Capiert Posted November 23, 2020 Share Posted November 23, 2020 (edited) wrt the initial_speed vi; & relativistic if you (only) swap (=reverse) its perspective from wrt earth's minimum_speed 0 to wrt earth's maximum_speed c. Reversing the Kinetic_energy (perspective) KE=m*v*va (wrt Earth’s_speed), *(-1)' gives -KE'=m'*(-)*v'*va' (wrt light’s_speed) which is already relativistic. Please notice if I let c=v+u, u=c-v, u=-v', (prime_symbol ‘ is wrt light’s_speed) then the (negative) speed_difference (wrt light’s_speed), is -v'=-(vf'-vi')=vi'-vf' & the average_(accelerated)_speed (wrt light’s_speed), is va'=(vi'+vf')/2 for initial_speed vi' (=-0'=v_min, wrt light’s_speed, or v_max=c wrt Earth’s_speed) & final_speed vf' (wrt light’s_speed). -KE'=m'*(vi'-vf')*((vi'+vf')/2), combine brackets -KE'=m'*((vi'^2)-(vf'^2))/2, let vi’=-0’=c -KE'=m'*((vi'^2)-(vf'^2))/2, bring c^2 out from the brackets (c^2)/(c^2)=1/1=1 -KE'=m'*(c^2)*(1-(vf'^2)/(c^2))/2, let gamma’^2=(1-(vf'^2)/(c^2)) -KE'=m'*(c^2)*gamma’*gamma’/2. That equation has (=contains) "half" of (DePretto's 1903, vis_viva), Energy E=m*(c^2) & 2 (Fitzgerald_Lorentz, relativistic_contraction similar) coefficients (named) gamma_primed gamma’=(1-(vf’^2)/(c^2))^0.5 where the rest_mass m=m' is (the same, =identical) constant(_variable) for either perspective. I.e. Conservation of mass com (wrt Earth’s_speed) m=m' (wrt light’s_speed). Speeds are the variables (instead). 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1419 PS Wi_(stripped).pdf 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1356 PS Wi__with old _Excel_ formula text_(mix).pdf Edited November 23, 2020 by Capiert Link to comment Share on other sites More sharing options...

swansont Posted November 23, 2020 Share Posted November 23, 2020 Yes, we know that KE is relative, since it depends on v, which is relative. 32 minutes ago, Capiert said: wrt the initial_speed vi; & relativistic if you (only) swap (=reverse) its perspective from wrt earth's minimum_speed 0 to wrt earth's maximum_speed c. Reversing the Kinetic_energy (perspective) KE=m*v*va (wrt Earth’s_speed), *(-1)' gives -KE'=m'*(-)*v'*va' (wrt light’s_speed) which is already relativistic. The term in KE is v^2, which is v*v. It is not two different speeds multiplied together. "wrt light's speed" is nonsensical. Speed is relative to some frame of reference, and light does not have an inertial frame of reference. It is also not inherently tied into the earth's speed. KE is relativistic if you (need to) use the relativistic form of the equation 32 minutes ago, Capiert said: Please notice c=v+u, No, it's not (further nonsense deleted; there's no point) Link to comment Share on other sites More sharing options...

