthermore, R lies in this plane as well, and the length of R is D, as measured by a ruler at rest in frame S. Let R lie on the positive y axis of frame S. Thus, D is the perpendicular distance from the line of travel of the rocket, and a point not on that line (the center of inertia of the moon). Here is the Euclidean construction for R... Construction for the shortest path from a point to a line In fact, we can now write R in terms of the j^ vector of reference frame S. [math] \vec R = D \hat j [/math] Now, let the X axis of reference frame S be parallel to the line of rectilinear motion of the center of inertia of the rocket. Now, let the i^ vector of reference frame S point in the direction of motion of the center of inertia of the rocket. I wish I could draw this. Here is what I will do, I will use Euclid's drawing found here. Let the center of inertia of the moon be permanently located at the point C, in Euclid's figure. As I previously stipulated, the point H in Euclid's figure is a distance D away from the point C in Euclid's figure. When the laser on board the rocket is located at point H, the laser is going to fire a photon in the direction of the vector which points from H to C (in Euclid's figure). Also, the vector R points from C in Euclid's figure, to H in Euclid's figure. Now, in Euclid's drawing, let the direction of motion of the rocket be from A to B. Thus, the motion of the center of inertia of the rocket lies entirely along the infinite straight line through points A,B. In fact, I will introduce a third frame, which is at rest in S, which I will refer to as the Euclidean frame. The origin of the Euclidean frame is the point H in Euclid's figure, and the point C lies on the positive y axis of the Euclidean frame, and the line through AB coincides with the x axis of the Euclidean frame, and the i^ direction of the Euclidean frame points from A to B. Since S is an inertial frame, and at least three noncollinear points of the Euclidean frame are at rest in S, it must be the case that the Euclidean frame is also an inertial reference frame. Let the i^ vector of S point in the same direction as the i^ vector of the Euclidean frame. Now, the origin of frame S` is permanently attached to the center of inertia of the rocket. Let the j^ vector of frame S` point in the same direction as the j^ vector of the Euclidean frame, and let the i^ vector of frame S point in the same direction as the i^ vector of the Euclidean frame. At this point, we should be able to determine the path of the photon in frame S, as well as frame S`. Since the object being fired is a photon, this complicates things quite a bit. For either one will hold that the speed of a photon is c in all inertial frames, or another will hold that the speed of a photon c is relative to the emitter. But in this problem here, the emitter is being subjected to an external force, prior to the laser being turned on. Hence the rest frame of the emitter is a non-inertial reference frame. To complicate matters, there is no general agreement on whether or not the mass of a photon is zero. The general concensus is that the "rest mass" of a photon is zero, but that the gravitational mass of a photon is non-zero. For now, let us suppose that the photon has mass. In fact, let the mass of the photon fired be quite substantial. In other words, think of the photon which is fired towards the center of the moon, as having been fired by a cannon. Thus, there will be an impulse on the ship at the moment in time at which the center of inertia of the ship is located at the point H in Euclid's figure. And that impulse will be in the direction of R, which is the same direction as the -j^ vector of the Euclidean frame.

You previously showed that it is relevent to the problem. Choose a simple model for Fe

You know, this is never adequately explained in texts, this shift from three dimensional form, to one dimensional form or vice versa. What is the correct way to visualize the wave? Do you know?

I thought so. Yes, you are absolutely right. So what is the correct way to analyze this problem? For sure the acceleration of the center of inertia of the pebble in this frame is zero, that is plain to see. I don't want to write W=mg, I would rather write: [math] \vec W = -\frac {GMm}{r^2} \hat r [/math] Is this correct... [math] -\frac {GMm}{r^2} \hat r + \vec F_e = 0 [/math] Where Fe is the electric repulsion force ?? How would you write Fe ??? Which are the electrons which participate in the interaction?

You better explain this one more. Also, in a previous post in this thread, you stated the speed of light isn't constant in the frame in question. Could you say why please?

Thank you both. PS: That notation was very good Dave.

