Chapter 4. Dynamics and geometry of successive collisions

Mechanics gets difficult for many collisions, dimensions, or masses. A single one-dimensional two-mass (1D-2-body) collision occupies Ch. 2-3. Now we do more dangerous things such as an X2-super bouncer from Project Ball, a 1969 class project. (Am. J. Phys. 39, 656 (1971)) See product liability disclaimer in Fig. 4.1.

 

Fig. 4.1 The X2-pen launcher with product liability disclaimer.

 

            At first, the X2 looks like a 1D-2-body device. A superball(©™Whammo Corp.) of mass M₁ =70gm launches a ballpoint pen of mass M₂ =10gm. But, it has a 3^rd body, bounce plate mass-M_O=10kg shown by a rectangle in Fig. 4.1. Actually the third body most responsible for this experiment is good old Mother Earth of mass. (Earth mass and solar mass are good-to-2-figure numbers for astrophysicists to remember. More precisely: and .)

            Collisions of very large with very small masses beg thorny questions (Like, “What IS mass?” or how do we deal with it?) As a mass ratio M₁/ M₂ approaches zero or infinity the slope of the P-conservation line in (V₁,V₂)-space (Recall Fig. 3.2.) approaches infinity or zero, respectively, as drawn in Fig. 4.2(a-b).

Geometric construction in Fig. 4.2a of final velocity for an elastic collision is a vertical reflection thru the COM point (V₁=V₂) on the P-line if M₁>> M₂ or else a horizontal reflection in Fig. 4.2b if M₁<< M₂. Inelastic final points approach the COM point more closely if inelasticity is significant. (Recall Fig. 3.2.)

            You should understand how a relatively large mass may give huge momentum to a smaller one but transfer only tiny amounts of energy. Each P-line in Fig. 4.2 is part of a KE-ellipse. In the COM frame (where the COM point is at origin) the P-line sits on top of an entire E-ellipse as the ratio M₁/ M₂ approaches (a) infinity or (b) zero. I visualize COM P-lines as ultra-thin ellipses between I₀ and F₀ and other P-lines in Fig. 4.2 as segments of a KE-ellipse that has (a) a huge V₂-axis or (b) a huge V₁-axis .

Fig. 4.2 Extreme mass-ratio collisions (a) M₁/ M₂ approaches infinity. (b) M₁/ M₂ approaches zero.

 

            Fig. 4.2a reflects our common experience of a bouncy ball of mass M₂ hitting the Earth of mass with velocity –V₀(point I₀) and being reflected with velocity +V₀(point F₀). While standing in the Earth frame, one is very nearly in the COM frame, too. Earth’s COM velocity is a tiny fraction of the apparent ball velocity V₀. For super-balls of mass M₂=60gm, the fraction  is 0.06/(6·10²⁴)=10^-26.

Bounce momentum absorbed by Earth is 2 M₂V₀ (or M₂V₀ if the ball goes “Ka-runch!”) but Earth absorbs at most a tiny KE of , that is, a fraction 10^-26 of ball KE: . Moreover, for elastic collisions, Mother Earth returns all the KE to M₂ but she absorbs double momentum P=2 M₂V₀.

However, common experience does not prepare us for X2 easily rebounding M₂ with more than twice its drop velocity in Fig. 4.3. (As we’ll see that means M₂ rises to more than four times its drop height!)

 

Fig. 4.3 n-Body collision experiments. (a) X-2 drop. (b) Independent collision model. (c) Ball towers.

 

 

Independent collision  models (ICM)

To compute final velocities of M₁ and M₂ it helps to idealize the collision of three bodies M₁, M₂, and as a sequence of two separate 2-body collisions that are completely determined by P and KE conservation. First M₁ bounces off Earth. Only then does M₁ knock M₂ to a faster speed as in Fig. 4.3b. The first collision is labeled Bang-1₍₀₁₎ in Fig. 4.4a followed by Bang-2₍₁₂₎ in Fig. 4.4b. The first Bang-1₍₀₁₎ between Earth and M₁ has a horizontal line like the I₀F₀ line in Fig. 4.2b. The second Bang-2₍₁₂₎ between mass M₁ and M₂ has a line of slope -M₁/ M₂ =-7 for a M₁ =70gm and M₂ =10gm (that of a superball and pen, respectively). The Bang-2₍₁₂₎ line is like the IF line in Fig. 3.1 or Fig. 3.2.

Fig. 4.4 (V₁-V₂)-plot of 2-Bang collision. (a) M₁ bounces off floor. (b) M₁ hits M₂ head-on.

 

This approximation is called an independent collision model (ICM) and is one secret to analyzing such 1D-3-body bang-up that otherwise has too many unknown velocities to be found by just two equations ÄP=0 and ÄKE=0 alone. ICM is exactly true if we initially separate M₁ and M₂ so three M₁, M₂, and never collectively bargain for available momentum and energy. ICM also applies to n-ball towers in Fig. 4.3c. They give very high-energy ejections and serve as classical models for supernovae. (N-body bangs are in Ch.8.)

