Now that we have derived the Euler-Lagrange equation from the principle of stationary action, Let’s see how it can be used to derive Newton’s second law. We will also make an important observation of how the concepts of momentum, force, and energy, which are familiar to use from Newtonian mechanics, make their way into the Lagrangian formalism. $\require{physics}$

2.3.1 — Connection to Newton’s second law

First, let’s apply the Euler-Lagrange equation for single particle of mass $m$ with one degree of freedom. In classical mechanics, we have $L=T-U$. For one particle of mass $m$, recall that the kinetic energy is $T(\dot q) = \frac{1}{2}m(\dot q)^2$ and is only a function of the generalized velocity, so $\pdv{T}{q}=0$. Assume the particle is experiencing a potential energy $U(q)$ that is only a function of the generalized coordinate, so $\pdv{U}{\dot q}=0$ (this is not a crucial assumption, but it makes our life easier).$^1$ Keep in mind that the generalized coordinates and velocities are independent from each other (so $\pdv{q}{\dot q}=0$ and $\pdv{\dot q}{q}=0$). This means that the partial derivative operator $\pdv{}{q}$ allows us to treat $\dot q$ as a constant.

We have

\begin{equation}
\begin{alignedat}{1}
\pdv{L}{q} &=\pdv{}{q}(T-U) \\
&= -\pdv{U}{q}
\end{alignedat}
\end{equation}

as well as

\begin{equation}
\begin{alignedat}{1}
\frac{d}{dt}\pdv{L}{\dot q} &= \pdv{}{\dot q}(T-U) \\
&= \frac{d}{dt}\pdv{T}{\dot q} \\
&= \frac{d}{dt}\pdv{}{\dot q}(\frac{1}{2}m\dot q^2) \\
&= m\frac{d}{dt}\dot q \\
&= m\ddot q
\end{alignedat}
\end{equation}

The Euler-Lagrange equation $\pdv{L}{q}=\frac{d}{dt}\pdv{L}{\dot q}$ then tells us

\begin{equation}
-\pdv{U}{q}=m\ddot q
\end{equation}

This is Newton’s second law! To be more clear, let’s say the particle is simply moving along the $x$-axis in the Cartesian plane, so that the generalized coordinate $q$ is $q=x$ and the acceleration is $\ddot q = a$. The term on the left-hand side is the force $F=-\pdv{U}{x}$ associated with the potential $U$. This gives us Newton’s second law in its most famous form.

\begin{equation}
F=ma
\end{equation}

$^1$ There do exist potential energies $U(q, \dot q, t)$ that are a function of velocities. For instance, the Lorentz force in electromagnetism has a velocity-dependent potential energy $U(\vb{r},\vb{v}, t) = q_e(\phi(\vb{r}, t)-\vb{v}\cdot\vb{A}(\vb{r}, t))$ where $q_e$ is the electric charge.

2.3.2 — Generalized forces and momenta

The most general form of Newton’s second law is $F=\frac{dp}{dt}$, where the net force on an object is equal to the rate of change in the object’s momentum $p$. It turns out that this form of Newton’s second law has an unexpectedly elegant connection to the Lagrangian formalism and the Euler-Lagrange equations. Our preceding derivation of Newton’s second law from the Lagrangian showed us that the force $F$ came from the $\pdv{L}{q}$ part of the Euler-Lagrange equation. This is the connection between the Lagrangian formalism and the Newtonian concept of forces.

So taking a derivative of of the Lagrangian gives us a force! But wait, there’s more. If $\pdv{L}{q}$ gives us a generalized force, what does $\pdv{L}{\dot q}$ give us? Let’s compare Newton’s second law side-by-side with the Euler-Lagrange equation:

\begin{equation}
\begin{alignedat}{1}
F &= \frac{d}{dt}p \\
\pdv{L}{q} &= \frac{d}{dt}\pdv{L}{\dot q}
\end{alignedat}
\end{equation}

The right-hand side of both equations involve the time derivative of some quantity (the momentum $p$ in Newton’s second law, and the quantity $\pdv{L}{\dot q}$ in the Euler-Lagrange equation). Indeed, the momentum associated with a degree of freedom comes from the $\pdv{L}{\dot q}$ part of the Euler-Lagrange equation.

