Johnny5

Gravitational Boost It's a trick used by NASA to save money in rocket fuel, during launch, or as you can see in the lower diagram, a large body can be used to alter the trajectory of a ship or what have you.

Ok, I have read it again, so now I have to ask you, "What is an absolute reference frame?"

No. And I can show you what I did explicitely. I still have my paper. It is interesting that you primed things, without knowing the event which I analysed.

They start off in the same frame, with no relative speed. In order to have the relative speed v between them be greater than zero, one of them must be subjected to an external force, in order so that the relative speed become greater than zero. And so if we start with them at rest in some frame, then once the external force acts upon one of the twins, that twin will have acceleration a=dv/dt in that frame. Let that applied force last for time delta T. Then turn off the engines, and then he is coasting at constant relative speed v to the other twin. But beforehand, he must accelerate at least for a little while.

This is a good question. Well for one, my mind seems to have no tolerance for space bending, and from what I can tell, GR says that space can bend. Also, I have found an error in the equivalence principle, using logic (and electrodynamics). So that explains this. As for SR, the same thing goes. The barn and ladder paradox... They can continue to use whatever makes them happy, that's inconsequential to me. There are so many things that physicists of today do not know, that I am sure to spot one or two of them in my time. Kind regards

I am reminded of Galileo, who after being excommunicated from the catholic church for maintaining that the earth moves around the sun, was heard to utter "and yet it moves." I know you were joshing me about the empiricism remark, but, in the spirit of Galileo, I would like to add this to my original comments... "And yet it's false." But what am i talkin about? Kind regards PS: Oh and one more thing, I have not the aptitude to redo everything from scratch, nor the lifespan. That's why I work on the part that's wrong.

Both measurements of what the amount of elapsed time? Yes, that is what relativity says. I understand relativity to the point where I know that is what it says. What assumption?

It depends upon what they think time travel is. We can always move one moment in time, into the future. But once we are there it becomes the present. It doesn't matter what people think. Let them think what they want, that doesn't alter what is true. Reality is objective not subjective. If you are just fooling around, instead of trying to learn that is one thing. But, this is a question which physics is supposed to be able to answer. The laws of physics involve conservation laws. Conservation of energy. Conservation of matter. Conservation of charge. Conservation of linear momentum. Conservation of angular momentum. Conservation of Baryon number. The list goes on and on. At least one of them is true. This is no argument, but as I say there are so many to choose from, all leading to the same conclusion.

Oh... why should something so obvious astonish you? Here is why, because you are as good as anyone else out there, that's why you would be the one to say. Regards

Explain what I meant about what?

As a matter of fact, you are the one to say. But you need to construct a proof. It is impossible for you to construct a proof, until you define what is meant by saying "travel through time." One such argument, might be that the location of the center of mass of the universe would instantaneously change, which is impossible. Regards

Yes but deriving the spherical harmonics of a hydrogen atom isn't a straightfoward fact. If the person already knows spherical coordinates, they could follow the derivation. Otherwise, at least instruct them on what to go learn, and return when they know it. Google isn't the answer to a question that complex. I was going to derive them for him. But right now I am working on several problems which you have given me on gravity.

