I wonder if the kinetic energy written as $\frac{d\mathbf x}{dt}\cdot d\mathbf p$ is related to chemical potential? I ask because if I use $\mathbf p = m \frac{d\mathbf x}{dt}$ as a constitutive equation for linear momentum, I get

$$d\mathbf p = dm \frac{d\mathbf x}{dt} + m d\frac{d\mathbf x}{dt}$$

\begin{align} \frac{d\mathbf x}{dt}\cdot dm \frac{d\mathbf x}{dt} &= dm \frac{d\mathbf x}{dt}\cdot \frac{d\mathbf x}{dt} \\ &= d(N \hat m)\frac{d\mathbf x}{dt}\cdot \frac{d\mathbf x}{dt} \\ &= dN \hat m\frac{d\mathbf x}{dt}\cdot\frac{d\mathbf x}{dt}+ N d\hat m \frac{d\mathbf x}{dt}\cdot \frac{d\mathbf x}{dt} \\ &= dN \hat m\frac{d\mathbf x}{dt}\cdot\frac{d\mathbf x}{dt}, \end{align}

if it is assumed that a specific kind of particle doesn't change it mass (e.g. an electron mass $\hat m_{electron}$ is constant), which is very similar to $dN\mu$ with $\mu = \hat m \frac{d\mathbf x}{dt}\cdot \frac{d\mathbf x}{dt}$. Finally

$$\int \frac{d\mathbf x}{dt}\cdot md\frac{d\mathbf x}{dt} = \frac{1}{2}\int md\left(\frac{d\mathbf x}{dt}\cdot\frac{d\mathbf x}{dt}\right)=\frac{1}{2}m\frac{d\mathbf x}{dt}\cdot \frac{d\mathbf x}{dt}$$

looks like the common expression for kinetic energy.

  • $\begingroup$ Here, presumably $\mathbf{x}$ is supposed to be the centre-of-mass coordinate of a multi-particle system? Do you think it makes sense for the chemical potential to vanish for any system whose centre-of-mass velocity is zero? $\endgroup$ Sep 15, 2015 at 10:02
  • $\begingroup$ Maybe there are other contributions? This being one of them? I don't know - I just wonder how the term should be interpreted. N here signifies one kind of particle by the way. $\endgroup$
    – Emil
    Sep 15, 2015 at 10:06
  • $\begingroup$ Yes but for $N\neq 1$ you are describing a multi-particle system (one that has multiple particles). In that case there are actually $6N$ variables specifying the state of the system, so you should think carefully about, and try to specify in the question, what your $\mathbf{x}$ and $\mathbf{p}$ are supposed to mean. $\endgroup$ Sep 15, 2015 at 10:36
  • $\begingroup$ I haven't constructed any model. Maybe N is a real number approximating the number of particles, maybe dN is the annihalation or creation operator and this should be a sum over all particles and N should be 0 or 1, or maybe something else entirely. $\endgroup$
    – Emil
    Sep 15, 2015 at 11:29
  • $\begingroup$ OK. It sounds a bit like you are trying to combine symbols so that they look like definitions of other symbols. If you want to find and understand a relation between physical quantities you should probably have at least some vague idea of a physical system in mind. In particular, the chemical potential is a property of many-body systems in thermal equilibrium so you must have $N\gg 1$ for the expression $\mu dN$ to mean what you want it to mean. $\endgroup$ Sep 15, 2015 at 11:43

1 Answer 1


I think the answer is yes to some extent, but have you neglected potential energy?

For example in the reaction

$$\mathrm{H-H + F \rightarrow H + H-F}$$

one of the electrons in the $\mathrm{H-H}$ molecule ends up in a significantly lower energy state than it was in before the reaction so that the potential energy of the system is lower on the right hand side.

Apologies that this is not a mathematical answer and does not address your equations.

In terms of chemical potential. The left hand side of the equation has a higher chemical potential than the right hand side, but due to the release of the potential energy I anticipate that the right hand side would have higher kinetic energy – another way of thinking about this is that it is an exothermic reaction and the heat generated is kinetic energy of the products.

  • $\begingroup$ Not sure how to get a dN term from other things, say the influence from the Lorentz force, but maybe it is possible. $\endgroup$
    – Emil
    Sep 15, 2015 at 10:17
  • $\begingroup$ Maybe electromagnetism can be included for each particle kind like this: $x^j= (ct,\mathbf x), A^j = (\phi / c, \mathbf A), dE?=ng_{ij}(\hat m\frac{1}{2}d[\frac{dx^i}{d \tau}\frac{dx^j}{d \tau}]+\hat e \frac{dx^i}{d \tau}dA^j) + dng_{ij}\frac{dx^i}{d \tau}(\hat m\frac{dx^j}{d \tau} + \hat eA^j)$ $\endgroup$
    – Emil
    Sep 15, 2015 at 16:03

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