            Velocity geometry suggests a family of X2 solutions as shown in Fig. 4.5 for a range of mass ratio M₁/M₂. This is an advantage of geometric solutions. Just a few points in Fig. 4.5a show all elastic (V₁-V₂) points lie on the 45°-line CPL. Extreme or optimal cases are located in Fig. 4.5b.

 Extreme and optimal cases

First, the upper limit for elastic final velocity is V₂=3·V₀ at pt-I for infinite mass ratio M₁/M₂. If no energy is lost, a particle of dust on a superball could be ejected three times the speed that the ball hits the floor. (And, it could go nine (9=3²) times the drop height. However, the elastic ICM model is not so good for tiny M₂ due to weak molecular forces. So bouncing balls don’t embed dust in ceilings. (But in a vacuum...!)

Second, an optimal performance case is shown by pt-M where the collision achieves a 100% transfer of energy to projectile M₂. The M-point is the intersection of the CPL line with the V₂-axis on which the M₁-ball velocity is zero. (V₁=0) There mass ratio is M₁/M₂=3.0, the slope of the M-line.

 

 

Fig. 4.5 X2-Final (V₁,V₂) (a) Final point locus. (b) Infinite ratio pt. I and maximum transfer pt. M.

 

            Another singular point U is for unit ratio M₁/M₂=1, a familiar ratio for players of billiards or pool. U undergoes inversion of velocities (+1,-1)-> (-1,+1). (Its COM point lies at origin.) If the U-line is boosted by (-1) to (0,-2)-> (-2,0) it is like a straight elastic pool shot. A 100% of KE transfers from a moving ball to an equal sized ball that was stationary. The same process at half that speed is (0,-1)-> (-1,0) shown by the Galileo-shifted line U₁-> U₂ in the lower left hand side of Fig. 4.5b.

            Points D between U and M have ball M₁ knocked to negative velocity by the down-coming M₂. Then M₁ hits the floor (Earth) at velocity –v to rebound at +v. For unit ratio case U, M₁ and M₂ rebound quite like a rigid body. Below U, ball M₁ rebounds at a speed faster than M₂ to hit M₂ again. In cases of low mass ratio, (M₁/M₂<<1) mass M₁ must hit M₂ many times to turn it around. We will study this effect shortly.

 Integrating velocity plots to find position

It is important to see how velocity values of Fig. 4.4b are turned into space-time position plot lines. Consider the first collision (Bang-1₍₁₀₎) in Fig. 4.6a and corresponding space-time paths in Fig. 4.6b. Initial velocity V_y1(0)=-1.0 gives a slope (distance)/(time) of an M₁ path but doesn’t tell where is the path or particle. The same for velocity V_y2(0)=-1 of M₂ in Fig. 4.6a. The paths need location, location,…

Initial position values such as (y₁(0)=1, y₂(0)=3) locate the paths as shown in Fig. 4.6b. Each path keeps its slope until a collision (Bang-1₍₁₀₎) between M₁ and the floor occurs at y₁(t=1) where its path and the floor intersect. Then, according to Fig. 4.6a, M₁ bounces its slope from V_y1=-1 up to V_y1=+1. Meanwhile, the upper path (M₂) maintains its down slope of V_y2=-1 until it intersects the rising path of M₁.

 

Fig. 4.6 Plots of 1^st collision (Bang-1₍₁₀₎). (a) Velocity-velocity plot. (b) Space-time plot.

 

            At time (t=2) there is an intersection of paths and the 2^nd collision (Bang-2₍₁₂₎) between M₁ and M₂ at space-time point (y₁(2)=1, y₂(2)=3). This gives V_y1=0.5 and V_y2=2.5 in Fig. 4.4b or in Fig. 4.7a-b below. Then to keep M₂ from flying away we install an elastic ceiling at y=7.

            The game becomes more interesting as Bang-3₍₂₀₎ between the ceiling (part of Earth ) is shown in Fig. 4.7b by a vertical arrow (like an IF line in Fig. 4.2a) reflecting M₂ to speed V_y2=-2.5. Then M₂ has Bang-4₍₁₂₎ between M₁ and itself that sends it back to the ceiling at a blistering speed of V_y2=+2.7  as M₁ returns more slowly toward the floor with velocity V_y1=-0.5.

The high speed of M₂ lets it go to the ceiling for Bang-5₍₂₀₎ and return to knock M₁ down once more (Bang-6₍₁₂₎) before M₁ hits the floor at V_y1=-0.9. (Bang-7₍₁₀₎) Then M₂ having lost speed to V_y2=+1.5 hits the ceiling (Bang-8₍₀₂₎) and returns for Bang-9₍₁₂₎ with M₁ rising at V_y1=+0.9.

Masses are treated as point-masses moving along straight lines between collisions in space-time plots. This is an ideal gravity-free ICM approximation with only straight lines in VV-plots. So we may derive motion without having to integrate the kinetic equations at the end of Ch. 3.

Fig. 4.7 Collision sequence. (a-b) Up to Bang-4₍₁₂₎. (c-d) Up to Bang-9₍₁₂₎.

