4.2 The Lagrangian from Hamilton’s Principle

Hamilton’s Principle is based on the idea that a physical system is fully described by a function of, \(L(q_1,\dots ,q_n,\dot{q}_1,\dots ,\dot{q}_n,t)\) that depends only on the generalized position and velocities of the particles in the system. The evolution of the system is determined by fixing the value of \(L\) at two positions in configuration time and requiring that the functional:

\begin{align} S=\int_{t_1}^{t_2} L(q_1,\dots ,q_n,\dot{q}_1,\dots ,\dot{q}_n,t) dt \tag{4.17}\end{align}

(called the action) is stationary. We have seen that if \(L\) is the Lagrangian, then Hamilton’s principle is equivalent to D’Alembert’s Principle and, thus, Newton’s Second Law. From variational calculus, we know that Hamilton’s principle leads to the Euler-Lagrange equations of motion for each coordinate \(q_i\):

\begin{align} \delta S=0\to\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i}&=0 \tag{4.18}\end{align}

The Lagrangian for a free particle

We can now ask ourselves what the function \(L\) can look like for describing a free particle. Our main requirement is that \(L\) should be a function that does not depend on the inertial frame of reference that we choose to describe the system. Given two frames of references, moving with a velocity \(\vec{V}\) with respect to each other, the position, \(\vec{r}\) and velocity of a particle, \(v\), must transform according to:

\begin{align} \vec{r'}=\vec{r}+\vec{V}t\\ \vec{v'}=\vec{v}+\vec{V} \tag{4.19}\end{align}

Note that we implicitly assume that time is “absolute” and independent of reference frame. This of course is not true in Special Relativity.

The Lagrangian should be such that applying Hamilton’s Principle is independent of the inertial frame of reference that is chosen. This means that the Lagrangian cannot depend on the absolute position of the particle. For a free particle, it should not matter where in space the particle is; it should always be described in the same way. We thus require:

\begin{align} \frac{\partial L}{\partial \vec{r}}=0 \tag{4.20}\end{align}

Similarly, the Lagrangian cannot depend on the direction of the velocity. Space is isotropic and the behaviour of the free particle will not depend on which direction it travels. The Lagrangian can thus only depend on its magnitude, \(v^2\):

\begin{align} L&=L(v^2)\tag{4.21}\\ \therefore \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}} \right)&=0 \tag{4.22}\end{align}

Lagrange’s equations thus requires that \(\frac{\partial L}{\partial \dot{q_i}}\) is a constant in time, thus requiring that the magnitude of \(v\) be constant in time. Of course, that is the result that we expect for a free particle. However, we don’t know the form for the Lagrangian, in principle it could be any function of \(v^2\), such as a polynomial, \(L=av^2+b(v^2)^2+\dots\).

Let’s consider how things are affected when we transform the Lagrangian to a frame of reference that is moving with an infinitesimal velocity \(\vec{\epsilon}\) with respect to the original inertial frame of reference. The particle thus has a velocity \(\vec{v'}=\vec{v}+\vec{\epsilon}\)

\begin{align} L(v'^2)&=L((\vec{v}+\vec{\epsilon})\cdot(\vec{v}+\vec{\epsilon}))\\ &=L(v^2+2\vec{v}\cdot\vec{\epsilon}+\epsilon^2) \tag{4.23}\end{align}

We can expand this as a Taylor series of \(L\) centered at \(v^2\), neglecting terms of order \(\epsilon^2\) and higher:

\begin{align} L(v'^2)&=L(v^2)+\frac{\partial L}{\partial v^2}2\vec{v}\cdot\vec{\epsilon}+\dots\\ &=L(v^2)+\frac{\partial L}{\partial v^2}x \tag{4.24}\end{align}

where \(2\vec{v}\cdot\vec{\epsilon}\) is just some constant number that we call \(x\) (it is constant since \(\vec{v}\) and \(\vec{\epsilon}\) are constants). Applying Hamilton’s Principle and using the Euler-Lagrange equations:

\begin{align} \frac{d}{dt}\left(\frac{\partial L(v'^2)}{\partial \dot{q}_i} \right) - \frac{\partial L(v'^2)}{\partial q_i}&=0\\ \frac{d}{dt}\left(\frac{\partial L(v'^2)}{\partial \dot{q}_i} \right)&=0\\ \frac{d}{dt}\left(\frac{\partial L(v^2)}{\partial \dot{q}_i} \right)+\frac{d}{dt}\left(\frac{\partial}{\partial \dot{q}_i}\frac{\partial L}{\partial v^2}x \right)&=0\\ \frac{\partial}{\partial \dot{q}_i}\frac{\partial L}{\partial v^2}&=k \tag{4.25}\end{align}

where we used the fact that the original Lagrangian does not depend on position, that the velocity, \(\vec{v}\) is constant, and we introduced \(k\) as another constant. For the last line to be true, \(L\) must be linear in the velocity squared:

\begin{align} L(v^2)=av^2 \tag{4.26}\end{align}

Note that for a single particle, the Lagrangian can be multiplied by any number and remain invariant (the number \(a\) just factors out). In principle, if we have a system of \(N\) non-interacting particles, they will each have their own value of \(a\), and only the ratios of the different values of \(a\) are relevant. We choose a new constant instead of \(a\) and call it \(\frac{1}{2}m\), so that we have the Lagrangian:

\begin{align} L=\frac{1}{2}mv^2 \tag{4.27}\end{align}

and we call \(m\) the “mass” of the particle. Note that we found this particular form of the Lagrangian simply by requiring that the Lagrangian be invariant under transformations from one inertial frame of reference to another. More fundamentally, we can see that for a system of non-interacting particles, it is the ratio of their masses that matters, not their absolute value.