The Lagrangian function for unconstrained motion
Summary:
- Define something called the "Lagrangian" as the kinetic energy minus the potential energy of a system: $${\cal L} \equiv T-U.$$
- For a free particle moving in a potential energy field, you can write out the components of $\myv F=m\myv a$ in terms of the Lagrangian as (for example, for the $x$ component): $$\begineq F_x&=&m\dot v_x\\ \frac{\del {\cal L}}{\del x} &=& \frac{d}{dt} \frac{\del {\cal L}}{\del \dot{x}} \endeq$$
- The equations above look like the Euler-Lagrange equations which we'd use to find the path that minimizes this path-dependent quantity, "$S$": $$S=\int_{t_1}^{t_2} {\cal L} \,dt.$$ where $S$ is called the "action" for a particular path.
- This leads us to suspect that "minimizing the action" might be a more general strategy for finding the path, or trajectory, of a dynamic system--"Hamilton's principle".
Unconstrained motion
Consider:
- just one particle,
- moving through space,
- with no constraints (no frictionless walls around,
- subject to a conservative force.
Such a particle has an unchanging total energy.
Let's write $\myv F=m\myv a$ in terms of $T$ and $U$
For a conservative force, the force is the negative gradient of the potential energy: $$-\myv \grad U(\myv r) = \myv F(\myv r).$$ The $x$ component of this vector equation is: $$-\frac{d}{dx}U=F_x.$$
We can relate $m\myv a$ to the particle's kinetic energy, using:
- for a particle of unchanging mass, $m$, we have $\frac{d}{dt}(\myv p)=\frac{d}{dt}(m\myv v)=m\dot{\myv v}$.
$$\begineq m\myv{a}=m\dot{ \myv{v}} =& \frac{d}{dt}(mv_x\uv x+mv_y\uv y+mv_z\uv z)\\ =&\frac{d}{dt}\left(\frac{d}{dv_x}\frac12 mv^2_x\uv x+\frac{d}{dv_y}\frac12 mv^2_y\uv y +\frac{d}{dv_z}\frac12 mv^2_z\uv z\right)\\ =&\frac{d}{dt}\left(\frac{\del}{\del v_x}T\uv x+\frac{\del}{\del v_y}T\uv y +\frac{\del}{\del v_z}T\uv z\right)\\ =&\frac{d}{dt}\left(\frac{\del}{\del \dot x}T\uv x+\frac{\del}{\del \dot y}T\uv y +\frac{\del}{\del \dot z}T\uv z\right)\\ \endeq $$ So, writing out the $x$ component of $\myv F=m\myv a$: $$\begineq F_x =& ma_x\\ -\frac{\del}{\del x} =&m\dot{v_x}\\ -\frac{\del}{\del x} =&\frac{d}{dt}\frac{\del}{\del \dot x} T. \endeq $$
Consider this combination of potential and kinetic energy, which we call the "Lagrangian": $$\cal L\equiv T-U.$$ Since $U=U(x,y,z)$ is a function of position alone, and $T=T(\dot x, \dot y, \dot z)$ is a function of velocity alone, we could re-write our $F_x=ma_x$ equation above in terms of this combination as... $$\frac{\del}{\del x}\cal L = \frac{d}{dt}\frac{\del}{\del \dot x} \cal L.$$ And we can write similar equations for the $x$ and $y$ components. This looks like an Euler-Lagrange equation for an "accounting function" $\cal L=\cal L(x,y,z,\dot x, \dot y, \dot z)$.