 

For comparison, a force-law simulation using BounceIt of the bang sequence of Fig. 4.7 is shown in Fig. 4.8. It has finite radius balls instead of ideal point particles, yet compares quite well. (So far as it goes.)

                       

Fig. 4.8 BounceIt x-vs.-t simulation to compare with Fig. 1.4.7d up to Bang-6. (V₁/V₂ =7/1.)

 

Fig. 4.7c and BounceIt V₁-V₂ simulations in Fig. 4.9 build an ellipse out of multiple IF lines. (This is a quite non-traditional ellipse construction!) Ellipse radii (a,b) follow from KE conservation equation (3.7b).

 

As time increases (Fig. 4.9a to Fig. 4.9c) the ellipse may fill with IF-lines that are dense (ergodic) or else just retrace sets of paths as in Fig. 4.9b. (Ch. 5 treats non-ergodic paths.) High sensitivity-to-initial-conditions-or-parameters (STICOP) means tiny ICOP variation has big effects. Extreme STICOP gives stochasticity or chaos.

Vector notation and space-space plots

Balance equation (3.4) concisely sums up preceding constructions or plots of elastic collisions.

             or:             (3.4)_repeated

More concise notation uses vector equations or arrays.

   (4.1)

It saves writing two (=)’s and two (-)’s. Also, each column vector may be labeled by a “fat” letter.

                             (4.2)

The Gibbs vector form of equation (3.4) or (4.1) uses fat-v and/or over-arrow-.

             v^FIN = 2 V^COM – v^IN ,               or:                   .                             (4.3)

           

Fig. 4.9  BounceIt V₁-V₂ simulation up to (a) Bang-15 (b) Bang-150 and (c) beyond.

     

Fig. 4.10 Vector collision velocity diagrams (After equation (4.3) and virtually identical to Fig. 3.3.)

Note vector V^COM bisecting the (v^IN+ v^FIN)-parallelogram diagonal as per T-symmetry relation from (3.12a) and Fig. 3.3. Here vectors v=(v₁, v₂) denote two particles each in one-dimension. More common is vector v=(v_x, v_y) (or v=(v_x, v_y, v_z)) for one particle in two-dimensions (or three dimensions).

Fig. 4.11 shows how velocity v(n) vectors find results of Bang-1₍₀₁₎ and Bang-2₍₁₂₎ collisions in Fig. 4.7. What’s new is a space-space y₂ vs. y₁ or position-vector y(n)-plot whose paths are spatial-trajectories or just plain trajectories. Space-time paths are found in Fig. 4.6 and Fig. 4.7 by transferring velocity slopes over to the space-space or space-time plot, but vectors in Fig. 4.11 simplify this process. Again, ideally small masses called point masses are assumed.

 

As the construction steps in Fig. 4.11 show, one easily transfers each velocity vector v(n) from the V₂ vs.V₁ plot so it points away from start point y(n) in the y₂ vs. y₁ plot. Step-0 does this by drawing initial velocity v(0)=(-1,-1)  to point away from our given initial position y(0)=(1,3). Then you extend that v-vector until it hits the floor (as v(0) does at y(1)=(0,2)), or else hits the collision line (y₂=y₁) (as v(1) does at y(2)=(1,1)), or else hits the ceiling (as v(2) does at y(3)=(2.2,7).). Each such “hit” is a Bang, Bang-1₍₀₁₎ at y(1), Bang-2₍₁₂₎ at y(2), or Bang-3₍₂₀₎ at y(3). Then from each Bang-n position point y(n) is drawn the next v(n)-velocity vector from the V₂ vs.V₁ plots. This process continues in exercises that lead to Fig. 4.12 and beyond.

Fig. 4.11 Vector collision velocity diagrams with Velocity-Velocity space and space-space.

Fig. 4.12 Vector collision diagrams continued with velocity-time and space-time plots added.

 

Help! I’m trapped in a triangle.

The trajectory in these figures is confined to the triangle above the 45°-collision line. Our model keeps m₂ above m₁. The right-hand “ceiling” in the figures never is hit because m₁ always is knocked down by m₂ before it touches the ceiling, and m₂ never sees the floor because m₁ is in the way. (Modern physicists beware! Quantum theory doesn’t encourage this feature. Quantum objects pass easily through each other! )

Two balls in 1D vs. one ball in 2D

For ball-Earth collisions involving ceiling or floor, the paths bounce in the space-space plot as though they’re inside a box. Only one component V₁ or V₂ changes each time and only by changing ±sign. Off the floor: (V₁ ,V₂) changes to (-V₁,V₂) , off of ceiling: (V₁,V₂) changes to (V₁,-V₂). It is like a single particle bouncing around a pool table. Here (V₁,V₂) acts like (V_X ,V_Y) in two dimensions, so two particles in one-dimension use graphs similar to one particle in two dimensions, an interesting analogy in quantum theory.