Sriman Dutta Posted November 23, 2020 Share Posted November 23, 2020 41 minutes ago, Capiert said: wrt the initial_speed vi; & relativistic if you (only) swap (=reverse) its perspective from wrt earth's minimum_speed 0 to wrt earth's maximum_speed c. Reversing the Kinetic_energy (perspective) KE=m*v*va (wrt Earth’s_speed), *(-1)' gives -KE'=m'*(-)*v'*va' (wrt light’s_speed) which is already relativistic. Please notice c=v+u, u=c-v, u=-v', (prime_symbol ‘ is wrt light’s_speed) the (negative) speed_difference (wrt light’s_speed), is -v'=-(vf'-vi')=vi'-vf' & the average_(accelerated)_speed (wrt light’s_speed), is va'=(vi'+vf')/2 for initial_speed vi' (=-0'=v_min, wrt light’s_speed, or v_max=c wrt Earth’s_speed) & final_speed vf' (wrt light’s_speed). -KE'=m'*(vi'-vf')*((vi'+vf')/2), combine brackets -KE'=m'*((vi'^2)-(vf'^2))/2, let vi’=-0’=c -KE'=m'*((vi'^2)-(vf'^2))/2, bring c^2 out from the brackets (c^2)/(c^2)=1/1=1 -KE'=m'*(c^2)*(1-(vf'^2)/(c^2))/2, let gamma’^2=(1-(vf'^2)/(c^2)) -KE'=m'*(c^2)*gamma’*gamma’/2. That equation has (=contains) "half" of (DePretto's 1903, vis_viva), Energy E=m*(c^2) & 2 (Fitzgerald_Lorentz, relativistic_contraction similar) coefficients (named) gamma_primed gamma’=(1-(vf’^2)/(c^2))^0.5 where the rest_mass m=m' is (the same, =identical) constant(_variable) for either perspective. I.e. Conservation of mass com (wrt Earth’s_speed) m=m' (wrt light’s_speed). Speeds are the variables (instead). 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1419 PS Wi_(stripped).pdf 24.61 kB · 2 downloads 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1356 PS Wi__with old _Excel_ formula text_(mix).pdf 27.01 kB · 0 downloads Suggestion 1: Know what you are talking Suggestion 2: Know physics Link to comment Share on other sites More sharing options...

Capiert Posted November 23, 2020 Author Share Posted November 23, 2020 (edited) 1 hour ago, swansont said: Yes, we know that KE is relative, since it depends on v, which is relative. The term in KE is v^2, which is v*v. It is not two different speeds multiplied together. If v=vf-vi & KE=m*((vf^2)-(vi^2))/2 & ((vf^2)-(vi^2))/2=(vf-vi)*(vf+vi)/2, & va=(vi+vf))/2 then why is NOT ((vf^2)-(vi^2))/2=v*va? & How can KE possibly be m*v*v/2? When should be KE=m*(((v^2)/2)+v*vi), instead. Quote "wrt light's speed" is nonsensical. Speed is relative to some frame of reference, Does a (virtual) moving_frame need (to have) mass? Quote and light does not have an inertial frame of reference. I did NOT declare an inertial frame, I used a (virtual, non_inertial) perspective. (I do NOT have to touch a(ny) thing.) I simply stated a speed_difference u=c-v between the maximum possible speed c & the moving_point's speed v. u is only the (complementary_) speed needed to add to v (in order) to be(come) c. That is simple algebra. No magic. Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. I see no problems & get astounding results (for constant_speeds). It's values are different from yours by a factor of -1/2. What do you mean by "inertial"_frame? 1_stone (1905) calculated the photons mass m=KE/(c^2). Quote It =? (light? or speed?) Quote is also not inherently tied into the earth's speed. If you mean speed, I was only using the earth's speed as an example. Quote KE is relativistic if you (need to) use the relativistic form of the equation Considering my results, I don't (believe) I need your SR form. Perhaps you can convince me otherwise? What I wanted to say is: Please notice (If I let) c=v+u Quote No, it's not Why NOT? Quote (further nonsense deleted; there's no point) I can't comment your deletions. My point is "virtual". Standard definition: no size. & moving (at constant speed). The (so_called) Rest_mass is inherent (=included) in (my) KE's opposite_perspective (-KE'); & (thus it) does NOT have to be added extra (by haphazard random guessing). Edited November 23, 2020 by Capiert Link to comment Share on other sites More sharing options...

