I'm trying to derive Hamilton's Principle from Lagrange's Equations, as I've heard they're logically equivalent statements, and am stuck on a final step. For simplicity, assume we're dealing with a system with only one generalized coordinate $q$, so that there is only one equation of motion of the system:

$$\frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) - \frac{\partial L_q}{\partial q} \circ \Lambda_t = 0 \quad,$$

where $L_q:\mathbb{R}^3 \to \mathbb{R}$ is the Lagrangian of the system expressed in terms of $q$, $\dot{q}$ and time $t$, $\Lambda_t(t) = (q_t(t), \dot{q_t}(t),t)$, where $q_t:\mathbb{R} \to \mathbb{R}$ is an actual trajectory, so a solution to the differential equation.

Working back from the steps taken to arrive at this equation from the variational principle, I did the following: let $\delta q_t:\mathbb{R} \to \mathbb{R}$ be a deviation from the actual trajectory, so a smooth function that satisfies $\delta q_t (t_1) = 0, \delta q_t(t_2) = 0$, for some given time instants $t_1$ and $t_2$. Multiplying the differential equation through by $\delta q_t$ and integrating from $t_1$ to $t_2$, we get

$$\int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t - \frac{\partial L_q}{\partial q} \circ \Lambda_t \phantom{,} \delta q_t \right] = 0 \quad.$$

Since by hypothesis $\delta q_t (t_1) = 0, \delta q_t(t_2) = 0$, the following holds

$$\int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \delta q_t \right) = 0$$

$$\Rightarrow \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t = \int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \delta q_t \right) - \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t \right] = \int_{t_1}^{t_2} \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \dot{\delta q_t} \quad.$$

So the full integral becomes

$$\int_{t_1}^{t_2} \left[ \frac{d}{dt} \left( \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \right) \delta q_t + \frac{\partial L_q}{\partial \dot{q}} \circ \Lambda_t \phantom{,} \dot{\delta q_t} \right] = 0 \quad,$$

which we can recognize as being the statement that the variation of some functional of the form

$$A(q_t) = \int_{t_1}^{t_2} L_q(q_t(t), \dot{q_t}(t), t) \phantom{,} dt$$

must be equal to zero for all admissible $\delta q_t$, so

$$\delta A(q_t, \delta q_t) = 0 \quad.$$

So starting from Lagrange's Equation for the system, we logically arrive at the implication that the variation of the functional $A$ (the action functional) must be zero at the system's true trajectory, for such $\delta q_t$. But this doesn't on its own imply that $q_t$ is in fact an extremum of $A$, which is what Hamilton's Principle states, and from which $\delta A(q_t, \delta q_t) = 0$ follows. Where's the missing piece that lets us conclude that $q_t$ is in fact an extremum of $A$, and not only a point at which its variation vanishes? Is it a mathematical fact about the form of the functional itself that allows us to conclude that?


3 Answers 3


Hamilton’s principle is often sloppily stated. It is not a “least-action” principle, nor is it a “principle of extremal action”. It is a principle of “stationary” action.

For example, on the round sphere, take two non-antipodal points, and consider the longer arc of the great circle which joins these two points (i.e the longer geodesic). This path is a saddle point for the length functional on the sphere (the length functional is the action in this context).

  • $\begingroup$ Oh, that wraps it up. I had read that Hamilton's Principle is often wrongly stated as a >least<-action principle, so I took that to mean that it the action could also be a maximum. Even my professor stated it as an extremal principle, not simply a stationary one. So my derivation is sufficient? $\endgroup$
    – jvf
    May 27 at 19:36
  • $\begingroup$ Yes it is sufficient. $\endgroup$
    – peek-a-boo
    May 27 at 20:06

In general, a solution of the Euler-Lagrange equation is not a local extremum of the action. This is why you were having a tough time.

Take for example the humble harmonic oscillator: $$ \begin{align} L&=\frac{\dot q^2-q^2}{2}\\ S&=\int_0^T Ldt\\ 0&= \ddot q +q \end{align} $$

The function is quadratic in $\delta q$, so you you just need to check the homogeneous quadratic part to see whether the stationary trajectory is an extremum. It is diagonalisable using the expansion: $$ \delta q=\sum_{n=1}^\infty \delta q_n \sin\left(\frac{\pi nt}{T}\right) $$ and has the spectrum: $$ \sigma= \left(\frac{\pi n}{T}-1\right)_{n\in\mathbb N^*} $$ which is positive iff $T<\pi$. In other words, if $T<\pi$, the harmonic trajectory is a global minimum and when $T>\pi$ it is a saddle point.

This confirms that the true trajectory is not necessarily an extremum (even local). However, the example shows that if $T$ is small enough, then it is a minimum. This is actually pretty general which is why it is rightfully called the principle of least action.

