# Are these motion equations equivalent in Newtonian mechanics?

Let's say we're working with a particle being subjected to a conservative force $$\vec{F}$$. Using Newtonian mechanics, we know that the equation of motion $$r(t)$$ for the particle is found using Newton's Second Law: $$m\ddot{\vec{r}} = \vec{F}\left(\vec{r}\right) = - \nabla U\left(\vec{r}\right) \tag{1}$$ where we used that the force is conservative, and hence, we can write it as the gradient of the one-dimensional potential.

On the other hand, by conservation of energy, we can say that $$E = K + U\left(\vec{r}\right)$$, but recalling that $$K = \frac{1}{2}m \lvert \dot{r}\rvert^2$$ we can establish the following relationship as well $$\frac{1}{2}m \big\lvert \dot{\vec{r}}\big\rvert^2 = E- U\left(\vec{r}\right) \tag{2}$$

Both $$(1)$$ and $$(2)$$ are differential equations for the same function $$\vec{r}(t)$$, so my question is, are both of these equations equivalent? And if so, how could I show that these equations are equivalent?

No, they are not!

What are our unknowns? They are three real-valued functions of time: $$\bigl(x(t), y(t), z(t) \bigr) =: \pmb{r}(t)$$.

The vector equation $$m \ddot{\pmb{r}} = -\pmb{\nabla}U[\pmb{r}]$$ is a collection of three real differential equations: \left\{\begin{aligned} m\ \ddot{x}(t) &= -(\partial_x U)[x(t), y(t), z(t)]\\ m\ \ddot{y}(t) &= -(\partial_y U)[x(t), y(t), z(t)]\\ m\ \ddot{z}(t) &= -(\partial_z U)[x(t), y(t), z(t)]\\ \end{aligned}\right. From them and given values at, say, $$t=0$$ of $$\pmb{r}(t)$$ and $$\dot{\pmb{r}}(t)$$ we can find the three functions.

The equation $$\tfrac{1}{2}m \lvert\dot{\pmb{r}}(t)\rvert^2 + U[\pmb{r}(t)] = E$$ is one real-valued differential equation: $$\tfrac{1}{2}m\ [\dot{x}(t)^2 + \dot{y}(t)^2 +\dot{z}(t)^2] + U[x(t), y(t), z(t)] = E$$ From it we cannot find three real-valued functions.

As Qmechanic wrote, this latter equation is implied by the former vector equation, but not vice versa. It can be used to find the three real-valued functions only if additional information (equivalent to two more equations), such as symmetries, is given.

When we consider thermodynamic systems, the former vector equation is not sufficient anymore, the energy equation (which will include work and heating terms) is not derivable from it and must be included as an independent fourth equation. The reason is that we have an additional fourth unknown real-valued function, namely the temperature $$T(t)$$ or equivalently the internal energy $$\epsilon(t)$$.

### References

For what it's worth $$(1)\Rightarrow (2)$$ but they are only equivalent in 1D.

(E.g. in 3D there could be a magnetic Lorentz force that (2) doesn't know about.)

It's quite easy, Let's see the only $$x$$-component $$m\ddot{x}=F_x=-\frac{\partial U}{\partial x}$$

From energy conservation $$\frac{1}{2}m(\dot{x}^2+\cdots )+U=E$$ Differentiate with respect to $$t$$ $$m\dot{x}\ddot{x}+\frac{\partial U}{\partial x}\dot{x}=0$$ $$\rightarrow m\ddot{x}=-\frac{\partial U}{\partial x}$$ As desired!

• Ohhhhh. It's so obvious now that you explained it! Thank you very much. Commented Jun 12, 2021 at 4:13
• There is a flaw in this reasoning. The "$\dotso$" parts do not disappear when you take the time derivative, so you cannot obtain the next-to-last equation. Commented Jun 12, 2021 at 9:38