swansont Posted November 23, 2020 Share Posted November 23, 2020 Just now, Capiert said: If v=vf-vi & KE=m*((vf^2)-(vi^2))/2 & ((vf^2)-(vi^2))/2=(vf-vi)*(vf+vi)/2, & va=(vi+vf))/2 then why is NOT ((vf^2)-(vi^2))/2=v*va? You noted va is "relative to earth's speed" and that's nonsense. I can find the KE in any frame. It is, as you note, relative. The earth is not inherent to the equation. Just now, Capiert said: & How can KE possibly be m*v*v/2? dW = Fdx for a constant force Integrate that and you get 1/2 mv^2 The work done is the change in kinetic energy. energy is conserved. Just now, Capiert said: When should be KE=m*(((v^2)/2)+v*vi), instead. Why "should" it be that? What valid physics principle is this based on? Just now, Capiert said: Does a (virtual) moving_frame need (to have) mass? A frame of reference is just coordinate systems. Just now, Capiert said: I did NOT declare an inertial frame, If you are applying Newton's laws you need to be in an inertial frame. Just now, Capiert said: I used a (virtual, non_inertial) perspective. (I do NOT have to touch a(ny) thing.) I simply stated a speed_difference u=c-v between the maximum possible speed c & the moving_point's speed v. u is only the (complementary_) speed needed to add to v (in order) to be(come) c. That is simple algebra. No magic. No valid physics, either. Just now, Capiert said: Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. We already have a solution. It's special relativity. Just now, Capiert said: I see no problems & get astounding results (for constant_speeds). It's values are different from yours by a factor of -1/2. Then your answers are wrong. Just now, Capiert said: What do you mean by "inertial"_frame? Non-accelerating. One in which Newton's laws would work. Just now, Capiert said: 1_stone (1905) calculated the photons mass m=KE/(c^2). No, that's the mass equivalent of the energy. Photons are massless. Just now, Capiert said: =? (light? or speed?) If you mean speed, I was only using the earth's speed as an example. Then it should not appear in a general equation. Just now, Capiert said: Considering my results, I don't (believe) I need your SR form. Perhaps you can convince me otherwise? Does your formulation give correct answers? If I drop a mass from some height, how fast will it be moving when it hits the floor? You can solve this with energy, or with kinematics. You can also compare to experiment. The factor of 1/2 will matter. Link to comment Share on other sites More sharing options...

