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# 13.4: The Lagrangian Equations of Motion

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

This section might be tough – but do not be put off by it. I promise that, after we have got over this section, things will be easy. But in this section I do not like all these summations and subscripts any more than you do.

Suppose that we have a system of $$N$$ particles, and that the force on the $$i$$th particle ($$i=1$$ to $$N$$) is $$\bf{F}_{i}$$. If the $$i$$th particle undergoes a displacement $$\delta\bf{r}_{i}$$, the total work done on the system is $$\sum_{i}\bf{F}_{i}\cdot\partial\bf{r}_{i}$$. The position vector $$\bf{r}$$ of a particle can be written as a function of its generalized coordinates; and a change in $$\bf{r}$$ can be expressed in terms of the changes in the generalized coordinates. Thus the total work done on the system is

$\sum_{i}\bf{F}_{i}\cdot\sum_{j}\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}\delta q_{j} \label{13.4.1}$

which can be written

$\sum_{j}\sum_{i}\bf{F}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}\delta q_{j}. \label{13.4.2}$

But by definition of the generalized force, the work done on the system is also

$\sum_{j}P_{j}\cdot\delta q_{j}. \label{13.4.3}$

Thus the generalized force $$P_{i}$$ associated with generalized coordinate $$q_{i}$$ is given by

$P_{j}=\sum_{i}\bf{F}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}. \label{13.4.4}$

Now $$\bf{F}_{i}=m_{i}\ddot{r}_{i}$$, so that

$P_{j}=\sum_{i} m_{i}\ddot{r}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}. \label{13.4.5}$

Also

$\dfrac{d}{dt} \left(\dot{\bf{r}}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}\right)=\ddot{\bf{r}}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}+\dot{\bf{r}}_{i}\cdot\dfrac{d}{dt}\left(\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}\right). \label{13.4.6}$

Substitute for $$\ddot{\bf{r}}_{i}\cdot\dfrac{\partial\bf{r}_{i}}{\partial q_{j}}$$ from Equation $$\ref{13.4.6}$$ into Equation $$\ref{13.4.5}$$ to obtain

$P_{j}=\sum_{i}m_{i} \left[\dfrac{d}{dt} \left(\dot{\bf{r}}_{i}\cdot\dfrac{\partial \dot{\bf{r}}_{i}}{\partial q_{j}} \right)-\dot{\bf{r}}_{i}\cdot(\dfrac{\partial \dot{\bf{r}}_{i}}{\partial q_{j}}) \right]. \label{13.4.8}$

Now

$\frac{\partial \mathbf{r}_{i}}{\partial q_{j}}=\frac{\partial \dot{\mathbf{r}}_{i}}{\partial \dot{q}_{j}} \label{assert1}$

and

$\frac{d}{d t}\left(\frac{\partial \mathbf{r}_{i}}{\partial q_{j}}\right)=\frac{\partial \dot{\mathbf{r}}_{i}}{\partial q_{j}}. \label{assert2}$

Therefore

$P_{j}=\sum_{i} m_{i}\left[\frac{d}{d t}\left(\dot{\mathbf{r}}_{i} \cdot \frac{\partial \dot{\mathbf{r}}_{i}}{\partial \dot{q}_{j}}\right)-\dot{\mathbf{r}}_{i} \cdot\left(\frac{\partial \dot{\mathbf{r}}_{i}}{\partial q_{j}}\right)\right] \label{conclusion}$

You may not be immediately comfortable with the assertions in Equations \ref{assert1} and \ref{assert2} so I’ll interrupt the flow briefly here with an example to try to justify these assertions and to understand what they mean.

Consider the relation between the coordinate $$x$$ and the spherical coordinates $$r,\theta, \phi$$:

$x=r\sin\theta\cos\phi \label{A1}\tag{A1}$

In this example, $$x$$ would correspond to one of the components of $$\bf{r}_{i}$$, and $$r, \theta, \phi$$ are the $$q_{1},q_{2},q_{3}$$.

From Equation $$\ref{A1}$$, we easily derive

\begin{align*} \dfrac{\partial x}{\partial r}&=\sin\theta\cos\phi \tag{A2.1} \\[4pt] \dfrac{\partial x}{\partial\theta} &=r\cos\theta\cos\phi \tag{A2.2} \\[4pt] \dfrac{\partial x}{\partial \phi}&=-r\sin\theta\sin\phi \label{A2}\tag{A2.3} \end{align*}

and differentiating Equation $$\ref{A1}$$ with respect to time, we obtain

$\dot{x}=\dot{r}\sin\theta\cos\phi+r\cos{\theta}\dot{\theta}\cos\phi-r\sin\theta\sin\phi\dot{\phi} \label{A3}\tag{A3}$

And from this we see that

\begin{align*} \dfrac{\partial\dot{x}}{\partial\dot{r}} &=\sin\theta\cos\phi \tag{A4.1} \\[4pt] \dfrac{\partial\dot{x}}{\partial\dot{\theta}} &=r\cos\theta\cos\phi \tag{A4.2} \\[4pt] \dfrac{\partial\dot{x}}{\partial\dot{\phi}} &=-r\sin\theta\sin\phi \label{A4}\tag{A4.3} \end{align*}

Thus the first assertion is justified in this example, and I think you’ll see that it will always be true no matter what the functional dependence of $$\bf{r}_{i}$$ on the $$q_{j}$$.