Angle of incidence=Angle of  reflection (or NOT)

When paths bounce off the floor and ceiling in the space-space plot, the angle of incidence equals the angle of reflection just as light rays reflect off mirrors. (Newton imagined little light corpuscles bouncing around.) It is customary to measure path angles from the normal or perpendicular to a mirror so a normal bisects the angle between the incident and reflected paths.

For m₁-m₂ Bangs off the 45°-collision line, the bisecting line has the slope -M₁/M₂=-7. It is like having mirror facets at slope M₂/M₁=1/7 along the 45°-collision line. For equal-mass-(M₁=M=M₂) balls, or one ball in two dimensions, the bisecting line slope at the 45°-collision line is –1 or -45° and the collision line acts like a unit-slope mirror on a triangular billiard table. It is not quite that simple if.

Consider the two collisions Bang-3₍₂₀₎ and Bang-4₍₁₂₎ in Fig. 4.12. Velocity v(2) bounces off the ceiling in Bang-3₍₂₀₎ into v(3), whose velocity slope is close to the mass-ratio M₁/M₂ which is 7:1 here. So the next collision Bang-4₍₁₂₎ bounces v(3) off the diagonal into v(4) which is close to –v(3). It’s followed by another ceiling bounce Bang-5₍₂₀₎ into v(5) heading down for another collision Bang-6₍₁₂₎.

Bang force

Lower Fig. 4.12 has a velocity vs. time plot next to a space-time plot. (A y-t plot in gray is under the V-t plot, too.) Each Bang means a change in velocity for any particle involved in the collision. By Newton’s 2^nd law (3.10c) each change in velocity, v to v+Äv, or better, each change in momentum, mv to m(v+Äv), requires a force impulse F·Ät= m(Äv) on each mass that changes. Shortly, we study ways to deal with this F.

Kinematics versus Dynamics

The velocity-velocity (v₁,v₂) plots, such as the left side of Fig. 4.12, fall in a category known as kinematics, or momentum analysis, which is concerned with how things are going, where they’re headed, or what is their velocity or momentum and energy. (kinos means movement.)

In contrast, the space-time plots, such as the right side of Fig. 4.12, fall in a category known as dynamics, or coordinate analysis, which is concerned with how things are located, where they are, or what are their coordinate or position and time schedules. (dynos means change.) We introduced the space-space (x₁,x₂) plot, another geometric or trajectory representation of dynamics.

Before going on, let’s compare how kinos and dynos play out in classical Newtonian physics versus their corresponding roles in quantum physics. This is a preview for later Unit 4, Unit 7, and Unit 8.

Dynos and Kinos: Classical  vs. quantum theory

In Newtonian physics, a precise position plot (y_k vs. time) lets you find a precise velocity plot, too, and, a velocity plot (V_k vs. time) lets you find a position plot if you know starting position values. (We did just that in Fig. 4.7 and Fig. 4.11.) In calculus, finding position from velocity values is called integration, and finding velocity from position values is called differentiation. Of the two, the latter is formally easier but numerically and experimentally more sensitive to imprecision and noise.

In quantum physics, having a precise velocity plot renders a position plot meaningless and vice-versa! Werner Heisenberg was the first to state this quantum idea, now known as Heisenberg’s Principle. If you know momentum exactly, that means a uniform wave is everywhere, and all positions are equally possible. If you know position exactly, that means every momentum is possible, implying a “wave-bomb” about to blow up the universe! (Neither of these extremes really exist and fortunately so for the last one.)

All this sounds crazy to most of us who are born-and-bred Aristotelean-to-Newtonian students. It is difficult enough to go from Aristotle’s what-you-see-is-what-you-get (WYSIWYG) universe to Newton’s corpuscular one. A quantum universe is yet another step removed on the WYSIWYG scale.

A way to see the quantum universe (Perhaps, it is the way.) is to learn about wave kinematics and dynamics without Newtonian corpuscles and see how waves mimic corpuscles and do so quite cleverly. The quantum universe is a WYDAWYG (waves-you-don’t see-are-what-you-get) world!

So our plan is to cast classical Newtonian kinematics and dynamics in a form that carries over into vibration and wave kinematics and dynamics. It is done by analogy with classical waves such as sound waves, water waves, and (most important) light waves. Many classical wave analyses invoke corpuscles (including, for Newton, light waves) so these analogies, like any analogy, need critical use of an Occam’s razor that must be sharpened. Above all, symmetry (and same-try) principles must be taken seriously.

 IF-ellipse geometry of Ch. 3 relates velocity, momentum and energy, and Ch. 4 derives space-time paths. Later this relates Lagrangian and Hamiltonian mechanics and finally leads to geometries of relativity and quantum mechanics. Then space-space and space-time plots relate to modern physics in subtle ways.

 

 

 

Exercise 1.4.1: Construct a history of a 4:1 mass ratio bounce. x₁(0)=1.5, x₂(0)=3.0, v₁(0)=-1, v₂(0)=-1

Ceiling height=7.0.(For bottom row: Ceiling height=6.0 ) The 4:1 mass ratio case is surprisingly periodic.

