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From: PD on 18 May 2010 11:01 On May 17, 5:47 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: In physics there is no requirement that > "work" (force x distance moved) be done before a force can be > replied. Of course. You can apply a nonzero force and generate no displacement, and there will be no work. You can have a displacement without applying any force, and there will be no work. But if you apply a net force, even with no resistance, AND there is displacement in that direction, then there will be work. An example of the last is an object in free-fall, in which there is a force (gravity), no resistance, and a displacement in the direction of gravity. So an object in free fall is one example where work is done, and that's where the kinetic energy of the falling object comes from, and why it grows as the velocity squared, as we derived earlier in the thread. 7th graders have no problem understanding these three cases. Why is it so confusing to you? > A two year old kid pushing on a square marble block as tall > as the kid, is applying a FORCE to the block, even though there is > ZERO displacement. Objects which are being held in some high place > prior to being dropped in an experiment, are actually already being > acted upon by the steady force of gravity that's = the object's > weight. The latter static weight is my contribution to understanding > momentum-variant kinetic energy. My correct formula, which replaces > KE = 1/2mv^2, and E = mc^2 / beta, is: KE = a/g (m) + v / 32.174 > (m). An object falling for one second will have a KE (force in > pounds) of two weight units; two seconds, three weight units; three > seconds, four weight units; four seconds 5 weight units. Those values > were confirmed in TWO KE experiments, including that $40.00 one which > you could easily do yourself. But since your motive is to disagree, > you'll never accept any science truth. NoEinstein > > KE = 1/2mv^2 is disproved in new falling object impact test.http://groups.google.com/group/sci.physics/browse_thread/thread/51a85... Here's another simple derivation for you, so you can see how it is inescapable. There is a nice kinematic relationship that you will find in engineering texts, as well as high school texts: (v_final)^2 - (v_initial)^2 = 2*a*x, where x is the displacement, a is the acceleration, and v_final and v_initial are final and initial velocities. I can show you where this relationship comes from, if you like. Now, we have the definition of work: Work = Net force x displacement. But we also know from Newton's 2nd law, which was written in the 1600's, Net force = mass x acceleration So by arithmetic (please don't tell me this confuses you) Work = mass x acceleration x displacement. But we can recognize that (acceleration x displacement) from the kinematic relation I mentioned above -- there it is on the right hand side, next to the factor of 2. So we also know acceleration x displacement = (1/2)(v_final)^2 - (1/2) (v_initial)^2 And so again by arithmetic Work = (1/2)(mass)(v_final)^2 - (1/2)(mass)(v_initial)^2 So you see? The right hand side is the change in the kinetic energy, and here we have the law of conservation of energy: any change in the kinetic energy is due to the work done. All we used in this derivation were three things, which you surely must accept: 1. The kinematic relationship mentioned above 2. Newton's 2nd law: F=ma 3. Work = force x displacement and besides that, just the eensiest bit of arithmetic. Awfully hard to find fault with it, isn't it? > > > > > On May 14, 3:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 7, 6:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > PD hasn't quoted any authoritative source showing that WORK is in any > > > way involved in calculating KE. > > > Oh, yes, I have, John. You don't seem to remember anything that was > > told to you the day before. > > Do you like easy to read pages? Here's one for students at West > > Virginia University:http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/WorkEngergyTheor... > > "The energy associated with the work done by the net force does not > > disappear after the net force is removed (or becomes zero), it is > > transformed into the Kinetic Energy of the body. We call this the Work- > > Energy Theorem." > > > > And he hasn't quoted any > > > authoritative source saying that "work" can be done simply by > > > COASTING, against no resistance! > > > The definition of work is in high school books, John. > > > > And he certainly can't explain how > > > 'gravity' could possibly 'know' the velocities of every falling object > > > (like hail from varying heights) and add the exact semi-parabolic KE > > > increase to each. > > > Doesn't have to, John. The force is not solely responsible for the > > increase in energy. The work is. The work is the product of both the > > force and the displacement. That's how the work increases in each > > second. It's simple, John. Seventh graders can understand it. I don't > > know why you're so much slower than the average 7th grader. > > > > In short, PD is a total, sidestepping FRAUD! And > > > 95% of the readers know that he's a fraud! NoEinstein > > > > > On May 7, 3:16 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 