# Capiert

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486

1. ## Gravity g's, missing term(?)

That surprises me because I can read the following (from above) The fallen height (taken from a graph), h=vi*t+g*(t^2)/2 is only 2 terms: the constant (initial_)speed term vi*t; & the accelerating term g*t*t/2. & also in my posted (above, 2nd of 2) .pdf's (above). https://www.scienceforums.net/applications/core/interface/file/attachment.php?id=23252 Surely you will be able to determine when that (equation) was posted. Good! (for that product). But now for the quotient -2*vi/t in g (alone). Someone to assist me in getting the WRONGLY strikedthrough text (above) to be NOT strikedthrough, anymore. Would you help? But to answer your question: Again, clearly -2*vi/t was NOT present in g (before my threads). I have NOT seen that term in books; but algebraically it is correct (as to how I gave it to you). Science should confirm itself, but it seems (to me) you (might) probably prefer NOT that the g equation be intact, when rearranged (e.g. if you leave out that term -2*vi/t)? (Is that possible?) I have only moved the acceleration g to the left side, & all other terms to the right side. That should give the correct g (equation). (Otherwise, the (general) equality is either destroyed, or distorted.) It'(=What I have done i)s NOT complicated.
2. ## Gravity g's, missing term(?)

From g. You obviously disagree, otherwise you would NOT have asked. Why NOT? It works. (As far as I know) I have shared the equation: the fallen height (taken from a graph), h=vi*t+g*(t^2)/2. Typical is NOT all cases, but instead most (e.g. a majority). For me it was an unknown that I wanted to find, e.g. experimentally. Can someone please help me with unstrikethrough my text (above)?

4. ## x+j*x#r

I received the telegraph (message) instructions: "go east 3 strides, then go 4 strides north. Where are you (now)?" Answer: 5 strides east. Signed Virtual.

6. ## Kinetic_Energy is already relative

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. I agree. I only used it (=Earth) as an (easy, simple) example to "try" & get the message across. The Work_energy's element F multiplied by the distance element dx. 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. 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. 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. 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? I'( wi)ll assume, (your answer is) no. (=constant_speed) 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. 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. 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? Capiert said: 1_stone (1905) calculated the photons mass m=KE/(c^2). What do you mean by that? Newton believed photons were particles. That means small (=less) mass; NOT NO_mass. 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. 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. Sorry. 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. Answer: the final_speed is vf=((vi^2)+g*h*2)^0.5 .
7. ## Kinetic_Energy is already relative

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. Does a (virtual) moving_frame need (to have) mass? 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). =? (light? or speed?) If you mean speed, I was only using the earth's speed as an example. 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 Why NOT? 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).
8. ## Kinetic_Energy is already relative

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

I'm glad you see it too. Thanks.

It does NOT have to be a car. Surely simpler contraptions exist to test. Confirmation. Yes (& reliability). I'm interested in at least 1 formula I could rely on. From there I could adapt & fine_tune to a general formula if possible. How many things did Newton (have to) go thru til he could (finally) settle on F=m*a? "The proof is in the pudding." I'm surprised this non_linear acceleration (formula, stuff) is so rare (for something (that is) (supposed to be) so universal). Do you tell students to learn (useless) linear_acceleration in your universities because it's NOT universal? (Surely) I hope NOT. But that's what's taught. Free_fall's ("linear" acceleration) is pretty common throughout the whole universe. So linear acceleration is NOT so uncommon. Even if you don't dare to call it universal. I'm just curious about how non_linear (acceleration) is proportioned to linear_acceleration, & the methods of confirmation. Is my curiousity justified, in a dark energy dilemma?

I want a reliable formula that I can equate to for comparisons. I can make up anything I want, but it doesn't mean it's accurate. Your team surely has better experience there, for real measurements. (I'm just guessing.)

(Yes) but please give me a formula example so I can get feel for the number_values.

Then I suppose your syntax v is my syntax vf. Is that true? Please elaborate, if not. Just attempting to be thorough(ly defined), which is NOT what I could claim everybody does. Capiert: I thought that (linear acceleration) was understood, that everybody knows that. It's clearly stated in the 3rd line (at the intro). Surely you have overseen my definition.? I gave you a specific example: the linear acceleration. But now you have made me curious. Please tell me more, e.g. an example (or more), for further (kinds of acceleration).

