What is the experimentally measurable difference between rest mass and the 'relativistic mass' of the photon ??!!
in [Relativity]
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From: Tony M on 22 Apr 2010 09:56 Artful, PD, 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? Furthermore, how would one determine the center of mass for this system? Which individual masses of the particles should one consider when doing this? Are photons taken into account for the center of mass? Assuming now there are nuclear reactions taking place in this system (which we defined as closed => all products of these reactions are considered part of the system), how would these reactions affect the invariant mass, center of mass, momentum and total energy of this system?
From: PD on 22 Apr 2010 10:26 On Apr 22, 8:56 am, Tony M <marc...(a)gmail.com> wrote: > Artful, PD, > > 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! 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. > > Furthermore, how would one determine the center of mass for this > system? Which individual masses of the particles should one consider > when doing this? Are photons taken into account for the center of > mass? It's not necessary to choose the center of mass frame for this system, as the invariant mass is invariant, independent of frame. > > Assuming now there are nuclear reactions taking place in this system > (which we defined as closed => all products of these reactions are > considered part of the system), how would these reactions affect the > invariant mass, center of mass, momentum and total energy of this > system?
From: Tom Roberts on 22 Apr 2010 11:23 PD wrote: > 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. The way to explain this is to note that mass is not really the important physical quantity. For instance, mass is not an extrinsic property of systems -- the mass of a collection of objects is in general NOT the sum of their individual masses. [Though that is often said to be so when speaking loosely, using a common PUN that interprets "[total] mass of a system" differently from "mass of an object".] In our current theories, the important physical quantity is 4-momentum. It is an extrinsic quantity -- the 4-momentum of a collection of objects is the sum of their individual 4-momenta. This then simplifies the answer: the mass of any object is the norm of its 4-momentum. This applies to the individual constituents of a system, and to the system as a whole considered as a sum of its parts. For a collection it yields the formula PD gave above (his p is 3-momentum). Tom Roberts
From: Y.Porat on 22 Apr 2010 12:16 On Apr 22, 4:26 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Apr 22, 8:56 am, Tony M <marc...(a)gmail.com> wrote: > > > Artful, PD, > > > 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! > > The way to find the invariant mass is very simple: > m^2 = (sum[E])^2 - (sum[p])^2 ------------------ very simple (:-)indeed but i showed you that p=mc AND NOTHING THERE TO MULTIPLY IT BY ZERO !! NORE GAMMA FACTOR TO MAKE IT DIFFERENT *QUANTITATIVELY* FROM REST MASS ??!! so now you **invented **different masses in different circumstances but we see that **quantitatively* you have no base for your new invention ... yet bingo i stated to understand you !!: while one sort of mass is going to a wedding it has a happy face and while i t is going to a funeral it gets a sad face so we got here a new kind of chameleon mass ---- --- A** CHAMELEON MASS **!! from the *chameleon* school directed by PD !! what are those CIRCUMSTANCES ?? it is ONLY for the genius (Shakespear )PD to decide !! i just wonder what else and how many new circumstances and the related kinds of mass !! (:-) Y.Porat --------------------------------- SO WHAT IS THE DIFFERENC EBETWEEN THE TWO ??!! > 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. > > > > > Furthermore, how would one determine the center of mass for this > > system? Which individual masses of the particles should one consider > > when doing this? Are photons taken into account for the center of > > mass? > > It's not necessary to choose the center of mass frame for this system, > as the invariant mass is invariant, independent of frame. > ----------------------- > > > > Assuming now there are nuclear reactions taking place in this system > > (which we defined as closed => all products of these reactions are > > considered part of the system), how would these reactions affect the > > invariant mass, center of mass, momentum and total energy of this > > system? ---------------------------
From: PD on 22 Apr 2010 12:26
On Apr 22, 11:16 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > On Apr 22, 4:26 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > On Apr 22, 8:56 am, Tony M <marc...(a)gmail.com> wrote: > > > > Artful, PD, > > > > 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! > > > The way to find the invariant mass is very simple: > > m^2 = (sum[E])^2 - (sum[p])^2 > > ------------------ > very simple (:-)indeed > but i showed you that p=mc No, Porat, if you'll look, I've told you that p=mc describes nothing in our universe. This formula for momentum is flat wrong. It does not work for *anything*. > AND NOTHING THERE TO MULTIPLY IT BY ZERO !! > NORE GAMMA FACTOR TO MAKE IT > DIFFERENT *QUANTITATIVELY* FROM REST MASS > > ??!! > so now you **invented **different masses > in different circumstances > but we see that **quantitatively* you have no base for your new > invention ... > yet > bingo i stated to understand you !!: > while one sort of mass is going to a wedding > it has a happy face > and while i t is going to a funeral > it gets a sad face > so we got here > a new kind of chameleon mass ---- > > --- A** CHAMELEON MASS **!! > from the *chameleon* school directed by PD !! > what are those CIRCUMSTANCES ?? > it is ONLY for the genius (Shakespear )PD to decide !! Nonsense, Porat. It's all pretty basic stuff, and it's all in pretty low level textbooks. I've not made it up, and it's not a new invention of any kind, and it isn't really up for a vote or an argument on a newsgroup. You either learn it or you don't. > i just wonder what else and how many new > circumstances and the related kinds of mass !! (:-) > > Y.Porat > --------------------------------- > > SO > WHAT IS THE DIFFERENC EBETWEEN THE TWO ??!! > > > > > 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. > > > > Furthermore, how would one determine the center of mass for this > > > system? Which individual masses of the particles should one consider > > > when doing this? Are photons taken into account for the center of > > > mass? > > > It's not necessary to choose the center of mass frame for this system, > > as the invariant mass is invariant, independent of frame. > > ----------------------- > > > > Assuming now there are nuclear reactions taking place in this system > > > (which we defined as closed => all products of these reactions are > > > considered part of the system), how would these reactions affect the > > > invariant mass, center of mass, momentum and total energy of this > > > system? > > --------------------------- |