In terms of the particle's Cartesian coordinates, its kinetic energy is: $$\begineq T=&\frac{1}{2}mv^2 = \frac{1}{2}m \dot{\myv{r}}\cdot \dot{\myv{r}} \\ =& \frac{1}{2}m[\dot{x}^2 + \dot{y}^2+\dot{z}^2] \\ =&T(\dot{x},\dot{y},\dot{z}),\endeq$$
Since any forces are conservative, the particle has a potential energy which depends on position (alone): $$U=U(\myv{r}) = U(x,y,z).$$
The Lagrangian function (which is not the mechanical energy $E$) is defined as:
$${\cal L} \equiv {\color{red}T}-{\color{blue}U}= {\cal L}({\color{red}\dot{x}, \dot{y}, \dot{z}}, {\color{blue} x, y, z, t}) $$
- The derivative of this Lagrangian, ${\cal L}$, wrt $x$ is: $$\frac{\del {\cal L}}{\del x} = -\frac{\del U}{\del x}=F_x.$$
- The derivative of the Lagrangian wrt to $\dot x$ is: $$\frac{\del {\cal L}}{\del \dot{x}} = \frac{\del T}{\del \dot{x}} = m\dot{x} = p_x$$
According to Newton's law, $F_x = \frac{d}{dt}p_x$ in an inertial frame, we have the following relationship between these derivatives of the Lagrangian: $$\frac{\del {\cal L}}{\del x} = \frac{d}{dt} \frac{\del {\cal L}}{\del \dot{x}}$$
And similarly for the other coordinates $$\frac{\del {\cal L}}{\del y} = \frac{d}{dt} \frac{\del {\cal L}}{\del \dot{y}};\ \frac{\del {\cal L}}{\del z} = \frac{d}{dt} \frac{\del {\cal L}}{\del \dot{z}}$$
These equations, together, look just like the Euler-Lagrange equations for minimizing $\int {\cal L}\,dt$, and so we might work backwards to:
Hamilton's principle
The actual path which a particle follows between two points 1 and 2 in a given time interval, $t_1$ to $t_2$ is such that the action, $S$, given by:
$$S=\int_{t_1}^{t_2} {\cal L} \,dt = \int_{t_1}^{t_2} {\cal L}(\dot{x,} \dot{y,} \dot{z,} x, y, z, t)\,dt$$
is a minimum (or more precisely, it's stationary) when calculated along the actual path of the particle.
This is also known as the principal of least action.
Other coordinate systems
Consider the general situation of converting Cartesian coordinates $(x, y, z)$ into some other set of "generalized" coordinates, $(q_1, q_2, q_3)$. For example, if we're converting to spherical polar coordinates... $$q_1 \equiv r=\sqrt{x^2+y^2+z^2},$$ $$q_2\equiv \theta=\arccos(z/r),$$ $$q_3\equiv \phi=\arctan(y/x).$$
Is the coordinate transformation from Cartesian to the new system a one-to-one relationship? That is, are both of the relations below "functions" in the sense of giving a unique value of $q_3$--for example $\phi$--for a given $x, y,$ and $z$? $$q_i = q_i(x, y, z)\ \ \text{and}\ \ \myv{r} = \myv{r}(q_1, q_2, q_3)$$
How could we fix these up to make them into one-to-one relations?
If the coordinate transforms are one-to-one, then we can uniquely re-write ${\cal L} = \frac{1}{2}m \dot{\myv{r}}^2 - U(\myv{r})$ in terms of the generalized coordinates, and the action becomes $$S = \int_{t_1}^{t_2} {\cal L}(\dot{q}_1, \dot{q}_2, \dot{q}_3, q_1, q_2, q_3, t)\,dt.$$
From a mathematical point of view, this is just a change of variables. And a change of variables does not change the value of a definite integral.
The same path in space (just in terms of different coordinates) is still the stationary solution that minimizes this integral.
This means that we could run through the same arguments about finding the least action path in this new coordinate system, and we'd once again come up with the Euler-Lagrange criteria for minimizing the action, namely:
$$\frac{\del {\cal L}}{\del q_i} = \frac{d}{dt}\frac{\del {\cal L}}{\del \dot{q}_i},$$ for each of the new "generalized" coordinates $q_i$.
This whole argument depends only on having a one-to-one transformation from Cartesian coordinates to the new coordinate system. So...
You can find the trajectory of a particle (or system of particles) by minimizing the action -- solving the E-L differential equations for each coordinate (parameterized in terms of the time)--in any coordinate system you like.
[*] A further condition is that you must always write the kinetic and potential energy in an inertial frame.
[**] But once you have written the Lagrangian in an inertial frame, you are free to do a coordinate transformation to a system in which the the coordinates are time dependent (even accelerating!).
