Is the concept of rest mass correct? Is the concept of rest mass correct?
All these years, we (me, and my classmates of Undergraduate 1st Year) have been accustomed to the concept of rest-mass, and the relativistic transformation of mass at speeds of the order of c. We had been introduced to this concept during our school-years, we were comfortable with it for quite a long period of time, until yesterday, when our teacher declared with a convincing degree of confidence, that it is not mass that changes relativistically, but rather, it is the momentum possessed by the body, that does so. Mass, as he says, is an invariant. P.S., he even declared that the conception that we had so long "cherished" in our hearts, is a wrong-one. Naturally, the whole class was wonder-struck. I am looking here for views on the subject from other, reliable sources, because I am (rather, we all are) in rather a big-dilemma whether to accept our teacher's view, or to reject it. You should know, that our undergrad. classes have started just 3 days ago, and at the very onset, we have faced a heavy-weight punch.
 A: Your teacher is correct.
The link you give, gives the correct definitions of the various "mass" concepts in special relativity.

m_0 here characterises an entity that is being described by a four momentum vector in the special relativity framework.
In vector spaces, the vector has an invariant length, (otherwise mathematically it would not be a vector space).
Elementary particles are the building blocks of all matter and energy that we see. They have a very specific rest/invariant/proper mass that characterizes them, together with the other quantum numbers.
In vector algebra, vectors can be added vectorially. The proton, for example is composed of three quarks and a soup of gluons. The four vectors of the internal constituents are bounded by the mass of the proton and when added vectorially the "length" is the mass of the proton. From the formula we see that if the momentum is zero the system has only its rest/proper/invariant mass.
The confusion comes from the famous E=m*c^2 formula , where m is the relativistic mass of an object moving with velocity v.That is variable and depends on the velocity of the particle under observation. 

It comes from the classical definition of momentum. Momentum relativistically is


which is the ordinary definition of momentum with the mass replaced by the relativistic mass. 

Even though the concept of relativistic mass gives a continuity between classical Galilean transformations and the framework of very high relativistic velocities, particle physicists have stopped using the term. Mass in particle physics is the rest mass, and all formulae refer to the four vectors of the problem ( E, px,py,pz) and use the rest masses only, to avoid such confusions.
A: It's a matter of terminology. You can define an object's mass to be proportional to its energy in the reference frame you observe it from, which includes kinetic energy in the mass, or you can define it to be proportional to the object's energy in its own rest frame, which excludes kinetic energy. Or you can do both, which is what physicists originally did shortly after relativity was first developed: the former definition of mass, which depends on speed, was called the relativistic mass, and the latter definition was called the rest mass.
In modern times, the convention has largely shifted, and most physicists will just use the word "mass" to refer to what used to be called rest mass: the energy of an object in its own rest frame, which does not depend on the object's motion. In this convention, mass is an invariant, and momentum and energy are not. What you used to call "mass" will now be known as the total energy. I would strongly recommend that you adopt this convention because it's what most other people in the physics community are using.
That being said, you're not going to find an authoritative source saying that one usage is correct and the other is not, because strictly speaking, neither one can be correct - it's a conventional choice.

Let me also refer you to this answer of mine which says more or less the same thing in more words.
