Energy Methods: Sect. 2.6 Consider a point particle under the influence of a conservative force in 1d (for simplicity). Conservation of mechanical energy:  E = T+U = (½)mv 2 + U(x) = constant Rewrite this, using v = (dx/dt) as: v(t) = (dx/dt) =  [2(E - U(x))/m] ½  v(x) Clearly, this requires (for real v): E  U(x) For general U(x), & given initial position x o at t = t o, formally solve the problem by integrating (get t(x), rather than x(t)): (limits: x o  x) t - t o =  ∫dx[2(E - U(x))/m] -½ v as a function of x! 

t(x) - t o =  ∫dx[2(E - U(x))/m] -½ Given U(x), (in principle) integrate this to get t(x) & (in principle) algebraically invert t(x) to get x(t), which is what we want! –In future chapters, we will do (in detail): 1. The harmonic oscillator: U(x) = (½)kx 2 2. Gravitation: U(x) = -(k/x) t as a function of x! 

General U(x): Learn a lot about particle motion by analyzing plot of U(x) vs. x for different energies E: T = (½)mv 2  0  Any real physical situation requires E = T + U(x)  U(x)

T = (½)mv 2  0  E = (½)mv 2 + U(x)  U(x) Consider motion for different E. E = E 1 : Bounded & periodic between turning points x a & x b. Bounded  Particle never gets out of the region x a  x  x b. Periodic: Moving to left, will stop at x a, turn around & move to right until stops at x b, & turn around again, repeating forever. x a & x b are called Turning Points for obvious reasons. Turning Points: v = 0, T= 0, E 1 = U(x). Gives x a & x b E = T + U = constant, but T & U change throughout motion.

T = (½)mv 2  0  E = (½)mv 2 + U(x)  U(x) Consider motion for different E. E = E 2 : Bounded & periodic between turning points x c & x d and separately between turning points x e & x f. Bounded  The particle never gets out of region x c  x  x d or out of region x e  x  x f. Periodic: Goes from one turning point to another, turns around & moves until stops at another turning point. Repeats forever. Particle is in one valley or another. Can’t jump from one to another without getting extra energy > E 2 (but, in QM: Tunneling!) Turning points: v = 0, T= 0, E 2 = U(x). Gives x c, x d, x e & x f

T = (½)mv 2  0  E = (½)mv 2 + U(x)  U(x) Motion for different E: E = E 0 : Since E = U(x), T= 0 & v = 0. Particle doesn’t move. Stays at x 0 forever. x 0 is determined by E 0 = U(x 0 ) E = E 3 : If the particle is initially moving to the left, it comes in from infinity to turning point x g, stops, turns around, & goes back to infinity. Turning point x g determined by v = 0, T=0, E 3 =U(x). E = E 4 : Unbounded motion. The particle can be at any position. Its speed changes as E - U(x) = T = (½)mv 2 changes.

Motion of particle at energy E 1 : This is similar to the mass- spring system. Approximate potential for x a  x  x b is a parabola: U(x)  (½)k(x-x 0 ) 2 x 0 = equilibrium point (Ch. 3!)

For the motion of a particle at energy E: If there are 2 turning points, x a & x b, the situation looks like the figure. The Approximate Potential for x a  x  x b is a parabola: U(x)  (½)k(x-x 0 ) 2 where x 0 is the Stable Equilibrium Point x0x0 xbxb xaxa

Equilibrium Points Equilibrium Point  Point where the particle will stay & remain motionless. –Stable Equilibrium Point  If the particle is displaced slightly away from that point, it will tend to return to it. (Like the bottom of parabolic potential well). –Unstable Equilibrium Point  If the particle is displaced slightly away from that point, it will tend to move even further away from it. (Like the top of an upside down parabolic barrier.) –Neutral Equilibrium Point  If the particle is displaced slightly away from that point, it will tend to stay at new point. (Like a flat potential).

Assume that the equilibrium point is at x = 0. In general, expand U(x) in a Taylor’s series about the equilibrium point [(dU/dx) 0  (dU/dx) x0 ] U(x)  U 0 + x(dU/dx) 0 + (x 2 /2!)(d 2 U/dx 2 ) 0 + (x 3 /3!)(d 3 U/dx 3 ) By definition, if x = 0 is an equilibrium point, the force = 0 at that point:  F 0  - (dU/dx) 0 = 0 We can choose U 0 = 0 since the zero of the potential is arbitrary. So: U(x)  (x 2 /2!)(d 2 U/dx 2 ) 0 + (x 3 /3!)(d 3 U/dx 3 )

For a general potential U(x), not far from an equilibrium point, keep the lowest order term only: U(x)  (x 2 /2!)(d 2 U/dx 2 ) 0 Or: U(x)  (½)kx 2 where k  (d 2 U/dx 2 ) 0 Equilibrium conditions: 1. k = (d 2 U/dx 2 ) 0 > 0 : Stable equilibrium. U(x)  Simple harmonic oscillator potential 2. k = (d 2 U/dx 2 ) 0 <0: Unstable equilibrium. 3. k = (d 2 U/dx 2 ) 0 = 0: May be neutral equilibrium, but must look at higher order terms.

Example 2.12 A string, length b, attached at A, passes over a pulley at B, 2d away, & attaches to mass m 1. Another pulley, with mass m 2 attached passes over string, pulls it down between A & B. Calculate the distance x 1 when system is in equilibrium. Is the equilibrium stable or unstable? Work on the board!

Potential: U(x) = -Wd 2 (x 2 +d 2 )/(x 4 +8d 4 ) Sketch this potential & discuss motion at various x. Is it bounded or unbounded? Where are the equilibrium positions? Are these stable or unstable? Find the turning points for E = -W/8. Example 2.13

Limitations of Newtonian Mechanics Sect. 2.7 Implied assumptions of Newtonian Mechanics: –r, v, t, p, E are all measurable (simultaneously!) –All can be specified with desired accuracy, depending only on the sophistication of our measuring instruments. True for MACROSCOPIC objects! Not true for MICROSCOPIC (atomic & smaller) objects! –Quantum mechanics is needed for these! Heisenberg uncertainty, for example tells us that ΔxΔp  (½) ħ  We cannot precisely know the x & p for a particle simultaneously! –Quantum mechanics   Newtonian mechanics as size of the object increases.

Newtonian mechanics also breaks down when the speed v of a particle approaches a significant fraction of the speed of light c. –Need Special Relativity for these cases (Ch. 14) –The is no concept of absolute time. –Simultaneous events depend on the reference frame. –There is time dilation. –There is length contraction. –Light speed c is limitation on speed of objects.

Practical limitation to Newtonian mechanics: –It is impractical when dealing with systems of huge numbers of particles  –Even with the most sophisticated & powerful computers, we cannot simultaneously solve this many coupled differential equations! –For such problems we need the methods of Statistical Mechanics (Physics 4302). –This uses the methods of probability & statistics to compute average properties of the system.