For the second assertion, consider

$\dfrac{\partial x}{\partial r}=\sin\theta\cos\phi \tag{A5}$

and hence

$\dfrac{d}{dt}\dfrac{\partial x}{\partial r}=\cos\theta\dot{\theta}\cos\phi-\sin\theta\sin\phi\dot{\phi}. \label{A5}\tag{A6}$

From Equation $$\ref{A3}$$ we find that

$\dfrac{\partial \dot{x}}{\partial r}=\cos\theta\dot{\theta}\cos\phi-\sin\theta\sin\phi\dot{\phi}, \tag{A7}$

and the second assertion is justified. Again, I think you’ll see that it will always be true no matter what the functional dependence of $$\bf{r}_{i}$$ on the $$q_{j}$$.

The kinetic energy $$T$$ is

$T=\sum_{i}\dfrac{1}{2}m_{i}\dot{r}_{i}^{2}=\sum_{i}\dfrac{1}{2}m_{i}\dot{\bf{r}}_{i}\cdot\dot{\bf{r}}_{i} \label{13.4.9}$

Therefore

$\dfrac{\partial T}{\partial q_{j}}=\sum_{i}m_{i}\dot{\bf{r}}_{i}\cdot\dfrac{\partial \dot{\bf{r}}_{i}}{\partial q_{j}} \label{13.4.10}$

and

$\dfrac{\partial T}{\partial \dot{q}_{j}}=\sum_{i}m_{i}\dot{\bf{r}}_{i}\cdot\dfrac{\partial \dot{\bf{r}}_{i}}{\partial \dot{q}_{j}}. \label{13.4.11}$

On substituting these in Equation $$\ref{conclusion}$$ we obtain

$P_{j}=\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{q}_{j}}-\dfrac{\partial T}{\partial q_{j}}. \label{13.4.12}$

This is one form of Lagrange’s equation of motion, and it often helps us to answer the question posed in the last sentence of Section 13.2 – namely to determine the generalized force associated with a given generalized coordinate.

### Conservative Forces

If the various forces in a particular problem are conservative (gravity, springs and stretched strings, including valence bonds in a molecule) then the generalized force can be obtained by the negative of the gradient of a potential energy function – i.e. $$P_{j}=-\dfrac{\partial V}{\partial q_{j}}$$. In that case, Lagrange’s equation takes the form

$\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{q}_{j}}-\dfrac{\partial T}{\partial q_{j}}=-\dfrac{\partial V}{\partial q_{j}}. \label{13.4.13}$

In my experience, this is the most useful and most often encountered version of Lagrange’s equation.

The quantity $$L=T-V$$ is known as the lagrangian for the system, and Lagrange’s equation can then be written

$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{q}_{j}}-\dfrac{\partial L}{\partial q_{j}}=0. \label{13.4.14}$

This form of the equation is seen more often in theoretical discussions than in the practical solution of problems. It does enable us to see one important result. If, for one of the generalized coordinates, $$\dfrac{\partial L}{\partial q_{j}}=0$$ (this could happen if neither $$T$$ nor $$V$$ depends on $$q_{j}$$ – but of course it could also happen if $$\dfrac{\partial T}{\partial q_{j}}$$ and $$\dfrac{\partial V}{\partial q_{j}}$$ were nonzero but equal and opposite in sign), then that generalized coordinate is called an ignorable coordinate – presumably because one can ignore it in setting up the lagrangian. However, it does not really mean that it should be ignored altogether, because it immediately reveals a constant of the motion. In particular, if $$\dfrac{\partial T}{\partial q_{j}}$$, then $$\dfrac{\partial V}{\partial q_{j}}$$ is constant. It will be seen that if $$q_{j}$$ has the dimensions of length, $$\dfrac{\partial L}{\partial \dot{q}_{j}}$$ has the dimensions of linear momentum. And if $$q_{i}$$ is an angle, $$\dfrac{\partial L}{\partial \dot{q}_{j}}$$ has the dimensions of angular momentum. The derivative $$\dfrac{\partial L}{\partial \dot{q}_{j}}$$ is usually given the symbol $$p_{j}$$ and is called the generalized momentum conjugate to the generalized coordinate $$q_{j}$$. If $$q_{j}$$ is an “ignorable coordinate”, then $$p_{j}$$ is a constant of the motion.

In each of Equations $$\ref{13.4.12}$$, $$\ref{13.4.13}$$ and $$\ref{13.4.14}$$ one of the $$q$$s has a dot over it. You can see which one it is by thinking about the dimensions of the various terms. Dot has dimension T-1.

So, we have now derived Lagrange’s equation of motion. It was a hard struggle, and in the end we obtained three versions of an equation which at present look quite useless. But from this point, things become easier and we rapidly see how to use the equations and find that they are indeed very useful.