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.

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    $\begingroup$ 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? $\endgroup$
    – user4552
    Commented Nov 28, 2018 at 1:28
  • $\begingroup$ @BenCrowell yeah spherically symmetric cf fields have zero curl. But im interested in showing $dE/dt =0$ $\endgroup$ Commented Nov 28, 2018 at 1:29
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    $\begingroup$ A constant with respect to time does not have a vanishing time derivative? What is the distinction you are trying to make? $\endgroup$ Commented Nov 28, 2018 at 2:10
  • $\begingroup$ @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 . $\endgroup$ Commented Nov 28, 2018 at 4:17
  • $\begingroup$ This term is correct? $(\dot{r}\,\dot{\varphi})^2$ or this one $({r}\,\dot{\varphi})^2$ $\endgroup$
    – Eli
    Commented Nov 28, 2018 at 8:27

2 Answers 2


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 $$E\big(x(t), \dot{x}(t)\big) = E(x_0, v_0) = E_0$$

  • $\begingroup$ 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? $\endgroup$
    – Kyle Kanos
    Commented Nov 28, 2018 at 11:14
  • $\begingroup$ @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 ... $\endgroup$ Commented Nov 28, 2018 at 12:34
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    $\begingroup$ 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. $\endgroup$
    – Kyle Kanos
    Commented Nov 28, 2018 at 12:41
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    $\begingroup$ @Futurologist great answer . Helped me a lot . Thanks. Keep posting such detailed answers . Great job. $\endgroup$ Commented Nov 28, 2018 at 15:32

We start with the LAGRANGE-Function

\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*}


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