What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
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From: Inertial on 28 Apr 2010 09:02 "Y.Porat" <y.y.porat(a)gmail.com> wrote in message news:3972780d-41a6-4326-8af7-a06878f703de(a)12g2000yqi.googlegroups.com... > On Apr 27, 10:59 pm, Darwin123 <drosen0...(a)yahoo.com> wrote: >> On Apr 27, 11:08 am, Darwin123 <drosen0...(a)yahoo.com> wrote:> <Porat>> do >> you agree withme that if you use the formula >> >> > > E ^2 == mc^2^2 PLUS !!! (pc) ^2 >> >> > If "m" is the rest mass, "E" the total energy, "p" the momentum and >> > "c" the speed of light. In the equation written here, "m" is not the >> > "relativistic mass." It is the "rest mass." >> > You may disagree as to the validity of the equation. However, you >> > clearly agree that the units of this equation are correct. >> >> > > THEN THE PC ^2 MUST AHBE EXACTLY THE SAME DIMENSION AS mc^2 >> > > (even if they have different scalr multipliers !!?? >> >> > No. Clearly false. You are mathemetaically illiterate. >> >> Sorry. You are mathematically illiterate. I misspell things sometimes, >> too. >> However, I do know that the units of "mc^2" are not the same as >> those of "mc^2^2". If you don't address this point, then you are a >> liar in addition to being an idiot. > > ------------------- > if mc^2 has the same units as pc It does > then their second degree ^2 > is as well the same But that's not what you wrote You wrote mc^2^2 .. which means the same as m((c^2)^2) ... which is NOT the same as (mc^2)^2 You also got (pc)^2 confused with pc^2, and mc^2 confused with (mc^2)^2) .. basically , you stuffed up the formulas completely > you are not even a fucken mathematician If you knew any maths you would see how wrong you are. .. but as with physics, you don't know enough to even begin to discuss either subject That you replied as you did above and criticized darwin123 as you did just further illustrates your ignorance for all to see.
From: Darwin123 on 28 Apr 2010 13:26 On Apr 19, 2:28 pm, PD <thedraperfam...(a)gmail.com> wrote: > Photons don't have a rest mass, and they don't have a relativistic > mass. And relativistic mass is an antiquated notion that has been > largely abandoned because it confuses amateurs and some structural > engineers. Yes, but who cares about them!? Just kidding. I am not sure what you mean by "abandoned." I have talked with other physicists on topics related to relativity. We sometimes used the concept of "relativistic mass," with the understanding that it has limited applicability. Since Einstein actually used the concept in some of his papers, and since his use of it was correct in that specific context, the concept of relativistic mass shouldn't be totally abandoned. In biology, the concept of "species" leads to all sorts of confusion among amateurs and zoo keepers. The idea of "species" really isn't valid far beyond a certain limited context. However, I don't think total abandonment is advisable. Some useful concepts need to be flagged, not abandoned. BTW: The relativistic mass that I am talking about is "m" in the equation: m=m_0 /sqrt(1-v^2/c^2) "m_0" is the rest mass, "v" is the relative velocity, and "c" is the speed of light. Although it leads to problems among beginners, I don't think it should be discarded. It is still taught in introductory classical mechanics courses.
From: PD on 28 Apr 2010 13:31 On Apr 28, 12:26 pm, Darwin123 <drosen0...(a)yahoo.com> wrote: > Although it leads to problems among beginners, I don't think it > should be discarded. It is still taught in introductory classical > mechanics courses. Read these two sentences a few times and this may explain why a good number of physicists have slapped their foreheads "D'oh!" and decided to stop perpetuating the madness.
From: Darwin123 on 28 Apr 2010 19:19 On Apr 28, 1:31 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 28, 12:26 pm, Darwin123 <drosen0...(a)yahoo.com> wrote: > > > Although it leads to problems among beginners, I don't think it > > should be discarded. It is still taught in introductory classical > > mechanics courses. > > Read these two sentences a few times and this may explain why a good > number of physicists have slapped their foreheads "D'oh!" and decided > to stop perpetuating the madness. Thank you! I reread the two sentences that I wrote several times, on your suggestion. I am beginning to understand.
