Physics 430 Lecture 10 Energy of Interaction Dale E Gary NJIT Physics Department 4 9 Energy of Interaction of Two Particles Up to now we have been discussing the energy of a single particle Now let s look at the energy of two particles Imagine two isolated particles alone in the universe no external forces interacting with each other Particle 2 at position r2 from the origin exerts a force F12 on particle 1 while the particle 1 at position r1 exerts an equal and opposite force F21 F12 on particle 2 The distance between the two particles is then r r1 r2 as shown in the figure For definiteness consider a gravitational interaction so that as usual Gm1m2 Gm1m2 r r 2 3 r r From the definition of r and the vector difference of the two position vectors we can rewrite this Gm1m2 as r1 r2 F12 3 r1 r2 F12 2 r r 1 r2 1 r2 r1 O The fact that this depends only on the difference of the two position vectors means that the force is translationally invariant October 3 2008 Energy of Interaction of Two Particles 2 This fact of translational invariance the fact that the force of interaction depends only on the distance between two particles and not on their absolution position allows us to greatly simplify the discussion by choosing a point r2 say as the origin even though the position of particle 2 is changing i e the origin need not be a fixed point With this choice our earlier discussion of force on a single particle applies For example if the force F12 on particle 1 is conservative then 1 F12 0 where the operator 1 x y z x1 y1 z1 with respect to the coordinates r1 x1 y1 z1 When the curl above is zero i e the force is conservative then we can define a potential energy F12 1U r1 Note that U r1 is defined throughout space but the force at particle 1 is the gradient of U r1 evaluated at the point where particle 1 is i e at r1 October 3 2008 Energy of Interaction of Two Particles 3 The foregoing is specific to the case when particle 2 is at the origin but we can always translate to some other origin such that particle 2 is at position r2 so that F12 1U r1 r2 1 The trick is that we do not have to modify the operator because x1 a derivative like is unchanged by adding a constant to x1 We can then find the force on particle 2 by particle 1 as F12 1U r1 r2 F21 2U r1 r2 U r1 r2 What this says is that we can consider a single potential energy for the interaction between particles 1 and 2 and evaluate its gradient at the position of each particle to find the force on each Force on particle 1 1U particle Force on particle 2 2U Now consider the two particles moving through space under their mutual interaction so that particle 1 moves through a distance dr1 under the force F12 and particle 2 moves through a distance dr2 October 3 2008 under the force F21 Total Energy of Interaction By the work KE theorem there will be a change in kinetic energy of each of the particles dT1 F12 dr1 and dT2 F21 dr2 We simply add these to find the total work done Wtot F12 dr1 F21 dr2 F12 dr1 F12 dr2 where the last equality is because again F21 F12 This can be rewritten W dr dr F d r r 1U r1 r2 tot 1 12 1 2 If we rename r1 r2 as 2r we finally have Wtot this dr U r dU that if the force of Let s pause to appreciate result It says interaction between two particles is conservative then we can define a potential energy throughout space As the particles move the potential energy changes but the change is just the negative of the change in kinetic energy that results from the interaction force Obviously then the total energy is conserved i e Notice that there areEtwo one for each particle but T kinetic U T1energies T2 U constant only one potential energy arising from the configuration of the particles October 3 2008 Elastic Collisions Elastic collisions are ones that take place through conservative forces so that no energy is lost to heat or other mechanism A collision between a proton and an electron for example occurs through the conservative electrostatic force Collisions between two billiard balls is largely elastic because the balls are made to act like stiff springs when they collide Example 4 8 An Equal Mass Elastic Collision Statement of the problem Consider an elastic collision between two particles of equal mass m1 m2 m as shown in the figure Prove that if particle 2 is initially at rest then the angle between the two outgoing velocities is 90o Solution mv1 mv1 mv 2 Conservation of momentum gives The fact that1 mv the2 collision 2 v 2 1 mv 2 is1 elastic mv 2 means v 2 vthat 2 1 2 1 2 2 Squaring the Comparing these we see that v1 v 2 0 1 1 1 2 1 2 v1 2 2 momentum equation after2 1 1 1 2 2 v v 2 v v v eliminating m gives 2 v v1 and v2 are 90 degrees apart October 3 2008 2 v 1 4 10 Generalizing to Multiple Particles In the case of many particles the text goes through the argument explicitly for 4 particles and then for N particles the same thing holds Taking the system of N particles interacting through conservative forces one can consider the internal potential energy of the system U int and any external potential energy U ext imposed from outside The total potential energy is then int ext ext U U U U U where the sums are over all of the particles and the in the second of the double sum ensures that we do not double count an interaction e g U12 and U21 The total kinetic energy is much simpler just One can calculate the force on particle by meaning calculate the gradient of U at the position of the particle The total energy F U E T U is conserved T T 12 m v 2 October 3 2008 Rigid Bodies The subtlety of U U int U ext U U ext can be illustrated with an example of rigid body rotation For a macroscopic rigid body the number of particles atoms may be vast However because the body is rigid the interaction forces between atoms is constant and the particles do not shift position appreciably to change the internal potential energy Thus the internal potential energy is constant and we can completely ignore it However if the body is NOT rigid and the shape or configuration of atoms changes then there may be an appreciable change in internal kinetic energy that DOES have to be taken into account An example is a star that may expand or contract during its lifetime …