# Does $\frac{d}{dt}E=0$ always yield unambiguous equations of motion?

If you have a conservative system, one way you can derive the equations of motion is by using the fact that the total energy $$E$$ of the system is conserved, i.e.:

$$\frac{d}{dt}(T+V)=0$$

For example consider a simple textbook-pendulum. The total energy is:

$$T + V = \frac{1}{2}ml^2\dot\varphi^2 + mgl(1-\cos\varphi)$$

Differentiating once with respect to time leads to the following condition, which is trivially true if $$\dot\varphi = 0$$ or if the terms within the brackets are zero. The latter case leads directly to the familiar equation of motion for a simple pendulum.

$$0 = \dot\varphi (ml^2\ddot\varphi + mgl\sin\varphi)$$

$$\Rightarrow \ddot\varphi + \frac{g}{l}\sin\varphi = 0$$

## Now to the fun stuff

Consider a system where a pendulum is attached to a mass which itself is attached to a spring. This is a system which has two independent degrees of freedom $$x$$ and $$\varphi$$.

The total energy of the depicted system is the following mess:

$$T + V = \frac{1}{2}m_1\dot x^2 + \frac{1}{2}m_2(\dot x^2 + 2l\dot\varphi \dot x \cos\varphi + l^2 \dot\varphi^2) + m_2gl(1-\cos\varphi) + \frac{1}{2}kx^2$$

Differentiating this mess with respect to time leads to an even bigger mess:

$$\dot x \ddot x(m_1 + m_2 + k) + m_2l\left(\ddot \varphi \dot x \cos\varphi + \dot\varphi \ddot x \cos\varphi - \dot\varphi^2\dot x \sin\varphi + l\dot\varphi\ddot\varphi - g\dot\varphi\sin\varphi\right)$$

Since the system has two degrees of freedom, we should get two coupled equations of motion. And we can indeed get two such equations if we do the same thing we did above, namely factoring out $$\dot x$$ and $$\dot\varphi$$. There is just one slight inconvience here. This term exists.

$$-m_2l\dot\varphi^2\dot x \sin\varphi$$

Here we can factor out both, $$\dot x$$ and $$\dot\varphi$$ which effectively gives two possible sets of equations of motion. Namely these two:

$$0 = \ddot x (m_1 + m_2) + m_2l(\ddot\varphi\cos\varphi - \dot\varphi^2\sin\varphi) + kx$$

$$0 = m_2l(\ddot x \cos\varphi + l^2\ddot\varphi - g\sin\varphi)$$

or these two:

$$0 = \ddot x (m_1 + m_2) + m_2l\ddot\varphi\cos\varphi + kx$$

$$0 = m_2l(\ddot x \cos\varphi + l^2\ddot\varphi - g\sin\varphi - \dot\varphi \dot x \sin\varphi)$$

How can we know which way of factoring and which equations are correct? Are both possibilities equaly true? Or is this method just not "valid" for systems with multiple degrees of freedom?

• FWIW: Energy conservation gives 1 EOM. This is not enough if there are more than 1 DOF. Commented Mar 13, 2021 at 21:23
• $\dot E=0$ is one equation, so it yields a single equation of motion. For 1d systems with a single degree of freedom, $\dot E=0$ yields every equation of motion (i.e., one equation). For systems with more degrees of freedom, $\dot E=0$ does not yield all the equations, but only one of them. Commented Mar 13, 2021 at 21:23
• Thank's! So I basically need some kind of second "constraint" in addition to $\dot E = 0$, to describe this system completely (or I could just use Lagrange, but where is the fun in that? :D) If so, what kind of constaint could that be? Commented Mar 13, 2021 at 21:32
• – user87745
Commented Mar 13, 2021 at 21:59
• If you're already working with the energy, then the Hamiltonian equations of motion are likely what you're looking for. Turning conservation equations into equations of motion is essentially the purpose of transforming to action-angle variables. Commented Mar 14, 2021 at 6:23