This definition of momentum in Lagrangian (and Hamiltonian) mechanics is sometimes referred to as the conjugate momentum or canonical momentum associated with the coordinate $q$. This definition of momentum is more general than the kinematic momentum $\vb{p}=m\vb{v}$ that we are used to in Newtonian Mechanics (so the generalized momentum $p$ may not be equal to $p=m\dot q$ in some cases).

We are now ready to make an important observation about conserved quantities in Lagrangian mechanics. Recall from Newton’s second law $F=\frac{dp}{dt}$ that if the force $F$ in a particular direction on a particle is zero, then the momentum of the particle is conserved in that direction. The analogous statement in the Lagrangian formalism is this: if the generalized force $F=\pdv{L}{q}$ is zero for a generalized coordinate $q$, then the generalized momentum $p=\pdv{L}{\dot q}$ conjugate to $q$ is conserved. This is just the conservation of momentum popping up within the Lagrangian formalism.

So consider a Lagrangian $L$ for a particular system with multiple degrees of freedom $\{q_i\}$. If we observe by eye that a certain coordinate $q_i$ does not appear in the Lagrangian, then $\pdv{L}{q_i}=0$. However, this implies that its conjugate momentum $p_i=\pdv{L}{\dot q_i}$ is a conserved quantity (it is a constant and does not change with time). We can summarize this with some new vocabulary: if a generalized coordinate $q_i$ does not explicitly appear in the Lagrangian, then we say $q_i$ is a cyclic coordinate, and its conjugate momentum $p_i=\pdv{L}{q_i}$ is a constant of motion.

This is where the efficiency of Lagrangian mechanics starts becoming clear. Taking just one derivative of the Lagrangian ($\pdv{L}{q}$) gives us a force. Taking a different derivative of the Lagrangian ($\pdv{L}{\dot q}$) gives us a momentum. Lastly, noticing by eye that a particular coordinate $q$ is absent in the Lagrangian tells us that the quantity $\pdv{L}{\dot q}$ is a constant.

2.3.3 — Energy and the Hamiltonian

For a system consisting of one degree of freedom $q$, consider the new quantity $H$ defined as

\begin{equation}
H \equiv p\dot q- L
\end{equation}

where $p=\pdv{L}{q}$ is the canonical momentum and $L$ is the Lagrangian. Plugging in $L=T-U$, $T=\frac{1}{2}m\dot q^2$, and $U=U(q)$ gives

\begin{equation}
H=\pdv{L}{\dot q}\dot q-L
\end{equation}

\begin{equation}
H=\pdv{(T-U)}{\dot q}\dot q- (T-U)
\end{equation}

\begin{equation}
H = \pdv{}{\dot q}(\frac{1}{2}m\dot q^2)-\frac{1}{2}m\dot q^2 + U
\end{equation}

\begin{equation}
H = m\dot q^2 – \frac{1}{2}m\dot q^2 + U
\end{equation}

\begin{equation}
H=\frac{1}{2}m\dot q^2 + U
\end{equation}

\begin{equation}
H=T+U
\end{equation}

It turns out that, in this case, this mysterious quantity $H$ is equal to the total mechanics energy $E=T+U$ that we are familiar with from Newtonian mechanics!

This quantity $H$ is known as the Hamiltonian.

We have way more to say about this object in Chapter 3, which is entirely dedicated to Hamiltonian mechanics. The point of introducing it here is because of the deep connection the the energy $E$ has with the Hamiltonian, and by extension, the Lagrangian. The Hamiltonian $H$ is usually equal to the energy $E=T+U$, but not always. For systems that have a potential $U=U(q, \dot q)$ that depends on the velocities, or systems that have rheonomic constraints (constraints in which time appears as an explicit variable), the Hamiltonian is not equal to the total energy $H\neq E$. However, for all other systems, we shall take eq.\eqref{eqn:hamiltonianenergy} as the definition for the total mechanical energy in Lagrangian and Hamiltonian mechanics, so that

\begin{equation}
E = \sum_{i=1}^{i=N}\pdv{L}{\dot q_i}\dot q_i- L
\end{equation}

To investigate the conservation of total energy, let’s take the time-derivative of $E=\pdv{L}{\dot q}\dot q-L$ for the case of a single degree of freedom.

\begin{equation}
\frac{dE}{dt}=\frac{d}{dt}(\pdv{L}{\dot q}\dot q)-\frac{dL}{dt}
\end{equation}

