E=MC2 ... or not? Problem of massless stable particles.?

Question

OK. I'm no physicist, but the formula and the physical evidence somehow don't add up.

Most sources indicate that the photon is "massless" (M=0). A table in Zukov's "Dancing Wu Li Masters", compiled by Lawrence Berkeley Laboratory, adds graviton, electron neutrino and muon nutrino to the list of M=O particles. Well, if in these cases M=0, applying the simple math of the E=MC2 formula, E (not C?) would ALSO equal 0 for these particles. However, for the photon, at least, clearly there IS some energy involved.

So ... is there any conventional explanation for this apparent contradiction ?

Any professional physicists out there with a ready answer?

Also ... if some particles have such small mass that they are simply "deemed" to be zero ... how are any meaningful calculations carried out on them? Doesn't the problem remain ... how to apply E=MC2 to such particles AND get a non-zero result for E and/or C?

Hmmm. ... momentum w/o mass? That's even more strange.

Answer 1

The "m" in E = mc^2 is the rest mass of the object.
Further more, E = mc^2 is not the complete equation, the object's momentum component is often omitted. Although photons have no rest mass, they do have momentum since they have energy (Per the Plank's equation, E = hv).

A photon of light's momentum is given as,
E = pc,
where E is the photon's energy, p is its momentum, and c is the speed of light in a vacuum.

Photons, by their vary nature, are never at rest.
Photons are the particle nature of light, hence they ALWAYS are traveling at the speed of light. In ANY and EVERY frame of reference, the speed of light is ALWAYS measured to be the same.

A graviton is still a theoretical particle, it has not yet been confirmed to exist.
As for neutrinos, experimental evidence has shown that these particles have mass, but it is just extremely small...so small, it is easier to measure it in energy equivalent terms (electron volts, eV).

Answer 2

First, the E=mc^2 formula applies to massive particles, not massless particles. It tells you that if all the mass of a particle was converted to energy the amount of energy that it would be is mc^2. So there is no contradiction. I've heard of some people giving the photon a "relativistic mass," but I don't believe that's proper. Eitherway, physicists never talk about relativistic mass.

Second, new evidence suggests (and I think this is the consensus now) that the neutrinos (there are 3 of them, not two - and maybe more in the form of "sterile neutrinos") have a small mass on the order of an eV (note that an eV is an energy unit. To get mass, divide energy by c^2). The list you looked at sounds like it is out of date.

So photons are definitely massless (as far as we can measure, and according to theory) and neutrinos have mass. Gravitons and gluons are also massless, though gravitons have not been experimentally confirmed.

So to answer your first question, again, there is no contradiction because the equation only applies to massive particles (massless particles transform in a different way - they can travel at c).

Your second question can be answered very simply: Not all equations in physics require a mass. As a matter of fact, a significant number of them don't. The mass of a particle basically tells us how that particle couples with spacetime via Inertia and Gravity. It requires something called the Higgs Boson that may or may not exists (we should know within 10 years through the Large Hadron Collider).

Answer 3

E=MC2

The important thing to note here is that mass and energy ARE equivalent - i.e. the same entity in different forms.

Light is pure energy. Its rest mass is completely converted to energy. The formula predicts both the light's energy potential and its total rest mass. Once the mass is completely converted to energy, an amount the formula predicts, the formula no longer applies - i.e. there is no M to be converted any longer.

A simple DC electrical analogy would be E=IR. Where E is the voltage, R is the resistance and I is the current. If the resistance is lowered to zero (i.e. a short circuit), according to the formula, the voltage would be zero and, hence, the current would also be zero. That DEFINITELY isn't what takes place when you short circuit the load resistance !
Again, the formula only holds true if we are working with values above zero.

Answer 4

I am not a professional physicist, but I have taken college-level physics as an engineering major.

A photon is in fact massless. The E=mc2 equation describes how energy can be converted to mass and vice-versa. A photon is pure energy but that energy could be converted to mass. Likewise, the explosion of a nuclear weapon produces a great amount of energy by converting a small amount of mass to energy.

That's my understanding, but college was 10 years ago, so my memories are a little rusty.

Answer 5

The easy answer for me is that neither general or specific relativity pretend to address quantum mechanics in any way. More than that - I cannot say except that I don't believe in M=0 particles. I believe in M=0 virtual particles, but those only have a probability of existing in the first place, so how can they probably have mass? That isn't intuitive for me. Like yourself, I am not a physicist, but I read a lot, and what I read tells me that all conventional "particles" have at least some mass.

Answer 6

The energy (E), mass (m), and momentum (p) of any particle are related as the sides of a right triangle in Pythagoras' Theorem:
E² = (mc²)² + (pc)².
This is the general form for the total energy in Special Relativity, and it is true for every particle. It reduces to E = pc for the case of mass-less particles (as a photon of light), and to E = mc² for "momentum-less" particles, that is, things at rest. The total energy is always greater than zero because mass-less particles are never at rest.

Answer 7

M = 0 particles got their energy from somewhere else. Electrons with more then 30,000 electron volts, striking a tungsten target 1 x 1 mm give up some of their energy in the form of X ray photons. This is how an X-ray machine works.

Answer 8

You are applying a formula with no knowledge of where it came from or what it was meant to cover.

"E=mc^2" is a rather old way of writing things. It confuses rest mass-energy with kinetic energy. The more usual way is "sqrt(E^2-P^2c^2)=Mc^2" which expresses the length of the energy-momentum 4-vector in terms of the rest mass M. Notice that when M=0, E=cP (energy is proportional to momentum). Light always has momentum--as does anything which travels at the speed of light (because there is no rest frame for it).

Answer 9

The photon is massless, but it does have energy according to Plank's Law E=h*f, where f=frequency of the light, and h = Plank's constant. According to relativity theory that energy will act like mass in a gravitaional field and in conservation of momentum stiuations (collisions). Therefore it will be subject to deflection in a gravitational field (as has been observed).

Answer 10

E=mc^2 is a expression for rest energy, ie energy a particle would have at v=0. All this energy is due to its mass. Photons are massless but they are not at rest.