What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
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From: Y.Porat on 19 May 2010 10:15 On May 19, 3:31 pm, "Paul B. Andersen" <paul.b.ander...(a)somewhere.no> wrote: > On 18.05.2010 17:16, Inertial wrote: > > > > > "Paul B. Andersen" <paul.b.ander...(a)somewhere.no> wrote in message > >news:hsu4au$2ald$2(a)news01.tp.hist.no... > >> On 12.05.2010 15:55, Tony M wrote: > >>> On May 12, 8:49 am, "Paul B. Andersen"<paul.b.ander...(a)somewhere.no> > >>> wrote: > >>>> On 12.05.2010 07:02, Koobee Wublee wrote: > > >>>>> Let's bring back the two > >>>>> equations describing energy, mass, and momentum below. > > >>>>> *1* E^2 = m^2 c^4 + p^2 c^2 > >>>>> *2* E^2 = m'^2 c^4 > > >>>>> Where > > >>>>> ** m = Rest mass > >>>>> ** m' = Observed mass > >>>>> ** p = Observed momentum > > >>>>> These equations are actually identical, but there are always bevies of > >>>>> Einstein Dingleberries who would aloofly swear to their god Einstein > >>>>> the nitwit, the plagiarist, and the liar that equation *1* is the only > >>>>> one that is any valid. > > >>>> They are identical for massive objects only. > >>>> What does *2* say the energy of a photon is? > >>>> Is it correct? > > >>>> -- > >>>> Paul > > >>>>http://home.c2i.net/pb_andersen/-Hide quoted text - > > >>>> - Show quoted text - > > >>> It says the energy of a photon is equal to its observed (or > >>> relativistic) mass (not the rest mass) multiplied by c^2. > >>> And yes, it is very correct. > > >> A massless particle has no 'relativistic mass'. > >> The 'mass equivalent' of its energy, h*nu/c^2 > >> is not the same as 'relativistic mass'. > > > Sorrry. .but that appears not to be correct. Please specify your source > > for those statements, > > It depend on how you define 'relativistic mass', > it's not a very well defined entity. > One definition is m' = E/c^2, which is applicable > for all kind of particles, even massless ones. > > With this definition, you are right, then > the 'relativistic mass' of a photon is indeed h*nu/c^2 > > But I find this definition rather pointless. > Remember that it originally stems from Einstein > (even if he didn't use that term in 1905); since > momentum is mv, the mass has to increase at high velocities. > This increased mass was later called the 'relativistic > mass', as opposed to rest mass. > > The definition E/c^2 came much later. > > The trend now is, as I am sure you know, not to > use the term 'relativistic mass' at all. It is > confusing and utterly unnecessary. > > It's only one kind of mass, the invariant mass. > The photon mass is zero. ====================== so ??!! zero ??!! > p = m*gamma*v for massive objects, > p = h/lambda for photons so if there is just one kind of mass why do you say that for the photon it is zero--- according to you P photons is h/lambda it is as well m c see the Plank momentum !! so were the hell do you see A GAMMA FACTOR TO MAKE IT 'RELATIVISTIC '???!!! ( what is that gamma factor wile V=c ??!!!) TIA Y.Porat -----------------------------------
From: Y.Porat on 19 May 2010 10:27 On May 19, 2:55 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 18, 11:57 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > > On May 18, 10:09 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 18, 2:19 pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > > > > > Present the evidence; all you need do is argue your case logically, > > > > based on acceptable axioms, and I'll accept what you say. > > > > When one teaches, two learn. > > > > I'm quite willing to listen, but I'll jump down your throat and rip your > > > > lungs out (metaphorically, of course) if you try to bullshit me. > > > > > Able to follow? > > > > Sorry, but you're talking mathematics and philosophy, not physics. > > > Cases in physics are not made by arguing logically from mutually > > > acceptable axioms. > > > If this is how you'd like to proceed with physics, then you're barking > > > up the wrong tree. > > > > Physics does things a little differently. The proposer puts together a > > > model that includes some axioms, whether the axioms or the model are > > > considered acceptable or not. Quite often, the axioms are considered > > > outlandish and contrary to conventional wisdom. So what? From that > > > model and those *assumed* axioms, certain conclusions are deduced from > > > them, including a number of conclusions that are distinct from other > > > models and which can be directly or indirectly tested in experimental > > > measurement. When a significant number of those distinguishing > > > conclusions are found in accord with measurement, then this > > > provisionally *forces* the acceptance of the model and the axioms, in > > > that the model has demonstrated itself to be superior to competing > > > models on the grounds of experimental confirmation. > > > > If you think that's a sucky way to do business, and you're not about > > > to go along with that game plan, then that's fine, but you're no > > > longer doing physics. > > > > PD > > > -------------------------- > > crook PD > > one sign that you are talking a lot about how physics should be > > tells us the less physicist you are > > not a physicist you are a crook politician!! > > a parrot politician!! > > > there is just one kind of mass !! > > no matter how do you call it > > > it is shown easily by the momentum > > of the photon: > > > P (photon momentum) =m c !!! > > That's not the momentum of the photon. It is not the momentum of any > object known, in fact. > We've already discussed this. Case closed. > > > > > see the Plank momentum > > Pl momentum = Plank mass times c !!! > > > now > > in that m c > > > THERE IS NOTHING 'RELATIVISTIC' > > (or anything else ) > > EVEN IF YOU WILL STAND ON YOUR BLOCKED > > crooked HEAD > > no gamma factor non of the slightest sign that > > that mass is something else than just **mass!!