7, 9:12 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD, you are a LIAR! Never ONCE have you explained why KE = 1/2mv^2 > > > > > isn't in violation of the Law of the Conservation of Energy. Until > > > > > you do (and you CAN'T) everyone will know that you are just an air- > > > > > head FRAUD! NoEinstein > > > > > Oh, but I have. If you really need to have it explained again, I ask > > > > you this time to print it out. > > > > > The law of conservation of energy says that any change in the energy > > > > of a system must be due solely to the work done on the system. > > > > > The work is the force acting on the object times the displacement of > > > > the object. So any change in energy of the object must be due solely > > > > to this work. > > > > > In the case of a falling body released from rest, we'll look at the > > > > increase in the kinetic energy, which must be due to the work done by > > > > the only force acting on the body -- gravity. If the increase of > > > > kinetic energy the body has at any time is accounted for by the work > > > > that was done on the body during that time, then we know that the law > > > > of conservation of energy has been respected. > > > > > In the first second, the body will fall 16 ft. In the next second, it > > > > will fall an additional 48 feet. In the third second, it will fall an > > > > additional 80 feet. During these first three seconds, the force has > > > > remained constant, so that it is the same in the first second, the > > > > second second, the third second. The speed increases linearly, so that > > > > it is falling at 32 ft/s after the first second, 64 ft/s after the > > > > second second, and 96 ft/s after the third second. > > > > > Now, let's take a look at the work. The work done since the drop, > > > > after the first second, is the force of gravity times the > > > > displacement. This is mass x g x (16 ft). So this is how much kinetic > > > > energy the object has after one second. Now, in the second second, > > > > we'll add more work, in the amount mass x g x (48 ft), since that's > > > > the displacement for the next second. This increases the kinetic > > > > energy of the body, so that it now has kinetic energy mass x g x (16 > > > > ft + 48 ft) = mass x g x (64 ft), and that number is four times bigger > > > > than it was after the first second. Now, in the third second, we'll > > > > add more work, in the amount mass x g x (80 ft), since that's the > > > > displacement for the next sentence. Since energy is conserved, this > > > > added energy must add to the kinetic energy of the body, so that it > > > > now has kinetic energy mass x g x (64 ft + 80 ft) = mass x g x (144 > > > > ft), and that number is nine times bigger than it was after the first > > > > second. > > > > > Now, it should be plain that the kinetic energy is conserved, since > > > > the only thing that has been contributing to it is the work done in > > > > subsequent seconds. We lost nothing, and we added only that which > > > > gravity added. The energy is conserved. > > > > > It should also be apparent that the kinetic energy is increasing in > > > > the ratios 1:4:9. > > > > Meanwhile, the velocities are increasing linearly, in the ratios > > > > 1:2:3. > > > > > Now, any fourth grader can see that we've completely conserved energy, > > > > losing track of nothing, and yet the kinetic energy is increasing as > > > > the square of the velocity. 1:4:9 are the squares of 1:2:3. > > > > > There is no violation of conservation of kinetic energy, and yet KE is > > > > proportional to v^2. > > > > > Now, don't you feel silly that a 4th grader can understand all of > > > > this, but you've never understood it? > > > > > > > On May 6, 8:54 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > OH? Then please explain, PD, how a UNIFORM force inputthe static > > > > > > > weight of the falling objectcan cause a semi-parabolic increase in > > > > > > > the KE. Haven't you heard?: Energy IN must = energy OUT! > > > > > > > NoEinstein > > > > > > > I have explained this to you dozens of times. I gather that you do not > > > > > > remember any of those posts, and you do not know how to use your > > > > > > newsreader or Google to go back and find any of those dozens of times > > > > > > when it has been explained to you. > > > > > > > I surmise that you are slipping into dementia, where each day begins > > > > > > anew, with any lessons learned the previous day forgotten. > > > > > > > I don't think it's a good use of my time to explain the same thing to > > > > > > you each day, only to have you retire at night and forget it by > > > > > > morning, do you? > > > > > > > PD- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > >
From: PD on 18 May 2010 11:02 On May 17, 9:31 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > > > On May 14, 3:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 7, 6:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > PD hasn't quoted any authoritative source showing that WORK is in any > > > way involved in calculating KE. > > > Oh, yes, I have, John. You don't seem to remember anything that was > > told to you the day before. > > Do you like easy to read pages? Here's one for students at West > > Virginia University:http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/WorkEngergyTheor... > > "The energy associated with the work done by the net force does not > > disappear after the net force is removed (or becomes zero), it is > > transformed into the Kinetic Energy of the body. We call this the Work- > > Energy Theorem." > > > > And he hasn't quoted any > > > authoritative source saying that "work" can be done simply by > > > COASTING, against no resistance! > > > The definition of work is in high school books, John. > > And the definitions of Parasite; Dunce; and Imbecile are in any > dictionary. Ha, ha HA! NE Hey, John, you asked for a direct quote from a reliable reference and I gave you one. If you don't like getting exactly what you ask for, then why do you ask for it? > > > > > > And he certainly can't explain how > > > 'gravity' could possibly 'know' the velocities of every falling object > > > (like hail from varying heights) and add the exact semi-parabolic KE > > > increase to each. > > > Doesn't have to, John. The force is not solely responsible for the > > increase in energy. The work is. The work is the product of both the > > force and the displacement. That's how the work increases in each > > second. It's simple, John. Seventh graders can understand it. I don't > > know why you're so much slower than the average 7th grader. > > > > In short, PD is a total, sidestepping FRAUD! And > > > 95% of the readers know that he's a fraud! NoEinstein > > > > > On May 7, 3:16 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 7, 9:12 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD, you are a LIAR! Never ONCE have you explained why KE = 1/2mv^2 > > > > > isn't in violation of the Law of the Conservation of Energy. Until > > > > > you do (and you CAN'T) everyone will know that you are just an air- > > > > > head FRAUD! NoEinstein > > > > > Oh, but I have. If you really need to have it explained again, I ask > > > > you this time to print it out. > > > > > The law of conservation of energy says that any change in the energy > > > > of a system must be due solely to the work done on the system. > > > > > The work is the force acting on the object times the displacement of > > > > the object. So any change in energy of the object must be due solely > > > > to this work. > > > > > In the case of a falling body released from rest, we'll look at the > > > > increase in the kinetic energy, which must be due to the work done by > > > > the only force acting on the body -- gravity. If the increase of > > > > kinetic energy the body has at any time is accounted for by the work > > > > that was done on the body during that time, then we know that the law > > > > of conservation of energy has been respected. > > > > > In the first second, the body will fall 16 ft. In the next second, it > > > > will fall an additional 48 feet. In the third second, it will fall an > > > > additional 80 feet. During these first three seconds, the force has > > > > remained constant, so that it is the same in the first second, the > > > > second second, the third second. The speed increases linearly, so that > > > > it is falling at 32 ft/s after the first second, 64 ft/s after the > > > > second second, and 96 ft/s after the third second. > > > > > Now, let's take a look at the work. The work done since the drop, > > > > after the first second, is the force of gravity times the > > > > displacement. This is mass x g x (16 ft). So this is how much kinetic > > > > energy the object has after one second. Now, in the second second, > > > > we'll add more work, in the amount mass x g x (48 ft), since that's > > > > the displacement for the next second. This increases the kinetic > > > > energy of the body, so that it now has kinetic energy mass x g x (16 > > > > ft + 48 ft) = mass x g x (64 ft), and that number is four times bigger > > > > than it was after the first second. Now, in the third second, we'll > > > > add more work, in the amount mass x g x (80 ft), since that's the > > > > displacement for the next sentence. Since energy is conserved, this > > > > added energy must add to the kinetic energy of the body, so that it > > > > now has kinetic energy mass x g x (64 ft + 80 ft) = mass x g x (144 > > > > ft), and that number is nine times bigger than it was after the first > > > > second. > > > > > Now, it should be plain that the kinetic energy is conserved, since > > > > the only thing that has been contributing to it is the work done in > > > > subsequent seconds. We lost nothing, and we added only that which > > > > gravity added. The energy is conserved. > > > > > It should also be apparent that the kinetic energy is increasing in > > > > the ratios 1:4:9. > > > > Meanwhile, the velocities are increasing linearly, in the ratios > > > > 1:2:3. > > > > > Now, any fourth grader can see that we've completely conserved energy, > > > > losing track of nothing, and yet the kinetic energy is increasing as > > > > the square of the velocity. 