You’ll need functional calculus to prove that though and you’d be leaving the realm of physics. Essentially, what saves you is the kinetic energy which scales up when $T\to0$ since it involves a time derivative. Typically, the statement is true for a Lagrangian of the form: $$ L=\frac{1}{2}\dot q^2-V(q) $$ with pretty loose assumptions on $V$. It can be further generalized to more general forms most notably to geodesics where the kinetic part is more complicated.

Even in peek-a-boo’s example, if the end points are sufficiently close, the geodesic is a true length minimizing curve.

Mathematically, being a minimum is pretty relevant since it allows the use of compacity arguments. This gives you powerful tools to prove the existence of trajectories.

Btw the moment the stationary solutions are not minima anymore is connected to caustics. In the harmonic example, $\pi$ it the time where trajectories start intersecting and you lose unicity.

Hope this helps.


(...) that $q_t$ is in fact an extremum of $A$, and not only a point at which its variation vanishes?

As discussed in the answer by contributor peek-a-boo: one should not pay too much attention to the customary name of the concept.

As you found out in the course of your exploration: the actual criterion is: find the point in variation space such that the derivative of Hamilton's action is zero.

So: a more descriptive name of the concept would be:
Hamilton's vanishing derivative

That raises the question: what is it about taking the derivative that makes it essential?

In order to access the necessary information the operation that is performed is taking the derivative with respect to the spatial coordinate(s).

The fact that taking a derivative (with respect to position) is necessary is correlated with the fact that potential energy does not have an intrinsic zero point. (Unlike, say, temperature, which does have an intrinsic zero point.) The choice of zero point of potential energy is arbitrary; whenever a potential energy is evaluated it is in terms of difference of potential.

To access the necessary information you take the derivative of the potential energy with respect to the spatial coordinate.

When you are using generalized coordinates: the derivative of the potential energy with respect to that generalized coordinate $q$ is referred to as a 'generalized force'.

Quoting the discussion of generalized forces by Richard FitzPatrick:

(...) a generalized force does not necessarily have the dimensions of force. However, the product $Q_i\,q_i$ must have the dimensions of work. Thus, if a particular $q_i$ is a Cartesian coordinate then the associated $Q_i$ is a force. Conversely, if a particular $q_i$ is an angle then the associated $Q_i$ is a torque.

The (generalized) force is a quantity that does have an intrinsic zero point.

Generalizing the relation between force and second time derivative of position

In cartesian coordinates:

In polar coordinates:
Torque = moment of inertia $\times$ second time derivative of angle

For current through an inductor (LC-circuit):
Electromotive force = inductance $\times$ change of current

The general pattern is:

  • There is an entity that tends to cause change of state, in mechanics referred to as 'force'.
  • There is opposition to change of state, in mechanics referred to as 'inertia', in an inductor referred to as 'Inductance', etc.
  • The relation between the above two is in terms of a second time derivative of the state, in mechanics referred to as 'acceleration'.

Integration of force/torque/etc. with respect to position coordinate gives work done, and potential energy is defined as the negative of work done.

Integral of acceleration with respect to position:

(Using the relations $ds=v \ dt$ and $a \ dt =dv$ to change the differential, with corresponding change of limits.)

$$ \int_{s_0}^s a \ ds = \int_{t_0}^t a \ v \ dt = \int_{t_0}^t v \ a \ dt = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 $$

The pattern is:
Take a second derivative wrt position (here: acceleration), and integrate with respect to position. The result is an expression that is quadratic in the first time derivative of position: ($\tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2$)

Conversely, taking the derivative of $\tfrac{1}{2}v^2$ with respect to position recovers the acceleration.

As we know: the true trajectory has the property that throughout the trajectory the derivative with respect to time of the sum of potential and kinetic energy is zero.

$$ \frac{d(E_k + E_p)}{dt} = 0 \tag{1} $$

There is another derivative-is-zero property of the true trajectory: derivative with respect to the position coordinate $s$:

$$ \frac{d(E_k - E_p)}{ds} = 0 \tag{2} $$

Of course: the striking thing is:
In (1) the potential energy has a plus-sign, and in (2) the potential energy has a minus-sign.

The plus sign being replaced with a minus sign is because of the following:
With respect to time the two energies (potential and kinetic), are counter-changing, such that the sum remains the same value.
When sweeping out variation of the trial trajectory: with respect to the position coordinate the two energies (potential and kinetic), are co-changing. Hence (1) features a plus-sign and (2) features a minus sign.

The thing is: (1) and (2) are not independent equations. (1) and (2) are two ways of expressing the same physical property.

The effect of applying the Euler-Lagrange equation

In every situation where you apply the Euler-Lagrange equation you are applying the following operation: differentiation with respect to the position coordinate.

By differentiating with respect to the position coordinate you are accessing the information that you need.

The fact that Hamilton's stationary action holds good goes back to the property that is expressed by (1) and (2).


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