What I am trying to do' date=' is learn something I don't already know. I will leave it as Fe, and keep in mind that it is a constant. Ok so that means this... [math'] \vec W =- \vec F_e [/math] The cause of the weight is due to gravity: [math] \frac{GMm}{r^2} \hat r =- \vec F_e [/math] And as swansont pointed out, the value of each side is constant in time, whilst the hand is holding the pebble at a constant height h above the surface of the earth. That means that the derivative of either side of the equation with respect to time is zero. So that I can do this... [math] \frac {G \hat r}{r^2} [ m \frac{dM}{dt} + M \frac{dm}{dt} ] = 0 [/math] I have assumed that the Newtonian gravitational constant, doesn't vary in time. The center to center distance between the center of inertia of earth, and the center of inertia of the pebble was stipulated to be constant in this particular problem, so that in the rest frame of the earth dr/dt=0. And the direction of the gravitational force is towards the center of the earth. Oh and I neglected that the graviational force is attractive. So now is the time to specify the direction of r^ very clearly. I think the right thing to do is to use gravitational field theory. So what I will do is this. I will define the gravitational field of an object with mass M as follows: [math] \vec \Gamma = - GM \frac{\hat r}{r^2} = - GM \nabla (\frac{-1}{r}) [/math] So now the question is, what frame are we talking about. The gravitating object has a center of inertia somewhere. That center of inertia is going to be the origin of the frame. Now, if the gravitating body is spinning, that spin will define a unique axis, which I will call the spin axis. We can then use that spin axis to begin defining a rectangular coordinate system. For now let us assume the gravitating object isn't spinning. So pick a fixed point on the surface of the gravitating body. In fact, let the gravitating body have the shape of a sphere, so that we can intuitively know where the center of inertia of the object is, it is at the center of the sphere. So now, just have set up a three dimensional rectangular coordinate system, with origin located at the center of the shere, and one of the axes passes through one of the fixed points on the sphere. Now, if we don't fix another of the axes, then it isn't clear that we have defined a frame in which the axes aren't spinning. So now, pick a second fixed point on the sphere, and fix the y axis. Therefore, there sphere isn't spinning in the X,Y,Z frame. Now, r is a vector from the origin of this frame, to a field point. So that - r is a vector from the field point to the origin. Gravitational force must come out attractive, so that the weight of an object of mass m (whose own gravitational field is currently being neglected), in the gravitational field of an object of mass M, is to be along the center to center line, and is given by: [math] \vec F = m \vec \Gamma [/math] So using Gamma as defined above, we have: [math] \vec F = -mGM \frac{\hat r}{r^2} [/math] Which is the familiar form of Newtonian gravity. So now we have this: [math] m \vec \Gamma = - \vec F_e [/math] And the magnitude of Fe is constant in time, as swansont pointed out. At least whilst the center to center distance is constant. The LHS of the formula above is the weight of the object whose mass is m, and has a direction towards the center of the object of mass M. The quantity Fe is the electric repulsion between the hand and the pebble. And has a direction antiparallel to the direction of the weight of the pebble. So if we differentiate both sides with respect to time we have this: [math] \frac{dm}{dt} \vec \Gamma + m \frac{d \vec \Gamma}{dt} = 0 [/math] Now we have: [math] \vec \Gamma = GM \nabla (\frac{-1}{r}) [/math] So that we have this: [math] \frac{dm}{dt} GM \nabla (\frac{-1}{r}) + m \frac{d}{dt}(GM \nabla (\frac{-1}{r}) ) = 0 [/math] Now the value of G is constant in time, and it is certainly nonzero, so we can divide both sides of the formula above to obtain: [math] \frac{dm}{dt} M \nabla (\frac{-1}{r}) + m \frac{d}{dt}(M \nabla (\frac{-1}{r}) ) = 0 [/math] Now, in spherical coordinates, the gradient is just partial with respect to r r^... so... [math] \frac{dm}{dt} M \frac{\partial}{\partial r} (\frac{-1}{r}) \hat r + m \frac{d}{dt}(M \frac{\partial}{\partial r} (\frac{-1}{r}) ) \hat r = 0 [/math] Dividing both sides by r^ we have: [math] \frac{dm}{dt} M \frac{\partial}{\partial r} (\frac{-1}{r}) + m \frac{d}{dt}(M \frac{\partial}{\partial r} (\frac{-1}{r}) ) = 0 [/math] Now, we are not to assume that the mass of the objects is constant in time. I have my reasons. So... For whatever it is worth, I think I need access to more powerful mathematics. The next step is to compute the following... [math] \frac {d}{dt}( M \frac{\partial}{\partial r} (\frac{-1}{r}) ) [/math] Ultimately, dr/dt =0 since the center to center distance is stipulated to be constant in time. So therefore we have: [math] \frac {d}{dt}( M \frac{\partial}{\partial r} (\frac{-1}{r}) ) = \frac{dM}{dt} \frac{1}{r^2} [/math] This leads to... [math] \frac{dm}{dt} M \frac{\partial}{\partial r} (\frac{-1}{r}) + m \frac{dM}{dt} \frac{1}{r^2} = 0 [/math] From which follows that: [math] \frac{dm}{dt} M \frac{1}{r^2} + m \frac{dM}{dt} \frac{1}{r^2} = 0 [/math] Multiplying both sides by r^2 we have: [math] \frac{dm}{dt} M + m \frac{dM}{dt} = 0 [/math] Which leads to this... [math] \frac{dm}{dt} M = -m \frac{dM}{dt} [/math] This is certainly true if it is impossible for the inertial mass of an object to vary in time, then each side would be equivalent to zero. Hence the statement would be true. But, if the inertial mass M can vary, the above relation would still have to be true. Is there any theory out there where gravitational mass can vary? Is there any theory out there where inertial mass can vary?