1:4:9 are the squares of 1:2:3. > > > > > There is no violation of conservation of kinetic energy, and yet KE is > > > > proportional to v^2. > > > > > Now, don't you feel silly that a 4th grader can understand all of > > > > this, but you've never understood it? > > > > > > > On May 6, 8:54 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > OH? Then please explain, PD, how a UNIFORM force inputthe static > > > > > > > weight of the falling objectcan cause a semi-parabolic increase in > > > > > > > the KE. Haven't you heard?: Energy IN must = energy OUT! > > > > > > > NoEinstein > > > > > > > I have explained this to you dozens of times. I gather that you do not > > > > > > remember any of those posts, and you do not know how to use your > > > > > > newsreader or Google to go back and find any of those dozens of times > > > > > > when it has been explained to you. > > > > > > > I surmise that you are slipping into dementia, where each day begins > > > > > > anew, with any lessons learned the previous day forgotten. > > > > > > > I don't think it's a good use of my time to explain the same thing to > > > > > > you each day, only to have you retire at night and forget it by > > > > > > morning, do you? > > > > > > > PD- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > >
From: xxein on 19 May 2010 19:20 On May 18, 11:02 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 17, 9:31 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 14, 3:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 7, 6:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > PD hasn't quoted any authoritative source showing that WORK is in any > > > > way involved in calculating KE. > > > > Oh, yes, I have, John. You don't seem to remember anything that was > > > told to you the day before. > > > Do you like easy to read pages? Here's one for students at West > > > Virginia University:http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/WorkEngergyTheor... > > > "The energy associated with the work done by the net force does not > > > disappear after the net force is removed (or becomes zero), it is > > > transformed into the Kinetic Energy of the body. We call this the Work- > > > Energy Theorem." > > > > > And he hasn't quoted any > > > > authoritative source saying that "work" can be done simply by > > > > COASTING, against no resistance! > > > > The definition of work is in high school books, John. > > > And the definitions of Parasite; Dunce; and Imbecile are in any > > dictionary. Ha, ha HA! NE > > Hey, John, you asked for a direct quote from a reliable reference and > I gave you one. > If you don't like getting exactly what you ask for, then why do you > ask for it? > > > > > > > > > And he certainly can't explain how > > > > 'gravity' could possibly 'know' the velocities of every falling object > > > > (like hail from varying heights) and add the exact semi-parabolic KE > > > > increase to each. > > > > Doesn't have to, John. The force is not solely responsible for the > > > increase in energy. The work is. The work is the product of both the > > > force and the displacement. That's how the work increases in each > > > second. It's simple, John. Seventh graders can understand it. I don't > > > know why you're so much slower than the average 7th grader. > > > > > In short, PD is a total, sidestepping FRAUD! And > > > > 95% of the readers know that he's a fraud! NoEinstein > > > > > > On May 7, 3:16 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 7, 9:12 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > PD, you are a LIAR! Never ONCE have you explained why KE = 1/2mv^2 > > > > > > isn't in violation of the Law of the Conservation of Energy. Until > > > > > > you do (and you CAN'T) everyone will know that you are just an air- > > > > > > head FRAUD! NoEinstein > > > > > > Oh, but I have. If you really need to have it explained again, I ask > > > > > you this time to print it out. > > > > > > The law of conservation of energy says that any change in the energy > > > > > of a system must be due solely to the work done on the system. > > > > > > The work is the force acting on the object times the displacement of > > > > > the object. So any change in energy of the object must be due solely > > > > > to this work. > > > > > > In the case of a falling body released from rest, we'll look at the > > > > > increase in the kinetic energy, which must be due to the work done by > > > > > the only force acting on the body -- gravity. If the increase of > > > > > kinetic energy the body has at any time is accounted for by the work > > > > > that was done on the body during that time, then we know that the law > > > > > of conservation of energy has been respected. > > > > > > In the first second, the body will fall 16 ft. In the next second, it > > > > > will fall an additional 48 feet. In the third second, it will fall an > > > > > additional 80 feet. During these first three seconds, the force has > > > > > remained constant, so that it is the same in the first second, the > > > > > second second, the third second. The speed increases linearly, so that > > > > > it is falling at 32 ft/s after the first second, 64 ft/s after the > > > > > second second, and 96 ft/s after the third second. > > > > > > Now, let's take a look at the work. The work done since the drop, > > > > > after the first second, is the force of gravity times the > > > > > displacement. This is mass x g x (16 ft). So this is how much kinetic > > > > > energy the object has after one