What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
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From: Tony M on 3 May 2010 11:28 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.
From: PD on 3 May 2010 12:41 On May 3, 10:07 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > On May 3, 4:55 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 1, 4:38 pm, "hanson" <han...(a)quick.net> wrote: > > > > ------- AHAHAHAHA... ahahaha.... AHAHAHA ------- > > > > Paul Draperbegin_of_the_skype_highlighting end_of_the_skype_highlighting: "PD" <thedraperfam...(a)gmail.com> wrote:> In <http://groups.google.com/group/sci.physics/msg/3e4af12502a33a7b> > > > > Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > > > >> PD <thedraperfam...(a)gmail.com> wrote: > > > > Koobee Wublee wrote: > > > > > > ** E^2 = m^2 c^4 + p^2 c^2 ------- [1] > > > > > > ** E = m c^2 ------- [2] > > > > hanson wrote: > > > > E^2 = m^2 c^4 --------- [3] = [2]^2 > > > m^2 c^4 = m^2 c^4 + p^2 c^2 ---------- [3] = [1] > > > p^2 c^2 = m^2 c^4 - m^2 c^4 > > > p^2 = m^2 c^2 - m^2 c^2 > > > p^2 = (m^2 - m^2) * c^2 ---------- or > > > p^2/c^2 = m'^2 - m^2 ---------- [4] > > > > Already here in [4] it is seen that momentum p happens > > > to be a (function of) mass, no matter what one calls it... > > > A rose by any other name is still a rose... Any momentum > > > needs mass to be present > > > Lovely. So let's see if I understand this game. > > PV=nRT, > > so R = PV/nT > > So the gas *constant* R is clearly a function of volume, according to > > you. > > > Stick to hyena-cackling, ahahahahanson, you are safer there than when > > you try to do something serious. > > > >... or to quote Y. Porat: > > > ----------- No mass --- No physics ---------- > > > with Porat implying that the photon has mass... etc, etc... > > ----------------- > V (Volume) is not a basic dimension!! > of the M K S > and we have as well that n there !! > so again > PD is a master of obfuscations > and a very sore looser !! > Y.Porat > ------------------------- Very well, then R(n/PV) = T, where T is a basic dimension, and one then concludes, following Hanson (and you) that R is a function of temperature. PD
From: Y.Porat on 3 May 2010 12:52 On May 3, 5:28 pm, Tony M <marc...(a)gmail.com> wrote: > 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 said before as well before that the gamma factor does not apply to the photon so ?? Y.P ----------------------
From: hanson on 3 May 2010 12:56 ==== AHAHAHAHA... AHAHAHAHAHAHAHAHA ===== Anti-Semite Paul Fraper "PD" <thedraperfamily(a)gmail.com>, in his Pedestrian Display, cranked himself grievously as he wrote: > >"hanson" <han...(a)quick.net> wrote: > ------- AHAHAHAHA... ahahaha.... AHAHAHA ------- Paul Draper wrote: == Paul Draper begin_of_the_skype_highlighting http://www.urbandictionary.com/define.php?term=skype Entry #7: lable for a megabitch, and of the jewish race she never thinks of anyone but herself. She's a skype. == end_of_the_skype_highlighting: > > "PD" <thedraperfam...(a)gmail.com> wrote: > > in http://groups.google.com/group/sci.physics/msg/3e4af12502a33a7b > > Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > > > PD <thedraperfam...(a)gmail.com> wrote: > > Koobee Wublee wrote: > > > > ** E^2 = m^2 c^4 + p^2 c^2 ------- [1] > > > > ** E = m� c^2 ------- [2] > > hanson wrote: > E^2 = m�^2 c^4 --------- [3] = [2]^2 > m�^2 c^4 = m^2 c^4 + p^2 c^2 ---------- [3] = [1] > p^2 c^2 = m�^2 c^4 - m^2 c^4 > p^2 = m�^2 c^2 - m^2 c^2 > p^2 = (m�^2 - m^2) * c^2 ---------- or > p^2/c^2 = m'^2 - m^2 ---------- [4] > > Already here in [4] it is seen that momentum p happens > to be a (function of) mass, no matter what one calls it... > A rose by any other name is still a rose... Any momentum > needs mass to be present. > The Pedestrian Display by Paul Draper who wrote: Lovely. So let's see if I understand this game. PV=nRT, so R = PV/nT So the gas *constant* R is clearly a function of volume, according to you. Stick to hyena-cackling, ahahahahanson, you are safer there than when you try to do something serious. > hanson wrote: ahahahaha... AHAHAHAHA... Paul, what kind of bizarre thought train has derailed your generally normal mentation? ... Porat is right, when he says that your MO is "obfuscation", when you get caught with your pants down, like you just were... ahaha... Get a hold of yourself, Paul. Your students are chuckling behind your back and refer to you as a senile old fart. Thanks for the laughs though... ahahaha... ahahahahanson > hanson cited and wrote: >... or to quote Y. Porat: > ----------- No mass --- No physics ---------- > with Porat implying that the photon has mass... etc, etc... > ------------ -----Paul, here is the entire post again, for your benefit---- > ------- AHAHAHAHA... ahahaha.... AHAHAHA ------- > Paul Draper: "PD" <thedraperfamily(a)gmail.com> wrote: > In < http://groups.google.com/group/sci.physics/msg/3e4af12502a33a7b> > Koobee Wublee <koobee.wub...(a)gmail.com> wrote: >> PD <thedraperfam...