A The equations of variable mass systems

[wrobel]

The equations of variable mass systems are usually deduced from some very informal argument. It is so at least for the books I know.

So I tried to construct a formal proof based on the continuous media equations.

Criticism, remarks etc are welcomed.

Let $D\subset \mathbb{R}^q$ be an open and bounded domain with $C^1-$smooth boundary $\partial D$. The cases $q=1,2,3$ are physically reasonable.

The space $\mathbb{R}^q$ is endowed with an inertial frame of reference $Ox^1,\ldots,x^q$, where $x=(x^1,\ldots,x^q)$ are the standard Cartesian coordinates:
$$\boldsymbol{x}=x^i\boldsymbol{e}_i\in \mathbb{R}^q.$$ A theorem we formulate below has an invariant form but we use the Cartesian coordinates just for convenience.

Let $(\cdot,\cdot)$ denote the standard inner product in $\mathbb{R}^q$.

Note also that in the Cartesian frame there is no need to distinguish co- and contravariant components of tensors.

Let $$\boldsymbol w(t,x)=w^i(t,x)\boldsymbol e_i,\quad w^i\in C^1(\mathbb{R}^{q+1})$$ stand for a vector field in $\mathbb{R}^q$ and $g^t_{t_0}(x)$ be its flow:
$$\frac{d}{dt}g^t_{t_0}(x)=\boldsymbol w(t,g^t_{t_0}(x)),\quad g^{t_0}_{t_0}(x)=x.$$ We assume that $g^t_{t_0}(x)$ is defined for all real $t,t_0$ and for all $x\in \mathbb{R}^q$.
Introduce a notation $$D(t)=g^t_{t_0}(D).$$
Loosely speaking $\boldsymbol w$ is a velocity of the volume $D(t)$.

Assume that the domain $D(t)$ is filled with a continuous media with a mass density $\rho(t,x)$ and a flow velocity field $\boldsymbol{v}(t,x)=v^i(t,x)\boldsymbol{e}_i$;
$$v^i,\rho\in C^1(\overline\Omega),\quad \Omega=\{(t,x)\mid x\in D(t),\quad t\in\mathbb{R}\}\subset\mathbb{R}^{q+1}.$$
The local conservation of mass equation is
$$\rho_t+\mathrm{div}(\rho\boldsymbol{v})=0$$ or
$$
\rho_t+\frac{\partial (\rho v^i)}{\partial x^i}=0.\qquad(1)$$
Let $$\boldsymbol{F}(t,x)=F^i(t,x)\boldsymbol{e}_i,\quad F^i\in C(\overline\Omega)$$ stand for a force per unit mass. So that the whole matter in $D(t)$ experiences a net force $$\boldsymbol{G}(t)=\int_{D(t)}\rho(t,x)\boldsymbol{F}(t,x)dV,$$ where $dV$ is the volume element.
We denote the stress tensor by $p^{ij}(t,x)$. The boundary $\partial D(t)$ experiences the following external contact net force
$$\boldsymbol{P}(t)=\int_{\partial D(t)}\boldsymbol{p}_n dS,\quad \boldsymbol{p}_n=p^{ij}n_j\boldsymbol{e}_i,$$
where $\boldsymbol{n}=n^i\boldsymbol{e}_i$ is the outer unit normal to $\partial D(t)$; $dS$ is the area element of the surface;
$$p^{ij}\in C^1(\overline\Omega).$$
The local equation of linear momentum balance is
$$
\rho\Big(v^k_t+\frac{\partial v^k}{\partial x^i}v^i\Big)=\rho F^k+\frac{\partial p^{kj}}{\partial x^j}.\qquad (2)$$
Let
$$\boldsymbol{Q}(t)=\int_{D(t)}\boldsymbol{v}(t,x)\rho(t,x) dV$$stand for the linear momentum of $D(t)$.

THEOREM. The equation of linear momentum balance for the volume $D(t)$ is as follows
$${\boldsymbol{\dot Q}}=\boldsymbol{G}+\boldsymbol{P}+\boldsymbol{R},$$where
$$\boldsymbol{R}=-\int_{\partial D(t)}\boldsymbol{v}(\boldsymbol{v}-\boldsymbol{w},\boldsymbol{n})\rho dS.$$PROOF. By the well-known theorem from the integral calculus and due to (2) it follows that
$$\frac{d}{dt} Q^k=\int_{D(t)}\big( v_t^k\rho+ v^k\rho_t+\mathrm{div}\,(\rho v^k\boldsymbol w) \big)dV$$
$$=\int_{D(t)}\Big[ \Big(F^k-\frac{\partial v^k}{\partial x^j}v^j\Big)\rho+\frac{\partial p^{kl}}{\partial x^l}+ v^k\rho_t\Big] dV$$
$$+\int_{D(t)}\mathrm{div}\,(\rho v^k\boldsymbol w)dV.$$
Integration by parts gives
$$\int_{D(t)}\frac{\partial v^k}{\partial x^j}v^j \rho dV=\int_{\partial D(t)} v^kv^j n_j\rho ds-\int_{D(t)} v^k\mathrm{div}(\rho\boldsymbol{v})dV$$
$$\int_{D(t)}\frac{\partial p^{kl}}{\partial x^l} dV=\int_{\partial D(t)} p^{kl}n_ldS,$$
$$
\int_{D(t)}\mathrm{div}\,(\rho v^k\boldsymbol w)dV=\int_{\partial D(t)}\rho v^kw^jn_jdS.$$
To finish the proof it remains to employ (1).

The Theorem is proved.

Other theorems (the angular momentum balance and the energy balance) can be deduced by the same way.

