What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
in [Relativity]
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From: Y.Porat on 3 May 2010 10:16 On May 3, 3:53 pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > "Tony M" <marc...(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? > > Guys, what is this nonsense about spinning wheels having mass but no energy? ------------------- exactly!! thank you Andro!! Y.Porat ----------------------
From: Y.Porat on 3 May 2010 10:35 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 -----------------------
From: PD on 3 May 2010 10:51 On May 1, 3:36 pm, glird <gl...(a)aol.com> wrote: > On May 1, 10:49 am, PD wrote: > > >< The first equation [e = mc^2] stems directly from the expression for the norm of the 4-momentum. > > > Assuming that "the norm of the 4 momentum" means "the line > perpendicular to the momentum in x,y,z,t" and that momentum = mv, Um, no. Good heavens, no. Please google "4-vector", "norm of 4-vector" and "4-momentum". If you don't know what words mean, please look them up rather than just guessing. You may also be interested to know that p=mv where m is associated with a quantity of matter is NOT a good definition for momentum, even though you will find that in the first few chapters of an introductory physics text. Later in the same text, you will usually see an apology for the oversimplification and an amendation to the more correct answer. > please show how "The first equation stems directly from the expression > for the norm of the 4-momentum". > > Along the way, note that in the expression "mv" the m is the mass > (quantity of matter) of > the object that moves at v; and explain how the > m in e = mc^2 -> m = e/c^2 is that same quantity of matter. > > glird
From: PD on 3 May 2010 10:55 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... >
From: Y.Porat on 3 May 2010 11:07
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 ------------------------- |