Download Sects. 2.6 & 2.7

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 as a function of x!

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
t 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)
• General U(x): Learn a lot about particle motion by
analyzing plot of U(x) vs. x for different energies E:
T = (½)mv20

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 massspring 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 the
Stable Equilibrium Point
xb
xa
x0
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 2.12
A 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!
Example 2.13
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.
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.