15. ## Period versus (inverse) rotation_speed?

Period versus 1/(vc), frequency versus vc ? I find it very confusing that rotation_speed vc=2*Pi*r/t looks (very similar, &) proportional to frequency f (cycles per second; or (1) cycle (e.g. circle) per time (in units of seconds, or less than a second)); & inverted_rotation_speed 1/(vc)=t/(2*Pi*r) (the time "per" cycle_or_circle_(circumference)) looks proportional to Period T=1/f which is inverted_frequency! (Thus units are suppose to be inverted.) So (confusing) that I must start with what we know. In fact I am so confused, that I know nothing except the rotation_speed vc=2*Pi*r/t (which is obvious), & the frequency f=1/T inverse_period relation (definition). The rest seems (to me) like non_sense! Thus I can derive everything (I need) from what we know: The circumferential_speed vc=cir(cle)/t(ime) vc=2*Pi*r/t. ~f. That looks like cycles (=circles) "per" second =[cps]=[c/s]. (Units in square brackets.) I.e. That looks (very much) like frequency! (Although you guys are only interested in the inverse(=1_divided_by)_time 1/t for the "number" value of f (without units, =excluding units [c/s]). We also know the "inverse" circumferential_speed 1/(vc)=t(ime)/cir(cle) 1/(vc)=t/(2*Pi*r). ~T. That looks like the time_(in seconds)_"per"_cycle(=circle) =[spc]=[s/c]. I.e. That looks (very much) like the Period! (Although you guys are only interested in time t for the "number" value of T (without units, =excluding units [s/c]), because you have defined frequency f=1/T as inverse_period. So the time ("for" a cycle) is t=(2*Pi*r)/(vc) time=(circle (or cycle)) "per" rotation_speed). That is the (number) value for what you guys call period (T). Your units are seconds (because all circles cancel). It's inverse 1/t=vc/(2*Pi*r) inverted_time=rotation_speed "per" (1) circle (or cycle), gives the number (value) for what you (guys & gals) call frequency (f). In other words, stripped of the cycle(s). You now have units [Hz] Herz (=hurts! Ouch!)

study & discuss their (great?) work, energy WE=m*a*d using linear_acceleration a (=v/t, =2*((vi/t)+(d/(t*t)), =F/m) & the force(d) distance d (=va*t) of a mass m (=F/a). That (linear accelerated) force is F=m*a, where the average (accelerated) speed (velocity) va (=(vi+vf)/2), & speed (velocity, difference) is v=vf-vi, for final_speed vf, minus initial_speed vi. ((Even) although it would also be possible to use factoring (an initial unit_speed of 1 m/s), instead). The (moving) kinetic_energy is KE=m*v*va, (pronounced key), or KE=m*((v^2)/2)+v*vi) using initial_speed vi, or KE=m*((v^2)/2)-v*vf) with (respect to) final_speed vf. The potential_energy PE=m*g*h (pronounced pea, as in pee) is a mass m g accelerated (fallen, multiplied by) height h=d distance. PE=Wt*h is (the force F=) weight Wt (=m*g) multiplied by height h. (No distance fallen_able, is no potential_energy.) Equating (both energies) PE=KE (pronounced peek) m*g*h=m*v*va without mass (divided from both sides) is g*h=v*va is only linear_acceleration a*d=v*va a*d=(vf-vi)*((vi+vf)/2) of the distance d, a*d=((vf^2)-(vi^2))/2, *2 a*d*2=(vf^2)-(vi^2), (pronounced add too). That's standard mechanics. (Anything wrong there?) Swapped sides (vf^2)-(vi^2)=a*d*2, +(vi^2) (vf^2)=(vi^2)+a*d*2, ^0.5 vf=((vi^2)+a*d*2)^0.5 is the final_speed (velocity) (of linear_acceleration). The speed difference (velocity) v=vf-vi, vf=((a*d*2+(vi^2))^0.5) v=((a*d*2+(vi^2))^0.5)-vi is final_speed (velocity) vf, minus the initial_speed (velocity) vi. Momentum('s impulse) mom=m*v is mass m multiplied by (the accelerated) speed difference (velocity) v. mom=m*(((a*d*2+(vi^2))^0.5)-vi) mom=m*((a*d*2+(vi^2))^0.5)-m*vi is the final_momentum momf=m*vf (pronounced mumf, as in eating fast), minus the the initial_momentum momi=m*vi (pronounced mommy). The average momentum moma=m*va (pronounced mama) is the mass m multiplied by the average (accelerated) speed (velocity) va=(vi+va)/2. Any questions? Since KE & mom only use mass & speeds, & all energy can be equated to KE, then it seems imaginable to equate all energies into impulse(s) of mom=F*t. Mom=E/va. Mother nature (pronounce Eva (the German Eve); or else Elva from James Cameron's Avatar).
17. ## Why does this ball go faster than c?

