What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
in [Relativity]
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From: BURT on 21 Apr 2010 21:18 On Apr 21, 5:15 pm, artful <artful...(a)hotmail.com> wrote: > On Apr 22, 5:24 am, Tony M <marc...(a)gmail.com> wrote: > > > YP, > > > Do not mix discussions about photons and other particles. While > > photons are particles, not all particles can be treated equally. > > Particles are divided into massive and massless. That grouping has > > nothing to do with how light or heavy particles are, and it doesnt > > mean that some particles have no mass. > > Actually .. it does. > > But that does not mean that the energy of such a particle cannot have > an mass-equivalence (ie an amount of mass that that energy could > possibly be converted to) > > > All particles have both mass > > and energy. > > They all have a combination of mass and energy. Photons, for example, > are little packets of energy and are all energy (no mass). > > > Massive particles can only travel at v<c and have non- > > zero rest mass. Massless particles exist only at c and have no rest > > mass. > > When we say 'mass' we generally mean the invariant / proper / rest > mass, unless the context implies we are talking about something else. > > > Therefore, one must be careful about the form of the equations > > one uses to describe the two types of particles, especially when using > > m, gamma or v. > > Yeup Light is a spread out electrtic energy wave. Matter is point particle of infinite density of energy. Mass is infinitely dense energy. Mitch Raemsch
From: Y.Porat on 23 Apr 2010 20:40 On Apr 23, 4:33 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 22, 9:12 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > > On Apr 22, 10:24 pm, PD m make quantities equal. > > > > > > There is no object that has momentum mc. > > > > > > I cannot believe that you do not know the first thing about units.. > > > > > ------------------- > > > > > that is a new level of impertinanace !! > > > > > mc^2 and pc are > > > > COMBINED WITH THE PLUS SIGN > > > > ON THE SAME LINE OF THE FORMULA > > > > WE AR E NOT TALKING ABOUT THEIER > > > > QUANTITIES > > > > WE ARE TALKING ABOUT THEIR > > > > PROPEERTIES THAT MAKES IT POSSIBLE TO COMBINE THEM > > > > IN THE SAME KLINE OF THE FORMULA > > > > WITH A LUS SIGN ON THE SAME LINE !! > > > > COULD YOU COMBINE THEM INCASE FOR INSTANTANCE THAT > > > > ONE UNIT WAS MISSING THERE ?? > > > > That's exactly what I said. Having the same dimensions does not make > > > the quantities equal. > > > ----------------- > > shameless crook!! > > > if you have a formula > > > C= A plus B > > > then even a dumb mathematician agreed that > > A and B > > must have exactly the same ***dimensionally !!** > > Yes, but this does not mean that A and B are *equal*. > So while mc^2 and pc have the same *dimensions* (and they do), ----------------------- THANK GODNESS THAT YOU UDERSTOOD IT UNDER MY PRESSURE ... (youhad no choice but to admit it unless you are a complete idiot ! now lets see about honesty ,,(:-) ------------- this > does NOT mean that you can set mc^2 = pc and deduce that p=mc. ---------------- and here goes the cheating of Paul Draper !! DDI I SAYED THAT mc^2 is pc ????!!! common draper it is time for you to know well that to say HALF THE TRUTH IS CHEATING !!! (i dont mention here Josef Goebbels ...) i SAID EXPLICITLY ABOVE THAT mc^2 if like pc ONLY DIMENSIONALLY !!! IE THEY MUST HAVE **ALL THE WAY** THE exact SAME DIMENSIONS no one missing no one additional and no one of one system unit and and the 'same the same unit ' from another system unit !!) no rest mass and no PD or other cheating mass just the exactly same mass with no double private interpretations from the new school of Paul Draper or others crooks and suckers school and other fucken mathematician as circumstantial mass'!! 2 I SAID ABOVE ***EXPLICITLY** THAT mc^2 AND pc HAVE THE same units dimensions (ie energy) -------BUT THEY HAVE DIFFERENT ******** SCALAR MULTIPLIERS ***** GOT IT PARROT ???!!!! keep well Y.Porat --------------------------
From: Inertial on 26 Apr 2010 08:55 "Y.y.Porat" <y.y.porat(a)gmail.com> wrote in message news:1207b72d-584e-43fe-b08b-e66a7d83c986(a)z11g2000yqz.googlegroups.com... > just tell me how do you see it?? > do you agree with me that > if you say that there are different kinds of mass > while is any physics formula defined by the mks system there is JUST > ONE KIND of N MASS > YOU YOU CLAIM OTHER KINDS OF MASS > IN THE MKS SYSTEM IE > > M K1 K2 K3 S SYSTEM > th e formula ddint hear about your > 3 explanations for it !!... > b > do you agree withme that if you use the formula > > E ^2 == mc^2^2 PLUS !!! (pc) ^2 > > THEN THE PC ^2 MUST AHBE EXACTLY THE SAME DIMENSION AS mc^2 > (even if they have different scalr multipliers !!?? > (i swear you never could dee > suc argument as mind just above > IN ANY OF YOUR Parroting BOOKS !! > (:-) > and that is because i am a thinking physicist and not a parrot NOBODY is saying there is more than one dimension of mass .. I've not seen *anyone* in this newsgroups claim that there is. The only one to mention it is you when you go on and on asking people if there is and arguing against something that noone claims.
From: Koobee Wublee on 26 Apr 2010 14:32 On Apr 22, 7:26 am, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 22, 8:56 am, Tony M <marc...(a)gmail.com> wrote: > > If you dont mind, lets elaborate on the invariant mass of a closed > > system of particles (which includes both massive and massless > > particles <photons>). These particles would have arbitrary velocities > > (not applicable to photons of course) and directions within the > > system. How do we define the invariant mass of such a system? > > First, what it clearly is not is the sum of the rest masses of the > constituents of the system, which in this case would be zero, since > the rest masses of the photons are zero. What may not be obvious, but > is also true, is that the invariant mass is not necessarily the sum of > the relativistic masses of the photons either! Ah, disclaimer statements with double meanings. I bet if the newsgroup allows you to use fine prints, you would have done so. <shrug> > The way to find the invariant mass is very simple: > m^2 = (sum[E])^2 - (sum[p])^2 > where the sum[E] is the scalar sum of the energies of the > constituents, and > sum[p] is the vector sum of the momenta of the constituents. 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> 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>
From: PD on 26 Apr 2010 15:04
On Apr 26, 1:32 pm, Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > On Apr 22, 7:26 am, PD <thedraperfam...(a)gmail.com> wrote: > > > On Apr 22, 8:56 am, Tony M <marc...(a)gmail.com> wrote: > > > If you dont mind, lets elaborate on the invariant mass of a closed > > > system of particles (which includes both massive and massless > > > particles <photons>). These particles would have arbitrary velocities > > > (not applicable to photons of course) and directions within the > > > system. How do we define the invariant mass of such a system? > > > First, what it clearly is not is the sum of the rest masses of the > > constituents of the system, which in this case would be zero, since > > the rest masses of the photons are zero. What may not be obvious, but > > is also true, is that the invariant mass is not necessarily the sum of > > the relativistic masses of the photons either! > > Ah, disclaimer statements with double meanings. I bet if the > newsgroup allows you to use fine prints, you would have done so. > <shrug> I don't know what you find suspicious in what I said. > > > The way to find the invariant mass is very simple: > > m^2 = (sum[E])^2 - (sum[p])^2 > > where the sum[E] is the scalar sum of the energies of the > > constituents, and > > sum[p] is the vector sum of the momenta of the constituents. > > 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. > > 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? |