# Proof of total energy is constant in a central force field

I saw many proofs of total energy is constant in a central force field. But all the proofs end up showing this formula $$m[{\dot r}^2 + r ^2{\dot \theta}^2 ]+ \int f(r)dr = E$$ is constant. But nobody showed $$dE/dt = 0$$ can somebody proof that or give any reference. And suppose $$f(r)$$ is proportional to $$1/r^2$$ if its needed for the proof. For any f(r) somebody can prove energy is constant that is subject to extra credit.

• Maybe I'm not understanding something about the question, but doesn't this follow simply because a central force is conservative, i.e., has zero curl?
– user4552
Commented Nov 28, 2018 at 1:28
• @BenCrowell yeah spherically symmetric cf fields have zero curl. But im interested in showing $dE/dt =0$ Commented Nov 28, 2018 at 1:29
• A constant with respect to time does not have a vanishing time derivative? What is the distinction you are trying to make? Commented Nov 28, 2018 at 2:10
• @CosmasZachos im asking if E is a constant throughout a particular trajectory then it's differential wrt time should also be zero . As it never changes with time . Commented Nov 28, 2018 at 4:17
• This term is correct? $(\dot{r}\,\dot{\varphi})^2$ or this one $({r}\,\dot{\varphi})^2$
– Eli
Commented Nov 28, 2018 at 8:27

Assume you are given an inertial coordinate system $$O\vec{e}_1\vec{e}_2\vec{e}_3$$ and denote by $$x = x_1\,\vec{e}_1 + x_2\,\vec{e}_2 + x_3\,\vec{e}_3$$ the position vector of a point with respect to the given system. Then, we can also write $$x =\begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} \, \in \, \mathbb{R}^3$$ to simplify notations

In this shorthand notation, a central force field is a vector function of the form $$F(x) = - \, f(\,\|x\|\,)\,x = - \, f\Big(\, \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 }\, \Big)\begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix}$$ where $$f \, : \, \mathbb{R} \to \mathbb{R}$$ is usually a differentiable function and $$\|x\| = \sqrt{(x\circ x)} = \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 }$$ The notation $$(x\circ y)$$ is for the dot product between the two vectors $$x$$ and $$y$$.

A particle of mass $$m$$ moves in such a force field following the system of differential equations $$\frac{d^2 x}{dt^2} = - \,\frac{1}{m}\,f(\|x\|)\, x$$ or component-wise \begin{align} &\frac{d^2 x_1}{dt^2} = -\,\frac{1}{m}\, f\Big(\, \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 }\, \Big) \, x_1\\ &\frac{d^2 x_2}{dt^2}= - \,\frac{1}{m}\, f\Big(\, \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 }\, \Big) \, x_2\\ &\frac{d^2 x_3}{dt^2} = - \,\frac{1}{m}\, f\Big(\, \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 }\, \Big) \, x_3 \end{align} I think you get the picture of shorter vector notations. From now on I will also write derivative as upper dots, i.e. $$\dot{x} = \frac{dx}{dt}$$ and $$\ddot{x} = \frac{d^2x}{dt^2}$$.

Proposition 1. Every central force field, as defined above, is conservative, i.e. there exists a potential function $$U : \mathbb{R}^2 \to \mathbb{R}$$ such that $$\nabla U(x) = f(\|x\|)\, x$$

Proof: Define the following function $$V(r) = \int_{0}^{r}\, \rho\,f(\rho) \, d\rho$$ for $$\rho > 0$$ as long as the integral is convergent at $$\rho = 0$$, which puts some restriction on the type of functions $$f(r)$$ you are allowed. They should behave well near $$r=0$$, i.e. they may have singularity, but it should be of a certain type. Anyway, this is a technicality.