 

Exercise 1.4.2: Complete Fig. 4.7 and Fig. 4.11 by constructing more steps using same ceiling height=7.0.

Continue until you reach the “gameover” point of last possible M₁-M₂ collision assuming the floor is open after Bang-1 so both masses fall thru indefinitely. When and where do they last collide?

 

Note, position y(n)-vectors of the Bang-n points are not drawn in Fig. 4.12 to avoid clutter.

 

 

 

Chapter 5 Multiple collisions and operator analysis

Analysis of many collisions with very different masses requires an advanced kind of geometry and algebra involving matrices and symmetry operators. Similar analysis is needed for quantum theory so this is a good opportunity to learn about these concepts using a more “down-to-Earth” classical bang physics. 

Doing collisions with matrix products

Fig. 5.1 shows a big mass m₂=49 bang a little mass m₂=1 more than ten times off the ceiling before being halted. This tests our collision precision! To check our results we use our previous vector equation (4.1) to make a matrix equation in (5.1) with  and total mass M = m₁+ m₂.

  (4.1)_repeated                 (5.1a)

(Let andhere.) Vector equation (5.1a) is converted to matrix equation in (5.1b).

                                                                                                        (5.1b)

Each IN-to-FIN bang is a operation (5.2a). Matrix product (5.4b) is bang-M following bang-N.

     (5.2a)                                    (5.2b)

Matrix M operates column-by-column on another matrix N as it does on a vector v. The off-the-ceiling matrix C = changes (v₁, v₂) to (v₁, -v₂) (Odd-n Bang-n₍₀₂₎) A 2-ball collision matrix M (Even-n Bang-n₍₁₂₎) and ceiling bang C act p-times in matrix products  to give Fig. 5.1.

       (5.3)

(5.4) shows (p=5) double-bangs  following a floor-bounce  or 11 bangs in all.

(5.4)

 

Even after 9 bangs, big m₁ still has a small upward velocity v₁=0.2925.

After Bang-11₍₀₂₎ big m₁ is nearly stopped and little m₂ is coming down at v₂=-7.071 with all the energy!

                                                                  (5.5)

Look out below! As m₁ turns back it crosses v₁=0 axis in Fig. 5.1a. The greatest curvature (acceleration and force) for the path of m₁ is between Bang-8 and Bang-14 in Fig. 5.1b just when m₂ is busiest.

Fig. 5.1 Multiple Bangs of the m₁=49 and m₂=1 superball system. (a) V vs V plot. (b) Y vs time.

 

Big m₁ is repelled down by repeated m₂ hits and gains speed as m₂ loses it. If no floor intervenes to rebound m₁ there comes a final bang that leaves m₂ slower than m₁ who falls away so m₂ can’t hit it again.  (We’ll call this a game-over point. As an exercise, you will be asked find it for various cases.)

However, if a floor intervenes, then a 2^nd floor-bounce matrix F=changes (v₁, v₂) to (-v₁, v₂) and bounces ball-m₁ back up to start the whole process over again. Ball-m₁ does another graceful up-then-down time trajectory very much like the one shown on the right-hand side of Fig. 5.1.

Except for floor bounces, the m₁-ball in Fig. 5.1 experiences a smoother flight than in Fig. 4.7 where a more massive m₂-ball jerks it severely. A smaller mass m₂ has less momentum-per-bang and gives a quasi-continuous force field for m₁. We will derive a funny kind of force and potential field theory from this.

Rotating in velocity space: Ticking around the clock

Here is an example of geometry and slope ratios being helpful. If you view the ellipse in Fig. 5.1a lower-edge-on (and do the exercise to finish it!) you may see it as a circular clock with each double-bang (odd-bangs 1,3,5,…) rotating the v-vector like a clock hand ticking equal-angle jumps around a dial.

You can make an energy ellipse (2E=m₁v₁²+ m₂v₂²) like Fig. 5.1(a) sketched in Fig. 5.2(a) into an energy circle (2E =V₁²+V₂²) like Fig. 5.2(b) by rescaling velocity (v₁, v₂) to (V₁ = v₁·√m₁, V₂ = v₂·√m₂).

            V₁=v₁·√m₁,      V₂=v₂·√m₂     where: 2E=m₁v₁²+ m₂v₂²=V₁²+V₂²                          (5.6)

Big-V variables replace little-v’s by setting (v₁ =V₁/√m₁, v₂ =V₂/√m₂) in matrix relation (5.1).

(5.1)_repeated                         (5.7)

Clearing scale factors √m_k gives the following big-V matrix relations to replace (5.1) above.

    (5.8)             (5.9)

The trick is to notice a Pythagorean relation x²+y²=1 for the circular bang-matrix components.

                                                                                       (5.10a)

The matrix can be defined using sinè  and cosè shown for m₁=49 and m₂=1 and angle è =16.26° in Fig. 5.2(c).

                                                                     (5.10b)     A 1-Bang matrix is a reflection by è. Our 2-Bang matrix is a rotation by angle -è =-16.26° in V space.

