Newton's law of gravitational attraction
in [Relativity]
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From: PD on 8 Dec 2009 15:13 On Dec 8, 2:02 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > If you will agree that both E = 1/2mv^2 and E = mc^2 / [1 - v^2 / c^2] > ^1/2 go to infinity close to velocity 'c', then you will be > acknowledging that the OUTPUT energy exceeds the INPUT energy > (velocity). The issue here is the terminal energy, not the disparity > in the profile of the two curves. NoEinstein > Well, I'm glad to see that you replaced E=mc^2 with E=mc^2/(1-v^2/ c^2). At least now you're not talking about rest energy. And it's pretty obvious that E=mc^2/(1-v^2/c^2) goes to infinity when v approaches c. The term in the denominator becomes zero, and you know what happens when you divide by zero. On the other hand, I wouldn't dream of saying that E=(1/2)mv^2 goes to infinity when v approaches c. Any child with a 5th grade math education will see that it doesn't. I assume you've at least had a 5th grade math education. Of course, E=(1/2)mv^2 is known to be an approximation that only works when v is much smaller than c, and can't be used reliably at all with v anywhere near c. However, the fact that energy goes to infinity when v approaches c doesn't violate energy conservation at all. It is true, just as it has always been true, that any energy increase is due to work done by an exerted force, and this is what energy conservation requires. It is also true that for any v at all less than c, the energy is not infinite. Big is not infinite. And as we calculated before, the energy required to get a proton up to 0.9999995 c is actually pretty small, about comparable to dropping a nickel from your pants pocket. Moreover, particles that actually do travel at c do so without accelerating them from any lower speed, and so no big amount of energy is required to do that either. PD
From: Autymn D. C. on 9 Dec 2009 00:34 On Dec 3, 8:00Â am, Raymond Yohros <b...(a)birdband.net> wrote: > you are smart pd but what i dont understand is > why you talk so much. > a troll will simply never learn > no matter how hard you try to > make them understand. a troll â them 1 â 2
From: Autymn D. C. on 9 Dec 2009 05:13 On Dec 8, 12:13 pm, PD <thedraperfam...(a)gmail.com> wrote: > Moreover, particles that actually do travel at c do so without > accelerating them from any lower speed, and so no big amount of energy > is required to do that either. They do not, and shed Rntgen and Chèrèncov heat along the way.
From: NoEinstein on 10 Dec 2009 21:59 On Dec 8, 3:13 pm, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD, the... should I still call him the Parasite Dunce?: For ONCE you are acknowledging that SR goes to infinity at 'c'. Thanks for realizing that fact! Pardon me for not copying that God d. Lorentz divisor every time I mention SR. Most encyclopedias only show E = mc^2 when talking about relativity. My bet is that Einstein had no idea whatsoever that Lorentz had screwed up his simple minded notion: "Energy can be converted to mass; and mass can be converted to energy." In no way is the velocity of the velocity of light part of the measure of the total energy within a given mass. Einstein was also totally clueless as to just what kinds of energy will convert to mass. Unfortunately for Einstein, VELOCITY can never be converted to mass! When I say "exponential" I mean at a rate of increase greater than linear. "Quadratic" applies specifically to parabolic rates of increase, like d = t^2. I am well aware that Coriolis's equation, KE = 1/2mv^2 doesn't have the KE going to infinity at velocity 'c'. What I was wanting you to grasp is that both equations increase exponentially (non-linearly). Thus, both equations violate the Law of the Conservation of Energy. That's because the two equations "input" velocity uniformly, but "output" the E or KE exponentially (non- linearly). Effectively, both Coriolis and Einstein created energy in their own minds that was never being imparted by Nature! Coriolis's equation was semi-quadratic, not quadratic. I suspect Einstein dropped the... "1/2" so that people wouldn't realize that he had copied Coriolis. Einstein had no idea whatsoever how much energy resides inside of a given mass. He would leave that up to future generations of scientists to figure out. I hope you feel better for beginning to side with science truths, not just the errant science status quo. The problem for both of us, now, is that those in academia couldn't care less about science truths so long as they can keep repeating their boring memorized lessons, and keep getting NSF grant money to research anything and everything about... Einstein's notions. The desire of young minds to 'understand' relativity has been a cash cow for universities. If I had the power, I would fire 75% of the physics teachers at universities and increase the entry requirement so as to reduce the number of ALL students by 50%. There is far too much emphasis being put on obtaining worthless college degrees. The USA is loosing its Work Ethic > > On Dec 8, 2:02 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > If you will agree that both E = 1/2mv^2 and E = mc^2 / [1 - v^2 / c^2] > > ^1/2 go to infinity close to velocity 'c', then you will be > > acknowledging that the OUTPUT energy exceeds the INPUT energy > > (velocity). The issue here is the terminal energy, not the disparity > > in the profile of the two curves. NoEinstein > > Well, I'm glad to see that you replaced E=mc^2 with E=mc^2/(1-v^2/ > c^2). At least now you're not talking about rest energy. > > And it's pretty obvious that E=mc^2/(1-v^2/c^2) goes to infinity when > v approaches c. The term in the denominator becomes zero, and you know > what happens when you divide by zero. > > On the other hand, I wouldn't dream of saying that E=(1/2)mv^2 goes to > infinity when v approaches c. Any child with a 5th grade math > education will see that it doesn't. I assume you've at least had a 5th > grade math education. > > Of course, E=(1/2)mv^2 is known to be an approximation that only works > when v is much smaller than c, and can't be used reliably at all with > v anywhere near c. > > However, the fact that energy goes to infinity when v approaches c > doesn't violate energy conservation at all. It is true, just as it has > always been true, that any energy increase is due to work done by an > exerted force, and this is what energy conservation requires. > > It is also true that for any v at all less than c, the energy is not > infinite. Big is not infinite. And as we calculated before, the energy > required to get a proton up to 0.9999995 c is actually pretty small, > about comparable to dropping a nickel from your pants pocket. > Moreover, particles that actually do travel at c do so without > accelerating them from any lower speed, and so no big amount of energy > is required to do that either. > > PD
From: PD on 11 Dec 2009 11:24
On Dec 10, 8:59 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Dec 8, 3:13 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the... should I still call him the Parasite Dunce?: For ONCE > you are acknowledging that SR goes to infinity at 'c'. Well, "SR" doesn't go to infinity. There is a Lorentz factor gamma that WOULD go to infinity at c. However, no massive particle ever gets to c, so no massive particle ever sees a Lorentz factor gamma of infinity, and the Lorentz factor gamma does not apply to massless particles. > Thanks for > realizing that fact! Pardon me for not copying that God d. Lorentz > divisor every time I mention SR. Most encyclopedias only show E = > mc^2 when talking about relativity. And this shows the problem with trying to understand relativity by reading encyclopedia articles. It may be of value to you to learn read something of more depth and quality if you want to really understand it better. > My bet is that Einstein had no > idea whatsoever that Lorentz had screwed up his simple minded notion: > "Energy can be converted to mass; and mass can be converted to > energy." In no way is the velocity of the velocity of light part of > the measure of the total energy within a given mass. Einstein was > also totally clueless as to just what kinds of energy will convert to > mass. Unfortunately for Einstein, VELOCITY can never be converted to > mass! Well, you may want to consider how two particles as light as electrons can collide at high velocities and produce a particle that is 90,000 times heavier than the two electrons combined. And yes, this has been done millions of times. > > When I say "exponential" I mean at a rate of increase greater than > linear. Then this is a terminology problem, because that is definitely not what "exponential" means to mathematicians and scientists. This is part of the thing that I've been reminding you of -- that certain words have very specific meanings and you've been using them in a much looser and inaccurate way. > "Quadratic" applies specifically to parabolic rates of > increase, like d = t^2. ANY function that goes as the square of another quantity is a quadratic function. So, for example, (1/2)mv^2 is quadratic in v. > I am well aware that Coriolis's