* > > the only mass that exists in that M K S system !! > > (it is the K there !!) > > > FULL STOP !! > > NO NEED FOR LONGER MUMBLINGS !!!! > > keep well > > Y.Porat > > --------------------------- so why should it be called (by scientists ) Plank momentum ?? does Plank momentum is by principle another kind of photon momentum?? may be you can prove that it is a new kind of momentum?? while its dimensions are the same as your dimensions of photon momentum!! it cant be by definition other dimensions !! including mass times c P is not zero c is not zero so were do you see any zero !!!! AND WERE ANOTHER GAMMA ?? 2 see my answer to Anderson where the hell you see a gamma factor in** his** (your** ) formula for photon momentum ??! what happens to gamma while v= c !!!! ATB Y.Porat -------------------
From: Paul B. Andersen on 19 May 2010 16:20 On 19.05.2010 16:15, Y.Porat wrote: > On May 19, 3:31 pm, "Paul B. Andersen"<paul.b.ander...(a)somewhere.no> > wrote: >> >> The trend now is, as I am sure you know, not to >> use the term 'relativistic mass' at all. It is >> confusing and utterly unnecessary. >> >> It's only one kind of mass, the invariant mass. >> The photon mass is zero. > ====================== > > so ??!! zero ??!! >> p = m*gamma*v for massive objects, >> p = h/lambda for photons > > so if there is just one kind of mass > why do you say that for the photon it is zero--- Because it is. A photon is a massless particle. > > according to you > P photons is h/lambda Quite. This is the magnitude of the momentum. Since momentum is a vector, a better expression is p = hbar k, where k is the wave vector since |k| = 2pi/lambda, |p| = h/lambda You must be pretty ignorant of physics since you didn't know this. > > it is as well m c No. > see the Plank momentum !! 'Planck momentum' is a unit, equal to 6.52485 kg m/s. This is about the momentum of a rifle bullet, a gigantic momentum compared to that of a photon, which for visible light is in the order of 10^-27 kg m/s. The units 'Planck mass' and 'Planck momentum' are related through the equation p_planck = m_planck * c So what about the Planck momentum? > so > > were the hell do you see > > A GAMMA FACTOR TO MAKE IT 'RELATIVISTIC '???!!! > ( what is that gamma factor wile V=c ??!!!) Why would you like to have a 'gamma factor' in the photon's momentum? The frame dependency is taken care of through the lambda. It transforms 'relativistic' if you use the relativistic Doppler shift. -- Paul http://home.c2i.net/pb_andersen/
From: spudnik on 19 May 2010 23:04 depends upon which frame of reference, not neccesarily one of A's or B's; I know, that is totally elementary, dood. what ever, the one thing that is not approachable is the speed of light -- not encroachable, at any rate, unless you happen to be a photon with no momentum. > What kind of lunacy prompted Einstein to say > the speed of light from A to B is c-v, > the speed of light from B to A is c+v, > the "time" each way is the same? thusNso: which has more decimal places: the integer value of Avagadro's No., or the surfer's canonical value of pi? thusNso: "infinite descent & more" is just a contradiction of some sort, which assuredly is plausible for such a negative conjecture/theorem. why is unique factorization problematic for these non-allowed integral values (assuming, Fermat was correct, for once ?-) whether mod arithmetic Day One is inadequate, I don't know enough of it to say. > no problem with quadratic reciprocity, though. thusNso: twins are always of the form, 6n plus and minus one? thusNso: on the wayside, please, attempt to "save the dysappearance" of Newton's God-am corpuscular "theory," by not using them in equations with "momentum (equals mass times directed velocity)." thusNso: actually, receding glaciers are probably better for rafting, compared to advancing ones, iff there's more water. thusNso: can one tell a priori that a black surface will absorb more infrared, since it is invisible in the first place, invoking, perhaps, blackbody curves (and, there are "line spectra" for both absorption & emmission) ?? I wish folks like Y'know and y'Know would at least *try* to write their syllogistical theories in terms of, "There Are No Photons?" just this afternoon, a lecturer showed a slide with a graph of "phonons from 0 to over 1 teracycles;" is that the sound of light? http://www.xs4all.nl/~johanw/PhysFAQ/General/LightMill/light-mill.html thusNso: I like all three of those; note that there is a raw infinity of trigona, two of whose edges are perpendicular to the other edge, as far as spherical trig goes, and I really like those "half lunes." --y'know dot the surfer's value of pi dot com period semicolon & I mean it! http://\\:btty
From: spudnik on 19 May 2010 23:06
I looked at your website; are you sure it's not steganogrphy? I don't know, becuase I have never been able to see one of those -- don't tell me, how! Hide quoted text - > > - Show quoted text - |