E=Mc^2

So if I were to continue adding energy to a particle (say a proton, mass 1 atomic mass unit) some of the energy would become matter and the particle would become heavier. This has been proven in particle accelerators.

When this energy is converted to matter, what type of particle is it converted into? Are quarks created, giving you a proton consisting of a greater number of quarks and thus, a greater mass? Could somebody please explain

Here's the way I understand it:

Every different kind of particle has a particular rest mass (m[sub]0[/sub]), which is an inherent property of that particle. This is the mass measured by an observer in a frame of reference at rest with respect to the particle.

Now when you "add energy" to the particle, such as is done in an accelerator, what you are doing is adding kinetic energy associated with its motion, which according to Mr. Einstein is equivalent to increasing its mass as measured by an observer past whom the particle is whizzing. At relativistic speeds the increase in mass can be appreciable.

Yet none of this changes the inherent rest mass of the particle, and hence it keeps its identity unless it collides with and interacts with other particles. So making, say, a proton (composed of three quarks) heavier by accelerating it doesn't change its constituent makeup.

As you may know, no physical theory has yet been devised to explain the exact rest masses of the various particles of matter. These values have to be measured experimentally.

This equation relates to electro-magnetic rays, and those alone.

The mass represented in this equation by the 'm' is the change in mass, or the difference in mass. Or, phrased differently, the amount of mass used up.

So, basically the variable quantity (practically) is not really the mass but the energy!

Hope I've helped,
Pin~Jinx

The complete form of the equation is E² = m₀²c⁴ + p²c², where m₀ is rest mass and p is relativistic momentum.

For objects at rest, p = mv = 0, so the above simplifies to E = mc².

For electromagnetic waves, m₀ = 0, which gives E = pc, and p = E/c. If we let p = mc, we get m = E/c², which means that due to its momentum a photon has relativistic mass. However, as Professor points out above, this "mass" is bound up in kinetic energy.

The m in E = mc² is not a difference in masses in any sense of the term. It refers to an object's rest mass, which is an absolute value.

MT: Thanks for the edifying clarification On notation: It sounds like "m-sub-zero" refers only to photons or other massless particles, rather than to rest mass in general, is that right?

Also, how did you make the HTML subscript? Does it have to do with using angle brackets instead of square brackets?

Mack is fully capable of answering on his own, and I mean no presumption in answering in his stead. The notation m₀ is used to denote rest mass of any kind, that of fermions or bosons equally, and also that of very large aggregate masses in connection with gravity, e.g.

Yes, using angled brackets instead of square brackets is the trick.

Thanks, Maiku.