- #1

binbagsss

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- 11

So I am aiming to derive the continuity equation using the fact that phase space points are not created/destroyed.

So I am going to use the Leibiniz rule for integration extended to 3-d:

## d/dt \int\limits_{v(t)} F dv = \int\limits_{v(t)} \frac{\partial F}{\partial t} dV + \int\limits_{A(t)} F \vec{u_{A}}.\vec{n} dA ##,

##\vec{n}## the normal unit vector, where ##F=F(x_{1},x_{2},x_{3},t) ##

##\vec{u_{A}} ## is the surface velocity.

So let ##N_{D}## denote the number of phase points in a volume element ##D##, ##N## the total number of phase points.

Conservation of phase points =>

## \frac{d}{dt} N_{D}(t) =0 ##

## = - N \frac{d}{dt} \int\limits_{D} dp dq \rho (p,q,t) ##

where ##\rho = \rho (p,q,t) ## is the density of function.

So by Leibniz rule above I get a term describing the change due to the changing surface of ##A## - ## \partial D ## here

## = - N \int\limits_{\partial D} \rho \vec{V}.\vec{n} d\sigma ##, which is fine ,

where ##\vec{V}## is the velocityMY QUESTION:

The corresponding term to ##\int\limits_{v(t)} \frac{\partial F}{\partial t} dV ## is zero here.

I.e the contribution due to the change of ##\rho## within ##V##. I'm just having a little trouble understanding this conceptually... so my book states that:

"Since phase points cannot be created the only way that phase points can be added or subtracted from the volume V is to flow across its boundary, S. "

That's fine, and seems to make sense to me. But, if points are flowing across the boundary then surely the number inside V is changing. So looking at

## \int\limits_{v(t)} \frac{\partial \rho}{\partial t} dV ##,

##\frac{\partial \rho}{\partial t} \neq 0 ## if there is either a change in the number of phase points or a change in the volume, doesn't saying that the number is changing across the surface => number is changing within the volume and so ##\neq 0 ## for a given volume due to the number of phase points changing?

( I am fine with the rest of the derivation just stuck in this step)

Many thanks in advance.