A very general discussion-Not specific to a system:
The internal energy, $U$, of a system is a function of state, which means that its value only depends on the thermodynamic variables ($P, V, T)$ for example, at a given state (this means for a given set of values of these variables).
Let us make this more concrete: Imagine the system is in a thermodynamic state where the thermodynamic variables have the values ($P_i, V_i, T_i$) ($i$ stands for initial). At these values of the thermodynamic variables the internal energy has a value:
Internal energy at the initial state $i$: $U(P_i,T_i,V_i)$.
You can think of a gas at pressure, volume and temperature condition ($P_i, V_i, T_i$). Now imagine you change the thermodynamic variables to these ones ($P_f, V_f, T_f$) ($f$ stands for final). The internal energy now has a new value
Internal energy at the initial state $f$: $U(P_f,T_f,V_f)$.
In this process you have changed the internal energy of the system by an amount:
Change in U: $\Delta U= U(P_f,T_f,V_f)- U(P_i,T_i,V_i)$
I hope it is clear to observe that the system could have followed an infinitely large set of $(P,V,T)$-points, along an infinitely large number of different paths in order to go from state $i$ to state $f$. However, these are not, in any way, influencing by how much $U$ will change, you can take which ever path you please to go from state $i$ to state $f$. So the system has no memory of the intermediate states.
In mathematical terminology, this means that the differential change, $dU$, is a perfect differential and this is stated by the simple mathematical expression
$\oint_C dU=0$
It is very similar to the gravitational potential of the Earth, for example, which tells us that the amount of energy we need to spend to lift an object by 3m, does not depend whether we bring it straight vertically up or we follow some other path.
I am tremendously confused about the exact difference between thermal energy and internal energy
Internal energy $E$ is all contained energy. And there are many ways to contain/store energy - one of them is thermally:
- In chemical bonds, when atoms join and form molecules
- In different phases of matter - ice at $0^\circ$ contains less internal energy than water at at $0^\circ$, because some energy has broken then bonds. Here a phase transition energy also called latent heat is the amount of energy needed to melt or freeze to do this phase change.
- As kinetic energy when particles move within a system
- As potential energies within the system when an object / objects is out of equilibrium (a book on a book-shelf wants to fall, or a compressed spring wants to jump out)
- As thermal energy, which is giving objects their temperature.
- Etc.
So no, internal energy and thermal energy is not at all the same. Internal energy is everything and includes thermal energy.
BUT when talking about ideal gases, their are no chemical interaction, no complex phase structure (because of no interaction), (usually) no significant potential energy. And the kinetic energy of each of the atoms is just the same as thermal energy or temperature. So here the inernal energy equals thermal energy - but only for an ideal gas.
Some websites loosely state that it is the "jiggliness" of the atoms/molecules that constitute the thermal energy of the system. What are they trying to say?!
Let's make this a bit clearer. Thermal energy is what we measure as temperature. So on the macroscale, if is felt as warmth/coldness.
But on the microscale, thermal energy is simply vibrating atoms. In a solid, they vibrate at their spot and more and more violently, as they are heated up. Soon they are heated so much that they vibrate so much that they rip themselves loose from the structure - and the materials is now melting.
This is why the word "jiggliness" is used. Thermal energy and temperature is in fact just microscale kinetic energy of the particles.
Best Answer
Your definition is a bit funky. I'll explain.
Explanation
As much as I like that you explicitly express internal energy $U$ as a function of state, for clarity I'm going to drop $\boldsymbol{R}$ for now.
We can't directly measure the absolute internal energy of a system, we can only infer its change by understanding some defined process. Consider the change in internal energy between two arbitrary states. By definition this is:
$\Delta U_{1 \rightarrow 2} = U_2 - U_1$
This applies equally to the case where we are considering internal energy relative to some reference state. We can express this as
$\Delta U = U - U_\text{ref}$
It doesn't matter what our reference state is, we can set $U_\text{ref} = 0$ so that
$\Delta U = U$
For a given state, internal energy is always relative to the internal energy of some other (reference) state. In practice, analysis of a non-trivial system will involve multiple states, we could choose any as the reference or pick an appropriate standard state; in any case setting the value to zero or dealing exclusively with $\Delta U$ has a similar effect.
For some arbitrary state of interest (rather than processes), I'd argue setting $U_\text{ref} = 0$ is not only convenient, it's semantically the most appropriate thing to do. We make it zero by definition rather than making it, say $12.7$ (which would mean $U = \Delta U + 12.7$). There's no such thing as the "true" reference state, only the one you define.
Visualisation
Here's an illustration to underline why the absolute value of the reference state is unimportant.
This is a generic visualisation relevant to two arbitrary states of interest.
The relative positioning of $\boldsymbol{R}_\text{ref}$ doesn't affect $\Delta U_{1 \rightarrow 2}$.
If we set $\Delta U_1 = 0$, we're defining $\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$
$\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$ implies $\Delta U_{1 \rightarrow 2} = \Delta U_2 = U_2$