Capiert Posted November 23, 2020 Author Share Posted November 23, 2020 (edited) 4 hours ago, swansont said: You noted va is "relative to earth's speed" and that's nonsense. Sorry, but I only live here (on Earth); I'm going to no other planet. So Earth was my best example. va is relative to the initial_speed vi which for this example is the earth's(_speed, which could be anything (reasonable)). So if we are traveling at the same speed as the Earth then that initial_speed vi=0 is zero for at rest wrt on Earth. Quote I can find the KE in any frame. It is, as you note, relative. The earth is not inherent to the equation. I agree. I only used it (=Earth) as an (easy, simple) example to "try" & get the message across. The Work_energy's element Quote dW = Fdx for a constant force F multiplied by the distance element dx. Quote Integrate that and you get 1/2 mv^2 How do we know that acceleration a=x/(t^2)? If the momentum mom=m*v & mom^2=(m*v)^2. I DON'T see the coherence. Quote The work done is the change in kinetic energy. energy is conserved. WE=F*d. I would prefer to say the mass*Energy m*E is conserved, instead. NOT the energy. Capiert said: Which should be KE=m*(((v^2)/2)+v*vi), instead. Quote Why "should" it be that? Algebra. The average_speed is va=(vi+vf)/2. *2 2*va=vi+vf, swap sides vi+vf=2*va, -vi vf=2*va-vi. The speed_difference is v=vf-vi, swap sides vf-vi=v, +vi vf=v+vi. Both vf's can be equated also vf=vf v+vi=2*va-vi, -vi v=2*va-2*vi v=2*(va-vi), /2 2*v=(va-vi), +vi 2*v+vi=va, swap sides va=2*v+vi. Or expanding the kinetic_energy KE=m*((vf^2)-(vi^2))/2, ((vf^2)-(vi^2))=(vf-vi)*(vf+vi) KE=m*(vf-vi)*((vf+vi)/2), (v=vf-vi & thus) vf=v+vi (Notice: KE=m*v*va, for v=(vf-vi) & va=(vi+vf)/2 but continue from above for below) KE=m*(v+vi-vi)*((v+vi+vi)/2), vi-vi=0 & vi+vi=vi*2 KE=m*(v)*((v+vi*2)/2), KE=m*(v*v+v*vi*2)/2, expand KE=m*((v*v/2)+(v*vi)). I derived (=equated) that above (for you also further above) via substitution, & I see no error in my equations, thus I must conclude they are correct. They are simple, algebra. Quote What valid physics principle is this based on? Mechanics: Gravity's (free_fall, linear_acceleration a=g) fallen_height h=hf-hi=vi*t+g*t*t/2 for time t & initial_speed vi gives final_speed vf=((vi^2)+g*h*2)^0.5 Capiert said: Does a (virtual) moving_frame need (to have) mass? Quote A frame of reference is just coordinate systems. I'( wi)ll assume, (your answer is) no. Quote If you are applying Newton's laws you need to be in an inertial (=constant_speed) Quote frame. No valid physics, either. Sorry, my mistake I used a (virtual, constant_speeed=inertial) perspective, instead of non_inertial=accelerating frame. Thanks for clearing that. Capiert said: Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. Quote We already have a solution. It's special relativity. It looks questionable (=doubtful) to me. The author tried to get rid of it (e.g. avoid it) in 1920 chapter 22. He didn't recommend it (anymore); NOR did the Nobel committee give him a prize for Relativity. (Why would they NOT if it were really important for Physics?) I guess they also had their doubts. Capiert said: I see no problems & get astounding results (for constant_speeds). It's values are different from yours by a factor of -1/2. Quote Then your answers are wrong. How do you know that? How do you know if SR is correct? E.g. When its author discouraged its usage, 1920 chapter 22. Capiert said: What do you mean by "inertial"_frame? Quote Non-accelerating. One in which Newton's laws would work. Capiert said: 1_stone (1905) calculated the photons mass m=KE/(c^2). Quote No, that's the mass equivalent of the energy. What do you mean by that? Newton believed photons were particles. Quote Photons are massless. That means small (=less) mass; NOT NO_mass. Quote No, that's the mass equivalent of the energy. That'( energy i)s what he used for a photon's mass in 1905. What do you mean by "No"? Are you trying to say E#m*(c^2). Mass is NOT (a kind of) energy? To me it seems rather obvious what things are, with an equation (=equality). I do NOT understand why you shirk (~avoid) from using math for physics, to get to the bottom of things. What prevents the (mass versus energy) connection? Light's_speed squared c^2? I.e. Mass & energy are NOT identical. Quote Photons are massless. If you are implying (=saying) NO_mass, then how do you know that? Or are you implying photos are (sort of like) energy, but not yet mass? How can you possibly have energy, without mass? Mass is only a construct, a coefficient (=factor). Capiert said: =? (light? or speed?) If you mean speed, I was only using the earth's speed as an (easy) example. Quote Then it should not appear in a general equation. Sorry. Quote Does your formulation give correct answers? I suspect yes. & I can convert to your SR answers with the factor -2; & visa versa from SR values to mine with the factor -1/2. Quote If I drop a mass from some height, how fast will it be moving when it hits the floor? Answer: the final_speed is vf=((vi^2)+g*h*2)^0.5 . Quote You can solve this with energy, or with kinematics. You can also compare to experiment. The factor of 1/2 will matter. Edited November 23, 2020 by Capiert Link to comment Share on other sites More sharing options...

MigL Posted November 23, 2020 Share Posted November 23, 2020 Your posts give me headaches, Capiert. And its not just the formatting ... 1 Link to comment Share on other sites More sharing options...