I'm sorry, I did not make myself clear enough. I hope you do not make random answers when you answer. (=wide) (which they were) 'any' speed except c, but its true Michelson's sketch is an exageration. Was it v=c/10? I thought the point was (for) the M&M experiment of 1888, fig 2. Yes, so why aren't we (still) talking about the real experiment? Yes, any reasonable speed. Does that mean trial & error? (E.g. decide (=arbitrate) from random values.) E.g. It's (=the mirror has) got to be wide (=big) enough.? They were 5 cm wide (& made of metal). Quite the contrary, (I'm interested in specific details). If the mirror is made too small, wouldn't a small (& coherent enough) light beam get cutoff? E.g. Wouldn't a (vertically) 90 degree incident ray (either) partially (or never) hit a too small mirror b that (mirror b) is moving away (to the right) at v? How can you change 90 degree x,y,z orthogonality, by changing reference systems? E.g. Formula?
18. ## Why does this ball go faster than c?

I'm sorry that did not answer my questions, e.g. y/n's.
19. ## Why does this ball go faster than c?

In which case, wasn't that angle decided by the draftsperson (that was) sketching? (because in that case a new angle would be needed for every different speed v ?); or else if the upper mirror is long enough to be hit at 90 degrees, & reflection occurs in a wavefront, wouldn't the diagonal beam, where it's twice Michealson's (drafted) reflected angle, hit the semi_mirror in perfect sync with the horizontal path? i.e. identical delays for both (vertical & horizontal) paths? (I mean the (reflected) angle in question is <1 degree (difference, wrt incidence at 90 degrees) for the intended speed v (~c/1000). Surely a slur distortion, or front might account for that interception (fringe intensity) at the semi_mirror.?)
20. ## Elastic collision?

"optimism" denied. Because I had not recognized you were impling WE=F*d is elastic; instead of stored momentum might be the (2*a*d)^0.5 part of the momentum mom=m*(((vi^2)+2*d*a)^0.5)-vi.? Sorry, (it's vague, general) typo, reminder. That comment is (awkwardly) in the wrong place. Is that a bit better? So momentum is the unlieing (basis) principle for collision.? For all collisions (non_elastic, partialy elastic, & totally elastic)? Which means we need to integrate then? But does that work well for (all) non_linear accelerations? I think it is zero wrt to my (immediate) surroundings (on earth), but I know I'm moving fast (eastwards) as the earth rotates thru the day. Yes, but our (so_called static) immediate surroundings are a deception. You know the earth is rotating, with no motion you are claiming identical speed, an inherent motion (i.e. speed) must exist (if we automatically imply (identical) speed). The real question is which (moving) reference to choose, to quantify (how fast). No motion is (truely) absurd. All things are moving (in the universe. Thus inherent motion exists.) Or isn't it? (I doubt that you can convince me otherwise. If you're clever enough you might. I don't know the outcome. But as the wording stands, it makes no sense otherwise (to me). & it's Newtonian.) Then nothing is moving.? That's not a conceptual failure, it's a fact; or else bad wording.? Everything in this universe moves although we can not say an independent speed wrt no reference. Wrt light we are moving at -c. I do not see an error. Einstein said there is no preferred reference, they all work well. You're right, they do not produce a difference, so that is the marvelous advantage when studying (=observing) collisions; until we get down to atoms & sub_atomic particles where the music changes (becoming significant). Then it is a non_elastic collision; or else a partially elastic collision. Yes. I did not convert after deriving (probably assuming (the momentum energy relation mom=E/va is so obvious) I could, later; & forgot that it (the existing work_energy formula) still was energy). It would matter for your challenge, if I had a momentum equivalent for your (physicist's) WE. Or wouldn't?