That's not right. It's more complicated it's this... [math] -\frac{ \hbar^2}{2m} \nabla^2 \Psi + U \Psi = i \hbar \frac{\partial \Psi}{\partial t} [/math] Where [math] \Psi = \Psi (x,y,z,t) [/math] Now, if we let the position vector be denoted as follows... [math] \vec r = x \hat i + y \hat j + z \hat k [/math] then we can write: [math] \Psi = \Psi (x,y,z,t) = \Psi (r,t) [/math] In the case where the wavefunction is constant in time, the lowercase psi is usually used, that is: [math] \psi = \Psi (x,y,z) = \Psi ® [/math] You have some object with inertial mass m in some potential U. hbar is Planck's constant of nature divided by 2 pi. i is the square root of negative one. Psi is called the wavefunction. The fundamental postulate of quantum mechanics is that the momentum of an object is related to an inertial wavelength in the following way: [math] |\vec P| = \frac{\hbar}{\lambda} [/math] From classical mechanics we have: [math] |\vec P| = |m \vec v| = m |\vec v| [/math] The quantity |v| is the speed of the center of inertia of the object in some particular inertial reference frame. Speed is a strictly positive quantity, and is defined as distance traveled in the frame, divided by time of travel (as measured by a clock at rest in the frame). Now, there are two kinds of velocity associated with a quantum mechanical inertial wave. They are called the group velocity , and the phase velocity. Here is a somewhat more detailed discussion of group velocity Suppose that we equate the classical expression for momentum with the quantum mechanical expression for momentum. Then we arrive at this formula here: [math] m |\vec v| = \frac{\hbar}{\lambda} [/math] Now, suppose that the object with inertial mass m is moving through an inertial reference frame with nonzero speed |v|. Then we can divide both sides of the equation above to get this: [math] m = \frac{\hbar}{|\vec v| \lambda} [/math] So there is a problem introduced, if we try to analyze an object in a reference frame in which it's speed is zero. At any rate, the LHS of the formula above gives the inertia of the object in terms of it's speed in a frame, and it's quantum mechanical wavelength.

I didn't exactly understand this. Suppose you have the following function of x,y,z [math] f(x,y,z) = 3xy^3 + 7zxy - 2x [/math] The partial derivative of f with respect to x is given by [math] \frac{\partial f}{\partial x} = 3y^3 + 7zy - 2 [/math] During the differentiation process, the variables y,z were treated as constant. So I do know that. So they cannot be equivalent.

Sum of forces equals zero... yes i know that force down due to gravity... force of hand on pebble due to electrodynamic repulsion. Is this what you mean... [math] \vec W + \vec F_e = \frac{d\vec P}{dt} = 0 [/math] ??

In another thread, yourdadonapogos asked whether or not [math] \frac{\partial}{\partial t} = \frac{d}{dt} [/math] Dave didn't want to answer it in that thread, because that thread is devoted to ordinary derivatives, but I think the question is a good one. Does anyone know the answer? Regards