We will use the product rule for the first term, and since $L=L(q,\dot q,t)$, we will use the multi-variable chain rule for the second term $\frac{dL}{dt}=\pdv{L}{q}\frac{dq}{dt}+\pdv{L}{\dot q}\frac{d\dot q}{dt}+\pdv{L}{t}$.

\begin{equation}
\frac{dE}{dt}=\Big(\frac{d}{dt}\pdv{L}{\dot q}\Big)\dot q + \pdv{L}{\dot q}\frac{d\dot q}{dt}- \Big(\pdv{L}{q}\frac{dq}{dt}+\pdv{L}{\dot q}\dv{\dot q}{t}+\pdv{L}{t}\Big)
\end{equation}

The $\pdv{L}{\dot q}\frac{d\dot q}{dt}$ terms cancel each other out. For the first term, we can use the Euler-Lagrange equation $\pdv{L}{q}=\frac{d}{dt}\pdv{L}{\dot q}$, and for the third term we can recall our notation $\frac{dq}{dt}=\dot q$.

\begin{equation}
\frac{dE}{dt}=\pdv{L}{q}\dot q- \pdv{L}{q}\dot q-\pdv{L}{t}
\end{equation}

Clearly the first two terms cancel out, and we are left with

\begin{equation}
\frac{dE}{dt}=-\pdv{L}{t}
\end{equation}

This important result tells us when the total energy of a system is conserved in the Lagrangian formalism.

So, if time does not explicitly appear in the Lagrangian, then the total energy of the system is conserved. Remember that this is only true for systems in which we can define the total energy as in eq.$\eqref{eqn:hamiltonianenergy}$. We will encounter certain systems for which the Lagrangian has no explicit time dependence ($\pdv{L}{t}=0$), yet the total energy is not conserved. These systems are usually driven by an external force that is itself time-dependent (leading to a rheonomic constraint).

In summary, our discussion of generalized forces, generalized momenta, and the total energy hopefully shed some light on the connection between the Newtonian, Lagrangian, and Hamiltonian formalisms.

Exercises

2.5 — *Forces for velocity-dependent potentials. Most potential energies are only functions of position $U=U(\vb{r})$. Such a potential gives rise to a force $\vb{F}=-\pdv{U}{\vb{r}}$. However, there do exist potential energies that are a function of velocity $U=U(\vb{r},\vb{v})$. Using the Euler-Lagrange equation with $T=\frac{1}{2}m\vb{v}^2$, prove that such potentials are associated with a force

\begin{equation}
\vb{F}=-\pdv{U}{\vb{r}}+\frac{d}{dt}\pdv{U}{\vb{v}}
\end{equation}

Hint: prove the result in one dimension with $T=\frac{1}{2}m\dot x^2$ and $U=U(x,\dot x)$, and then argue for its generalization in three dimensions.

2.6 — *Canonical versus linear momentum. The Lagrangian for a charged particle with electric charge $q_e$ and mass $m$ in an external electromagnetic field is given by

\begin{equation}
L(\vb{r},\vb{v},t) = \frac{1}{2}m\vb{v}^2 + q_e\vb{v}\cdot\vb{A}-q_e\Phi
\end{equation}

where $\vb{A}(\vb{r},t)$ is the vector potential and $\Phi(\vb{r},t)$ is the scalar potential (note that the word potential in this context does not mean potential energy). These potentials give rise to the electric field $\vb{E}=\pdv{\Phi}{\vb{r}}-\pdv{\vb{A}}{t}$ and magnetic field $\vb{B}=\curl{\vb{A}}$. Show that the canonical momentum defined by $\vb{p}_\text{canonical}=\pdv{L}{\vb{v}}$ (this expression means that each component $p_i$ is given by $p_i=\pdv{L}{v_i}$) is related to the linear momentum $\vb{p}_\text{linear}=m\vb{v}$ by

\begin{equation}
\vb{p}_\text{canonical}=\vb{p}_\text{linear}+q_e\vb{A}
\end{equation}

Which momentum (canonical or linear) is necessarily conserved if a system with this Lagrangian satisfies $\pdv{L}{\vb{r}}=0$?

Hint: express the Lagrangian in terms of the components of the vectors $\vb{v}$ and $\vb{A}$ and prove the result one component at a time.