22. ## Elastic collision?

Then center of the big bang explosion as the reference instead. Well then I guess we can give up. You knew it (excluding your assumption of James Watt's horsepower history).
23. ## Elastic collision?

Your 1st sentence saying "It depends" gave an inkling of hope, that neither yes nor no were completely right. Thus I concluded a yes possibility also exists. I hope now you do. Ok. What I usually write as F=mom/t (so I can make the connection for myself, & don't forget.) F=m*v/t. Is my notation wrong? Force will change a mass's speed (velocity). But I still see the acceleration a=F/m of a mass m as the observable. Momentum changed is like a bank account. What goes into the mass m (as a speed change differencee v=vf-vi) can come out. If the speed was increased, then decreasing the speed can return the mass to its original speed. How we invoke & do the (speed) change is another story. But I think I'm off topic there, momentum_squared is the better concept instead of momentum & kinetic energy, there. We have no momentum wrt to each other, true. But I don't think we can say that for any other reference, because we are moving thru the universe (no doubt). I mean surely there is an underlieing inherent momentum for every mass, based on let us say the universe's center (from where you (=physicists) say the big bang happened). Ok. That makes things easier. i.e. less to think about. I guess according to the big bang (hypothesis, since I doubt we'll ever have a time machine or viewer to the past to prove it), everything is suppose to be flying away from the universe's center & being also gravitationally attracted (according to you physicists). That looks like a radial acceleration to me, e.g. a non_balanced force (setup) in ruffly 1 direction out from the core (center). Does that mean the earth on which the balance is fixed is not turning. Surely not.? Stationary is a static (d)illusion. No motion does not exist in the universe, everything is moving. Stationary (or static, at rest) only means both (the object (e.g. mass), & its reference sytem) are moving at the "same" speed. To say they do not move (at all) is absurd. Or do you disagree? Math or instrument machine? Ok. I said the balls deform like a spring, in order to decellerate (elastically, during the collision). That deformation needs time, (even if it's very quick, & the 2 balls are very hard, like steel or glass. Both materials still have (hard) elastic properties). I hope you understand me now.? (It seems you (also) did not register squared_momentum.? Have I said something wrong? Are you still trying to figure out what that means (because you have not commented on it so I assumed you are not sure). Please ask that I can try to fill in the blanks. Sometimes the mind automatically blocks things out too fast to notice.)
24. ## Elastic collision?

Yes, til you (helped improve that situation). Can I say it so?: I can trust energy when it is properly (isolated, &) incorporated in the momentum_squared equations. I guess I didn't recognize what you really wanted. In fact I know. That sounds like a possibility although you end up saying no. What example do you have in mind? That's so (=too) elegantly said that I'll need an example to picture it. I have to accept that. (But have difficulty & can't (quite).) But isn't every moving thing a storage of momentum? Non_recognizable (as static), when our surroundings are moving at our same speed.? When is that ever possible? (According to you, net gravity is from the universe's core.) But drop that example, & try, e.g. 2 weights balanced on a balance. Is potential momentum possible? (Similar to the potential energy PE=m*g*h=Wt*h.) mom=m*v v=((vi^2)+2*h*g)^0.5-vi where we have initial_speed vi height h free_fall acceleration g. I.e. The mass multiplied by the rooted 2*h*g (part). ? I'd be more than happy to to agree with you there. But you did not answer all of my post, leaving things: mom^2 (instead of KE), & Hooke's law. So it's difficult to orientate. Indirectly then, I have to assume they mean nothing for you here, & are thus an inadequate basis. You have brought me so far that (thru) the various threads I could incorporate Energy correctly into my momentum formulas. Thanks for helping me correct them. But if you do not comment on points (which I've based) adequately enough, then they are left in my head to rumble around, causing further disturbance. At present I can trust the momentum_squared formula, & in that (equation) is where I find (accurate) energy relations. If I have a seamless formula, then I will not hesitate (& be more than happy) to use it. e.g. mom_squared. I've been wrong on (so) many things. But that's not the important thing for me, getting it fixed=corrected is. If I have to make the effort to just believe (like christians often demand) (then that naive belief) is simply not enough. It won't stay (in), & that's valid for my technical mind. I must address the issues that make it (=my mind) tick so. (Me) evading those issues won't help me.