equation, KE > = 1/2mv^2 doesn't have the KE going to infinity at velocity 'c'. What > I was wanting you to grasp is that both equations increase > exponentially (non-linearly). Quadratic is not exponential. Neither of them is linear, but that doesn't make (1/2)mv^2 exponential. I'm asking you to learn how to use terms appropriately. > Thus, both equations violate the Law of > the Conservation of Energy. That's because the two equations "input" > velocity uniformly, but "output" the E or KE exponentially (non- > linearly). That's what ANY quadratic function does. For example, in d=(1/2)at^2, you input time uniformly and the output distance emerges non-linearly. That isn't a violation of anything. > Effectively, both Coriolis and Einstein created energy in > their own minds that was never being imparted by Nature! Nonsense. All the energy is delivered by work, which comes from force applied through a distance. There is no energy that comes from no place. It's all accounted for. > > Coriolis's equation was semi-quadratic, not quadratic. "Semi-quadratic" is a meaningless word. The area of a circle is quadratic in the diameter: A = (1/4)*pi*d^2. The presence of the constants out in front doesn't change that fact. A function can be linear in one variable and quadratic in another. For example, d=(1/2)a*t^2 is linear in acceleration a, and quadratic in the time t. All this is 6th grade arithmetic, John. Perhaps you have forgotten your 6th grade arithmetic. > I suspect > Einstein dropped the... "1/2" so that people wouldn't realize that he > had copied Coriolis. Einstein had no idea whatsoever how much energy > resides inside of a given mass. Please, John, stop making stuff up. It's transparent, and it makes you look bad. > He would leave that up to future > generations of scientists to figure out. > > I hope you feel better for beginning to side with science truths, not > just the errant science status quo. The problem for both of us, now, > is that those in academia couldn't care less about science truths so > long as they can keep repeating their boring memorized lessons, and > keep getting NSF grant money to research anything and everything > about... Einstein's notions. The desire of young minds to > 'understand' relativity has been a cash cow for universities. If I > had the power, I would fire 75% of the physics teachers at > universities and increase the entry requirement so as to reduce the > number of ALL students by 50%. There is far too much emphasis being > put on obtaining worthless college degrees. The USA is loosing its > Work Ethic You owe your livelihood to your college degree, John. You would not be able to work at your profession without it. > > > > > On Dec 8, 2:02 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > If you will agree that both E = 1/2mv^2 and E = mc^2 / [1 - v^2 / c^2] > > > ^1/2 go to infinity close to velocity 'c', then you will be > > > acknowledging that the OUTPUT energy exceeds the INPUT energy > > > (velocity). The issue here is the terminal energy, not the disparity > > > in the profile of the two curves. NoEinstein > > > Well, I'm glad to see that you replaced E=mc^2 with E=mc^2/(1-v^2/ > > c^2). At least now you're not talking about rest energy. > > > And it's pretty obvious that E=mc^2/(1-v^2/c^2) goes to infinity when > > v approaches c. The term in the denominator becomes zero, and you know > > what happens when you divide by zero. > > > On the other hand, I wouldn't dream of saying that E=(1/2)mv^2 goes to > > infinity when v approaches c. Any child with a 5th grade math > > education will see that it doesn't. I assume you've at least had a 5th > > grade math education. > > > Of course, E=(1/2)mv^2 is known to be an approximation that only works > > when v is much smaller than c, and can't be used reliably at all with > > v anywhere near c. > > > However, the fact that energy goes to infinity when v approaches c > > doesn't violate energy conservation at all. It is true, just as it has > > always been true, that any energy increase is due to work done by an > > exerted force, and this is what energy conservation requires. > > > It is also true that for any v at all less than c, the energy is not > > infinite. Big is not infinite. And as we calculated before, the energy > > required to get a proton up to 0.9999995 c is actually pretty small, > > about comparable to dropping a nickel from your pants pocket. > > Moreover, particles that actually do travel at c do so without > > accelerating them from any lower speed, and so no big amount of energy > > is required to do that either. > > > PD |