 

[ergospherical]

As an exercise I tried to determine the equation of angular momentum balance; I wondered if my working looks correct to you? The local equation of angular momentum balance is $p^{ij} = p^{ji}$. Let\begin{align*}
\boldsymbol{L}(t) &= \int_{D(t)} \boldsymbol{r} \times \boldsymbol{v}(\boldsymbol{r},t) \rho(\boldsymbol{r},t) dV \\

L^i(t) &= \int_{D(t)} \epsilon_{ijk} x^j v^k \rho dV
\end{align*}and also let\begin{align*}
\boldsymbol{M}(t) &= \int_{D(t)} \boldsymbol{r} \times \boldsymbol{F}(\boldsymbol{r},t) \rho(\boldsymbol{r},t)dV \\

M^i(t) &= \int_{D(t)} \epsilon_{ijk} x^j F^k \rho dV
\end{align*}Take the time derivative of $L^i$, using the local momentum balance equation as well as $\epsilon_{ijk} v^j v^k = 0$,\begin{align*}

\dfrac{d}{dt} L^i &= \int_{D(t)} \epsilon_{ijk} \left( v^j v^k \rho + x^j v^k_t \rho + x^j v^k \rho_t + \mathrm{div}(\rho x^j v^k \boldsymbol{w}) \right) dV \\

&= \int_{D(t)} \epsilon_{ijk} \left( x^j \left[ \left( F^k-\frac{\partial v^k}{\partial x^l}v^l\right)\rho+\frac{\partial p^{kl}}{\partial x^l} \right] + x^j v^k \rho_t \right) dV \\

&\hspace{35pt} + \int_{D(t)} \epsilon_{ijk} \mathrm{div}(\rho x^j v^k \boldsymbol{w}) dV \\ \\

&= \int_{D(t)} \epsilon_{ijk} \left( - x^j \frac{\partial v^k}{\partial x^l}v^l \rho + x^j\frac{\partial p^{kl}}{\partial x^l} + x^j v^k \rho_t \right) dV + M^i \\

&\hspace{35pt} + \int_{D(t)} \epsilon_{ijk} \mathrm{div}(\rho x^j v^k \boldsymbol{w}) dV

\end{align*}Using partial integration, and the relation $\rho_t + \mathrm{div}(\rho \boldsymbol{v})= 0$, we may write\begin{align*}
\int_{D(t)} \epsilon_{ijk} x^j\frac{\partial v^k}{\partial x^l}v^l \rho dV - \int_{\partial D(t)} \epsilon_{ijk} x^j v^kv^l n_l\rho ds &=-\int_{D(t)} \epsilon_{ijk} v^k \dfrac{\partial}{\partial x^l} ( x^j v^l \rho)dV \\

&= -\int_{D(t)} \epsilon_{ijk} v^k (v^j \rho + x^j \mathrm{div}(\rho\boldsymbol{v}))dV \\

&= -\int_{D(t)} \epsilon_{ijk} x^j v^k \mathrm{div}(\rho\boldsymbol{v})dV \\

&= + \int_{D(t)} \epsilon_{ijk} x^j v^k \rho_t dV

\end{align*}as well as\begin{align*}
\int_{D(t)} \epsilon_{ijk} x^j \dfrac{\partial p^{kl}}{\partial x^l} dV &= \int_{\partial D(t)} \epsilon_{ijk} x^j p^{kl} n_l ds - \int_{D(t)} \epsilon_{ijk} p^{kj} dV \\

&= \int_{\partial D(t)} \epsilon_{ijk} x^j p^{kl} n_l ds
\end{align*}where in the final equality the symmetry of $p^{ij} = p^{ji}$ was used. The time derivative of $L^i$ can hence be rewritten as\begin{align*}
\dfrac{d}{dt} L^i &=
\int_{\partial D(t)} \epsilon_{ijk} x^j (p^{kl} - v^k v^l \rho) n_l ds + M^i \\

&\hspace{35pt} + \int_{D(t)} \epsilon_{ijk} \mathrm{div}(\rho x^j v^k \boldsymbol{w}) dV
\end{align*}Does this equation look correct? It can probably be simplified further.

 

[wrobel]

I think you just missed $\rho$ and the formulas must be as follows

 

[ergospherical]

Thanks! I missed out a $\rho$ when writing my final equation [it is now corrected]. Then\begin{align*}

\dfrac{d}{dt} L^i &=

\int_{\partial D(t)} \epsilon_{ijk} x^j (p^{kl} - v^k v^l \rho) n_l ds + M^i \\
&\hspace{35pt} + \int_{D(t)} \epsilon_{ijk} \mathrm{div}(\rho x^j v^k \boldsymbol{w}) dV \\ \\&=
\int_{\partial D(t)} \epsilon_{ijk} x^j p^{kl} n_l ds + M^i \\
&\hspace{35pt} + \int_{D(t)} \epsilon_{ijk} \rho x^j v^k (w^l - v^l) n_l ds

\end{align*}Multiplying by $\boldsymbol{e}_i$ yields\begin{align*}
\dfrac{d}{dt} \boldsymbol{L} = \int_{\partial D(t)} \boldsymbol{r} \times \boldsymbol{p}_n ds + \boldsymbol{M} + \int_{\partial D(t)} \boldsymbol{r} \times \boldsymbol{v} (\boldsymbol{w} - \boldsymbol{v}, \boldsymbol{n}) ds
\end{align*}where $(\boldsymbol{p}_n)^i = p^{ij} n_j$.

 

[wrobel]

now we can turn back to https://www.physicsforums.com/threads/what-is-the-tension-of-the-rope.1004390/#post-6506923
:)