Sensei Posted November 23, 2020 Share Posted November 23, 2020 (edited) Physicists are making shortcut. "massless" means in the reality "having no rest-mass". Rest-mass is invariant mass. Mass which quantum object has in its rest-frame of reference. https://en.m.wikipedia.org/wiki/Rest_frame ps. Learn some physics... start from primary school physics textbooks.. Edited November 23, 2020 by Sensei Link to comment Share on other sites More sharing options...

swansont Posted November 23, 2020 Share Posted November 23, 2020 3 hours ago, Capiert said: Sorry, but I only live here (on Earth); I'm going to no other planet. Physics doesn’t care. It has to work under a wide range of conditions 3 hours ago, Capiert said: So Earth was my best example. va is relative to the initial_speed vi which for this example is the earth's(_speed, which could be anything (reasonable)). Then you aren’t doing physics, or anything particularly useful 3 hours ago, Capiert said: The Work_energy's element F multiplied by the distance element dx. How do we know that acceleration a=x/(t^2)? Anyone who can do calculus should be able to figure out, from the definition of acceleration, that x = 1/2 at^2 for something starting from rest. 3 hours ago, Capiert said: If the momentum mom=m*v & mom^2=(m*v)^2. I DON'T see the coherence. Momentum isn’t a necessary element of this analysis 3 hours ago, Capiert said: WE=F*d. I would prefer to say the mass*Energy m*E is conserved, instead. I don’t care what you prefer, if it has no basis in physics. 3 hours ago, Capiert said: & I see no error in my equations, thus I must conclude they are correct. I’ve pointed them out 3 hours ago, Capiert said: They are simple, algebra. Mechanics: Gravity's (free_fall, linear_acceleration a=g) fallen_height h=hf-hi=vi*t+g*t*t/2 for time t & initial_speed vi gives final_speed vf=((vi^2)+g*h*2)^0.5 Assume it starts from rest. How did you solve for t? 3 hours ago, Capiert said: Capiert said: Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. If v<<c, It works just fine 3 hours ago, Capiert said: How do you know that? How do you know if SR is correct? It matches experiments 3 hours ago, Capiert said: Newton believed photons were particles. That means small (=less) mass; NOT NO_mass. Photons require relativity 3 hours ago, Capiert said: That'( energy i)s what he used for a photon's mass in 1905. What do you mean by "No"? Are you trying to say E#m*(c^2). Mass is NOT (a kind of) energy? I never said this 3 hours ago, Capiert said: I do NOT understand why you shirk (~avoid) from using math for physics, to get to the bottom of things. Oh, FFS. I’m begging you to do math, which you’ve refused to do. I’m not making the idiotic claims. The burden of proof is on you. 3 hours ago, Capiert said: If you are implying (=saying) NO_mass, then how do you know that? Because relativity works. 3 hours ago, Capiert said: Or are you implying photos are (sort of like) energy, but not yet mass? How can you possibly have energy, without mass? Because relativity works Link to comment Share on other sites More sharing options...

