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)mv2 + 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 xo at t = to, formally solve the problem by integrating (get t(x), rather than x(t)): (limits: xo x)

t - to = ∫dx[2(E - U(x))/m]-½

v as a function of x!

t(x) - to = ∫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)kx2

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)mv20

Any real physical situation requires

E = T + U(x) U(x)

T = (½)mv2 0 E = (½)mv2 + U(x) U(x)• Consider motion for different E.

E = E1: Bounded & periodic

between turning points xa & xb.

Bounded Particle never gets out of the region xa x xb.

Periodic: Moving to left, will stop at xa, turn around & move

to right until stops at xb, & turn around again, repeating forever. xa & xb are called Turning Points for obvious reasons.

Turning Points: v = 0, T= 0, E1 = U(x). Gives xa & xb

E = T + U = constant, but T & U change throughout motion.

T = (½)mv2 0 E = (½)mv2 + U(x) U(x)

• Consider motion for different E.

E = E2: Bounded & periodic

between turning points

xc & xd and separately between

turning points xe & xf .

Bounded The particle never

gets out of region xc x xd or out of region xe x xf.

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 > E2 (but, in QM: Tunneling!) Turning points: v = 0, T= 0, E2 = U(x). Gives

xc , xd , xe & xf

T = (½)mv2 0 E = (½)mv2 + U(x) U(x)• Motion for different E:

E = E0: Since E = U(x),

T= 0 & v = 0. Particle

doesn’t move. Stays at x0 forever.

x0 is determined by E0 = U(x0)

E = E3: If the particle is initially

moving to the left, it comes in from infinity to turning point xg, stops, turns

around, & goes back to infinity. Turning point xg determined by v = 0, T=0, E3 =U(x).

E = E4: Unbounded motion. The particle can be at any position. Its speed changes

as E - U(x) = T = (½)mv2 changes.

• Motion of particle at energy E1: This is similar to the mass-spring system.

• Approximate potential for

xa x xb is a parabola: U(x) (½)k(x-x0)2

x0 = equilibrium point (Ch. 3!)

• For the motion of a particle at energy E: If there are 2 turning points, xa & xb, the situation looks like the figure.

The Approximate Potential for xa x xb is a

parabola:

U(x) (½)k(x-x0)2

where x0 is theStable Equilibrium Point

x0

xbxa

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) U0 + x(dU/dx)0 + (x2/2!)(d2U/dx2)0 + (x3/3!)(d3U/dx3)0 + ...

• By definition, if x = 0 is an equilibrium point,

the force = 0 at that point: F0 - (dU/dx)0 = 0

We can choose U0 = 0 since the zero of the potential is arbitrary.

So: U(x) (x2/2!)(d2U/dx2)0 + (x3/3!)(d3U/dx3)0 + ...

• For a general potential U(x), not far from an equilibrium point, keep the lowest order term only:

U(x) (x2/2!)(d2U/dx2)0 Or: U(x) (½)kx2

where k (d2U/dx2)0

• Equilibrium conditions: 1. k = (d2U/dx2)0 > 0 : Stable equilibrium.

U(x) Simple harmonic oscillator potential

2. k = (d2U/dx2)0 <0: Unstable equilibrium.

3. k = (d2U/dx2)0 = 0: May be neutral

equilibrium, but must look at higher order terms.

Example 2.12A string, length b, attached at A, passes over a pulley at B, 2d

away, & attaches to mass m1. Another pulley, with mass m2

attached passes over string, pulls it down between A & B.

Calculate the distance x1 when system is in equilibrium. Is the

equilibrium stable or unstable? Work on the board!

Potential: U(x) = -Wd2(x2+d2)/(x4+8d4) 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 1023.

– 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.