second. Now, in the second second, > > > > > we'll add more work, in the amount mass x g x (48 ft), since that's > > > > > the displacement for the next second. This increases the kinetic > > > > > energy of the body, so that it now has kinetic energy mass x g x (16 > > > > > ft + 48 ft) = mass x g x (64 ft), and that number is four times bigger > > > > > than it was after the first second. Now, in the third second, we'll > > > > > add more work, in the amount mass x g x (80 ft), since that's the > > > > > displacement for the next sentence. Since energy is conserved, this > > > > > added energy must add to the kinetic energy of the body, so that it > > > > > now has kinetic energy mass x g x (64 ft + 80 ft) = mass x g x (144 > > > > > ft), and that number is nine times bigger than it was after the first > > > > > second. > > > > > > Now, it should be plain that the kinetic energy is conserved, since > > > > > the only thing that has been contributing to it is the work done in > > > > > subsequent seconds. We lost nothing, and we added only that which > > > > > gravity added. The energy is conserved. > > > > > > It should also be apparent that the kinetic energy is increasing in > > > > > the ratios 1:4:9. > > > > > Meanwhile, the velocities are increasing linearly, in the ratios > > > > > 1:2:3. > > > > > > Now, any fourth grader can see that we've completely conserved energy, > > > > > losing track of nothing, and yet the kinetic energy is increasing as > > > > > the square of the velocity. 1:4:9 are the squares of 1:2:3. > > > > > > There is no violation of conservation of kinetic energy, and yet KE is > > > > > proportional to v^2. > > > > > > Now, don't you feel silly that a 4th grader can understand all of > > > > > this, but you've never understood it? > > > > > > > > On May 6, 8:54 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > OH? Then please explain, PD, how a UNIFORM force inputthe static > > > > > > > > weight of the falling objectcan cause a semi-parabolic increase in > > > > > > > > the KE. Haven't you heard?: Energy IN must = energy OUT! > > > > > > > > NoEinstein > > > > > > > > I have explained this to you dozens of times. I gather that you do not > > > > > > > remember any of those posts, and you do not know how to use your > > > > > > > newsreader or Google to go back and find any of those dozens of times > > > > > > > when it has been explained to you. > > > > > > > > I surmise that you are slipping into dementia, where each day begins > > > > > > > anew, with any lessons learned the previous day forgotten. > > > > > > > > I don't think it's a good use of my time to explain the same thing to > > > > > > > you each day, only to have you retire at night and forget it by > > > > > > > morning, do you? > > > > > > > > PD- Hide quoted text - > > > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - xxein: So the Bible (any interpretation) is reliable too?
From: NoEinstein on 20 May 2010 08:40 On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > KE, > > No, I didn't. I responded directly to your post about ether flow on > muons. > > > because you can't find any place in any text that states: "Work is > > being done even if there is no resistance. (sic) The only requirement > > to have work is that there be a displacement. (sic) Thus, if a hockey > > puck slides twice as far across the ice, twice as much work was done, > > and there is twice as much KE in the puck, even if the ice is > > frictionless. (sic)." > > You are not paying attention. > Remember, PD: I am King of the Hill in science. It isn't "my job" to pay attention to you. > > There is no work if there is no force present, even in the absence of resistance. > I correct you: 'There is no work if there is no force present', AND there is no corresponding resistance > There is work if there is a force present, even in the absence of resistance. > Dear PD: Newton's Laws of Motion require: "For every action there must be an equal and opposite reaction." It is IMPOSSIBLE to apply a force... UNLESS there is a corresponding resisting force! That is like applying the "force" of a skyscraper in a marsh. The maximum static force that can ever be applied is determined by the supporting capacity of the marsheffectively ZERO. My thesis in Architecture was: "Float Foundations for Poor Soils". Essentially, I created structural boats under buildings to support those by the bouyancy of the marsh. For the record, I made an 'A' on that thesis. NoEinstein > > > The resistance on electrons imposed by the ether IS the force being > > measured in those early Lorentz experiments. > > Sorry, what "Lorentz experiments"? > Lorentz experimented, extensively, with trying to measure the velocity and the mass of electrons. Those were inside vacuum tubes, and were speeded up by electromagnets. But, strangely, the electrons encountered exponentially more resistance the closer the velocity came to 'c'. The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years before the M-M experiment. Lorentz shoehorned such to also explain (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED in science. Expect bad results whenever that happens. > > > So, the very experiments > > you inquire about, only need the correct CAUSE, not a new set of > > experiments! > > A correct cause would be accompanied by calculations, indicating the > size of the effect expected due to this cause. Without those > calculations, you've got nothing. > PD, the drag on electrons due to the ether that is clumping in front is very close to Lorentz's Beta. The MATH is close to correct, but the cause is ether drag. Note: Ether can drag electrons and massive objects, but it never drags photons! NoEinstein