(5.11)                                      (5.12)

 

Fig. 5.2 Velocity-velocity clocks. (a) Energy ellipse (As in Fig. 5.1) (b-c) Energy bang-clock angles

            (d) Velocity-squared E-plot.  (e) Mass-scaled V-squared E-plot.  (f) Integral right triangles

 

Matrix (5.12) reduces N-double-bang chains like (5.4). N products of matrix (5.9) are done if è =16.26° in (5.12) is replaced by Nè =81.30° to give (5.13) below. (We take N=5 double-bangs to check against (5.5).)

               (5.13a)

Relate V’s to v’s by (V₁=v₁√m₁, V₂=v₂√m₂) gives (5.5). Note follows initial floor F: (v₁, v₂)=(1,-1).

 (5.13b)

Without a 2^nd floor-bounce-back operation F, this sequence ends near bang-21 our “game-over” point. (See exercise 5.1.) Matrix group products clarify collision sequences so they may be “engineered.”

Statistical mechanics: Average energy

If two balls of mass m₂=1 and m₁=7 bounce back and forth between wall the small ball goes faster on the average than the bigger one. How much faster? Let’s assume that arrows on the scaled velocity clock in Fig. 5.2(b) get uniformly distributed around its circle after many collisions. (Fig. 5.2(b) shows only m₁-m₂-bounce arrows. m₂-ceiling-bounce-arrows fill up the upper half.) A ball’s velocity and momentum must sum and average to zero otherwise it will not stay in the region between the floor and the ceiling.

 

But, what is average squared-velocity v² of each ball? An energy plot in the space (V₁)² vs (V₂)² of scaled velocity-squared helps to answer this. The result is a 45° line shown in Fig. 5.2(e). In other words points on the circle in Fig. 5.2(b) get mapped onto the 45° line in Fig. 5.2(e) by KE conservation.

                                      (V₁)² + (V₂)² = 2 KE = m₁(v₁)² + m₂(v₂)²

The average of all points on the 45° line is its bisector.

                          (V₁)² = KE = (V₂)²                or:                    m₁(v₁)² = KE = m₂(v₂)²   

This gives the average velocities or root-mean-square-speeds v₁^rms and v₁^rms of m₁ and m₂.

                                                                                                    (5.14)

Each ball, regardless of mass, gets equal share (50% if there are just two) of the total energy. So, if m₁ is 7 times m₂ then the mean speed of m₂ is √7=2.65 times faster than that of m₁. The 1^st bang in Fig. 4.4 gives 2.5.

Bonus: Rational right triangles

Geometry often offers interesting numerics. In this case, the general right triangle in Fig. 5.2(c) makes integer or rational fraction solutions to the Pythagorean sum a²+b²=c² such as the famous (a=3,b=4,c=5) right triangle. Perfect-square mass values (m₁ and m₂=1, 4, 9, 16, 25, 36, 49, 81, 100,…) will give integral valued right triangle altitude a=√(4 m₁·m₂), base m₁-m₂, and hypotenuse m₁+m₂. Examples in Fig. 5.2 are (a=14,b=48,c=50) for (m₁=49, m₂=1) and (a=12,b=5,c=13) for (m₁=9, m₂=4).

Reflections about rotations: It’s all done with mirrors

In 1843 Hamilton discovered his quaternion algebra {1,i,j,k}, a mathematical jewel. In 1930 Pauli found related spinor matrices {1,ó_X, ó_Y, ó_Z}. We label Pauli matrix ó_Z as sigma-A=ó_A (A for Asymmetric) and ó_X as sigma-B=ó_B (B for Balanced). They are Hamilton’s k and i with an imaginary factor attached.

                                              (5.15a)                                                 (5.15b)           

Other matrices, sigma-C=ó_C (C for Circular) and sigma-0=ó₀(0 for “Origin”) are products like ó_Aó_B or ó_A².

            (5.15c)                      (5.15d)

Hamilton’s {i,j,k} square to -1. (i²=j²=k²=-1) That is like . But, Pauli-ó’s square to +1. (1=ó_X²=ó_Y²=ó_Z².)

We now relate ó-matrices to simple super-ball collision reflections and rotations shown in Fig. 5.2. For example, the ó_A is our “ceiling bounce” C in (5.3) and our “floor bounce” F in (5.3) is just - ó_A.

= C                          (5.15a)                                     = F             (5.15b)           

A geometric view of ó_A (or - ó_A) is mirror reflection thru Cartesian x (or y) axes in Fig. 5.3a while ó_B (or - ó_B) is reflection thru mirror planes tilted at angle ð/4 (or −ð/4) between x-y axes in Fig. 5.3b. General reflection ó_ö thru a mirror plane tilted at angle ö/2 (Fig. 5.3c) is a sum (5.15c) of ó_A cosö and ó_B sinö. We now verify this.

                                                       (5.15c)

            Like all reflections, ó_ö must square-to-one. (ó_ö²=1) It does so because ó_A²=1=ó_B² and ó_Aó_B =-ó_Bó_A. We test ó_ö on unit vectors and  and see that matrix algebra checks with geometry in Fig.5.3c.

