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For Vlasov equation, we have $$\frac {\partial f}{\partial t}=-v\frac{\partial f}{\partial r}-a\frac{\partial f}{\partial v}.$$ This makes sense in the physical sense of the conservation law, which might be expressed as $$\frac{\partial (stuff)}{\partial{t}}=-\nabla \cdot{flux}.$$

However, mathematically speaking, if we forget all about the physical meaning of $f$ and just regard it as some function dependent on $\vec v$ and $\vec r$ which then depend on $t$ (reminiscent of how the Lagrangian relies on generalized coordinates and generalized velocity), we would find that, according to the chain rule for partial derivatives, $$\frac {\partial f}{\partial t}=\frac{\partial r}{\partial t}\cdot\frac{\partial f}{\partial r}+\frac{\partial v}{\partial t}\cdot \frac{\partial f}{\partial v}=v\frac{\partial f}{\partial r}+a\frac{\partial f}{\partial v}.$$ So completely opposite signs on the RHS? What happened?

Qmechanic
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Splinter_X
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2 Answers2

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The confusion is between full derivative and partial derivative. We have function $f(x,v,t)$, which depends on three variables $x,v,t$. Now, if we consider this function along a particle trajectory, paramerized by time, we have a function of one variable $F(t)=f(x(t),v(t),t)$, so that $$ \frac{dF(t)}{dt}=\frac{\partial f}{\partial x}\frac{dx(t)}{dt}+ \frac{\partial f}{\partial v}\frac{dv(t)}{dt}+ \frac{\partial f}{\partial t} $$ See Physical interpretation of total derivative for more background on differentials vs. partial derivatives.

Remark
an additional confusing factor is that that $F$ is often denoted by a lower case letter $f$, even though it is understood that functional forms of $f(x,v,t)$ and $f(x(t),v(t),t)$ are different (the former is a function of three variables, while the latter of only one variable.)

Roger V.
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0

$\def \b {\mathbf}$

$$~\frac{d}{dt}f(\b r~,\b p~,t)=0~$$ means that $$~f(\b r~,\b p~,t)-\rm constant=0$$ or $$~-f(\b r~,\b p~,t)+\rm constant=0$$.

which is equivalent

Example \begin{align*} &f(\,x(t),y(t)\,)=x^2+y^2-l^2=0\quad\Rightarrow\\\\ &\frac{d}{dt}\,f+\frac{\partial f}{\partial x}\,\dot x+ \frac{\partial f}{\partial y}\,\dot y=\frac{d}{dt}\,f+2\,x\dot x+2\,y\dot y=0\\\\ &\text{or}\\\\ &\frac{d}{dt}\,f-\frac{\partial f}{\partial x}\,\dot x- \frac{\partial f}{\partial y}\,\dot y=\frac{d}{dt}\,f-2\,x\dot x-2\,y\dot y=0\quad\Rightarrow\\ &-x^2-y^2+l^2=0 \end{align*}

Eli
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