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

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Rayleigh's dissipation function, defined as a sum involving the velocities of particles, is shown to be homogeneous of degree 2 in the generalized velocities. The transformation equations for generalized coordinates lead to the expression for the dissipation function in terms of these velocities. It is clarified that the function does not depend on the generalized velocities directly, which supports the proof of homogeneity. The coefficients in the dissipation function were initially overlooked but are essential for the correct formulation. Ultimately, it is confirmed that the dissipation function is indeed homogeneous of degree 2 in the generalized velocities.
Kashmir
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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}##?
 
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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.
 
vanhees71 said:
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?
 
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Kashmir said:
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
 
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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}##.
 
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wrobel said:
if ##\boldsymbol r_i## depends on t explicitly then it is not so
Yes, thank you. :)
 
vanhees71 said:
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. :)
 
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