What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
in [Relativity]
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From: whoever on 3 May 2010 21:32 "Tony M" <marcuac(a)gmail.com> wrote in message news:c088a6fa-58a1-4262-87c9-951eaf7d2313(a)d19g2000yqf.googlegroups.com... > On May 3, 10:35 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: >> On May 3, 3:27 pm, Tony M <marc...(a)gmail.com> wrote: >> >> >> >> >> >> > Guys, what is this nonsense about photons having energy but no mass? >> > Energy and mass are BOTH observer dependent quantities, so when we >> > discuss the two we MUST use the SAME frame of reference for BOTH. >> > Therefore, if we talk about the rest mass of a photon then we also >> > have to talk about the rest energy of the photon. Is there such a >> > thing as rest energy for a photon? There is no such thing (or you can >> > say it is zero). The same applies to its rest mass. >> > If we talk about photon energy in a frame where this energy is not >> > zero (photon is not at rest) then we MUST refer to the OBSERVED mass >> > in the SAME frame of reference, and that would NOT be zero either. We >> > can't mix non-rest energy with rest mass and say photons have energy >> > but no mass! >> > E=m c^2 applies, where BOTH E and m are OBSERVED quantities, >> > regardless of the frame of reference we choose, as long as it is the >> > SAME frame for both quantities. >> > Photon mass contributes to the invariant mass of a system which >> > contains photons. >> >> ------------------ >> quite right >> but i have some news for you >> in case you forgot!! >> THE GAMMA FACTOR DOES NOT APPLY TO THE >> PHOTON ENERGY!!! >> so you are right that we have to measure it at the original frame in >> which it was created (or emitted ) >> b >> if you measure it in aframe that is >> running away fast from the original one >> **you get the red shift phenomenon!! >> yet this i s rathre another prove that the photon >> energy (hf) >> is subdivided to smaller components >> iow >> hf >> IS NOT RIGHT DEFINITION OF THE SMALLEST PHOTON ENERGY !! >> >> and i devoted a special thread for it !!! >> but that is another Opera !! >> yet again >> it has nothing to do withthe question >> whether the photon has mass or not >> we proved above that it has mass >> AND ONLY ONE KIND OF MASS!!! >> (iow >> if you have two frames involved >> the mass ( or energy) >> can be divided or spread or shared >> between the two frames >> (while the overall energy and mass-- >> IS CONSERVED >> IN THOSE TWO FRAMES !!) >> >> AND THAT IS ANOTHER COPYRIGHT-- >> >> --INSIGHT ( OF MINE ) ABOUT THAT ISSUE !!! >> >> ATB >> Y.Porat >> ------------------------ Hide quoted text - >> >> - Show quoted text - > > The observed mass (or energy) of a photon or its value in different > frames of reference does not imply a gamma factor. The gamma factor > does not belong in photon equations. > Like I said in a different posting, f is frame dependent making E=hf > frame dependent. The observed mass of the photon is m=E/c^2, no gamma > involved. > Under no circumstance should the observed mass of a photon be > expressed as m = gamma x m0, where m0 is the rest mass. I've been explaining this to Porat from months (if not years). He doesn't get that m = gamma x m0 is indeterminate for anything travelling at c (and with m0 = 0) .. ie you cannot calculate a particular value for m from that equation. That does not mean that the laws of SR physics do not apply .. nor does it mean that m0 is other than zero. --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: whoever on 3 May 2010 21:40 "Tony M" <marcuac(a)gmail.com> wrote in message news:9e015792-b53a-428e-84e1-0ab7f4fe4ac1(a)o14g2000yqb.googlegroups.com... > Guys, what is this nonsense about photons having energy but no mass? No invariant mass. They do have a relativistic (frame dependent) mass measurement .. E = Mc^2, E = hf, so M = hf/c^2 > Energy and mass are BOTH observer dependent quantities, Depends on whether you mean relativistic mass or invariant mass. Just a the invariant (or proper) length of a rod does not change but its (relativistic, or frame-dependent) length is different for different observers Porat does not understand the difference between a frame dependent and an invariant measurement (of length of mass) > so when we > discuss the two we MUST use the SAME frame of reference for BOTH. Indeed they must. Though for invariant mass, the frame doesn't matter. > Therefore, if we talk about the rest mass of a photon then we also > have to talk about the rest energy of the photon. And as a photon is never at rest in any inertial frame, it is not meaningful. However invariant (frame-independent) mass does have meaning, and it must be zero for a photon. > Is there such a > thing as rest energy for a photon? There is no such thing (or you can > say it is zero). The same applies to its rest mass. But not to invariant mass. > If we talk about photon energy in a frame where this energy is not > zero (photon is not at rest) then we MUST refer to