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After all, energy conservation equation is a differential equation that can be solved to find the motion, but this is never done. It is alway considered equation of motion only the time derivative of energy conservation equation. Why? It's simpler? Consider for example the spring-mass system. I can write $$E= \frac{1}{2} m [x'(t)]^2 + \frac{1}{2} k [ x(t) - \bar{X}]^2 $$ This is a differential equation solved by $$ x(t) = \bar{X} + \sqrt{\frac{2E}{k}} \sin \left( \sin^{-1} \left( (x_0 - \bar{X}) \sqrt{\frac{k}{2E}} \right) +\sqrt{\frac{k}{m}} t \right) $$ We don't have the position as a function of $x_0$ and $v_0$, but as a function of $x_0$ and $E$, it is the same.

Being more general, consider $E=\frac{1}{2}m\dot{x}^2 + U$. If I do time derivative, if $\dot{x} \neq 0$ and exploiting $F=-\frac{dU}{dx}$ I can write $m \ddot{x} = F$. The reverse too can be done: the $m \ddot{x} = F$ can be written $m \frac{dv}{dt} + \frac{dU}{dx} = 0$. Integrating we have $m\int v dv + U =$ constant: call $E$ the constant and the job is done. Are energy conservation equation and equation of motion substantially equivalent of there is some reason to not use conservation equation as equation of motion?

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    $\begingroup$ What do you mean by it's ''never done''? Writting the energy conservation, separating variables and integrating is how I've mostly seen done the resolution of the Kepler problem, just to give a canonical example. Energy conservation is an equation of motion because it allows you to derive the motion of the system. You may need other equations if you have more degrees of freedom, of course. $\endgroup$
    – Gilgamesh
    Commented Apr 23, 2021 at 21:39
  • $\begingroup$ That's basically the root of hamiltonian formalism, which uses the total energy of a system to find the equations of motion. $\endgroup$
    – Triatticus
    Commented Apr 24, 2021 at 1:10
  • $\begingroup$ if the energy is conserved, you obtain the equation of motion with $\dfrac{dE}{dt}=0$ $\endgroup$
    – Eli
    Commented Apr 24, 2021 at 6:24

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Yes, energy conservation equations and Newton’s second law essentially can both be solved to yield the equations of motion for systems in which the forces involved are all conservative forces. (This is the same as requiring that a potential energy can be defined.) An example of a non-conservative force is the force of friction.

As you have found, you can rewrite the equations of motion in terms of a variety of different initial condition terms, and for any particular problem it is usually done in such a way that the equations look the most “intuitive” - which varies from problem to problem, and honestly from person to person. As far as which equation is easier to solve to obtain the equation of motion - again, it varies from problem to problem.

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Conservation of energy alone can only be used to solve for the motion of a particle in one space dimension. If you have more spatial dimensions, knowing $\tfrac{1}{2} m v^2$ does not necessarily tell you the direction of $\vec{v}$. In the Kepler problem, you have to actually use other symmetries to figure this out. You first restrict yourself to one 2D plane, and then use the conservation of angular momentum as well to pin down $\vec{v}$. So, for the Kepler problem (elliptic orbital motion) we were only able to solve the motion in this way because we had as many conserved quantities (energy and angular momentum) as dimensions. Another way to say this is that the Kepler problem is "integrable," i.e. it has enough conserved quantities to solve for the motion easily. In 1D, everything is integrable. However for higher dimensions and arbitary potentials $V(\vec{x})$ this will not be the case.

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To summarize what the OP says. Taking $E(x, \dot{x}, t)$, the energy in a general dependence of position, velocity and time. If we take the total derivative respect to time: $$ \frac{dE}{dt} = \frac{\partial E}{\partial x}\dot{x} + \frac{\partial E}{\partial \dot{x}} \ddot{x} + \frac{\partial E}{\partial t} \tag{1} $$ If system is conservative and energy does not depends explicitly upon time we have: $$\frac{\partial E}{\partial x}\dot{x} + \frac{\partial E}{\partial \dot{x}} \ddot{x} = 0 \tag{2}$$ You can write the energy in the case of conservative force like a sum of kinetic energy, quadratic in velocity and potential energy. $$ E = E_{kin}(\dot{x}^2) + U(x) \tag{3}$$ Using (2) and (3): $$ m\dot{x}\ddot{x} + \frac{\partial U}{\partial x}\dot{x} = 0 \tag{4}$$ Taking $\dot{x} $ not equal to zero, and dividing for it. $$ m\ddot{x} = - \frac{\partial U}{\partial x} \tag{5}$$ That is excactly the equation of motion in 1-D, for conservative force, where you can obtain force from gradient (in 1-D, simple derivative) of a function called potential energy.

When energy is a first integral and it's a one dimensional problem we know all the information we need to know. If energy is a constant of motion is constant along the trajectory, then energy constancy is a constrain that allow to write the solutions. All this talking resembles the Hamiltonian formulation of mechanics.

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You are correct that conservation of energy gives us equations of motion.

However, each time this works, notice that you impose some constraints. In your case, you imposed that the motion is 1-dimensional. Notice that conservation of energy gives you 1 equation. But, sometimes, we need a system of differential equations. For instance, consider a block on frictionless incline on frictionless ground. Both the block and incline will move, so one equation is insufficient.

The total energy of a closed system is actually the Hamiltonian, $\mathcal H = T+U$, and you can find the equations of motion using Hamiltonian mechanics.

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All mechanical equations can be formulated by using simple energy conservation, more information can be seen via https://iopscience.iop.org/article/10.1088/2399-6528/ac57f8 for discrete system, or

https://academic.oup.com/ptep/article/2022/12/123A01/6832290 for a continuous system.

Hamilton's principle uses energy and work to formulate equations of motion, but not uses energy conservation of energy.

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  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Hyperon
    Commented Jan 24, 2023 at 14:34

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