1
$\begingroup$

I have read some thermodynamical and statistical physics books but it feels like they always gloss over the definition of internal energy. In wikipedia they included terms like

  1. Kinetic energy. But each setting seems to have its own definition of kinetic energy (quantum physics, classical physics, special/general relativity, ...).

Also it seems to exclude things like

  1. Property X of the system as a whole. I have found no definition of what "property X of the system as a whole" means (in wikipedia they talk about the kinetic energy and the potential energy due to external forces as the properties). Does one subtract the expected value of X of the constitutents or what?

Sometimes the Hamiltonian is described as "the energy of the system". Is it different from the internal energy?

Is there any definition somewhere that actually defines all the terms, preferably using mathematics instead of words so the meaning is more clear?

$\endgroup$
1
$\begingroup$

Internal energy is a concept from thermodynamics, defined with help of the 1st law of thermodynamics. One formulation of this law, based on experience:

  • If a system is in a state of thermodynamic equilibrium $X_i$, addition of definite amount of heat to it with no work being done results in certain equilibrium state $X_f$. The same result may be accomplished by addition of corresponding amount of work while no heat is being transferred.

The system can change its state from initial $X_i$ to final $X_f$ via many different processes, which use different timing and amount of heat and work transferred.

The 1st law can be used to show that for any two such states of thermodynamic equilibrium, the sum of work and heat added while the system goes from $X_i$ to $X_f$ is always the same, irrespective of the process used. Mathematically, the integral

$$ \int_{t_i}^{t_f} \bigg(\frac{dW}{dt} + \frac{dQ}{dt}\bigg)\,dt $$

where $t$ is any parameter increasing monotonously with time, depends only on the two states $X_i, X_f$. It does not matter whether the process is reversible.

Now imagine a process where only work is being transferred; no heat is being transferred. The integral is

$$ \int_{t_i}^{t_f} \frac{dW}{dt}\,dt. $$

If we fix the state $X(t_i) = X_i$, the integral depends only on the final state $X(t_f)$, which we are free to choose from all the available states of the system.

Hence we can define function of the state $X$

$$ U(X) = \int_{t_i}^{t_f} \frac{dW(t)}{dt}\,dt, $$

where the process is chosen in such a way that the final state at time $t_f$ is $X$. This function is called internal energy in thermodynamics. In short, internal energy is work added to the system to get it from reference state to the current state with no heat transfer.

Because the reference state (for example, $T=373.15\,$K,$p=101325\,$Pa) is not unique, internal energy is not a unique function of the state variables; one can add any constant to it. The difference of $U$ for two states, which is measurable as work or heat, will be the same.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Now work and heat is undefined instead. You just moved the undefined-ness :) $\endgroup$ – Emil Jun 17 '15 at 20:45
0
$\begingroup$

Once the center of mass is defined, the total energy $E$ can be split as

$$E = K + U$$

with $K$ being the kinetic energy of the center of mass, and $U$ the internal energy. Non-relativistically

$$E = \frac{1}{2}Mv^2 + U$$

with $v$ the center of mass velocity. So one can define internal energy as

$$U \equiv E-K$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.