The $c^2$ factor is just a conversion factor from mass to energy? I have some questions:


*

*The $c^2$ factor is just a conversion factor from mass to energy. What does it mean? I know it's about $E=mc^2$ and so on, but really I need deep understanding of it.

*I know that the rest mass is used in undergraduate books and mass is just a number associated to the particle and the mass is really energy.
But does rest energy have something to do with rest mass?
 A: I'm assuming your first question is about the meaning of $c$ itself as opposed to $c^2$. In relativity, space and time are united in a geometric framework called $spacetime$. In the absence of gravity, the 'distance' between events in spacetime is given by the spacetime metric $\Delta s$,
$$(\Delta s)^2 = (c\Delta t)^2 - (\Delta r)^2,$$
where $\Delta t$ is the time between the two events and $\Delta r$ is the distance between them.
Even though they are united in the same framework, space and time are not the same, and if they are measured with different units then a universal constant conversion factor is necessary to convert between them. $c$ is that factor. 
Let's rephrase your second question: does rest mass have something to do with rest energy? Yes. The rest energy of an object is its energy measured in a reference frame in which it is at rest (or its center of mass is at rest). Einstein called it the Energieinhalt, the 'energy content'. It's an internal energy that includes the kinetic and potential energies (and masses!) of all the component parts of the object interacting via various fields at the atomic and subatomic levels. If you remove some quantity $\Delta E$ of this internal energy, the mass of the object will decrease by an amount $\Delta m = \frac{\Delta E}{c^2}$.
A: Maybe the most mathematically sound way to define this stuff is to do it like this:


*

*Define the proper time of an infinessimal spacetime interval to be


$$d\tau^2 = dt^2 - dx^2 - dy^2 -dz^2$$


*Define the four velocity of a particle to be


$$u^\alpha = dx^\alpha /d\tau)$$
where $\alpha$ is just an index that ranges from 0 to 4


*Define the four momentum of the particle to be


$$p^\alpha = m u^\alpha$$
$p^0$ would be called the energy $E$.
In the way I have done things (and the way most people do things now) the mass is just the scalar that we multiply the four velocity by to get the four momentum. Sometime it is called the "invariant mass," because anybody who measures the length of the four momentum in any reference frame will get the same mass.
"Proper mass" is a concept that is going out of style. In the past, people thought that
$$p= mv$$
So they wanted to change the definition of mass in order to make this equation still work. Particles can't travel faster than $c$, but they can still get infinitely high momentum.
The resolution was to say that mass is the proper mass times the gamma factor.
$$m = \gamma m_0$$
However, nowadays people have no problem writing
$$p=m\gamma v$$
for the $\alpha=1,2,3$ part of the four momentum. Here $m$ is the invariant mass.
The length of the four momentum is
$$m^2 = E^2 - p^2$$
or
$$E = \sqrt{m^2 + p^2}$$
That's the way most people do it pedagodically now.
A: A remarkable achievement of SR (special relativity) is the equivalence between energy and mass as expressed in the rest frame of a massive particle (three momentum nil), that is $E = m c^2$, where $m$ is the rest mass of the particle.  
As for the $c^2$ factor, it is just due to the units of measure. In natural units, $c = 1$, you simply read $E = m$. This clearly states that the rest energy of a massive particle is its rest mass.
