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I am trying to find hamiltonian for system described by EOM $$ \ddot{x}(t) + \beta \dot{x}^2(t)\sqrt{\dot{x}^2(t)+\dot{y}^2(t)} = 0, \\ \ddot{y}(t) + \beta \dot{y}^2(t)\sqrt{\dot{x}^2(t)+\dot{y}^2(t)} = 0. $$

I wanted to find Lagrangian first and use Legendre transformation but I have no idea how to even start. My first thought was to solve this analytially. Can somebody give me some hints how to start and how to do this?

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  • $\begingroup$ Related: physics.stackexchange.com/q/147341/2451 $\endgroup$ – Qmechanic Feb 24 at 12:54
  • $\begingroup$ Hi blahblah. Welcome to Phys.SE. Your above drag force (v2) is not necessarily antiparallel to the velocity. That seems physically wrong. $\endgroup$ – Qmechanic Feb 24 at 13:00
  • $\begingroup$ I got this form by expanding eq: $$ \ddot{x}(t) + \beta \dot{x}^2(t) = 0 $$ into two dimensions... Thank you, I have to try to do this again. $\endgroup$ – user281659 Feb 24 at 13:04
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Working in terms of velocity variables: $v_x=\dot{x}, v_y=\dot{y}$. The second suggestion redily reduces the order of the system: $$ \dot{v}_x + \beta v_x^2\sqrt{v_x^2+v_y^2}=0,\\ \dot{v}_y + \beta v_y^2\sqrt{v_x^2+v_y^2}=0 $$ One now has now multiple options available:

  • Switching to polar coordinates: $$ v_x=v\cos\phi, v_y=v\sin\phi $$
  • Solving for the trajectory and reducing the equatiosn to one variable $$ \frac{dv_x}{dv_y} $$
  • Solving for the kinetic energy: $$ \frac{d}{dt}K=\frac{d}{dt}(\frac{v_x^2+v_y^2}{2})=v_x \dot{v}_x + v_y\dot{v}_y $$
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  • $\begingroup$ I've already switched into velocity variables but I will try to work with polar coordinates, thank you! $\endgroup$ – user281659 Feb 24 at 13:00
  • $\begingroup$ I cannot judge beyond what is written in your question. I just see that you immediately reduce the order of the system by relabeling $\dot{x}$ and $\dot{y}$ and then it might be even without polar coordinates. $\endgroup$ – Roger Vadim Feb 24 at 13:03
  • $\begingroup$ @blahblah I expanded the answer for clarity $\endgroup$ – Roger Vadim Feb 24 at 13:09
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The Lagrangian is:

$$\mathcal L=\frac 12 (\dot x^2+\dot y^2)-x\,f_x-y\,f_y$$

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the Hamiltonian

$$H=\frac 12\left(\,{p_{{x}}}^{2}\,+{p_{{y}}}^{2}\right)+x{\it f_x}+y{\it f_y}$$

where $$ f_x= \beta \dot{x}^2(t)\sqrt{\dot{x}^2+\dot{y}^2} , \\ f_y=\beta \dot{y}^2(t)\sqrt{\dot{x}^2+\dot{y}^2} $$

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  • $\begingroup$ Now I noticed that I wrote wrong EOM... It should be without second square $$ \ddot{x}(t) + \beta \dot{x}(t)\sqrt{\dot{x}^2(t)+\dot{y}^2(t)} = 0, \\ \ddot{y}(t) + \beta \dot{y}(t)\sqrt{\dot{x}^2(t)+\dot{y}^2(t)} = 0. $$ How much does it change formula for $f_x$ and $f_y$? And can you give me some hints how to calculate this Lagrangian. And thank you very much for help! $\endgroup$ – user281659 Feb 26 at 15:03
  • $\begingroup$ I understand that, if $T=\frac{1}{2}(\dot{x}^2+\dot{y}^2},$ then when I want to get my EOM from E-L eq., $f_x$ and $f_y$ should be $$ x \cdot f_x = x \cdot \beta\dot{x}\sqrt{\dot{x}^2+\dot{y}^2}, \\ y \cdot f_y = y \cdot \beta\dot{y}\sqrt{\dot{x}^2+\dot{y}^2}. $$ But, is it other way to get this or not? $\endgroup$ – user281659 Feb 26 at 16:05
  • $\begingroup$ you can put inthe equations for $~f_x~,f_y~$ any force law . but I don't think that your radial damper force is correct. the radial damper force is: $~\beta\,\dot{r}\,\hat{\mathbf{e}}_r~$ with $~r=\sqrt{x^2+y^2}~$ and $~\mathbf e_r=x\mathbf e_x+y\,\mathbf e_y~$ $\endgroup$ – Eli Feb 26 at 16:08
  • $\begingroup$ first obtain the equations of motion with $~T=... -x\,f_x+y\,fy~$ put the force law in the equations of motion, not in the kinetic energy $\endgroup$ – Eli Feb 26 at 16:15
  • $\begingroup$ This should be your force law $\begin{align*} &\mathbf{F}_\beta= \left[ \begin {array}{c} {\frac {\beta\, \left( x{\dot{x}}+y{\dot{y}} \right) x}{{x}^{2}+{y}^{2}}}\\\\{\frac {\beta\, \left( x{\dot{x}}+y{\dot{y}} \right) y}{{x}^{2}+{y}^{2}}}\end {array} \right] \end{align*}$ $\endgroup$ – Eli Feb 26 at 16:20
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I don't think you got the right equations. In one of your comments above you said that basically you want the two dimensional version of the equation $$\ddot{x} \, + \, \beta\, \dot{x}^2 = 0$$ which is the equation for which the magnitude of the air drag force is proportional to the square magnitude of the velocity Well, the higher dimesional version, in vector form, should be $$\ddot{\vec{r}} \, + \, \beta \, \big|\, \dot{\vec{r}} \,\big| \, \dot{\vec{r}} \, = \, \vec{0} $$ In dimension two, the component-wise equations become \begin{align} &\ddot{x} \, + \, \beta \, \dot{x}\, \sqrt{\dot{x}^2 + \dot{y}^2} \, = \, 0\\ &\ddot{y} \, + \, \beta \, \dot{y}\, \sqrt{\dot{x}^2 + \dot{y}^2} \, = \, 0 \end{align} In vector form, your system is a different system: $$\ddot{\vec{r}} \, + \, \beta \, \big|\, \dot{\vec{r}} \,\big|^2 \, \dot{\vec{r}} \, = \, \vec{0} $$

