Forest UCM EnergyIntPart
Energy of Interacting particles
Translational invariance
Consider two particles that interact via a conservative force [math]\displaystyle{ \vec{F} }[/math]
Let [math]\displaystyle{ \vec{r}_1 }[/math] identify the location of object 1 from an arbitrary reference point and [math]\displaystyle{ \vec{r}_2 }[/math] locate the second object.
The vector that points from object 2 to object 1 may be written as
- [math]\displaystyle{ \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]\displaystyle{ \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]
- 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
- [math]\displaystyle{ \vec{F}_{12} = - \vec{\nabla}_1 U(\vec{r}_1) }[/math]
- [math]\displaystyle{ = - \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]\displaystyle{ = - \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) }[/math]
Newton's 3rd law requries that
- [math]\displaystyle{ \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]\displaystyle{ - \left (- \vec{\nabla}_2 U(\vec{r}_2) \right ) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2) }[/math]
or
- [math]\displaystyle{ \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]\displaystyle{ \vec{F}_{12} }[/math] as object 1 moves through [math]\displaystyle{ d\vec{r}_1 }[/math]
plus
- the work done by [math]\displaystyle{ \vec{F}_{21} }[/math] as object 1 moves through [math]\displaystyle{ d\vec{r}_2 }[/math]
This NET work can be determine by taking the derivative of the potential energy
Proof:
- [math]\displaystyle{ W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} }[/math]
- [math]\displaystyle{ = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot (-\vec{F}_{12}) }[/math]
- [math]\displaystyle{ = \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \vec{F}_{12} }[/math]
- [math]\displaystyle{ = \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \left ( - \vec{\nabla}U(\vec {r}_1 ) \right ) }[/math]
- [math]\displaystyle{ = -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}_1 - \vec{r}_2) \right ) }[/math]
- [math]\displaystyle{ = -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}) \right ) =-dU }[/math]
Total Mechanical Energy conservation
The work done by [math]\displaystyle{ \vec{F}_{12} }[/math] as object 1 moves through [math]\displaystyle{ d\vec{r}_1 }[/math] is given by the work energy theorem as
- [math]\displaystyle{ dT_1 = d \vec{r}_1 \cdot \vec{F}_{12} }[/math]
similarly for [math]\displaystyle{ \vec{F}_{21} }[/math]
- [math]\displaystyle{ dT_2 = d \vec{r}_2 \cdot \vec{F}_{21} }[/math]
- [math]\displaystyle{ 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]\displaystyle{ = dT = -dU }[/math]
Thus
- [math]\displaystyle{ dT + dU = 0 }[/math]
or
- [math]\displaystyle{ E_{\mbox{total}} = T_1 + T_2 + dU = }[/math]constant
Elastic Collisions
Definition
BOTH Momentum and Energy are conserved in an elastic collision
- Example
Consider two object that collide elastically
- Conservation of Momentum
- [math]\displaystyle{ \left ( p_1 + p_2 \right ) _{\mbox{initial}} = \left ( p_1 + p_2 \right ) _{\mbox{final}} }[/math]
- Conservation of Energy
- [math]\displaystyle{ \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]\displaystyle{ U_{\mbox{initial}} = U_{\mbox{final}} = 0 = }[/math] arbitrary constant
Example
Consider an elastic collision between two equal mass objecs one of which is at rest.
- Conservation of momentum
- [math]\displaystyle{ m \vec{v}_1 = m \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right ) }[/math]
- Conservation of Energy
- [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2 }[/math]
and you conclude that
- [math]\displaystyle{ 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 \;\;\;\; \Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime} }[/math]