              (5.16a)                     (5.16b)

Geometry Fig. 5.3d also shows that a product ó₂ó₁ of any two reflection matrices is a rotation matrix R.

In Fig. 5.3d  ó_öó_A is right-hand rotation R_+ö but ó_Aó_ö=R_−ö in Fig. 5.3e is left-handed. Rotation angle ö is twice the angle ^ö/₂ between mirrors. Direction of rotation ó₂ó₁ is from 1^st mirror (of ó₁) to 2^nd mirror (of ó₂).

       (5.17a)                   (5.17a)

For example, rotation ó_Bó_A is by +90° and ó_Aó_B is by -90°. Rotation ó_A(-ó_A)=(-ó_A) ó_A is by ±180°.

Through the clothing store looking glass

The rotation in V₁ vs V₂ space of Fig. 5.2b is a product of ceiling bounce and m₁-m₂ collision that are each a reflection. An even simpler example of paired-reflection rotation is a clothing store mirror in Fig. 5.4a. It lets you swing two mirrors like doors to view multiple images of yourself. If you set the angle between mirrors to ö/2=30° as in Fig. 5.3 d-e or to 60° as in Fig. 5.4a then you see yourself rotated by twice that angle. Images are turned 120° counter-clockwise in the right mirror and clockwise (-120°) in the left mirror of the latter.

            The sketches in Fig. 5.4a oversimplify the actual images shown by photos of a real mirror pair. The single reflections for ó_A are not shown in the sketch but clearly visible in photos where the ó_A and ó_ö images both have backwards text and a left hand image of the original right hand. This is corrected in the (-120°)-rotated ó_Aó_ö image and the (+120°)-rotated ó_öó_A image.

 

 

Fig. 5.3 Mirror-reflection geometry (a)±ó_A, (b) ±ó_B, (c) ó_ö. Right-and-left-handed rotation (e) ó_öó_A (f) ó_Aó_ö.

 

            A special case is rotation ó_A(-ó_A)=(-ó_A) ó_A by ±180° due to setting mirrors at exactly ö/2=90° as in Fig. 5.4b. The result is known as a corner-reflector image. Wherever you stand while viewing a 90° corner you see your image centered and rotated±180° to face you but it is not reflected. A 90° corner image is as others see you, complete with a readable monogram on your jacket and your right hand on the right side.

How fundamental are reflections?

A product of two reflections is a rotation R_ö=ó₂ó₁, but two rotations just give another rotation R_ö+è= R_öR_è and never a reflection. This makes reflections more basic and productive than rotations.

 

Fig. 5.4 Mirror reflections and their rotations with relative angle: (a) 60° (b) 90° (corner reflector images).

 

            On the other hand, you cannot do a reflection of a real solid object without entering an Alice-in-Wonderland looking-glass-world. Moving every atom in a classical object to a reflected position (without destroying it) is unthinkable! Yet, we easily rotate semi-solid objects (like your eyeballs while reading this).

            Waves, on the other hand, are very un-solid and do reflection effortlessly. Rotation takes twice the effort as seen in the looking glass images of Fig. 5.4. This is one reason reflection operations are so basic to the study of wave mechanics, quantum theory, and relativistic symmetry as we will see in later Units. They are elementary symmetry generators in a 1D world. A 1D translation by distance a is two reflections by 1D mirrors separated by distance a/2.

 

 

 

            Symmetry operation R or ó is defined by what it does to unit vectors and as ó_ö (5.16) is done in Fig. 5.3c. That matrix does that same operation to any and all vectors in the space.

                                             (5.18)

A way to distinguish rotation and reflection operators is by the determinant det|M| of their matrices.

                                                

A determinant of matrix M quantifies the space (area in this case) enclosed by vectors in M‘s rows or columns (u and v enclose a parallelogram in this case). 

            A rotation determinant is +1, but a reflection determinant is –1. Reflected area or angle in Fig. 1.3 is negative.

                                      

Determinants track the multiplication of matrices. The determinant of a product is a product of determinants.

                                                            det|M·N|= (det|M|)(det|N|)= det|N·M|

Thus, two reflections each with det|ó|=-1 form a product of det|ó₁ ó₂|=(-1)(-1)=+1, that of a rotation. This also shows a product of rotations cannot make a negative-det-matrix and so cannot be a reflection.

 

Exercise 1.5.1 Gameover 49:1

Complete Fig. 5.1 (for m₁:m₂=49:1) up to the gameover point where sequence would end without 2^nd floor bounce. Compare geometric results to analytic matrix analysis.

If the floor is open after the initial bounce of m₁, what mass-ratio near that of Fig. 5.1( m₁:m₂=49:1) would cause m₁ and m₂ to drop away with the same final velocity.

 

Exercise 1.5.2 Bigger bangs 100:1

(a) Construct a plot for m₁:m₂=100:1 to the “gameover” point after which the bigger ball would need a floor bounce to continue hitting the small one.  Such a large mass-ratio favors a rescaled (Estrangian) √m₁v₁ vs. √m₂v₂ circle–plot. You may use that instead of a plot like Fig. 5.1.

