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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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)

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  • $\begingroup$ 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

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