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

  • #1
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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Answers and Replies

  • #2
vanhees71
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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.
 
  • #3
Kashmir
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72
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?
 
  • #4
wrobel
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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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  • #5
vanhees71
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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}##.
 
  • #6
Kashmir
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if ##\boldsymbol r_i## depends on t explicitly then it is not so
Yes, thank you. :)
 
  • #7
Kashmir
437
72
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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