Yesterday, I was trying to find the trajectory of a photon fired at a right angle to the direction of the acceleration of a craft in the rest frame S` of the craft, as well as its path in the inertial rest frame of a moon, which the craft was flying by. As I recall, the problem isn't simple. The point I left off at, was right at the moment that the laser fired the photon at the center of mass of the moon. Here is the diagram that I was using: Euclid Book I, proposition 12 In the diagram C corresponded to the center of inertia of the moon, and the rest frame of the moon was considered to be an inertial reference frame, and I was referring to that frame as frame S. As I recall the rocket was moving along the line AB, and the direction of its acceleration a, in frame S was from A to B. The origin of S was the center of inertia of the moon. The point H was on the positive y axis of S. And the X axis of S was parallel to straight line AB. And the i^ vector of frame S pointed from left to right in Euclid's diagram. Then I defined another inertial frame which I was just referring as Euclid's frame. It was a frame in which at least three non-collinear points in it were at rest in frame S, from which we can also infer that Euclid's frame was also inertial. Keep in mind that frame S was stipulated to be inertial. The origin of Euclid's frame is the point H in Euclid's figure, the X axis of Euclid's frame is the line AB in Euclid's figure, the i^ direction of Euclid's frame was from A to B, and the point C lies on the positive y axis of Euclid's frame. And there was one other frame, S`, which is the rest frame of the rocket ship. Now, the speed of the rocket in frame S is changing in time, but the direction of motion of the rocket is along line AB. Now, there comes a moment in time when the rocket is at the point H in Euclid's figure. Right at that moment in time, the laser fires a photon right at the center of inertia of the moon, which is the point C in Euclid's figure. This is where I left off at yesterday. Now, if the photon is extremely massive, then there will be a substantial force on the rocket, which would cause frame S` to deviate off of it's straight line path AB. And that of course will affect the mathematical solution to the path of the photon in frame S`, after it is emitted from the laser. But, in reality, the inertial mass of the photon is negligible, in comparison to the mass of the rocket. In theory, the mass of the photon is literally zero, although there is no general consensus on this. But the reality is, that to a good approximation, the emission of a single photon by the laser, will hardly affect the direction of the acceleration of the rocket in frame S. So, for now, we can totally ignore this effect, and treat the motion of the center of mass of the rocket as constrained to move along line AB even after the photon is emitted from the laser (which is mounted to the rocket). But what is certainly true is this, at the moment that the laser emits the photon, that direction is at a right angle to the initial direction of the rocket through frame S. And the photon is emitted at a moment in time when the center of inertia of the rocket is at point H, in relation to the point C (the point C is permanently the center of inertia of a moon). So now the goal is to find the path of the photon in reference frame S, and reference frame S`. We need to keep in mind that S` is a non-inertial frame. Well think of things before the photon is emitted. Imagine that you are holding a baseball in your hand, as your rocket ship is accelerating along line AB. There is a force upon your hand. Then right when you pass the point H, release the baseball. The mass of the rocket + fuel just decreased by one baseball mass. But here is the thing to note. The moment that you release the baseball, the baseball goes from being in a non-inertial reference frame to being in an inertial reference frame. Can anyone else see that? The baseball will now be moving in a straight line at a constant speed, in reference frame S, in the direction from A to B. And had it been thrown at the center of inertia of the moon, instead of merely released, it would still be moving in a straight line at constant speed in frame S, only this time the direction would have been a diagonal line, in the first quadrant of the Euclidean frame. In other words, let v denote the velocity of the baseball in Euclid's frame at the moment it was thrown from the rocket straight (at point H) right at the center of inertia of the moon. [math] \vec V = v_1 \hat i + v_2 \hat j [/math] So now, the ship was accelerating at the moment when the baseball is thrown, and