Either way, you can solve analytically any system of the type $$\ddot{\vec{r}} \, + \, \beta \, \big|\, \dot{\vec{r}} \,\big|^m \, \dot{\vec{r}} \, = \, \vec{0} $$ where $m$ is a positive integer. First, since position is not explicitly present in the system, we can substitute $\vec{v} = \dot{\vec{r}}$. Then the vector equation becomes $$\dot{\vec{v}} \, + \, \beta \, \big|\, \vec{v} \,\big|^m \, \vec{v} \, = \, \vec{0} $$ or in another notation: $$\frac{d\vec{v}}{dt} \, = \, - \, \beta \, \big|\, \vec{v} \,\big|^m \, \vec{v}$$

One way to go on from here is to change the time-parameter $t$. Assume that $t$ itself can be written as a strictly increasing function $t = t(s)$ of a scalar parameter $s$, so that the function $t(s)$ satisfies the differential equation $$\frac{dt}{ds} \, = \, \frac{1}{\big|\vec{v}\big(t(s)\big)\big|^m}$$ where one can impose the initial condition $t(0) = 0$. Then $$\frac{d\vec{v}}{ds} \, = \, \frac{dt}{ds}\,\frac{d\vec{v}}{dt} \, = \frac{dt}{ds} \left(- \, \beta \, \big|\, \vec{v} \,\big|^m \, \vec{v}\right) \, = \, \frac{1}{\,\,\big|\vec{v}\big|^m} \, \left(- \, \beta \, \big|\, \vec{v} \,\big|^m \, \vec{v}\right) \, = \, - \, \beta\, \vec{v} $$ If you put the last two equations together you get the extended system \begin{align} &\frac{d\vec{v}}{ds} \, = \, - \, \beta\, \vec{v} \\ &\frac{dt}{ds} \, = \, \frac{1}{\,\,|\vec{v}|^m} \end{align} The first vector equation can be solved immediately: $$\vec{v} = \vec{v}(s) = e^{-\beta s}\,\vec{v}_0$$ where $\vec{v}_0$ is a constant vector (initial condition). Knowing $\vec{v}(s)$ as a function of $s$ allows us to substitute it in the second, scalar, equation: $$\frac{dt}{ds} = \frac{1}{\,\,\big|\, e^{-\beta s} \,\vec{v}_0 \,\big|^m} = \frac{e^{m \beta s}}{\,\,\big| \,\vec{v}_0 \,\big|^m}$$ The solution to the latter equation, with initial condition $t(0) = 0$, is $$t = t(s) = \frac{e^{m \beta s} - 1}{\,\,m \beta \,\big| \,\vec{v}_0 \,\big|^m}$$ When you solve for $s$, i.e. invert the function, writing $s = s(t)$ you get $$s = \frac{1}{m \beta}\,\ln\big( 1 + m\beta \, \big| \,\vec{v}_0\big|^m t \big)$$ Substitute the latter in $\vec{v} = e^{-\beta s}\,\vec{v}_0$ and you have the solution of the original vector equation $$\vec{v} = \left(\, e^{-\beta \frac{1}{m \beta}\,\ln\big( 1 + m\beta \, | \,\vec{v}_0|^m t \big)}\,\right)\vec{v}_0 = \big(1\, + \, m\beta \, \big| \,\vec{v}_0\big|^m t \big)^{- \, \frac{1}{m}}\, \vec{v}_0$$ Finally, you can rewrite it as $$\vec{v} \, = \, \left( \frac{1}{\sqrt[m]{1 \, + \, m\beta \big| \,\vec{v}_0\big|^m t \,\, }} \right) \vec{v}_0 $$ To find the position as a function of time $t$, integrate with respect to $t$ the last equation:

$$\vec{r} \, = \, \vec{r}_0 \, + \, \left(\int\, \frac{dt}{\sqrt[m]{1 \, + \, m\beta \big| \,\vec{v}_0\big|^m t \,\, }} \right) \vec{v}_0 $$ If $m > 2$, this integral does not look explicitly solvable (but I cannot bet on it). However, when $m=1$ or $m=2$ you get explicit solutions $$m=1: \,\,\,\,\, \vec{r} \, = \, \vec{r}_0 \, + \, \left(\frac{1}{\beta}\,{\ln\big(1 \, + \, \beta \big| \,\vec{v}_0\big|\, t \big)} \right) \frac{\vec{v}_0}{|\vec{v}_0|} $$ $$m=2: \,\,\,\,\, \vec{r} \, = \, \vec{r}_0 \, + \, \left(\frac{1}{2\beta}\,\sqrt{1 \, + \, 2\beta \big| \,\vec{v}_0\big|^2 t} \right) \frac{\vec{v}_0}{|\vec{v}_0|^2} $$

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