DrStupid said:
In another thread I demonstrated that relativistic mass is the same quantity as the mass as used in non-relativistic physics:
https://www.physicsforums.com/threads/relativistic-mass-still-a-no-no.892981/page-2#post-5620010
And relativistic mass is not mass in the modern sense of the word.
That's not gravitational mass but mass. The source of gravity in GR is the stress-energy-tensor. The fact, that this tensor may depend on mass only in special cases desn't turn mass into geravitational mass.
Well, there are many writings about these issues in popular-science websites and other non-peer-reviewed sources that are not up to date or even incorrect.
Again: To get the correct relation between Newtonian and special-relativistic mechanics one must not (I emphasize one must not!) use relativistic mass. The reason is simple: In Newtonian physics mass is frame independent, i.e., a scalar under Galileo transformations. If you use relativistic mass, the Newtonian limit, which you get by expanding in powers of ##v/c##, and there you see that the various relativistic masses depend on the speed of the particle at order ##\mathcal{O}(v^2/c^2)##, which is the same order you have to take for kinetic energy to get the correct non-relativistic limit.
The correct way to guess a relativistic equation of motion from a given Newtonian one (you cannot strictly derive it, because SR mechanics is more general than Newtonian mechanics), you have to go to an instantaneous inertial restframe of the particle, where Newtonian mechanics is supposed to hold as an approximation, and then write down the Newtonian equation of motion in terms of covariant quantities. Then you have a good guess for an SR-mechanics equation of motion.
For the kinematic quantities it's clear that to get a covariant expression you have to extend the position vector ##\vec{x}## by the time-poisition four-vector ##(x^{\mu})##. The time in the intantaneous rest frame defines the proper time of the particle. It's given in covariant form by
$$\mathrm{d} \tau=\frac{1}{c} \mathrm{d} s=\frac{1}{c} \sqrt{\mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \eta_{\mu \nu}}.$$
Thus the velocity and acceleration in Newton's Law is most conveniently extended to the corresponding four-velocity and four-acceleration
$$c u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}, \quad a^{\mu} = \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}.$$
The simplified equation of motion for a point particle in Newtonian physics ##\vec{F}=m \vec{a}## is thus generalized to
$$m a^{\mu}=K^{\mu},$$
where ##m## is the invariant mass of the particle, and it's the same mass as in Newtonian physics, i.e., a scalar (in Newtonian physics under Galileo transformations in SR under Poincare transformations).
Of course also in SR there are only 3 independent degrees of freedom. That's because by definition
$$u_{\mu} u^{\mu}=1 \, \Rightarrow u_{\mu} a^{\mu}=0.$$
The Minkowski four-force ##K^{\mu}## thus must fulfill the constraint
$$K_{\mu} u^{\mu}=0.$$
It implies that both ##a^{\mu}## and ##K^{\mu}## are space-like four-vectors.
Of course also energy and momentum easily are derived in this way. The three-momentum in Newtonian physics is ##\vec{p}=m \vec{v}##. This suggests to extend this definition to the relativistic four-momentum
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=m c u^{\mu}.$$
That indeed ##p^0=E/c## is an expression for the kinetic energy (up to an additive constant which is physically irrelevant) can be seen by expanding in powers of ##v/c=|\vec{u}|/u^0##. Note that ##\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t## are not (!) spatial components of a four-vector because one takes a derivative of space components of the four-vector ##x^{\mu}## with respect to the coordinate time ##t##. Of course we can use, for a moment, such non-invariant quantities to get the non-relativistic limit by expanding in powers of ##v/c##. For energy we have
$$E=mc^2 u^0=m c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau}=m c^2 \frac{1}{\mathrm{d} \tau/\mathrm{d} t}=m c^2 \frac{1}{\sqrt{1-\vec{v}^2/c^2}}=m c^2 \gamma.$$
Now expanding in powers of ##v/c##
$$E=m c^2 \left (1+\frac{v^2}{2 c^2} +\mathcal{O}(v^4/c^4) \right )=m c^2 + \frac{m}{2} \vec{v}^2 + \mathcal{O}(v^4/c^4).$$
This shows that in the non-relativistic limit we get, up to the constant rest-energy ##E_0=m c^2## the Newtonian kinetic energy. To include the rest energy in the energy of a relativistic particle is convenient, because as we've just shown, then you get a Lorentz-four vector and make energy and momentum co-variant quantities as they are components of the energy-momentum four-vector
$$(p^{\mu})=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}.$$
The mass is, by construction, a scalar obeying the "on-shell condition"
$$p_{\mu} p^{\mu}=m^2 c^2.$$
For a derivation from the point of view of Noether's theorem, see my SR FAQ article:
https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf