Johnny5 Posted April 3, 2005 Share Posted April 3, 2005 I wouldn't write it out as an electrostatic force. It's way too complicated and probably not relevant to the problem. You previously showed that it is relevent to the problem. Choose a simple model for F_{e} Link to comment Share on other sites More sharing options...

swansont Posted April 3, 2005 Share Posted April 3, 2005 You previously showed that it is relevent to the problem. Choose a simple model for F[sub']e[/sub] The presence of the force is relevant, but the model for it is irrelevant. It's a constant in the context of the problem. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted April 3, 2005 Share Posted April 3, 2005 You better explain this one more. Also' date=' 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?[/quote'] You chose the Earth as the frame. The Earth is spinning. You therefore have to account for the centrifugal "force" on your pebble even at "rest". While low speed newtonian physics works great in this frame on or near the Earth "corrections" must be made for greater precision. Pseudo forces such as the centrifugal force or coriollis force can be used to balance things out or correct for the fact that this is not an inertial frame. The speed of light is not constant in this frame. This would seem obvious if the Earth was spinning at much higher speed. Since it is not, the discrepancy is small locally but more obvious and much larger if you extended this frame out beyond the solar system. Go far enough and all matter, all stars etc. are all going much faster than 300,000 km/s. Of course this is far beyond the normal use of this frame. It is a very useful and common frame to use but you have to be aware of the limitations. It is a poor choice of frame to use to try and prove that SR or GR or the equivalence principle are wrong. They all require "corrections" in this frame. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 The presence of the force is relevant, but the model for it is irrelevant. It's a constant in the context of the problem. 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 ? Link to comment Share on other sites More sharing options...

swansont Posted April 4, 2005 Share Posted April 4, 2005 Also' date=' I have neglected the use of centrifugal force, as JC pointed out, so where does that fit in swansont? I can replace Fe by [b']C[/b] 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 ? How it fits in depends on what you are trying to do. You can rewrite it as C or just leave it as F_{e} Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 How it fits in depends on what you are trying to do. You can rewrite it as C or just leave it as F_{e} 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? Link to comment Share on other sites More sharing options...

geistkiesel Posted May 6, 2005 Share Posted May 6, 2005 If the mass of the object increases, it must also increase if you are moving, because you can't say which one of you is actually at rest. If the object under scrutiny had been accelerated, the force causing the measured relative motion exclusively (as in trains accelerating and train stations not accelerating) would this not be sufficient to establish "which object is moving with respect to another object"? An unaccelerated observer in a lab observing an electron acceleration process can tell from all the instrumentations, plus his lack of feeling any personal accelerations, unambiguously observes known forces accelerating the electron which is observed to undergo some "mass" perturbation. What do you mean the observer cannot determine if he is moving or if the electron is moving? I thought the "twin paradox" was resolved by an acceleration explanation: the accelerated twin is the slower aging wrt his twin which was unaccelerated; Of course this does pose a serious problem to the equivalence of inertial frames and the embedded "reciprocity" aspects of SRT anad the expected results from scrutinizing the inertial frames motion and effects of motion. Geistkiesel Link to comment Share on other sites More sharing options...

swansont Posted May 6, 2005 Share Posted May 6, 2005 If the object under scrutiny had been accelerated, the force causing the measured relative motion exclusively (as in trains accelerating and train stations not accelerating) would this not be sufficient to establish "which object is moving with respect to another object"? How do you know that the acceleration didn't bring the object to rest? Link to comment Share on other sites More sharing options...

geistkiesel Posted May 14, 2005 Share Posted May 14, 2005 How do you know that the acceleration didn't bring the object to rest? It is this kind of question that, by inference, describes the intrinsic flaws in SRT. Let us assume the electrons are produced from the ionization of hydrogen gas. The electrons are directed into an accelerating beam until a velocity of .99c is reached. The electron then exits the field and is detected by a cascade photomultiplier tube where the output current is proportional to the input enrgy of the electron. Let us perform this experiment with a wide specrum of accelrated electron velocities. The question is how would I know that the electron wasn't decerated to a state of rest? I would know from the reading of the detectors that the frist electron had an energy that was grossly elevated over the electrons that reached lower levels of velocity as described by the output describing all the accelerations. What would you do if asked the same question? Would you entertain giving a response that you really didn't know if the electron was accelerated or decelerated? Link to comment Share on other sites More sharing options...

swansont Posted May 14, 2005 Share Posted May 14, 2005 It is this kind of question that' date=' by inference, describes the intrinsic flaws in SRT. Let us assume the electrons are produced from the ionization of hydrogen gas. The electrons are directed into an accelerating beam until a velocity of .99c is reached. The electron then exits the field and is detected by a cascade photomultiplier tube where the output current is proportional to the input enrgy of the electron. Let us perform this experiment with a wide specrum of accelrated electron velocities. The question is how would I know that the electron wasn't decerated to a state of rest? I would know from the reading of the detectors that the frist electron had an energy that was grossly elevated over the electrons that reached lower levels of velocity as described by the output describing all the accelerations. What would you do if asked the same question? Would you entertain giving a response that you really didn't know if the electron was accelerated or decelerated?[/indent'] In this case you know the state of the hydrogen before the experiment started. What if you don't know the history of the particle? Plus, you are trying to measure the electron's energy in your frame. Not the "absolute energy" that it has (which would be contrary to SR). The energy you measure would be frame dependent. Link to comment Share on other sites More sharing options...

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