In Investigations on the Theory of the Browning Movement, on page 5, Einstein wrote:

of all atoms of the system), and if the complete system of the equations of change of these variables of state is given in the form $$\dfrac{\partial p_v}{\partial t}=\phi_v(p_1\ldots p_l)\ (v=1,2,\ldots l)$$ whence $$\sum\frac{\partial\phi_v}{\partial p_v}=0,$$

I assume it is an elementary result, since he gives no explanation on how to deduce it. How can I obtain this relation?

Attempt: I tried to consider $$\sum\frac{\partial \phi_v}{\partial p_v} ~=~ \sum\frac{\mathrm{d}t \phi_v}{\mathrm{d}t} \left(\partial_t p_v \right)^{-1} ~=~ \sum \frac{\partial_t \phi_v}{ \phi_v} \,,$$ but I couldn't go any further.

  • 1
    $\begingroup$ What is $\phi_v$ here? Make sure that your post is self-complete and it would be better to clear up all the terminologies. $\endgroup$ – user36790 Nov 2 '16 at 8:50

The variables

$$p^{\nu}, \qquad \nu=1,\ldots, \ell \tag{A}$$

are the phase space coordinates. The derivative $\frac{\partial p^{\nu}}{\partial t}$ in Einstein's paper is a total time derivative. The vector field

$$\phi~=~\sum_{\nu=1}^{\ell}\phi^{\nu}\frac{\partial }{\partial p_{\nu}} \tag{B}$$

generates time evolution. The divergence of a vector field

$$ {\rm div}\phi~=~ \frac{1}{\rho}\sum_{\nu=1}^{\ell}\frac{\partial (\rho\phi^{\nu})}{\partial p^{\nu}},\tag{C}$$

where $\rho$ is the density in phase space, which we will assume is constant

$$\rho={\rm constant} \tag{D}$$

(wrt. the chosen coordinate system). Apparently Einstein assumes that the vector field $\phi$ is divergencefree,

$$ {\rm div}\phi~=~0 .\tag{E}$$

We stress that not all vector fields are divergencefree.

Counterexample: The dilation vector field $$\phi~=~\sum_{\nu=1}^{\ell}p^{\nu}\frac{\partial }{\partial p^{\nu}}\tag{F}$$ is not divergencefree. The corresponding flow solution reads $$ p^{\nu}(t)~=~p^{\nu}_{(0)} e^t.\tag{G}$$

Assumption (D) and (E) follow e.g. in a Hamiltonian formulation because of (among other things) Liouville's theorem. Recall that Hamiltonian vector fields are divergence-free. See also this related Phys.SE post.

  • $\begingroup$ This answer is very different than the one above. What do you think about the argument $\sum \frac{\partial \phi_\nu}{\partial p_{\nu}} = \sum \frac{\partial^2 p_\nu}{\partial p_{\nu} \partial t} = \frac{\partial}{\partial t} \sum \frac{\partial p_\nu}{\partial p_{\nu}} = \frac{\partial}{\partial t} \sum 1 = 0$ It requires no divergencefree hypothesis. Is there a flaw in this argument? Cheers! $\endgroup$ – Conrado Costa Nov 3 '16 at 8:09
  • $\begingroup$ I gave a counterexample in my answer. $\endgroup$ – Qmechanic Nov 3 '16 at 8:17
  • $\begingroup$ if $p_\nu$ is $C^2$ then $\frac{\partial \phi_\nu}{\partial p_{\nu}}=\frac{\partial^2 p_\nu}{\partial p_{\nu} \partial t} = \frac{\partial}{\partial t} \frac{\partial p_\nu}{\partial p_{\nu}} = \frac{\partial}{\partial t} 1 = 0$? $\endgroup$ – Conrado Costa Nov 3 '16 at 8:43
  • $\begingroup$ $\uparrow $ No. $\endgroup$ – Qmechanic Nov 3 '16 at 13:11

In short : $$ \sum \frac{\partial \phi_\nu}{\partial p_{\nu}} = \sum \frac{\partial^2 p_\nu}{\partial p_{\nu} \partial t} = \frac{\partial}{\partial t} \sum \frac{\partial p_\nu}{\partial p_{\nu}} = \frac{\partial}{\partial t} \sum 1 = 0 $$ (Second equality comes from the fact that derivatives can be exchanged)

  • $\begingroup$ Nice! In the first equality instead of $\frac{\partial p_v}{\partial p_v \partial t}$ you mean $\frac{\partial^2 p_v}{\partial p_v \partial t}$? another question, do we need the sum here for this argument? $\endgroup$ – Conrado Costa Nov 2 '16 at 8:55
  • $\begingroup$ Comment to the answer (v2): Note that not all flows are divergencefree. $\endgroup$ – Qmechanic Nov 3 '16 at 8:21

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.