(a)gmail.com> wrote: > Koobee Wublee wrote: > > > ** E^2 = m^2 c^4 + p^2 c^2 ------- [1] > > > ** E = m� c^2 ------- [2] > hanson wrote: E^2 = m�^2 c^4 --------- [3] = [2]^2 m�^2 c^4 = m^2 c^4 + p^2 c^2 ---------- [3] = [1] p^2 c^2 = m�^2 c^4 - m^2 c^4 p^2 = m�^2 c^2 - m^2 c^2 p^2 = (m�^2 - m^2) * c^2 ---------- or p^2/c^2 = m'^2 - m^2 ---------- [4] > Already here in [4] it is seen that momentum p happens to be a (function of) mass, no matter what one calls it... A rose by any other name is still a rose... Any momentum needs mass to be present ... or to quote Y. Porat: ----------- No mass --- No physics ---------- with Porat implying that the photon has mass... etc, etc... > Now, depending on how the size of that difference [4] in [p] by (m' - m ) is mathematically treated, whether it is manipulated as a delta or as a differential, one will gain or loose a factor of 2 in the result, which is the loudly trumpeted **2** that appears to be the key argument in the analysis of the measurements of Mercury's precession and other experiments. So, it has nothing to do with any relativity... Newton still trumps and towers over Einstein, no matter how frantically the Einstein Dingleberries do twist, spin or bob in the breeze of the farts that come out of Albert's sphincter & do tremble in their angsts when KW says:" Einstein was a nitwit, plagiarist, and a liar. ? > Koobee Wublee wrote: > > > Where > > > ** E = Observed energy > > > ** m = Rest mass > > > ** m� = observed mass > > > ** p = observed momentum > > The second equation [2] can be directly derived as one > of the geodesic equations. In the meanwhile, the first > equation [1] can only be claimed after the second equation > is established. So, what is this absurd claim that the >second equation is not preferably accepted by the self- > styled physicists nowadays. It just does not make any sense. > So, good professor Draper, how do you derive at the first > equation without knowing the second equation? Paul Draper wrote Good grief, KW. Quit blustering and spluttering. The first equation stems directly from the expression for the norm of the 4-momentum. > hanson wrote: But Paul, you, as a pedagogue, (not necessarily a demagogue, as Porat would have it), should nevertheless not engage into the exact same behavior mode that you accuse KW of doing. To make this character flaw of yours more understandable let me quote my friend from Israel, Yehiel Porat, who stressed on 29-April, 6:16 the following,....... for your benefit, Paul: > Yehiel Porat wrote: === PD, 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 ?? 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 ?? 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 ?? === 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 ?? a btw question, can you have Gravitation without mass how long you was spending thinking about you way to cheat and obfuscate ??? === do you think that all the readers are idiot or suckers !!!??? Y.Porat > hanson wrote: Paul, try to understand that pigs are NOT kosher... ahahaha,, Thanks for the laughs, though, guys.... ahahahahanson --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---
From: Y.Porat on 3 May 2010 13:02
On May 3, 6:41 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 3, 10:07 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > > On May 3, 4:55 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 1, 4:38 pm, "hanson" <han...(a)quick.net> wrote: > > > > > ------- AHAHAHAHA... ahahaha.... AHAHAHA ------- > > > > > Paul Draperbegin_of_the_skype_highlighting end_of_the_skype_highlighting: "PD" <thedraperfam...(a)gmail.com> wrote:> In <http://groups.google.com/group/sci.physics/msg/3e4af12502a33a7b> > > > > > Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > > > > >> PD <thedraperfam...(a)gmail.com> wrote: > > > > > Koobee Wublee wrote: > > > > > > > ** E^2 = m^2 c^4 + p^2 c^2 ------- [1] > > > > > > > ** E = m c^2 ------- [2] > > > > > hanson wrote: > > > > > E^2 = m^2 c^4 --------- [3] = [2]^2 > > > > m^2 c^4 = m^2 c^4 + p^2 c^2 ---------- [3] = [1] > > > > p^2 c^2 = m^2 c^4 - m^2 c^4 > > > > p^2 = m^2 c^2 - m^2 c^2 > > > > p^2 = (m^2 - m^2) * c^2 ---------- or > > > > p^2/c^2 = m'^2 - m^2 ---------- [4] > > > > > Already here in [4] it is seen that momentum p happens > > > > to be a (function of) mass, no matter what one calls it... > > > > A rose by any other name is still a rose... Any momentum > > > > needs mass to be present > > > > Lovely. So let's see if I understand this game. > > > PV=nRT, > > > so R = PV/nT > > > So the gas *constant* R is clearly a function of volume, according to > > > you. > > > > Stick to hyena-cackling, ahahahahanson, you are safer there than when > > > you try to do something serious. > > > > >... or to quote Y. Porat: > > > > ----------- No mass --- No physics ---------- > > > > with Porat implying that the photon has mass... etc, etc... > > > ----------------- > > V (Volume) is not a basic dimension!! > > of the M K S > > and we have as well that n there !! > > so again > > PD is a master of obfuscations > > and a very sore looser !! > > Y.Porat > > ------------------------- > > Very well, then R(n/PV) = T, where T is a basic dimension, and one > then concludes, following Hanson (and you) that R is a function of > temperature. > > PD -------------------- T is an MKS dimension ??? 2 waht has all that got to do with our op question about mass of the *** photon ** 3 if we see that in ***your *** formula E^2 =mc^2)^2 plus (pc) ^2 there is nothing to multiply the mass by zero so mass in energy is non zero then you suggest or expect that in another VALID LEGITIMATE FORMULA you will find something contradictory to it ???!! (do you think now** that *your* formula is wrong in any aspect ??!! and not good enough to prove the above ?? Y.P ----------------------- |