Forest UCM EnergyIntPart

Energy of Interacting particles

Translational invariance

Consider two particles that interact via a conservative force $\vec{F}$

Let $\vec{r}_1$ identify the location of object 1 from an arbitrary reference point and $\vec{r}_2$ locate the second object.

The vector that points from object 2 to object 1 may be written as

$\vec r = \vec{r}_1 - \vec{r}_2$

The distance between the two object is given as the length of the above vector

If the force is a central force

$\vec F = \frac{k}{r^3} \vec r = \frac{k}{\left | r_1 - r_2 \right |^3} \left ( \vec{r}_1 - \vec{r}_2 \right )$

Notice

The interparticle force is independent of the coordinate system's position, only the difference betweenthe positions matters

If object 2 was fixed so it is not accelerating and we place the origin of the coordinate system on object 2

Then the force is that of a single object

One potential for Both Particles

Both forces from same potential

If the above force is conservative then a potential exists such that

$\vec{F}_{12} = - \vec{\nabla}_1 U(\vec{r}_1)$

$= - \vec{\nabla}_1 U(\vec{r}_1) = - \left( \hat i \frac{\partial}{\partial x_1} + \hat j \frac{\partial}{\partial y_1} + \hat k \frac{\partial}{\partial z_1} \right ) U(\vec{r}_1)$

$= - \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2)$

Newton's 3rd law requries that

$\vec{F}_{21} = - \vec{F}_{12} = - \left (- \vec{\nabla}_1 U(\vec{r}_1) \right ) = \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2)$

$- \left (- \vec{\nabla}_2 U(\vec{r}_2) \right ) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)$

or

$\vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)$

You can find the net external force on a body in the system once you have the potential for the system

Total work given by one potential

The total work is the sum of

the work done by $\vec{F}_{12}$ as object 1 moves through $d\vec{r}_1$

plus

the work done by $\vec{F}_{21}$ as object 1 moves through $d\vec{r}_2$

This NET work can be determine by taking the derivative of the potential energy

Proof:

$W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21}$

$= d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot (-\vec{F}_{12})$

$= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \vec{F}_{12}$

$= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \left ( - \vec{\nabla}U(\vec {r}_1 ) \right )$

$= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}_1 - \vec{r}_2) \right )$

$= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}) \right ) =-dU$

Total Mechanical Energy conservation

The work done by $\vec{F}_{12}$ as object 1 moves through $d\vec{r}_1$ is given by the work energy theorem as

$dT_1 = d \vec{r}_1 \cdot \vec{F}_{12}$

similarly for $\vec{F}_{21}$

$dT_2 = d \vec{r}_2 \cdot \vec{F}_{21}$

$W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} = dT_1 + dT_2$

$= dT = -dU$

Thus

$dT + dU = 0$

or

$E_{\mbox{total}} = T_1 + T_2 + dU =$constant

Elastic Collisions

Definition

BOTH Momentum and Energy are conserved in an elastic collision

Example

Consider two object that collide elastically

Conservation of Momentum

$\left ( p_1 + p_2 \right ) _{\mbox{initial}} = \left ( p_1 + p_2 \right ) _{\mbox{final}}$

Conservation of Energy

$\left ( T + U \right ) _{\mbox{initial}} = \left ( T + U \right ) _{\mbox{final}}$

When the initial and final states are far away fromthe collision point

$U_{\mbox{initial}} = U_{\mbox{final}} = 0 =$ arbitrary constant

Example

Consider an elastic collision between two equal mass objecs one of which is at rest.

Conservation of momentum

$m \vec{v}_1 = m \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right )$

Conservation of Energy

$\frac{1}{2} m v_1^2 = \frac{1}{2} m \left (v_1^{\;\prime} \right )^2 + \frac{1}{2} m\left ( v_2^{\;\prime} \right )^2$

Square the conservation of momentum equation

$\vec{v}_1 \cdot \vec{v}_1 = \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right ) \cdot \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right )$

$v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime}$

compare the above conservation of momentum equation with the conservation of energy equation

$v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2$

and you conclude that

$2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 \;\;\;\; \Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime}$

Forest_UCM_Energy#Energy_of_Interacting_Particles