How to prove conservation of energy from a central potential, taking Newton's laws as assumptions?

Question

This is an entirely mathematics question, but only with some trivial physics assumptions.

This is a two-body system made into a one-body system by using reduced mass.

$U(r) = -\int_{r_A}^{r}dr'\mathbf{F_1}(\mathbf{r'}).\mathbf{r'} + U(r_A) $

I know that $r_A$ is arbitrary. So,

$U(r_A) = U(r) +\int_{r_A}^{r}dr'\mathbf{F_1}(\mathbf{r'}).\mathbf{\hat{r}'}$ must hold for all $\mathbf{r}$ (or $|\mathbf{r}|$?).

Given it's a central force, $\mathbf{F_1}\propto\mathbf{r}$ and $|\mathbf{F_1}|=f(r)$ and therefore,

$U(r_A) = U(r) +\int_{r_A}^{r}dr'F_1(r')$.

But how does this (mathematically) lead to,

$E = \frac{1}{2}\mu\mathbf{\dot{r}}.\mathbf{{\dot{r}}}+U(r)$

given Newton's second law,

$\mathbf{F_1}=\mu\mathbf{\ddot{r}}$.

I'm looking for the explicit mathematically lines of reasoning and not physics.

Answer 1

You can actully derive it using the definition of work.

so we have a central force: $$ \vec{F}=-\frac{d U}{d r}\hat{r} $$ let's calculate the work: between to states $S_1$ and $S_2$: $$ W_{12}=\int_{S_1 \to S_2} \vec{F}\cdot \vec{\textrm{dr}}=-\int_{S_1 \to S_2} \frac{d U}{d r}\hat{r}\cdot \vec{\textrm{dr}} $$ which is obviously: $$ W_{12}=U(r_1)-U(r_2) $$ now let's do the same using newton second law $\vec{F}=\mu\frac{d\vec{v}}{dt}$: $$ W_{12}=\int_{S_1 \to S_2} \vec{F}\cdot \vec{\textrm{dr}}=\int_{S_1 \to S_2} \mu\frac{d\vec{v}}{dt}\cdot \vec{\textrm{dr}}= $$ $$ =\int_{S_1 \to S_2} \mu\frac{d\vec{v}}{dt}\cdot \frac{d\vec{r}}{dt}\textrm{dt}=\int_{S_1 \to S_2} \mu\frac{d\vec{r}}{dt}\cdot \vec{\textrm{dv}}=\frac{1}{2}\mu v_2^2-\frac{1}{2}\mu v_1^2 $$ (Remark: this demonstration of the work-energy principle can be done in much more mathematical detail in the lagrangian setting, this is mostly an intuitive proof, consult wikipedia:W-E Principle for more detail)

equate: $$ \frac{1}{2}\mu v_2^2-\frac{1}{2}\mu v_1^2 = U(r_1)-U(r_2) $$ rearrange so all variables of $S_1$ and $S_2$ are on opposite sides: $$ \frac{1}{2}\mu v_2^2+U(r_2) = \frac{1}{2}\mu v_1^2+U(r_1) = \textrm{constant} $$ so since $S_1$ and $S_2$ are arbitrary this holds for any state,thus we have obtained a constant of the motion, let's call it Energy: $$ E=\frac{1}{2}\mu v^2+U(r)=\frac{1}{2}\mu \dot{\mathbf{r}}\cdot\dot{\mathbf{r}}+U(r) $$

Answer 2

You will need the physical definitions of kinetic energy as $T = \frac{1}{2}m \dot{r}^2$ where $r = |\vec{r}|$, and total energy as $E = T + U$. Then you can use the fact that a central force is necessarily conservative, and so $\vec{F} = - \frac{\partial U}{\partial r}\hat{r}$, in this case. By Newton's Second Law, by the other hand, you have $\vec{F} = m\ddot{\vec{r}} $. Then all you need to do is to calculate $\frac{dE}{dt}$.