From: PD on 20 May 2010 09:42
On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > KE, > > > No, I didn't. I responded directly to your post about ether flow on > > muons. > > > > because you can't find any place in any text that states: "Work is > > > being done even if there is no resistance. (sic) The only requirement > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > puck slides twice as far across the ice, twice as much work was done, > > > and there is twice as much KE in the puck, even if the ice is > > > frictionless. (sic)." > > > You are not paying attention. > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > pay attention to you. :>) Just make sure you tell yourself that each day in the mirror. If you like, please add the line, "And I am the heir to the throne of the kingdom, by birthright." You may also consider adding, if you are feeling confident, "And I am irresistible to women." > > > There is no work if there is no force present, even in the absence of resistance. > > I correct you: 'There is no work if there is no force present', AND > there is no corresponding resistance There is no work done if there is no net force, regardless whether there is resistance or not. > > > There is work if there is a force present, even in the absence of resistance. > > Dear PD: Newton's Laws of Motion require: "For every action there > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > force... UNLESS there is a corresponding resisting force! Newton's 3rd law is not a statement about a resisting force. That is a common mistake by high school students that is corrected in virtually every textbook on the subject. For example, from the high school text that I've been quoting from, page 133: "One important thing to remember about action-reaction pairs is that each force acts on a different object. Consider the task of driving a nail into wood.... To accelerate the nail and drive it into the wood, the hammer exerts a force on the nail. According to Newton's third law, the nail exerts a force on the hammer that is equal to the magnitude of the force that the hammer exerts on the nail. "The concept of action-reaction pairs is a common source of confusion because some people assume incorrectly that equal and opposite forces balance one another and make any change in the state of motion impossible. If the force that the nail exerts is equal to the force the hammer exerts on the nail, why doesn't the nail remain at rest? "The motion of the nail is affected only by the forces acting on the nail. To determine whether the nail will accelerate, draw a free-body diagram to isolate the forces acting on the nail.... The force of the nail on the hammer is not included in the diagram because it does not act on the nail. According to the diagram, the nail will be driven into the wood because there is a net force acting on the nail. Thus, action-reaction pairs do not imply that the net force on either object is zero. The action-reaction forces are equal and opposite, but either object may still have a net force acting on it." > That is > like applying the "force" of a skyscraper in a marsh. The maximum > static force that can ever be applied is determined by the supporting > capacity of the marsheffectively ZERO. My thesis in Architecture > was: "Float Foundations for Poor Soils". Essentially, I created > structural boats under buildings to support those by the bouyancy of > the marsh. For the record, I made an 'A' on that thesis. > NoEinstein > > > > The resistance on electrons imposed by the ether IS the force being > > > measured in those early Lorentz experiments. > > > Sorry, what "Lorentz experiments"? > > Lorentz experimented, extensively, with trying to measure the velocity > and the mass of electrons. Those were inside vacuum tubes, and were > speeded up by electromagnets. But, strangely, the electrons > encountered exponentially more resistance the closer the velocity came > to 'c'. Reference, please. > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > before the M-M experiment. Lorentz shoehorned such to also explain > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > in science. Expect bad results whenever that happens. > > > > So, the very experiments > > > you inquire about, only need the correct CAUSE, not a new set of > > > experiments! > > > A correct cause would be accompanied by calculations, indicating the > > size of the effect expected due to this cause. Without those > > calculations, you've got nothing. > > PD, the drag on electrons due to the ether that is clumping in front > is very close to Lorentz's Beta. Prove that. > The MATH is close to correct, but > the cause is ether drag. Prove that. Show the derivation. Your bluff is called. > Note: Ether can drag electrons and massive > objects, but it never drags photons! NoEinstein |