Consider the problem of a classical pendulum whose state can be described by a function $\theta(t)$ where $\theta$ is measured from the line directly below. We then have that our pendulum's $\theta$ obeys the following differential equation

$$ \frac{d^2 \theta}{dt^2 } + \frac{g}{l}\sin \theta = 0 $$

Through the trick $ K = \frac{d\theta}{dt}, K \frac{dK}{d\theta} = \frac{d^2 \theta}{dt^2}$ we can re-write the above differential equation as a different one and then integrate it to find that there is a constant $Q_0$ such that

$$ \frac{1}{2} \left( \frac{d \theta}{d t} \right)^2 - \frac{g}{l} \cos(\theta) = Q_0 $$

It's fruitful to ask "what does this really mean?", what is that $Q_0$ actually supposed to be? and by multiplying both sides by $ ml^2 $ we find rather enlightening that we have the following:

$$ \underbrace{\frac{1}{2} ml^2 \left( \frac{d \theta}{d t} \right)^2}_{\text{Kinetic Energy}} + \underbrace{-mgl \cos(\theta)}_{\text{Gravitational Potential Energy}} = ml^2 Q_0 = E_0 $$

And now this is much less mysterious, it is clear this $Q_0$ is just a scaled version of $E_0$ the total energy of our system, which is constant as we should expect. Of course we can continue going forward here... Before we added the extra mass-length information the differential equation could have been re-written as:

$$ \frac{1}{\sqrt{2Q_0 + \frac{g}{l} 2\cos(\theta)}} \frac{d \theta}{d t} = 1$$

Again this can be integrated to yield another quantity...

$$ \sqrt{\frac{2}{Q_0 + \frac{g}{l}}} F \left[ \frac{\theta}{2} , 2 \frac{g}{l} \frac{1}{Q_0 + \frac{g}{l}} \right] = t + Q_1 $$

This suggests then that the following is true...
$$ \sqrt{\frac{2}{Q_0 + \frac{g}{l}}} F \left[ \frac{\theta}{2} , 2 \frac{g}{l} \frac{1}{Q_0 + \frac{g}{l}} \right] -t = Q_1 $$

I.E. there is some quantity $Q_1$ which does NOT vary with time, and can be found through that horrendous looking left hand side. What conserved Quantity is this $Q_1$ supposed to represent kinematically? It should be something akin to a "Second Energy" or "Momentum" of our pendulum but I can't figure what this thing is supposed to be and there doesn't seem to be any descriptions of it online. It does appear to be intimately related to the period. One could also theoretically verify it is conserved by measuring that LHS in an experiment and confirming it doesn't vary with time.

Some realization:

If you declare the state of your system to be $S$ at time $t=0$ then at any time thereafter you would also declare that "back in $t=0$ the state was $S$". The 'conservation' of $Q_1$ appears to be a restatement of just that.

  • $\begingroup$ Interesting. But I'd be surprised if in this simple model there were another conserved quantity besides energy. Could it be that that "horrendous looking left hand side" can actually be rewritten in terms of the energy only? $\endgroup$
    – fra_pero
    Commented Jun 10, 2020 at 7:18
  • $\begingroup$ Related: physics.stackexchange.com/q/8626/2451 $\endgroup$
    – Qmechanic
    Commented Jun 10, 2020 at 8:39

7 Answers 7



$Q_1$ is not a conserved quantity at all. It is just a parameter which depends on the initial conditions.


First of all, there's an error when you derived

$$\frac{1}{\sqrt{2Q_0 + \frac{g}{l} 2\cos(\theta)}} \frac{d \theta}{d t} = 1\tag{1}$$

You only took the positive square root, whereas you should have take both the possibilities of the RHS being $+1$ and $-1$. You can easily see that the equation $(1)$ never holds whenever $\theta$ is decreasing i.e. $\mathrm d \theta /\mathrm d t<0$. To correct this, we need to add a modulus around the $\mathrm d \theta/\mathrm dt$ term. Thus the corrected equation would be

$$\frac{1}{\sqrt{2Q_0 + \frac{g}{l} 2\cos(\theta)}} \left|\frac{d \theta}{d t}\right| = 1\tag{2}$$

I would advise you to re-integrate equation $(2)$ to find the correct solution which holds over the complete range of motion.