therefore the baseball had some initial speed v1 in the positive i^ direction, and the baseball was thrown right at the moon, so that at that same moment, the baseball was given some speed v2 in the positive j^ direction of Euclid's frame. So the path of the baseball in Euclid's frame, is going to be a diagonal line, in the first quadrant, and it's slope will be: [math] m = \frac {v2}{v1} [/math] Rise/run In other words, the dt's cancel out, and because of the way I have set up the problem we have: [math] m = \frac {v2}{v1} = \frac{dx}{dy} [/math] Now, what started me thinking about this, was the fact that swansont said the path of the emitted photon in the rest frame of the rocketship was curved. What I really want to do is find the formula for that curve, in the rocket frame, which I know is a non-inertial frame. Somewhere in this argument, I am going to fix the speed of a photon relative to its emitter as c in the center of mass frame. Since the ship is so much more massive then the photon, we are considering that the CM of the ship remains traveling on line AB, even after the photon is emitted. Ok swansont, I think I've worked out the basic idea of the path of the photon in the non-inertial reference frame of the ship. I know that once the photon is emitted, it's going to move in a diagonal line in frame S, and the Euclidean frame. But, the rocket is continually ejecting mass, and therefore accelerating along the line AB, and I've not yet specified the details of it. Consider the simplest case non-zero constant acceleration. Let the the rate of acceleration be g = 9.8 meters per second. So we can think of the rocket in free-fall, in the direction from left to right. Thus, in equal amounts of time, it's jumping larger and large distances, and the sum of those distances is the sum of squares... like this... X-X----X---------X----------------X-------------------------X And this change in distance is obeyed in the S frame, and Euclid's frame. So that means approximately this... In the rest frame of the ship, frame S`, the photon appears to have an acceleration g in the -i^ direction, and a constant speed v, in the j^ direction. So, if I'm not mistaken, the path of the photon after emission, in the rest frame of the rocket, frame S`, is a parabola, in the second quadrant of frame S`. Now I could have easily made a mistake, but I just wanted to check what Janus said, which was, let me go find the quote, so I get it right... Yes ok I agree with that, the path is a parabola. The next thing Janus says, is this: This I didn't understand Janus. The next thing Janus says is this: Here, it actually seems to me, as though you are arguing my point, which is that the time dilation formula is a false statement in reference frame S`. I do believe that I said this..., "If the time dilation effect occurs, then it is true in at most one of the two frames, not both." Janus then says this: Well yes there is a paradox created if you believe that the formula is true in both frames, which is what I told this person. The time dilation formula cannot be a true statement in both frames. If it is true in at least one of the frames, then it is true in at most one of the frames, in this twin paradox example. From which it follows that it is true in exactly one of the frames. But if SR is totally wrong, then it's true in neither of the frames. So the best thing to say is this... In the twin paradox problem, the time dilation formula is true in at most one of the two frames. This leaves open the possibility that it's false in both frames, but closes off the possibility that it's true in both frames. Lastly, you say this... You are going to confuse that person, because the time dilation formula cannot be true in both frames, that is impossible, because then this is what is happening... Since the formula is true in frame S in which twin one is at rest, Then after one year aboard the ship, twin one is ten years older than twin two. Suppose that when the trip begins each twin is age 30. Since the time dilation formula is true, there is a moment in time at which twin one is age 31, and twin two is 40. Now, suppose that the time dilation formula is true in frame S`, which is the non-inertial frame in which twin two is at rest. Since the time dilation formula is true, when twin two is age 31, it must be the case that twin 1 is aged forty. Therefore, twin one reaches his fortieth birthday before twin two reaches his fortieth birthday, and twin one does not reach his fortieth birthday before twin two reaches his fortieth birthday. So the time dilation formula cannot be true in both frames, which is all I told this person. Kind regards