Day: July 18, 2019

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201907181500 Homework 2 (Q3)

Two mass points, m_1 and m_2, move under the influence of a mutual central force, where the central force potential is given by U(\mathbf{r}_1,\mathbf{r}_2)=U(|\mathbf{r}_1-\mathbf{r}_2|). Assume the center of mass is at the rest, please find the equivalent one-body problem and show that the corresponding Lagrangian can be written as

\mathcal{L}=\displaystyle{\frac{1}{2}}\mu\dot{r}^2-U_{\textrm{eff}}

where r=|\mathbf{r}_1-\mathbf{r}_2| is the relative distance between the two mass points and \mu=\displaystyle{\frac{m_1m_2}{m_1+m_2}} is the reduced mass.

Solution.

(Reference: https://www.physics.rutgers.edu/~shapiro/507/book4(DOT)pdf)

Let \mathbf{R}\stackrel{\textrm{def}}{=}\displaystyle{\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}} and \mathbf{r}\stackrel{\textrm{def}}{=}\mathbf{r}_2-\mathbf{r}_1. Expressing \mathbf{r}_1 and \mathbf{r}_2 in terms of \mathbf{R} and \mathbf{r}, we write

\mathbf{r}_1=\mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r},

\mathbf{r}_2=\mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}

where M=m_1+m_2.

The kinetic energy T is computed as follows:

\begin{aligned} T & = \displaystyle{\frac{1}{2}m_1\dot{r}_1^2}+\displaystyle{\frac{1}{2}m_2\dot{r}_2^2} \\ & = \displaystyle{\frac{1}{2}m_1\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r} \bigg)\Bigg]^2}+\displaystyle{\frac{1}{2}m_2\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}\bigg)\Bigg]^2} \\ & = \displaystyle{\frac{1}{2}}(m_1+m_2)\dot{R}^2+\displaystyle{\frac{1}{2}\frac{m_1m_2}{M}}\dot{r}^2 \\ & = \displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2} \\ \end{aligned}

where \mu is the reduced mass \displaystyle{\frac{m_1m_2}{m_1+m_2}}.

From its formula above, T can be seen as the sum of the kinetic energy of the motion of the centre of mass, i.e., \displaystyle{\frac{1}{2}}M\dot{R}^2, and the kinetic energy of motion about the centre of mass, i.e., \displaystyle{\frac{1}{2}\mu\dot{r}^2}.

And from the fact that the Lagrangian \mathcal{L}=\displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r) is cyclic on R, the centre of mass is either at rest or in uniform motion.

Thus the equation of motion for r will not contain terms involving \mathbf{R} or \dot{\mathbf{R}}. We may hence ignore the first term, and what remains in the Lagrangian is

\mathcal{L}=\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r).

Now that we introduce a spherical coordinate system given by its equation of transformation from the Cartesian as:

\begin{aligned} x & =r\sin\theta\cos\phi \\ y & =r\sin\theta\sin\phi \\ z & =r\cos\theta \\ \end{aligned}

we can write the kinetic energy as

\begin{aligned} T & =\displaystyle{\frac{1}{2}}\mu (\dot{x}^2+\dot{y}^2+\dot{z}^2) \\ & = \displaystyle{\frac{1}{2}}\mu[(\dot{r}\sin\theta\cos\phi+\dot{\theta}r\cos\theta\cos\phi -\dot{\phi}r\sin\theta\sin\phi)^2\\ & \qquad +(\dot{r}\sin\theta\sin\phi+\dot{\theta}r\cos\theta\sin\phi+\dot{\phi}r\sin\theta\cos\phi)^2\\ & \qquad +(\dot{r}\cos\theta -\dot{\theta}r\sin\theta)^2]\\ & = \displaystyle{\frac{1}{2}}\mu [\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2]\\ \end{aligned}

(Note that the kinetic energy is cyclic on the coordinate \phi and the conjugate momentum P_\phi=\displaystyle{\frac{\partial\mathcal{L}}{\partial\dot{\phi}}}=\mu r^2\sin^2\theta\dot{\phi}=\textrm{constant}. Observe that r\sin\theta is the distance between the particle and the z-axis, and it can be easily seen that P_\phi is the z-component of the angular momentum \mathbf{L}.)

To simplify things, we choose the direction of angular momentum \mathbf{L} as the z-direction. It follows that \theta=\pi/2, \dot{\theta}=0, and L=\mu r^2\dot{\phi}.

I wish to obtain the Euler-Lagrange equation for r. Hence I compute the following:

\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial r}} & = \mu r\dot{\phi}^2- \partial_r U\\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{r}}} & = \mu\dot{r}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial \mathcal{L}}{\partial \dot{r}}}\bigg) & = \mu\ddot{r}\\ \end{aligned}

Then one can express the one-body problem as:

\mu\ddot{r}-\mu r\dot{\phi}^2+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0,

or,

\mu\ddot{r}-\displaystyle{\frac{L^2}{\mu r^3}}+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0,

or,

\mu\ddot{r}+\displaystyle{\frac{\mathrm{d}}{\mathrm{d}r}U_{\textrm{eff}}(r)}=0,

where U_{\textrm{eff}}(r)=U(r)+\displaystyle{\frac{L^2}{2\mu r^2}} is the effective potential.

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201907181434 Homework 1 (Q3)

Obtain the equation of motion for a particle falling vertically under the influence of gravity when the frictional forces obtainable from a dissipation function kv^2/2 are present. Integrate the equation to obtain the velocity as a function of time and show that maximum possible velocity for a fall from rest is v=mg/k.

Solution.

Write the Lagrangian \mathcal{L}=T-V by noting

T=\displaystyle{\frac{1}{2}m\dot{\mathbf{y}}^2} and V=-mg|\mathbf{y}|,

where the upward direction is taken to be positive. The frictional force is

\mathcal{F}=\displaystyle{\frac{k\dot{\mathbf{y}}^2}{2}}.

I wish to obtain the Euler-Lagrange equation, by computing the derivatives below:

\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial y}} & = mg \\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}} & =m\dot{y} \\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) & = m\ddot{y} \\ \displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & = k\dot{y}\\ \end{aligned}

Hence I obtain the E-L equation (with dissipation):

\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) -\displaystyle{\frac{\partial \mathcal{L}}{\partial y}}+\displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & =0\\ m\ddot{y}-mg+k\dot{y} & =0\\ \ddot{y}+\displaystyle{\frac{k}{m}}\dot{y} & =g\\ \end{aligned}

Treating u=\dot{y} as variable, I may obtain a first-order differential equation:

\dot{u}+\displaystyle{\frac{k}{m}u}-g=0

Solving it,

\begin{aligned} v(t) & =e^{\int \frac{k}{m}\mathrm{d}t}=e^{kt/m}\\ e^{kt/m}\dot{u}+\frac{k}{m}ue^{kt/m} & =ge^{kt/m}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(ue^{kt/m}) & = ge^{kt/m}\\ ue^{kt/m} & = \displaystyle{\frac{mg}{k}}e^{kt/m}+\textrm{constant }C\\ u(t)& =\displaystyle{\frac{mg}{k}}+Ce^{-kt/m}\\ v_{\textrm{max.}}=\dot{y} & =\displaystyle{\frac{mg}{k}}\qquad\qquad (e^{-\frac{kt}{m}}\rightarrow 0\enspace \textrm{as}\enspace t\rightarrow \infty )\\ \end{aligned}

In conclusion, it is proven that the maximum possible speed for a fall from rest is v=mg/k.