From: Y.Porat on 29 Apr 2010 09:16
On Apr 27, 4:18 am, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 26, 7:06 pm, Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > > > > > On Apr 26, 12:04 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Apr 26, 1:32 pm, Koobee Wublee wrote: > > > > It sounds like this so-called invariant mass is indeed your rest mass > > > > as indicated by your own mathematics above. <shrug> Yet, you are > > > > continuing to use twisted logics to spew up mysticism. > > > > > Lets see. The following represent indeed mathematical equivalence. > > > > The self-styled physicists want to promote mysticism. They are > > > > claiming the first equation is valid while the second one is out of > > > > favor. > > > > > ** E^2 = m^2 c^4 + p^2 c^2 > > > > ** E = m c^2 > > > > > Where > > > > > ** E = Observed energy > > > > ** m = Rest mass > > > > ** m = observed mass > > > > ** p = observed momentum > > > > > Clearly, both equations are indeed the same. Why do the self-styled > > > > physicists continue chanting about the nonsense of them being > > > > different? It makes one wonder if they understand the simple > > > > mathematics behind all that at all. <shrug> > > > > They're not different. However, you'll notice that m is *invariant*, > > > which means it has the same value in all frames, not just the frame in > > > which it is at rest. Numerically, invariant mass has the same value as > > > the rest mass. But the latter is the mass when measured at rest, by > > > definition. The former does not require it to be at rest to apply. > > > You'll also notice that m' is not invariant, and so yes, there are > > > reasons why m' is not so much used anymore, where m is. > > > So, both equations are correct. Why do you insist that only the first > > one is correct? So, if your structural engineer is able to apply the > > second equation to solve the problem, why are you making fun of that > > engineer using the second equation instead of the first? > > 1. What I said is that the second equation is used because it makes a > more consistent use of m. > 2. The structural engineer is not able to apply either equation to > solve the problem. > 3. The structural engineer is confusing himself with the meaning of > mass, which is different in each case. > > > > > The concept of the invariant mass is stupid and unnecessary since mass > > is always an observed quantity. The rest mass makes much more sense > > in any applications. <shrug> > > And is it your contention that mass is always observed at rest? > Particle physicists would strenuously disagree, and they observe mass > all the time. > > > > > > > > > OK? How do the self-styled physicists cope with the energy > > > > conservation in (E = m c^2) applied to gravitation? They dont. In > > > > desperation, they tossed around this gravitational wave nonsense to > > > > distract attentions. The answer to that question is indeed to claim > > > > the supposedly invariant rest mass not so invariant after all. This > > > > so-called invariant rest mass will now vary with the amount of > > > > curvature in spacetime. The more curved up spacetime is the less > > > > invariant mass it has as represented by the following equation > > > > derived through the geodesic equations of the Schwarzschild metric. > > > > > ** E^2 = m^2 c^4 (1 2 U) / (1 B^2) > > > > > Where > > > > > ** m = Rest mass in flat spacetime > > > > ** m sqrt(1 2 U) = Rest mass or invariant mass > > > > ** U = G M / c^2 / r > > > > ** B^2 c^2 = (dr/dt)^2 / (1 2 U)^2 + r^2 (dO/dt)^2 / (1 2 U) > > > > ** dO^2 = dLongitude^2 cos^2(Laitude) + dLatitude^2 > > > > > The above equation can simpler be written as the following. > > > > > ** E = m c^2 > > > > > Where > > > > > ** m = m sqrt(1 2 U) / sqrt(1 B^2) > > > > > Which one of the first two equations is more elegantly simpler? It > > > > should take no brainer to answer that question unless you are the one > > > > promoting mysticism in post-Aether physics where mysticism is the rule > > > > of the game. <shrug> > > > > Gee, why do you think that the equation that has fewer symbols in it > > > is the one that is fundamentally better? > > > Well, the second equation deals with the so-called invariant/rest > > mass. Although the first equation would lead to the same thing as the > > second one if the rest mass in variant under the curvature of > > spacetime, the second equation allows much easier concept of mass in > > which as Tony M pointed out that there is only one mass that is the > > observed mass. That observed mass in increased through the time > > dilation due to motion but decreased through time dilation due to > > gravitational time dilation. Oops! Did the cat get out of the bag? > > Does someone smell contradiction in GR? > > No, sorry, I don't sense any contradiction. Do you? -------------- nasty pig demagogue! lier ! why do you obfuscate the problem ?? we have a simple equation trhat you youself brought it is E^2 =(mc^2) plus (pc)^2 now nast pig PD if we have a physical formula withthe 3 dimjewnsions of M K S C = A +B does all of the above items MUST HAVE EXACTLY THE SAME DIMeENSIONS?? YES OR NO NASTY PIG ?? sow where do you see in those formula something like M K2 K2 K3 S DIMENSIONS ?? where do you see different kinds of mass!! as K1 or K2 ?? hey nasty pig ?? 2 you could say that in one item there is a sclalr multiplier of zero for the mass WHERE ( NASTY PIG ) YOU SEE A ZERO SCALAR MULTIPLIER ? FOR ANY OF THEM ?? 3 YOU COULD SAY THAT MAY BE THERE IS SOME RELATIVISTIC GAMMA IN IT SO WHERE (NASTY SHAMELESS DEMAGOGUE) YOU SEE A GAMMA FACTOR WHILE THE GAMMA FACTOR DOES NOT APPLY AT ALL TO THE PHOTON OR TO ENERGY ?? 4 why( nasty demagogue obfuscator) you mingle GR in it !! does your fucken GR create new kinds of mass !!??? while again he gamma factor does not apply to energy nor to mass or new kinds of ENERGY !!! other than the E^2 = ( m c^2)^ + (p c)^2 ????? is that formula that *you yourself brought at the beginning of this discussion ---- ** NOT VALID ANYMORE ** why ?? BECAUSE IT IS MORE DIFFICULT TO OBFUSCATE IT ?? 5 a btw question can you have Gravitation without mass 6 how long you was spending thinking about you way to cheat and obfuscate ??? 7 do you think that all the readers are idiot or suckers !!!??? Y.Porat ------------------------ |