Let us now define the potential $$U(x) = V(\|x\|)$$ Now we calculate the gradient of $$U(x)$$ by applying the chain rule. \begin{align} \nabla U(x) = & \nabla\big(\, U(x) \,\big) = \nabla\big(\, V(\|x\|)\, \big) = V'(\|x\|)\, \nabla \|x\| = V'(\|x\|)\, \nabla \big(\, (x \circ x)^{\frac{1}{2}}\, \big) = \\ = & \, V'(\|x\|)\, \frac{1}{2} (x \circ x)^{\frac{1}{2}-1}\, \nabla \big(\,(x \circ x)\, \big) = \, V'(\|x\|)\, \frac{1}{2} (x \circ x)^{- \, \frac{1}{2}}\, 2 \, x =\\ = & \, V'(\|x\|)\, \frac{1}{2} \, \frac{1}{(x \circ x)^{\frac{1}{2}}} \, 2\, x = \, V'(\|x\|)\, \frac{1}{2 \, \sqrt{(x \circ x)}}\, 2\, x = \\ = & \, V'(\|x\|)\, \frac{1}{\sqrt{(x \circ x)}}\, x = \, V'(\|x\|)\, \frac{1}{\|x\|}\, x = \\ = & \, \frac{V'(\|x\|)}{\|x\|}\, x \end{align} But recall that by Newton-Leibniz's theorem $$V'(r) = \frac{d}{dr} \, \int_{0}^{r} \, \rho \, f(\rho)\, d\rho = r \, f(r)$$ which implies $$V'(\|x\|) = \|x\|\, f(\|x\|)$$ Therefore \begin{align} \nabla U(x) &= \nabla\big(\, V(\|x\|)\, \big) = \, \frac{V'(\|x\|)}{\|x\|}\, x = \, \frac{\|x\|\, f(\|x\|)}{\|x\|}\, x =\\ &= \, f(\|x\|)\, x \end{align}

Define the total energy function $$E(x, \dot{x}) = \frac{m}{2} \|\dot{x}\|^2 + U(x)$$

Theorem. (Conservation of energy) Let $$x = x(t)$$ be the solutions to the initial value problem for the following system of differential equations \begin{align} &\ddot{x} = - \,\frac{1}{m}\, f(\|x\|)\, x \\ &x(0) = x_0\\ &\dot{x}(0) = v_0 \end{align} where $$x_0$$ and $$v_0$$ are two fixed vectors from $$\mathbb{R}^3$$. Then for any $$t \in \mathbb{R}$$ we have the following property $$E\big(x(t), \dot{x}(t)\big) = E(x_0, v_0)$$ i.e. the total energy $$E\big(x(t), \dot{x}(t)\big)$$ of the point particle moving in the central force field $$- \, f(\|x\|)\, x$$ at an arbitrary moment of time $$t$$ is the same as the energy $$E_0 = E(x_0, v_0)$$ of the point particle at the beginning of its motion, at time $$t=0$$. In other words, the total energy fo the particle always stays equal to $$E_0$$.

Proof: We know from calculus that $$E\big(x(t), \dot{x}(t)\big) = E(x_0, v_0)$$ for all $$t$$ if and only if $$\frac{d}{dt} E\big(x(t), \dot{x}(t)\big) = 0$$ By definition, $$E\big(x(t), \dot{x}(t)\big) = \frac{m}{2} \|\dot{x}(t)\|^2 + U\big(x(t)\big) = \frac{m}{2} \big(\dot{x}(t)\circ \dot{x}(t)\big) + U\big(x(t)\big)$$ To simplify notations, I will suppress the explicit notation for the argument $$t$$. Then $$E\big(x, \dot{x}\big) = \frac{m}{2} \big(\dot{x}\circ \dot{x}\big) + U\big(x\big)$$ Differentiate the latter identity with respect to $$t$$ \begin{align} \frac{d}{dt} E\big(x, \dot{x}\big) &= \frac{d}{dt} \left( \frac{m}{2} \big(\dot{x}\circ \dot{x}\big) + U\big(x\big)\right) = \frac{m}{2} \frac{d}{dt} \big(\dot{x}\circ \dot{x}\big) + \frac{d}{dt} U\big(x\big) = \\ &= \frac{m}{2} \, 2\, \big(\dot{x}\circ \ddot{x}\big) + \big( \, \nabla U(x) \circ \dot{x} \,\big) \end{align} Recall that by the system of differential equations (which are the mathematical manifestation of the second law of Newton) we have that $$\ddot{x} = -\, \frac{1}{m}\, f(\|x\|)\, x$$ and by the proposition from above $$\nabla U(x) = f(\|x\|)\, x$$ Plug these two expressions in the calculation for the derivative of the energy function \begin{align} \frac{d}{dt} E\big(x, \dot{x}\big) &= \frac{d}{dt} \left( \frac{m}{2} \big(\dot{x}\circ \dot{x}\big) + U\big(x\big)\right) = \frac{m}{2} \, 2\, \big(\dot{x}\circ \ddot{x}\big) + \big( \, \nabla U(x) \circ \dot{x} \,\big) = \\ &= \frac{m}{2} \, 2\, \Big(\, \dot{x}\circ \big( -\, \frac{1}{m}\, f(\|x\|)\, x\, \big) \, \Big) + \Big( \, \big( f(\|x\|)\, x \big) \circ \dot{x} \,\Big) = \\ &= \Big(\, \dot{x}\circ \big(\, - \, f(\|x\|)\, x\, \big) \, \Big) + \Big( \, \big( f(\|x\|)\, x \big) \circ \dot{x} \,\Big) =\\ &= - \, f(\|x\|)\, \big(\, \dot{x}\circ x\, \big) + f(\|x\|)\, \big( \, x \circ \dot{x} \,\big) = \\ &= - \, f(\|x\|)\, \big(\, {x}\circ \dot{x} \, \big) + f(\|x\|)\, \big( \, x \circ \dot{x} \,\big) = \\ &= 0 \end{align} Hence, there exists a constant $$E_0$$ such that $$E\big(x(t), \dot{x}(t)\big) = E_0$$ for all $$t \in \mathbb{R}$$. The constant cab be calculated by plugging for example $$t=0$$, which yields $$E\big(x(t), \dot{x}(t)\big) = E(x_0, v_0) = E_0$$