Here is are the trig identities I usually remember: sin (A+B) = sinAcosB+sinBcosA cos (A+B) = cosAcosB-sinAsinB And then I can figure out these, because I know that sine is an odd function, and cosine is an even function, so I don't have to memorize them: sin (A-B) = sinAcos(-B)+sin(-B)cosA = sinAcosB - sinBcosA cos (A-B) = cosAcos(-B)-sinAsin(-B) = cosAcosB+sinAsinB So cos (A+B) - cos (A-B) = (cosAcosB-sinAsinB)-(cosAcosB+sinAsinB) So.. cos (A+B) - cos (A-B) = -sinAsinB-sinAsinB = -2sinAsinB So 2sinAsinB = -cos (A+B) + cos (A-B) So sinAsinB = 1/2[cos (A-B)-cos(A+B)] There you go. While we're on it, there is a good way to derive the formulas using the exponential relationship... [math] e^{i \theta} = cos \theta + i sin \theta [/math] You do something, then equate real part to real part, imaginary part to imaginary part, and you get some trig relations. Let theta = A+B [math] e^{i (A+B)} = cos (A+B) + i sin (A+B) [/math] I'm trying to remember it. [math] e^{iA} e^{iB} = e^{iA+iB} = e^{i(A+B)} [/math] also [math] e^{iA} e^{iB} = (cosA+isinA)(cosB+isinB) [/math] Got it... Therefore: [math] (cosA+isinA)(cosB+isinB) = cos (A+B) + i sin (A+B) [/math] Now the LHS of the equation above is equivalent to: [math] cosAcosB+cosAsinB i + sinAcosB i - sinAsinB [/math] Where I have made use of the algebraic fact that: (A+B)(C+D) = AC+AD+BC+BD Now, grouping the real parts together, and the imaginary parts together we have: [math] (cosAcosB - sinAsinB) + (cosAsinB + sinAcosB) i [/math] And the expression just obtained, is equivalent to the following expression: [math] cos (A+B) + i sin (A+B) [/math] At which point we can immediately see the trig identities for the sum of two angles: [math] cos (A+B) = (cosAcosB - sinAsinB) [/math] [math] sin (A+B) = (cosAsinB + sinAcosB) [/math] To obtain the trig identities for the difference of two angles, just let B become -B in the formulas above, and use the fact that sine is an odd function and cosine is an even function, as I did earlier. Thus, we can derive most of the trig relations, if we already know the following formula: [math] e^{i \theta} = cos \theta + i sin \theta [/math] This formula is easily provable using the series representation of ex. It is: [math] e^x = \sum_{n=1}^{n= \infty} \frac{x^n}{n!} [/math] Let x = i times theta. [math] e^{i \theta} = \sum_{n=1}^{n= \infty} \frac{(i \theta)^n}{n!} [/math] In order to finish the derivation, you have to know the series expressions for sine and cosine. They are: [math] sin \theta = \sum_{n=1}^{n= \infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} [/math] and [math] cos \theta = \sum_{n=1}^{n= \infty} (-1)^n \frac{x^{2n}}{(2n)!} [/math] And of course these formulas can be obtained using a Taylor series expansion. Whew...

yes you are right dave, I will derive that later. Let's see what it was again: 2 sin (mx) sin (nx) = cos (m-n)x - cos (m+n) x That is what is to be derived. Well... 2 sin (mx) sin (nx) = sin (mx) sin (nx) + sin (mx) sin (nx) Therefore: sin (mx) sin (nx) + sin (mx) sin (nx) = cos (m-n)x - cos (m+n) x Let A = mx, let B = nx, therefore:

I'm sure there is, but I need to review integration by parts, it comes up like crazy during Fourier analysis. Well let me have a look at what you wrote... Oh I see, that's a trigonometric identity. Yeah that works good too. The only problem is I don't usually remember the product identities. But I have to know integration by parts, for other reasons. But yes that works, it is equally good. Thanks