[***] Well, *just about any* coordinate system. We'll eventually see some qualifications about "holonomic" systems.
[****] And we'll eventually find that this also works with constraint forces (like 'normal forces'), though not frictional forces.
Polar coordinates
Consider a particle that's free to move in two-dimensions. But we'll use polar coordinates, instead of Cartesian coordinates to describe the position of the particle. That is, the Lagrangian will be a function of... $${\cal L} = T - U = {\cal L}(\dot{r}, \dot{\phi}, r, \phi).$$
So we need to be able to write $T$ and $U$ in terms of the polar coordinates $r$ and $\phi$ and their time derivatives.
It appears that:
$$\begineq
\frac {\Delta \myv r}{\Delta t} =& \frac{\Delta r}{\Delta t}\uv r
+r\frac{\Delta \phi}{\Delta t}\uv \phi \\
\dot{\myv r} =&\dot{r} \uv r+r\dot{\phi} \uv \phi
\endeq$$
So, two orthogonal components of the velocity are $v_r = \dot{r}$ and $v_{\phi} = r \dot{\phi}$.
So, the kinetic energy is $$\begineq T=&\frac{1}{2}mv^2 = \frac{1}{2}m \myv{v} \cdot \myv{v} = \frac{1}{2}m (v_r^2 + v_{\phi}^2)\\ =& \frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2).\endeq$$
The Lagrangian is $${\cal L} = T-U=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2) - U(r, \phi )$$
- Write out the E-L equation for the $\phi$ coordinate: $$\frac{\del {\cal L}}{\del \phi} = \frac{d}{dt}\frac{\del {\cal L}}{\del \dot{\phi}}.$$
- Write out the E-L equation for the $r$ coordinate: $$\frac{\del {\cal L}}{\del r} = \frac{d}{dt}\frac{\del {\cal L}}{\del \dot{r}}.$$
- Re-write your results in terms of the following quantities:
$\myv{F} = -\myv{\grad} U$. Using the gradient in 2-d polar coords (see inside back cover of your textbook): $$\myv{\grad}U = \uv{r} \frac{\del U}{\del r} + \uv{\phi} \frac{1}{r} \frac{\del U}{\del \phi} =-F_r\uv{r} -F_\phi \uv{\phi}$$ The components of this vector equation give rise to the first two relations:
- $\frac{\del U}{\del \phi}=-rF_\phi=-\Gamma=$torque (relative to the origin)
- $\frac{\del U}{\del r}=-F_r$.
- the moment of inertia of a particle of mass $m$ is $I=mr^2$ (relative to the origin),
- angular momentum is $L=I\omega=I\dot{\phi}$.
- centripetal acceleration for a particle moving in a circle of radius $r$ is $a_c=mv_\phi^2/r=m(r\dot{\phi})^2/r=mr\dot{\phi}^2=mr\omega^2.$
- E-L equation for the $\phi$ component:
$$\begineq \frac{\del {\cal L}}{\del \phi}=&\frac{d}{dt}\frac{\del {\cal L}}{\del \dot{\phi}}\\
\frac{\del (-U)}{\del \phi}=&\frac{d}{dt}\frac 22mr^2\dot{\phi}\\
rF_{\phi}=&\frac{d}{dt}I\dot{\phi}\\
\Gamma=&\frac{d}{dt}L=\dot L\endeq$$
This is the familiar relationship between torque and change of angular momentum.
- E-L equation for the $r$ component:
$$\frac{\del {\cal L}}{\del r}=\frac{\del (-U)}{\del r}+
mr\dot{\phi}^2=F_r+ma_c.$$
$$\frac{d}{dt}\frac{\del {\cal L}}{\del \dot{r}}
=\frac{d}{dt}m\dot{r}=\frac{d}{dt}p_r=m\ddot{r}.$$
So putting these together: $$F_r+ma_c=m\ddot{r}.$$ Consider what this means when...
- Motion in the $\uv r$ direction only, when $\phi=$constant, and therefore $\dot{\phi}=0$.
- Motion in a circle when $r=$constant, and therefore $\ddot r=0$.
Homework
Chapter 7: 1, 2, 3, 4, 8, 18