What about the other equation

The final equation which you obtained

$$\sqrt{\frac{2}{Q_0 + \frac{g}{l}}} \left(F \left[ \frac{\theta}{2} , 2 \frac{g}{l} \frac{1}{Q_0 + \frac{g}{l}} \right]\right) -t = Q_1\tag{3}$$

only holds true for the cases where $\mathrm d \theta/\mathrm dt>0$, so for now, we'll only consider cases where the pendulum is going from left to right, but the insight provided below will also help you determine the physical meaning of the new constant that you would obtaing after integrating equation $(2)$. Also, the equation $(3)$ contains an incomplete elliptic integral of the first kind. One of the important properties of this function is that


where $k$ is any real number. Thus, substituting $\theta=0$ in equation $(1)$, we get

\begin{align} \sqrt{\frac{2}{Q_0 + \frac{g}{l}}} \left(F \left[ 0 , 2 \frac{g}{l} \frac{1}{Q_0 + \frac{g}{l}} \right]\right) -t_0 &= Q_1\\ 0-t_0&=Q_1\\ Q_1+t_0&=0\tag{4} \end{align}

where $t_0$ is the time when the pendulum passes throught its equilibrium position for the first time. And since we are only considering the case where $\mathrm d\theta /\mathrm dt>0$, thus the above equation is valid only for the cases where the pendulus comes from the left and goes to the right while passing through the equilibrium position.

Physical Significance

The physical significance of the constant $Q_1$ isn't as deep and profound as you expected. $Q_1$ is just a shifting constant applied to the time. This constant will change upon changing your definition of $t=0$. Thus, it's just a parameter which adjusts/shifts the time scale of the oscillation. It adjusts according to the initial conditions and doesn't give you any more information about the dynamical parameters of the system.

  • 26
    $\begingroup$ This reminds me of Landau's Mechanics, where in chapter two he notices that since the equations of motion are 2nd order differential equations, for every degree of freedom we have 2 constants of integration. So, for $d$ DOFs, the solution depends on $2d$ constants, which are effectively constants of motion. However, since time is translation-invariant, one of these constants always shifts the initial time. Hence, for $d$ DOFs we have always $2d-1$ independent constants of motion. For the pendulum $d=1$, so we have one constant of motion (energy), plus initial time shifting. $\endgroup$
    – DoeJohn
    Commented Jun 10, 2020 at 8:17
  • $\begingroup$ @HicHaecHoc Yeah, that's a great insight while solving problems. $\endgroup$
    – user258881
    Commented Jun 10, 2020 at 8:19
  • $\begingroup$ Why is $F$ an incomplete elliptic integral of the first kind instead of a function? $F[0,k]$ means taking the integral from zero to zero, which is, rather trivially, zero. This holds for any integral. $\endgroup$ Commented Jun 11, 2020 at 19:58
  • $\begingroup$ @descheleschilder No, the OP obtained the incomplete elliptic integral of the first kind after integrating their differential equation (which is correct from the POV of integration). $F$ isn't a generic function. It is a specific function used to represent elliptic integrals which can't be expressed in closed form using well known functions. $\endgroup$
    – user258881
    Commented Jun 11, 2020 at 20:20
  • $\begingroup$ @descheleschilder Yeah, so the property that $F(0,k)=0$ is more of a trivial result than a specific important property. Nonetheless, the answer still holds. $\endgroup$
    – user258881
    Commented Jun 12, 2020 at 10:59

Suppose that in any physical system, the solution to the equation of motion is $x(t) = f(t, x_0, v_0)$. Then $x - f = 0$, so it's conserved. In this way, you can manufacture a new conserved quantity for any physical situation. You can also add on any function $g(x_0, v_0)$ of the initial conditions, giving an infinite family of conserved quantities $x - f + g$. This is what you found.

For example, for a ball in freefall, you can easily check that $x - (x_0 + v_0 t - gt^2/2) + x_0$ is conserved, for this reason. But this isn't a new conserved quantity at all -- it's just a minor rewriting of the solution to the equation of motion, whose particular value is the initial position.

You can't use this idea to do anything. If you don't already know the general solution $f$ then you can't compute $x-f+g$, if you do know $f$ then you don't need it, and if you don't know $f$ but somehow know the numeric value of $x-f+g$, that just tells you about the initial conditions, which you already knew anyway.


The configuration space of a pendulum is 1D (in fact, a circle, $S^1$) so it's phase space is 2D (a cylinder, $S^1\times \mathbb{R}$). If there were two integrals of motion then we could label each point in the 2D phase space by those two values, and since they're meant to be conserved, the phase space dynamics would have to be trivial (i.e. position and momenta never change).

So whatever your $Q_1$ is it is either:

(a) Some function of $Q_0$ so not an independent integral of motion or

(b) A weaker kind of conserved quantity that is not just a function of phase space coordinates. For instance, the initial angle and angular velocities are strictly conserved quantities along a trajectory.

I suspect your $Q_1$ can be written in terms of the initial conditions, ie is of type (b).


I'm not sure that this represents an actual "conserved quantity." In order for it to, you'd need it to satisfy $$ \{ H, Q_1\} = \frac{\partial Q_1}{\partial t}. $$

Here, I am taking $H = Q_0$ to be the (rescaled) Hamiltonian which governs time evolution. Then the momentum is $p$ which, by Hamilton's equation, is $p = \dot \theta$.

The problem is you have defined $Q_1$ in terms of $Q_0$, which is really a function of $\theta$ and $p$. In order for this to be a true conserved quantity, it would have to satisfy the above equation on phase space (using the full definition of $Q_0(\theta, p)$ when plugged into $Q_1$) which I don't think it does, but I could be wrong.


