Forest UCM EnergyIntPart

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Energy of Interacting particles

Translational invariance

Consider two particles that interact via a conservative force [math]\vec{F}[/math]

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

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

[math]\vec r = \vec{r}_1 - \vec{r}_2[/math]

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

If the force is a central force

[math]\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 )[/math]

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

[math]\vec{F}_{12} = - \vec{\nabla}_1 U(\vec{r}_1) [/math]
[math] = - \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)[/math]
[math]= - \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) [/math]

Newton's 3rd law requries that

[math]\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) [/math]
[math]- \left (- \vec{\nabla}_2 U(\vec{r}_2) \right ) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)[/math]


[math] \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)[/math]

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 [math]\vec{F}_{12}[/math] as object 1 moves through [math]d\vec{r}_1[/math]


the work done by [math]\vec{F}_{21}[/math] as object 1 moves through [math]d\vec{r}_2[/math]

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


[math]W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} [/math]
[math]= d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot (-\vec{F}_{12}) [/math]
[math]= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \vec{F}_{12} [/math]
[math]= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \left ( - \vec{\nabla}U(\vec {r}_1 ) \right ) [/math]
[math]= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}_1 - \vec{r}_2) \right ) [/math]
[math]= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}) \right ) =-dU[/math]

Total Mechanical Energy conservation

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

[math]dT_1 = d \vec{r}_1 \cdot \vec{F}_{12} [/math]

similarly for [math]\vec{F}_{21}[/math]

[math]dT_2 = d \vec{r}_2 \cdot \vec{F}_{21} [/math]

[math]W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} = dT_1 + dT_2[/math]
[math]= dT = -dU[/math]


[math]dT + dU = 0 [/math]


[math]E_{\mbox{total}} = T_1 + T_2 + dU = [/math]constant

Elastic Collisions


BOTH Momentum and Energy are conserved in an elastic collision


Consider two object that collide elastically

Conservation of Momentum
[math]\left ( p_1 + p_2 \right ) _{\mbox{initial}} = \left ( p_1 + p_2 \right ) _{\mbox{final}}[/math]
Conservation of Energy
[math]\left ( T + U \right ) _{\mbox{initial}} = \left ( T + U \right ) _{\mbox{final}}[/math]

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

[math]U_{\mbox{initial}} = U_{\mbox{final}} = 0 =[/math] arbitrary constant


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

Conservation of momentum
[math] m \vec{v}_1 = m \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right )[/math]
Conservation of Energy
[math] \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[/math]

Square the conservation of momentum equation
[math] \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 )[/math]
[math] 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} [/math]

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

[math] v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2[/math]

and you conclude that

[math]2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 \;\;\;\; \Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime} [/math]