I know that energy is recognized through motion. Even in the mass-energy equivalence a velocity is present even though it is a rest-energy (Not really sure if this would count as a potential energy since there is no 'field' of acceleration that the mass is in)
So does kinetic and potential energy make up all other forms of energy by definition?
Originally even thermal energy was neither kinetic nor potential. Of course with the acceptance of the mechanical theory of heat we can interpret it as a manifestation of kinetic energy. Roughly, the distinction between kinetic and potential energy in classical mechanics reflected the difference between intrinsic energy of an object, and energy of interaction between objects, with the former reduced to the energy of mechanical motions (macroscopic or microscopic).
In this sense special relativity introduced a new form of energy based on the Einstein's mass-energy equivalence. In hindsight, the energy of electromagnetic field studied earlier by Lorentz and Poincare was a particular case. Another manifestation is energy released in nuclear reactions. One could say that as with the thermal energy this new energy microscopically "reduces" to the energy of subatomic motions and interactions, so it may not be that new. However, there is still a difference with classical statistical mechanics and thermal energy. When excited atom emits a photon we can not say that energy of some motion or interaction "in" the atom got "transferred" to the photon, which did not even "exist" before the emission, such classical parsing simply loses its meaning in quantum theory. Moreover, mass-energy equivalence assigns this internal energy even to truly elementary particles (with no parts that can move or interact) that are at rest.
Yes, at the fundamental level all energy terms are normally either kinetic or potential energy. The only demonstration of this that I know of requires a tool called the Lagrangian, which you might not be familiar with. But maybe you can at least get a flavor of how it goes.
The Lagrangian, very briefly, is a particularly useful way to represent all the possible dynamics of a system (in this case, a single particle). It can be used to find the equations of motion, and also the energy, which is what we'll do.
In general, the Lagrangian is a function of the position of a particle (conventionally called $q$) and all the possible time derivatives:
$L(q, \dot{q}, \ddot{q},...)$ .
This might seem like it has an indefinite number of possible forms, but actually there aren't that many:
-For starters, using Taylor expansion we can always write $L$ as a function of polynomials of these variables (okay, one could imagine a Lagrangian with a 'perverse' function that isn't Taylor-expandable, but I know of no such example that actually occurs in physics).
-Furthermore, it turns out that for a system with a minimum possible energy, only terms up to the first time derivative are possible. I refer you to an excellent previous question that discusses this.
-Now all the terms are like $\dot{q}^n q^m$, for $n,m$ as non-negative integers. However, many of these terms have no physical effect, because they can be made to vanish by a suitable integration by parts. This is true because it is actually the integral of the Lagrangian, $\int L dt$, that is physically significant (this is the action, which obeys the Principle of Least Action)*. Once this criterion is applied, only terms with $n=0$ or $m=0$ remain. (Note that this statement has been corrected)
Once you accept these arguments (which should take some time and careful thought!), the most general possible Lagrangian of a single particle is just:
$L=f(\dot{q})-V(q)+C$
Here $f(\dot{q})$ and $-V(q)$ are general functions of only velocity and position. I've also added explicitly a constant term, $C$, although this could have also been absorbed into the form for either $f$ or $V$. Now I will Taylor expand $f$:
$L=(\alpha \dot{q} + \beta \dot{q}^2+\gamma \dot{q}^3+\dots)-V(q)+C$
The prescription to find the energy of the particle from this (or, more precisely, the so-called Hamiltonian), is:
$H=p \dot{q} -L$, $p=\partial L/\partial \dot{q}$ .
So just plug this and simplify for:
$H=(\beta \dot{q}^2+2\gamma \dot{q}^3+\dots)+V(q)-C$
We end up with just a function of the velocity (with lowest term $\sim \dot{q}^2$) and the position. These can now be defined as the kinetic and potential energy. The constant term, again, could be grouped with either.
At sufficiently low velocity, we should expect that only the lowest order term in the kinetic energy is relevant. Then we have an energy like:
$H=\beta \dot{q}^2+V(q)-C$
Defining $\beta=m/2$, this recovers the normal non-relativistic form of the energy of a particle. $C$ has no effects on the dynamics in this limit, so it does not matter what value it is given.
However, for a relativistic particle the higher-order kinetic energy terms do matter, and one ends up with an energy like:
$$H=(\frac{1}{2}m\dot{q}^2+\frac{3}{8}m\frac{\dot{q}^4}{c^2}+\dots)+V(q)+mc^2$$ $$=\frac{mc^2}{\sqrt{1-\frac{\dot{q}^2}{c^2}}}+V(q)$$
As you can see, this exact form has the interesting aspect that the mass energy and kinetic energy end up having a combined expression, so it is somewhat natural to consider the mass energy as the constant part of the kinetic energy expression. But using the Taylor expanded expression one could justify grouping it with kinetic energy, potential energy, or as a separate category, so if you want to consider it as a third type of energy you can do that too.
This analysis was for a single particle, but field theories also show a similar division of the energy into terms involving derivatives of the field value and those involving the field value directly, which can be considered as generalizations of the kinetic/potential energy division.
*Note that I've made several (conventional) assumptions in these manipulations: most importantly, assuming that certain boundary terms can be neglected (which is generally true if the particle never reaches infinite distance in a finite time), and assuming that the Lagrangian does not depend directly on time, which corresponds to motion in a static field. The Lagrangian formulation as written above also can't handle a dissipative process like friction, but at the microscopic level dissipation is always just conservative coupling to a system with many degrees of freedom.
Edit: reading your comments to your question, I should emphasize that this is not a definition of energy, although this Lagrangian formalism turns out to be useful for that too. Take a look at this question for some interesting discussion about the best way to define energy.
There are only two forms of energy. Mass is one. The other is kinetic.
Potential energy is simply kinetic energy of a mass.
Heat energy is kinetic energy of individual molecules.
Electricity is the kinetic energy of an electron gas flowing through the metallic matrix.
Even light is the kinetic energy of a single photon.
Back to mass. Mass is a form of energy. When matter meets antimatter both are converted into various forms of kinetic energy such as heat and light and other radiation. All of which are kinetic.
Thus mass is equivalent to kinetic energy.
Proof of this is quite simple. Relativity says that when you accelerate a mass it's mass increases. As you get it near light speed then it's mass increases drastically. Where does this extra mass come from. That should be obvious. It comes from the kinetic energy of its velocity. Therefore mass and velocity (kinetic energy) are equivalent. Mass is a store of velocity.
So everything that exists is kinetic energy. Even matter is just energy. Kinetic energy.