For an ODE system with $N$ initial conditions, there are $N$ conserved quantities, say, $q_i(t)-f_i(q_1,q_2,...,q_N, t)=0$, where $q_i$ are coordinates and velocities of a general meaning and $f_i$ are the corresponding solutions of the ODE.

Any combination of conserved quantities is also a conserved quantity. You may add to the left-hand side and to the right-hand side any constant, for the simplest example.

And now you may combine them into complicated functions to be constants too.


Let's take a dimensional analysis approach:

You end up in the first part with:

$$ ml^2 Q_0 =E_0,$$

writing that it is clear that $Q_0$ is just a scaled version of $E_0$ (so also energy). Dimensional analysis shows that this can't be the case.

Insofar the mass $m$ is concerned one can write: $F=ma$, which translated in units gives: $N=kg\frac{m}{{sec}^2}$, with the result that mass has the unit $\frac{N{sec}^2}{m}$.
$l^2$ has obviously the unit $m^2$.
Combining both, $ml^2$ has the unit $Nm{sec}^2=J{sec}^2$.

It follows, because $E_0$ has the unit $J$, $Q_0$ must have the unit $\frac{1}{{sec}^2}$ (because $ml^2$ has the unit $J{sec}^2$). This means $Q_0$ is not energy, which implies it is not a scaled version of $E_0$ (which would be the case if $ml^2\gt 1$ was a dimensionless scaling factor, which it is not).

So the first quantity $Q_0$ is just $\frac{E_0}{ml^2}$, which is obviously conserved ($E_0$, $m$, and $l$ are constant for the pendulum).

When it comes to the second quantity $Q_1$ note that in your last equation,

$$ \sqrt{\frac{2}{Q_0 +\frac{g}{l}}}F\left[\frac{\theta}{2},2\frac{g}{l}\frac{1}{Q_0 + \frac{g}{l}}\right]-t = Q_1 ,$$

the square root part is constant and has the unit $sec$ ($Q_0$, $g$, and $l$ are all constants; both $Q_0$ and $\frac{g}{l}$ have the unit $\frac{1}{{sec}^2}$, so the square root of their inversed sum has the $sec$ as a unit). Let's call this point in time $t_2$.
Because we can't subtract two quantities with different units (the expression containing the square root and $F$, minus $t$), $F$ will have to spit out a real number without a unit.
Note that the second argument of $F$ has a constant value without a unit ($2\frac{g}{l}$ times the inverse of $Q_0 +\frac{g}{l}$, gives the unit $\frac{{sec}^2}{{sec}^2}=1$, i.e. no unit at all) for every dimensionless $\theta$ in the first argument, so the product with the square root has the unit $sec$, as it should. For every angle $\theta$, let's call the constant real number without a unit spit out $F$, so the product of the square root and $F$ gives $Ft_2$ It's clear that $Q_1$ has the $sec$ for a unit so it represents a point in time. Let's call this point $Ct_1$, in which $C$ is a constant without a unit. Altogether this gives

$$Ft_2 -t=Ct_1$$ so $$Ft_2-Ct_1=t$$

This means that $Q_1$ ($Ct_1$) is just the time that can be subtracted from a given time $Ft_2$ with as result the time $t$.


An Analogy

Let's examine the conceptually equal system the motion of a freely falling mass $m$ from height $h$ relative to the ground it falls on, which has one DOF (the vertical displacement), and is described by the same kind of differential equation:

$$ \frac{d^2 s}{dt^2 }-g = 0 $$

We can apply the same trick: $K=\frac{ds}{dt}$, $K\frac{dK}{ds}=K\frac{d(\frac{ds}{dt})}{ds}=K\frac{1}{dt}= \frac{d^2 s}{dt^2 }$, and integrate wrt to $s$:

$$\int_0^h(K\frac{dK}{ds}-g)ds=\frac{1}{2}K^2-gh=Q_0,$$ so, after multiplying both sides with the mass $m$ one gets $$\frac{1}{2}mv^2-mgh=E_0=mQ_0,$$ so $Q_0=\frac{E_0}{m}$ and this equation expresses the conservation of energy $E_0$.

So, $$v^2=2\frac{E_0}{m}+2gh,$$ from which it follows $$\frac{1}{\sqrt{\frac{2E_0}{m}+2gh}}\frac{ds}{dt}=1,$$ so $$\int_0^h\frac{1}{\sqrt{\frac{2E_0}{m}+2gh}}ds=t+C_1$$

Now both quantities under the square root sign have $\frac{m^2}{{sec}^2}$ so the inverse square root function has the unit $\frac{sec}{m}$. This means that the integral of this function over $ds$ (which has the value $Ch$ in which $C$ is the constant inverse square root) has the dimension time and so has $Q_1$. It's the difference between $Ch$ and a time somewhere on the trajectory of the freely falling mass.


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