Orthogonality relations, yes them. I am going to work them out right now. I know integration by parts, so... As a quick example, I will integrate the natural log of x, using integration by parts. [math] \int ln x \ dx = ? [/math] Underneath ln x, write the derivative with respect to x of ln x, and underneath dx, write the integral of dx. In other words: [math] \left \begin{array}{cc} ln x \ \ dx \\ \frac{1}{x} \ \ x \\ \end{array} \right [/math] We can now write the following true statement: [math] \int ln x \ dx = (ln x)(x) - \int (1/x)(x) dx [/math] Now look at the RHS. There are four entries in the array A11 A12 A21 A22 In this problem, A11 = ln x and A12 = dx A21 = 1/x A22 = x Using the integral by parts formula, we can show that the original integral is equivalent to: (A11)(A22) - integral of (A21)(A22) Finishing up we have: [math] \int ln x \ dx = xln x - \int dx = xln x - x + C [/math] Where C is an arbitrary constant of integration. Ok so that is an example of integration by parts. We can also use it to evaluate the integrals at the wolfram site. Let me have a look at the first integral we have to evaluate: [math] \int_{x=- \pi}^{x=\pi} sin (mx) sin (nx) dx [/math] In a problem like this, there is a quick way to do integration by parts. Make two colums... At the top of the first column, write sin (mx), and at the top of the second column write sin nx dx. Underneath sin (mx) write the derivative of sin (mx) {which is m cos (mx)}, and underneath sin (nx)dx write the integral of sin (nx)dx {which is -cos(nx)/n} Now do this one more time. Underneath m cos (mx) write the derivative with respect to x of m cos (mx) which is -m2 sin mx. [math] \left \begin{array}{cc} sin (mx) \ \ sin (nx) \\ m cos(mx) \ \ -\frac{cos(nx)}{n} \\ -m^2 sin (mx) \ \ -\frac{sin (nx)}{n^2} \\ \end{array} \right [/math] Notice that we have reached a row which is equivalent to the original row times a constant, so we can stop here. Now change the sign of every other row: [math] \left \begin{array}{cc} sin (mx) \ \ sin (nx) \\ -m cos(mx) \ \ -\frac{cos(nx)}{n} \\ -m^2 sin (mx) \ \ -\frac{sin (nx)}{n^2} \\ \end{array} \right [/math] So here is the array: A11 A12 A21 A22 A31 A32 Now, here is the pattern for integration by parts: The integral of (A11)(A12) = (A11)(A22)+(A21)(A32) - integral of (A31)(A32) So we can write out the answer as: [math] \int sin (mx) \ \ sin (nx) \\ dx = [/math]

Take your time, in the meantime I am going to refresh my memory on Fourier series. 2 days is about right for me to have it again, so it's just as well. Thank you Dapthar.

ok.

Take an object, like a pebble in your hand. Hold it at a constant distance above the center of the earth. Now, you have to choose a frame to define the acceleration of the object in. In the rest frame of the sun, for sure that stone has a nonzero acceleration. BUT, and here is the really important point. Frame switch to the rest frame of the earth. Somewhere inside the earth, is the center of mass of the earth. And let the pebble be roughly spherical, so that we can ensure that it's center of mass is inside of it. Ok so now there is a nonzero distance between the center of inertia of the earth, and the center of inertia of the pebble. And since you aren't raising or lowering your hand in the earth frame, this center to center distance is constant in time (at least for now). Now switch back to the rest frame of the sun. In that frame, the earth is spinning, so that the radial vector from the center of the earth to the center of the pebble is rotating. And that means that the center of the pebble is accelerating in the sun frame. Now switch to the earth frame where the radial vector is an axis of the frame. Therefore, the location of the center of inertia of the pebble isn't moving in this frame. That means this now... That means that the pebble is at rest in this frame. Certainly the pebble could be spinning in your hand, but the center of mass of that pebble is at rest in this frame. And when something is at rest in a frame, then its speed in the frame is zero. And this pebble is remaining at rest in this frame, hence it's speed isn't changing. Now here is the definition of acceleration: [math] \vec a = \frac{d\vec v}{dt} [/math] As you can see acceleration is a vector quantity. That means that it has magnitude, as well as direction. Now, dt is strictly positive, and it certainly is non-infinite, hence the only way the LHS (left hand side) of the equation above can be zero, is if the numerator of the RHS is zero. Now here is the numerator of the RHS: [math] d\vec v [/math] ok so That quantity is a differential of a velocity vector. Velocity is speed in a frame, times direction of motion in the frame. Right now, we are in in a frame in which the center of mass of earth is at rest. Now, there are many reference frames in which the center of mass of earth is at rest. In some of them, a fixed point on the surface of earth may be orbiting the CM of earth. But other frames we can concieve of have fixed points on the surface being at rest, in addition to the CM of earth being at rest. And this is the type of frame that the pebble is in, because it is hovering at a height h over the the same spot on the ground. So the point is, that the speed of the center of inertia of the pebble in this frame is zero, and hence the product of the speed of the pebble in this frame, and it's direction of motion in this frame must be zero, since its speed is zero, regardless of what it's direction of motion was previously, or later (since it's speed is zero over some interval of time in this frame, as measured by a clock at rest in this frame). With all of this keenly in mind, it must be the case that the acceleration of the pebble in this frame of reference is zero. It is only when you release the pebble, that it begins to fall, or using other words... it is only when you release the pebble that it begins to accelerate in this reference frame. My thoughts on this matter are improving. Now, once you release it, it has a measurable acceleration of g. But whilst you are holding that pebble in your hand, it has NO acceleration in this frame. On the other hand, you can feel the weight of the pebble at all moments in time for which the pebble has no acceleration. Now, suppose you use Newton's second law of motion to describe the weight of this pebble, as being measured by your hand (or equivalently a scale which you are holding, with the pebble on top). If you assume that the weight of the object is equivalent to its inertial mass, times its acceleration in this reference frame, then you will run into an immediate contradiction, and here's why: Suppose you write that force is equal to inertial mass, times acceleration, like this: [math] \vec F = M \vec a [/math] We are already clear on the fact that the acceleration of the object in this frame is zero, and it's weight is certainly nonzero. So if we set W equal to F in the formula above we get this: [math] \vec W = M \vec 0 [/math] In an earlier discussion with Tom Mattson, Mr Mattson showed me a simple argument proving that a vector quantity is equal to the zero vector if and only if its magnitude is zero. So let's write this: [math] W (dir \vec W) = (M)( 0 ) ( dir \vec 0 ) [/math] In the formula above, from left to right, W is what is measured by the scale, (and is certainly nonzero), dir W points towards the center of the earth, M is the inertial mass of the pebble, 0 is the magnitude of the acceleration of the center of inertia of the pebble in this frame, and ( dir \vec 0 ) is the same vector as (dir \vec W), since the statement is supposedly true in this frame. So dividing both sides by dir W, we obtain the following scalar equation: [math] W = M 0 [/math] It is an algebraic fact that zero times any finite quantity is equivalent to zero, and M is finite, hence M times zero is equal to zero, hence the RHS of the statement above is equivalent to zero. However, the LHS is the weight of the pebble as measured by the scale (or our hand), and is certainly not equivalent to zero, in fact it is positive. Therefore, we have reached a contradiction, W is, and isn't zero. Hence, it cannot be the case that the weight of the pebble is equivalent to its inertial mass times its acceleration in the frame. But Newton's law is not F = M a. Here is Newton's second law, expressed mathematically: [math] \vec F = \frac{d\vec P}{dt} [/math] In the formula above, P is the momentum of the pebble in this frame, note that the momentum of an object is a frame dependent quantity. In other words, it is not usually the case that the momentum of an object in one frame is the same as its momentum in another frame. Now, the definition of momentum is as follows: [math] \vec P = M \vec v [/math] In the formula above, M is the inertial mass of an object (which is considered to be a property of the object, and not a frame dependent quantity), and vector v is the velocity of the object in the frame (which is intuitively a frame dependent quantity). Now, using the differential calculus we can show that: [math] \vec F = \frac{d\vec P}{dt} = \frac{d(M\vec v)}{dt} = M \frac{d\vec v}{dt} + \vec v \frac{dM}{dt} [/math] We know that dv/dt =0, so let us continue our assumption that the weight of the pebble is given by Newton's second law. We then obtain: [math] \vec W = \vec v \frac{dM}{dt} [/math] Now, the LHS we know is nonzero, and on the RHS, there is multiplication by v, the speed of the pebble in this frame, which we know is zero, and remaining zero in this frame. Therefore, again we reach a contradiction. Therefore, it cannot be the case that the weight of the object satisfies Newton's second law in this reference frame. That is impossible.

You guys are great. I have to go now. This might just work... something...