# The Mean Value of a Mass Particle's Energy

1) Suppose the state of the system is determined by the variables $P_1, \ldots, P_n$ and let there be a sub-system, determined by the variables $p_1, \ldots, p_n$, which belongs to $P_1,\ldots, P_n$ but the energy of the whole system be divided into two parts -- one, energy $E$, which only depends on $p_1, \ldots, p_m$, while the rest is independent of $p_1,\ldots, p_m$. The rest of symbols have their usual meaning. Then, the question is, why is it that the probability $dW$ for anyone of these $p_i$'s to be within the region $dp_1\ldots dp_m$ should be given by $$dW = C e^{-\frac{E}{kT}}~ dp_1 \ldots dp_m. \tag 1$$ Isn't the energy $E$ the function of these $p_1,\ldots,p_m$ and, therefore, isn't also their change determining the $E$ and not vice versa? How is then the probability of the $p_1,\ldots, p_m$ being in the region $dp_1 \ldots dp_m$ determined by $E$? It must be somehow the opposite; that is, $E$ is determined by $dp_1 \ldots dp_m$?

2) Now, once this is understood, the question arises where this integral comes from $$\int_{dE} dp_1 \ldots dp_m = \omega(E)~dE.$$ The function $\omega(E)$ has the appearance of probability density but how exactly that whole thing came about, let alone that it caused the substitution in eq.(1) of $dp_1 ... dp_m$ by $$\omega(E)~dE:~ dW = C e^{-\frac{E}{kT}} \omega(E)~dE.\tag 2$$

3) If now one observes a linear harmonic oscillator, whose coordinates are wrt center of mass, then eq.(2) is written as $$dW = C e^{-\frac{E}{kT}}~ dx~ dv,\tag 3$$ where $x$ and $v$ are, resp., the coordinate and the velocity. Where does eq.(3) follow from, especially, when it is claimed that $\int{dx~dv} = \textrm{const}$, claiming further that this yields $\omega = \textrm{const}$, having in mind that, obviously, $E = ax^2 + bv^2$. So, that whole thing leads to $$dW = \textrm{const} \cdot e^{-\frac{E}{kT}}~dE.$$ How did this come about?

4) And, finally, it is claimed that $$\frac{{\displaystyle \int E e^{-\frac{E}{kT}}}~dE}{\displaystyle \int e^{-\frac{E}{kT}}~dE} = kT = \bar{E},$$ however, my best efforts to solve that quotient yield $E + kT$ and not just $kT$, let alone that the meaning of the quotient isn't quite clear. Can someone shed some light on these points? Thanks.

NOTE: Regarding point 1): Maybe the meaning of the probability $dW = C e^{-\frac{E}{kT}} dp_1 \ldots dp_m$ can be understood by considering that while, indeed, as defined, $E$ is a function of all $p_1 \ldots p_m$ (but not of their differentials $dp_1 \ldots dp_m$), a given $E_1$ may be the result of only the change of one of these $p_i$'s, the rest remaining the same, unchanged. Of course, such $E_1 = E_1(p_1 + dp_1 \ldots p_m)$ will differ from $E_m = E_m(dp_1 \ldots dp_m)$ and, therefore, if we write the whole set $dp_1 \ldots dp_m$ of differentials of these variables, then, if $E_1$ is our concern, the probability of all the $p_i's$ (determining $E_1$) to reside within $dp_1 \ldots dp_m$ will be $dW = C e^{-\frac{E_1}{kT}} dp_1 \ldots dp_m$; that is, there will be zero probability for any of the $p_i's$ to be within $dp_1 \ldots dp_m$, except for $dp_1$.