Dissipation function is homogeneous in ##\dot{q}## second degree proof

[Kashmir]

We have Rayleigh's dissipation function, defined as
$\mathcal{F}=\frac{1}{2} \sum_{i}\left(k_{x} v_{i x}^{2}+k_{y} v_{i j}^{2}+k_{z} v_{i z}^{2}\right)$

Also we have transformation equations to generalized coordinates as
$\begin{aligned} \mathbf{r}_{1} &=\mathbf{r}_{1}\left(q_{1}, q_{2}, \ldots, q_{3 N-k}, t\right) \\ & \vdots \\ \mathbf{r}_{N} &=\mathbf{r}_{N}\left(q_{1}, q_{2}, \ldots, q_{3 N-k}, t\right) \end{aligned}$

How can I prove that the dissipation function is homogeneous of degree 2 in $\dot{q}$?

 

[vanhees71]

You have $\vec{v}_i=\dot{\vec{r}}_i=\dot{q}^j \partial_j \vec{r}_i$ and thus
$$\mathcal{F}=\frac{1}{2} \sum_i \dot{q}^j \dot{q}^k (\partial_j \vec{r}_i) \cdot (\partial_k \vec{r}_i).$$
Sinc the $\vec{r}_i$ are functions of the $q^j$ and $t$ but not of $\dot{q}^j$ by assumption you have
$$\mathcal{F}(q,\lambda \dot{q},t)=\lambda^2 \mathcal{F}(q,\dot{q},t).$$
QED.

 

[Kashmir]

You have $\vec{v}_i=\dot{\vec{r}}_i=\dot{q}^j \partial_j \vec{r}_i$ and thus
$$\mathcal{F}=\frac{1}{2} \sum_i \dot{q}^j \dot{q}^k (\partial_j \vec{r}_i) \cdot (\partial_k \vec{r}_i).$$
Sinc the $\vec{r}_i$ are functions of the $q^j$ and $t$ but not of $\dot{q}^j$ by assumption you have
$$\mathcal{F}(q,\lambda \dot{q},t)=\lambda^2 \mathcal{F}(q,\dot{q},t).$$
QED.

Thank you. Why not a partial with time term in $\vec{v}_i=\dot{\vec{r}}_i=\dot{q}^j \partial_j \vec{r}_i$ also where did the kx, ky, kz term go?

 

[wrobel]

How can I prove that the dissipation function is homogeneous of degree 2 in ?

if $\boldsymbol r_i$ depends on t explicitly then it is not so

 

[vanhees71]

Indeed, I've overlooked this and also forgot the coefficients. So we must have
$$\vec{r}_i=\vec{r}_i(q)$$
and thus
$$\dot{\vec{r}_i}=\dot{q}^j \partial_{j} \vec{r}_i(q)$$
and
$$\mathcal{F}=\frac{1}{2} \sum_i \vec{v}_i^{\text{T}} \hat{K} \vec{v}_i = \frac{1}{2} \dot{q}^j \dot{q}^k g_{jk}$$
with
$$g_{jk} = \sum_i (\partial_j \vec{r}_i)^{\text{T}} \hat{K} \partial_k \vec{r}_i.$$
Now indeed $g_{jk}$ is a function of the $q$ only and don't depend on $\dot{q}$, and thus indeed $\mathcal{F}$ is a homogeneous function of degree 2 in the generalized velocities $\dot{q}$.

 

[Kashmir]

if $\boldsymbol r_i$ depends on t explicitly then it is not so

Yes, thank you. :)

 

[Kashmir]

Indeed, I've overlooked this and also forgot the coefficients. So we must have
$$\vec{r}_i=\vec{r}_i(q)$$
and thus
$$\dot{\vec{r}_i}=\dot{q}^j \partial_{j} \vec{r}_i(q)$$
and
$$\mathcal{F}=\frac{1}{2} \sum_i \vec{v}_i^{\text{T}} \hat{K} \vec{v}_i = \frac{1}{2} \dot{q}^j \dot{q}^k g_{jk}$$
with
$$g_{jk} = \sum_i (\partial_j \vec{r}_i)^{\text{T}} \hat{K} \partial_k \vec{r}_i.$$
Now indeed $g_{jk}$ is a function of the $q$ only and don't depend on $\dot{q}$, and thus indeed $\mathcal{F}$ is a homogeneous function of degree 2 in the generalized velocities $\dot{q}$.

Got it. Thank you very much. :)