Capiert Posted November 24, 2020 Author Share Posted November 24, 2020 10 hours ago, MigL said: Your posts give me headaches, Capiert. And its not just the formatting ... I sympathize with you. I've been wracking my brains on this stuff for years trying to make some sense of it. 10 hours ago, Sensei said: Physicists are making shortcut. "massless" means in the reality "having no rest-mass". I guess you missed the point Sensei. My (perspective_reversal) calculations show that the Rest_mass is inherent, (meaning it'( i)s) (already) in the KE('s reversed perspective). I don't need to add it (rest_mass) randomly because somebody thought it was forgotten & needed. 10 hours ago, Sensei said: Rest-mass is invariant mass. That's right, mass is conserved, conservation of mass com. The mass does NOT change (in my calculations), only the speeds (change). & I haven't done (=changed) anything; (except) only the perspective. & I get (Fitzgerald)_Lorentz_contraction similar results with (only) algebra. So I have to ask: Why do you think you (might) have to add rest_mass additionally (extra) ((randomly) out of the blue ((or a hat) (like a magician)))? (That's ridiculous.) Why should the rest_mass be in your relativistic equations "twice"? E.g. Originally (inherently); & then again (randomly, by you) because you thought you missed (=forgot) it, before. I DON'T need to add rest_mass to my KE calculations because it's already there. You however, think you do (need to do that). 10 hours ago, Sensei said: Mass which (a) quantum object has in its rest-frame of reference. https://en.m.wikipedia.org/wiki/Rest_frame The rest_frame is when the initial_speed vi=0 [m/s]. 10 hours ago, Sensei said: ps. Learn some physics... start from primary school physics textbooks.. I'( have) a bundle of them some read several times. But they DON'T solve my problems (=paradoxes). They (=those books) only "help" me solve them. Those answers are NOT in the books. In fact the books are (sometimes) misleading. Too much NONSENSE makes my brain shut off. (I CAN'T tolerate it.) So I have to take my time & unravel the puzzle. I have to try (new) alternatives; NOT (old) failures. All your textbooks only leed to dark_energy (=NONSENSE, errorful calculations). I'm sorry but I think you are behind the times judging me so. I'm searching for a recalibration to bring physics up to date, instead of the scattered mess it is in now. You guys (your team) is either going to help me, or not. Link to comment Share on other sites More sharing options...

Dagl1 Posted November 24, 2020 Share Posted November 24, 2020 @Capiert Could you do the calculations people ask you to? If you want to convince people here that your ideas are valid, show it with numerical examples. Show the formulas you use and the numbers you put in. Words are great, but eventually you will need to show math, and I really think this is that moment (or it was a few posts ago already). You make many claims, so why not just do the simple thing, use your own formulas/ideas to calculate some stuff, showing what you do and why. It shouldn't be hard. Oh and, if your answer is different from that of standard physics, please don't you know.... claim that all the regular physics must therefore be wrong, because we use the regular physics to make basically everything work, and it would be honestly absurd and extremely ignorant if you do do the calculations, find that your answer differs from that of regular formulas, and then just refute the regular formulas. I of course am sure that you won't do that, but I thought it would just say it anyway! Goodluck! Link to comment Share on other sites More sharing options...

joigus Posted November 24, 2020 Share Posted November 24, 2020 I join my voice to the chorus. And please tell me what a 'virtual reference system' is. Link to comment Share on other sites More sharing options...

Sensei Posted November 24, 2020 Share Posted November 24, 2020 (edited) 8 hours ago, Capiert said: In fact the books are (sometimes) misleading. Too much NONSENSE makes my brain shut off. ...."primary school natural science textbooks are nonsense"... ??? That (nonsensical) statement explains everything... ps. make photo of page were you think is written nonsense in your example science textbook and attach in the post in new thread with explanations why do you think it is nonsense and somebody here will explain to you why you are wrong... Edited November 24, 2020 by Sensei Link to comment Share on other sites More sharing options...

swansont Posted November 24, 2020 Share Posted November 24, 2020 8 hours ago, Capiert said: I'm searching for a recalibration to bring physics up to date, instead of the scattered mess it is in now. Your failure to understand physics is not evidence of a problem with physics. 8 hours ago, Capiert said: You guys (your team) is either going to help me, or not. ! Moderator Note We have tried, but you resist the help. People point out errors, and you insist they aren’t errors. But you never seem to be able to show how your physics actually applies to any problem, despite requests that you show this. It’s happened again in this thread. You shirk the responsibility of the burden of proof, and that’s not acceptable. Your approach of repeatedly asserting things without valid justification is against our rules (both in general and specifically for Speculations) I’m not going to let any more time and effort be wasted, as we’ve been down this path before. 1 Link to comment Share on other sites More sharing options...

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