Cham,
Spaceman Spiff,
there are so much misunderstanding and confusion in your response, I will not go into all the details here. This is just too discouraging.
I think I've made myself as clear as I can, notwithstanding I am wrong, and it's better to be clear and wrong than muddled, right or wrong. I strive for clarity so that my mistakes can be spotted and corrected. I note that you came in for more correction by Fridger than I, though I've had to voluntarily correct myself regarding gluon rest mass.
I'm certainly not throwing all you say into the 'bs' bin, but I also think my comments about your equations hold, and note that the point where I really got exasperated with you was about you thinking that Einstein didn't know about photons when he published SR.
Indeed, I find your writings (on this topic) very confused and muddled. That BlindedByTheLight had to re-ask several times might show this, and statements like these below really don't help me understand.
Cham wrote:Photon don't have a mass. But it has energy and momentum !
How does something have momentum but no mass?
Cham wrote:The formula [ E = mc?? ] still apply to a photon, but then, the "m" doesn't mean anything. It may be called "mass of the photon", but this is really arbitrary and doesn't have any experimental meaning without gravity.
Please explain further why the mass of the photon has no experimental meaning without gravity.
Cham wrote:I have to insist, the formula E = m c^2 has a clear physical meaning ONLY IN THE CASE OF MATTER (massive particles and objects). IT IS AMBIGUOUS for photons. You may use it, if you wish, to say that a photon has a mass if you write m = E/c^2, but this is an abuse of language and has no physical meaning.
Can I or can't I apply this equation to photon energy or not? Why is it only to be applied to rest mass? Well, you're repeating and insisting, but not explaining...
Cham wrote:Equation (1) doesn't have any physical meaning because it's simply a definition. It is not a RELATION between TWO observable things ("m" isn't observable, or measurable if you prefer). Equation (2) IS a relation between two independant entities, which can BOTH be subject to experiments. Just replace (1) into (2) so "m" vanish entirely from the equation. What remains are "E" and "m_0", which are the true observable things here.
Aren't
E and
m0 not observables? You have to calculate those from observed effects. In fact, I think energy (like momentum) is never an observable in physics, it's always derived from other observables. In this case, relativisitic mass is more observable than rest mass. You apply the Lorentz transform to what mass you measured, and get the rest mass.
OK, on with the rest...
Cham wrote:MASS DO NOT INCREASE WITH VELOCITY.
Ah, you're one of them what quibbles about the whether the Lorentz transform should only be allowed to apply as a time dilation only, or a space contraction only, but never as an equivalent change in mass. Or do you also want to ban the idea of time dilation and insist it's only space contraction, or
vice versa?
I think you'll find the quantity
m, as in 'relativistic mass', is just as observable as any other mass quantity in the usual way, the experiment transduces the effects of particle mass into a length or time measurement. One could think of the Lorentz transform as a correction factor due to the particle velocity
v relative to one's experiment at the time, because the relativistic particle's time dilation or space contraction causes it to interact with the experiment differently, and that causes us to measure a different mass. It's just an equivalence, but not meaningless. Meaningless is like dividing by zero.
Anyway, what I wrote about was what you find if you try and experimentally measure masses of speeding particles: you will find a higher mass which is the rest mass factored by the Lorentz transform. This site seems to back me up:
Mass-energy equivalence (
http://en.wikipedia.org/wiki/Mass-energy_equivalence ). It also points out that the kinetic energy put into the accelerated particle does indeed add mass to the particle.
Cham wrote:As I have shown previously, the "m" may be eliminated from all equations. And it even don't shows without ambiguity in the relativistic Newton equation :
But you are simply showing that what others conveniently call relativistic mass is just the rest mass factored by the Lorentz transform. This is just the 'mass dilation' way of thinking about it.
Some physicists point out that momentum doesn't really exist, and that it's not a meaningful physical quantity: it's just a useful book-keeping item for tracking motion. Some also claim energy doesn't really exist, and that it too is just an accounting tool in our theories... Some even say (as d.m.falk mentioned) that mass doesn't really exist, that everything must boil down to lengths or times. Fine, but if there is actually no such thing as mass, why do I displace my sofa when I sit on it?
Cham wrote:In any experiment, you cannot measure the particle's mass, while the particle is in motion. IT IS ITS KINETIC ENERGY. The only kind of mass you can measure, is the "rest mass" (or "proper mass", or whatever) while the particle is instantaneously at rest relative to the observer.
I think measuring the mass of a particle in motion is not an impossible case. One observes how one's experiment effects that very motion to extract the mass. How do they measure the masses of all those exotic sub-atomic particles flying though the bubble chambers at CERN? How do you actually directly measure the kinetic energy of something, rather than mass, or velocity? Surely, the relativisitic mass is measured, then the appropriate rest mass is worked out?
Cham wrote:About the photon : ... So you can't measure its "mass" without gravity. Even with gravity, the "m" does not have a clear physical meaning for a photon.
So I can't measure the momentum of any photon without gravity, or even with it? What's this
Compton scattering (
http://hyperphysics.phy-astr.gsu.edu/hb ... ptint.html ) stuff then? Where does that radiation pressure come from then?
Cham wrote:What the hell is "electromagnetic energy" anyway ? From Maxwell theory (which I know enough to teach it all year long), the total energy of some electromagnetic wave is given by the volumic integral of its electric and magnetic density over all volume of space (I will not write the equations here). IT IS KINEMATICAL AND POTENTIAL ENERGY.
Something like this:
http://en.wikipedia.org/wiki/Electromagnetic_energy.
Kinetic energy specifically refers to the energy stored in a massive body (as in finite, non-zero rest mass) due to motion, which is why the equation contains both
m and
v. It can be exchanged for potential energy in a gravitational field, or potential energy in an electrical field. What rest mass in a photon is moving in the oscillating electrical field so as to contain kinetic energy? (I'm not sure what you mean by kinematical, but I wonder if you mean kinetics, which does not consider mass, force or energy, only motion).
I did you a favour: I searched for the terms "photon kinetic energy" on Google. I got one result, not prominent, but try it:
Gamma Rays and Kinetic Energy (
http://www.newton.dep.anl.gov/askasci/p ... y00930.htm ).
Lo and behold, two people who say gamma rays have kinetic energy. But...
The first guy doesn't impress me. He supposes that if a photon slowed down by a tiny bit, it would lose all its kinetic energy, yet, it's travelling at near light speed. So, it has zero kinetic energy because it has no rest mass. This is what I say anyway! How does kinetic energy arise at the speed of light for a (rest) massless particle, and no other speed?
The second guy uses maths. Except he applies the
m of ??
mv?? to the photon's (non rest) mass (contradicting the first guy), and gets the
m from ...
m =
E/
c?? ! Then he finds that only half of the photon's energy is kinetic, and he isn't sure if the energy of the electric and magnetic field (what I referred to as electromagnetic energy) makes up the other half, or his application is invalid and electromagnetic energy makes up all the energy of the photon...
I'm not convinced.
Cham wrote:Timelessness has NOTHING to do with interference pattern.
You missed my point. You give the standard probability wave interference pattern argument. What I'm asking about is when there's only one photon.
Spiff.
P.s., yes, you can be exasperated at my idiocy too, but please do include clear explanations as to why if you can spare the time.
Just wanted to say thank you to everyone for chiming in (especially the professionals)... I will need to re-read this posting several times to make sense of it and let it sink in - as I am most definitely NOT a professional on these matters... just someone keenly interested.