• Why do you have an equals sign at the start and end of every line in the $\frac{d}{dt}E(x,\,\dot{x})$ equation array? Commented Nov 28, 2018 at 11:14
• @KyleKanos I do not know, I haven't thought about it too much. Just a habit, I guess. Like half of the equal sign stays on one line, the other half continues on the following line :D ... Commented Nov 28, 2018 at 12:34
• Okay, well there is no such thing as "half of the equal sign," so what you're doing is wrong. So you probably should kick that habit. Commented Nov 28, 2018 at 12:41
• @Futurologist great answer . Helped me a lot . Thanks. Keep posting such detailed answers . Great job. Commented Nov 28, 2018 at 15:32

\begin{align*} &L=T-V=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\varphi}^2\right)-V(r)\\ &\Rightarrow\\ &\text{Equation of motion,coordinate \varphi}\\ &p_\varphi=\frac{\partial L}{\partial \dot{\varphi}}=m\,r^2\dot{\varphi}\\ &\dot{p}_\varphi=\frac{d}{dt}\left(m\,r^2\dot{\varphi}\right)\\& \Rightarrow\\& m\,r^2\dot{\varphi} =\text{const=l}&(1)\\\\ &\text{for the coordinate r we obtain:}\\ &\frac{d}{dt}\left(m\,\dot{r}\right)-m\,r\,\dot{\varphi}^2+\frac{\partial V}{\partial r}=0\\ &\text{and with } -\frac{\partial V}{\partial r}=f(r)\quad \Rightarrow\\ &m\,\ddot{r}-m\,r\,\dot{\varphi}^2=f(r)&(2)\\ &\text{Equation (1) in (2) we obtain}\\ &m\,\ddot{r}-\frac{l^2}{m\,r^3}=f(r)\quad, m\,\ddot{r}+\frac{\partial}{\partial r}\left(\frac{1}{2}\frac{l^2}{m\,r^2}\right)=-\frac{\partial V}{\partial r}&(3)\\\\\\ &\text{The Energie equation is:}\\ &E=T+V=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\varphi}^2\right)+V(r)\\ &\text{with equation (1)}\\ &E=\left(\frac{1}{2}\,m\,\dot{r}^2+V+\frac{1}{2}\frac{l^2}{m\,r^2}\right)\\ \\&\text{multiply equation (3) with \dot{r} and rearange we get:}\\ &m\,\dot{r}\,\ddot{r}=-\dot{r}\,\frac{\partial}{\partial r}\left(V+\frac{1}{2}\frac{l^2}{m\,r^2}\right) &(4)\\ &\text{with}\quad m\,\dot{r}\,\ddot{r}=\frac{d}{dt}\left(\frac{1}{2}\,m\,\dot{r}^2\right)\quad \text{and}\quad \frac{d}{dt} V(r)=\frac{d V}{d r}\dot{r}\\ &\Rightarrow\quad\text{equation (4)}\\ &\frac{d}{dt}\left(\frac{1}{2}\,m\,\dot{r}^2\right)=- \frac{d}{dt}\left(V+\frac{1}{2}\frac{l^2}{m\,r^2}\right)\\\\ &\boxed{\frac{d}{dt}\underbrace{\left(\frac{1}{2}\,m\,\dot{r}^2+V+\frac{1}{2}\frac{l^2}{m\,r^2}\right)}_{=E}=0} \end{align*}