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The Lagrange equations

Classical mechanics

The Lagrange equations

flickr.com/Quinn Dombrowski

The Lagrange equations

We here derive the Lagrange equations in a step-by-step fashion, using the knowledge of generalized forces from our previous discussions.

In our discussion so far, we arrived at an expression for the generalized force in terms of time derivatives of the classical kinetic energy. We now make use of this concept to arrive at a remarkable set of equations called the Lagrange equations, which form an alternate route of describing the motion of classical objects.

With conservative forces

Rewriting the expression for the generalized force,

(1)   \begin{equation*} Q_j=\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}-\dfrac{\partial T}{\partial q_j}  \end{equation*}

We also know from the definition of generalized force,

    \[\sum_j Q_j \delta q_j=\sum_i\mathbf{F}_{i}.\delta \mathbf{r}_i\]

But from the transformation equations,

    \[\delta \mathbf{r}_i=\sum_j {\partial \mathbf{r}_i \over \partial q_j} \delta q_j\]

We have,

(2)   \begin{equation*} Q_j=\sum_i\mathbf{F}_{i}.{\partial \mathbf{r}_i \over \partial q_j}  \end{equation*}

As we mentioned earlier, let us consider conservative forces. These forces are special in the sense that work done by them in moving a mass in space is independent of the path taken and depends only on where you started and where you stopped (much like it is with electric forces). And so, we can express them as a gradient of a potential, \mathbf{F}_i=-\nabla_i V or \mathbf{F}_i=-\dfrac{\partial V}{\partial r_i}\mathbf{\hat{r}}_i. Also, since, \dfrac{\partial \mathbf{r}_i}{\partial q_j}=\dfrac{\partial r_i}{\partial q_j}\mathbf{\hat{r}}_i, equation(2) becomes,

    \[Q_j=-\sum_i\dfrac{\partial V}{\partial r_i}\mathbf{\hat{r}}_i.\dfrac{\partial r_i}{\partial q_j}\mathbf{\hat{r}}_i\]

And we are left with,

(3)   \begin{equation*} Q_j=-\dfrac{\partial V}{\partial q_j}   \end{equation*}

I.e., if the forces involved are conservative, we can also write the generalized force as the gradient of the potential with respect to the generalized coordinates.

Arriving at the Lagrange equations

Let us substitute the result (3) into equation (1), we have,

    \[-\dfrac{\partial V}{\partial q_j}=\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}-\dfrac{\partial T}{\partial q_j}\]

Or,

    \[\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}-\dfrac{\partial T}{\partial q_j}+\dfrac{\partial V}{\partial q_j}=0\]

We might as well combine the last two terms of the equation above (since differentiation is a linear operator) and write,

(4)   \begin{equation*} \frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}-\dfrac{\partial (T-V)}{\partial q_j}=0 \end{equation*}

If we further choose only potentials that do not depend on velocities, we have \dfrac{\partial V}{\partial \dot{q}_j}=0. So, no harm is done if we included this in the first term above,

    \[\frac{d}{dt} \dfrac{\partial (T-V)}{\partial \dot{q}_j}-\dfrac{\partial (T-V)}{\partial q_j}=0\]

Let’s name the difference (T-V)=L, the Lagrangian.
We then can write,

    \[\frac{d}{dt} \dfrac{\partial L}{\partial \dot{q}_j}-\dfrac{\partial L}{\partial q_j}=0\]

These are the all-famous Lagrange equations which form an alternate method of determining the equations of motion (besides the Newtonian method). To put them to use, all you need to know is the total kinetic energy and potential energy of the system. Take the difference of the energies, find the Lagrangian, L, and differentiate it according to this equation, and voila, you have your equation of motion! No need to even think of the constraint forces. No need to draw any force diagrams.

But keep in mind that we have derived these equations for forces that are conservative with a potential that is independent of velocities. For example, we can’t use the above form of the equations when forces such as friction or air drag or any other dissipative ones are acting on the body.

The Lagrangian

Now, focusing on the new function we introduced above – the Lagrangian. Since, it is just the difference of the kinetic and the potential energies, where the kinetic energy is a function of \dot{q} and t, and the potential energy of q and t; the Lagrangian must be a function of q, \dot{q} and t.

    \[L=L(q, \dot{q}, t)\]

As we know, there can be ambiguity regarding the potential function chosen, since we are free to chose any point in space as the point of zero potential. So, say, instead of a potential V that you chose, we choose a potential V+F, where F is also some function of q and t. How will the Lagrange’s equation turn out then?

    \[\frac{d}{dt} \dfrac{\partial (L+F)}{\partial \dot{q}_j}-\dfrac{\partial (L+F)}{\partial q_j}=0\]

Or,

(5)   \begin{equation*} \frac{d}{dt} \dfrac{\partial L}{\partial \dot{q}_j}+\frac{d}{dt} \dfrac{\partial F}{\partial \dot{q}_j}-\dfrac{\partial L}{\partial q_j}-\dfrac{\partial F}{\partial q_j}=0  \end{equation*}

We can rewrite the second term here as \dfrac{d}{dt} \dfrac{\partial F}{\partial \dot{q}_j}=\dfrac{\partial}{\partial \dot{q}_j}\dfrac{dF}{dt}.
And since F is a function of q and t, \dfrac{dF}{dt}=\sum_k \dfrac{\partial F}{\partial q_k}\dot{q_k}+\dfrac{\partial F}{\partial t}.
Which implies, \dfrac{d}{dt} \dfrac{\partial F}{\partial \dot{q}_j}=\dfrac{\partial}{\partial \dot{q}_j}\dfrac{dF}{dt}=\dfrac{\partial F}{\partial q_j}.
Hence, equation (5) becomes,

    \[\frac{d}{dt} \dfrac{\partial L}{\partial \dot{q}_j}+\dfrac{\partial F}{\partial q_j}-\dfrac{\partial L}{\partial q_j}-\dfrac{\partial F}{\partial q_j}=0\]

And we get back the Lagrange’s equation,

    \[\frac{d}{dt} \dfrac{\partial L}{\partial \dot{q}_j}-\dfrac{\partial L}{\partial q_j}=0\]

So, there seems to be no effect on the equations of motion, of adding the function F(q,t) to the Lagrangian. This means that we can construct a number of different Lagrangian functions (differing by this additional function F) all of which give the same, correct equations of motion.

We will next use the Lagrange equations for different classical systems and surprise ourselves with how simply we can obtain the same results with them, as compared with Newton’s way.

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Roshan Sawhil is a Physics postgraduate who rejoices both doing and explaining Physics. He also finds doing Philosophy as a leisure activity quite interesting. You can find and connect with him on Facebook and Twitter.

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