the OBSERVED mass > in the SAME frame of reference, Porat says that mass (the relativistic, or frame-dependent mass) doesn't exist > and that would NOT be zero either. Yeup .. but Porat says there is no such thing .. that an object can only have one mass measurement, and that value is not frame-dependent. He's in denial. > We > can't mix non-rest energy with rest mass and say photons have energy > but no mass! We CAN say it has a zero invariant mass. > E=m c^2 applies, where BOTH E and m are OBSERVED quantities, It is misleading to use 'm' there .. as 'm' usually means invariant mass. But assuming you are using 'm' for relativistic mass, then yes > regardless of the frame of reference we choose, as long as it is the > SAME frame for both quantities. > Photon mass contributes to the invariant mass of a system which > contains photons. You mean: Photon relativistic mass contributes to... Just saying 'mass' implies (by convention) invariant mass. I know its not consistent .. as it is the opposite convention that we use for length (say). --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: spudnik on 4 May 2010 20:12 it's just a "paradox" of assuming that it is a photon, when it is merely a wave; obviously, a thing with p=mv not equal to zero, can't have one of the terms being zero; so, it is not a particle or Newtonian corpuscle -- and did they *have* to give Einstein a Nobel, just to reify that foolishness of his? > There is no mass in the momentum of the photon. thus: it's probably just his grasp of English; don't you think? as for "Newton's law," its universality is actually due to Kepler; Hooke merely algebraized Kepler's orbital constraints, using some work of Huyghens (then, Knewton stole the inverse second-power thing from Hooke, and destroyed Hooke's portraits .-) thus: nor are most glaciers actually receding, although this fact is mainly unnoticed, because of a massive lack of historical data on virtually all glaciers. satellite telemetry has shown almost no change of Antarctic icesheets, but what else would one expect, considering that there is as much ice as can be accomodated, because "ice bergs do calve, period." > OK, NSIDC and NERSC don't agree. NERSC, who shows the years 2007, 2008, > 2009 and 2010, still shows 2010 with much more ice than 2007 and 2008. thus: I'm allowed to agree with Al Gore about one thing; am I not?... even though Mauna Loa is a weird place to measure CO2, it's still just one place, with a record since the '60s (I think). now, most of the effect of humans may not be the burning of Fossilized Fuels (tm), but the burning-up of soil biota & forests. (after all, oil comes out of the ground, by itself, under pressure -- even when we're pumping like crazy in the Gulf of Mexico and the Redondo Seep e.g.) thus: most of these things, you mention, are just theoretical interpretations from the Schroedinger's cat school of Copenhagen, "reifying the math" of the probabilities; they don't actually have any bearing on the correctness of QM or GR or SR or any thing, nor on your so-called theory. but, why do you say that conversation of momentum supposedly doesn't apply to a split quantum of light in some standard theory? and poor Nein Ein Stein believes that p = mv is a force and that F = ma is not, and some thing about Coriolis' force, merely from a didactic say-so of his (in some sort of pidgen English, which could be the whole problem). > - The future determines the past > - Virtual particles exist out of nothing > - Conservation of momentum does not apply to a downgraded photon pair --Light: A History! http://wlym.TAKEtheGOOGOLout.com
From: Y.Porat on 5 May 2010 14:06 On May 5, 7:03 pm, P > > > > and right !! > > > > RIGHT ?? > > > > the second step you will have to agree with me > > > > is another Golden historic rule of mine > > > > > NO MASS - NO REAL PHYSICS !! > > > > > (a few years later you will tell every body that it was > > > > your or your ---------------------- the theoretical above formula does not tell you tomultiply mc^2 by zero so where from you took it ?? 2 just tell me WHAT IS THE ERROR MARGINS OF YOUR TOOLS THAT SAY THAT THE PHOTON MASS IS **THEORETICALLY** ZERO ?? 3 IF ITS MASS IS SMALLER THAN CAN BE MEASURED BY IT HOW WPOD YOU KNOW THAT IT IS ZERO ?? 4 THE PHOTON eNERGY FORMULA IS E PHOTON = hf were do you see **in all the experiments that were done in which photon energy was found-- as nonzero **-- was there a scalar multiplier (in that experimental formula) that was multiplying the hf (and the m in it )by zero ?? while the dimensions of hf is mc^2?? hf== nonzero scalar times meter^2 /second ^2 do you deny that hf contain the dimension m ??!! and no scalar multiplier of zero to it Y.P ------------------------ i
From: spudnik on 5 May 2010 17:47
I agree with PD, but why are we arguing with a guy who is not quite literate in English (and perhaps not in any other language, like AP) ?? YP, get back to us, after you simply *try* to read Shakespeare -- which no-one can ever really completely comprehend, but ... |