(b) Compare the accuracy of your geometric results with an analytic calculation like (5.2) or (5.13).

 

 

Solutions to 1.5.1

 

 

Chapter 6 Introducing Force, Potential Energy, and Action

Analysis of force is one of the trickier parts of Newtonian mechanics and one that Aristotle seems to have not done so well. We, like Aristotle, feel we know force after being pushed and pulled around by it most of our conscious lives. Aristotle related force directly to mass and its motion. If he ever wrote equations then, perhaps, Aristotle’s equation would be F=MV.

NOT! MV is momentum, not force. Galileo and Newton may be the first to realize that force should be equated to a change in momentum. A famous equation F=Ma equates force to mass or inertia M times acceleration a, the rate of change of velocity. (It is called Newton’s 2^nd law or NEWTON-TWO. Recall (3.16d).)

                                                                                                     (6.0)

Fig. 6.1 Big mass m₁  feels “force field” or “pressure” of small ball rapidly bouncing to and fro.

MBM force fields and potentials

Motion of m₁ in Fig. 5.1b suggests a kinetic model and a potential force field. Boltzman uses this to derive gas force laws for volume, temperature, and pressure. As a big m₁-ball squeezes space (volume) for a tiny m₂-ball in Fig. 6.1, the speed v₂ and energy ¹/₂ m₂v₂² of m₂ increases. So does the momentum transfer rate or bang-force on m₁. Energy is related to temperature and bang-force is related to pressure. A furiously bouncing m₂ is like a single-atom gas getting hot when its Y-space is compressed as in Fig. 6.1b.

Fig. 6.1 Big mass-m₁ ball feeling “force-field” or “pressure” of small ball rapidly bouncing to-and-fro.

A “double-whammy” hits the m₁-ball as it closes in with velocity v₁ toward m₂ and wall (Y=0):

(1) Bang rate B with m₂ increases with shrinking distance 2Y traveled by m₂ between m₁ and wall.

(2) Increased velocity v₂ (due to v₁) increases momentum m₂v₂ and ÄP transferred to m₁ by each bang.

(3) Increased velocity v₂ (due to v₁) increases bang rate even more. It’s really a triple whammy!

            If m₁ is huge (say 1kg) compared to atom or molecule m₂ (say (2/3)·10^-27kg for an H-atom), the speed v₁ of the macro-mass m₁ may be negligible compared to typical atomic speeds v₂ of 10³ m/s. Then we ignore effects (2) and (3) due to tiny v₁ in a so-called isothermal model. An adiabatic model includes them.

Isothermal model force laws

Atom m₂ in Fig. 6.1 travels distance 2Y back & forth between m₁ and ceiling at Y for each bang m₁. If v₁ is slow, the time Ät between bangs is 2Y divided by velocity v₂ of m₂. Bang rate B is the inverse: B=1/ Ät.

            Ät = 2Y /v₂ (bangs per sec) (6.1a)                    B =1/Ät = v₂ /2Y (seconds per bang)   (6.1b)

Each head-on bang of big m₁ on small m₂ changes velocity of m₂ from −v₂ to +v₂^FIN as shown in Fig. 6.2.

            (for: m₁>>m₂):                        v₂^FIN = v₂+2v₁              (Å v₂ for: v₂>>v₁)                  (6.2)

Added speed for m₂ is 2v₁, twice that of incoming m₁. (V-V-plot Fig. 6.2 assumes large-m₁.) The change ÄP of momentum m₂v₂ is the difference between FIN value +m₂v₂^FIN and IN value −m₂v₂.

                        ÄP = (+m₂v₂^FIN)–(−m₂v₂)=2m₂v₂+2m₂v₁       (Å 2m₂v₂ for: v₂>>v₁)                        (6.3)

So, if “atomic” velocity v₂ is large compared to v₁ it gives a bang-force F=B· ÄP = ÄP/Ät on m₁.

                                                 BP= ÄP/Ät =F = 2m₂v₂(v₂ /2Y) = m₂v₂²/Y                             (6.4)

So a force field F=2·KE/Y on m₁ due to m₂ is proportional to KE=¹/₂m₂v₂² or temperature T of m₂. Boltzman’s constant k of proportionality (KE=kT) gives an isothermal force law FY=2kT. It is a 1-D version of Boyle’s ideal gas law: PV=2kT. Here a ceiling tries to keep energy or “temperature” of m₂ constant in spite of m₁.

 

Fig. 6.2 Large mass-ratio (m₁/m₂>>1) bounce sequence. (Compare to Fig. 4.2a.)

 Adiabatic model force laws

An elastic ceiling can’t give or take energy so each m₁ bang adds velocity 2v₁ to v₂ at rate B=v₂ /2Y (6.1). As m₁ closes at speed v₁ it reduces distance 2Y that m₂ travels. So bang rate B grows due to more v₂ and less Y.

   ,              y =v₁t=H-Y,                                            (6.5a)

We cancel time and v₁ to show this force is inverse-Y- cubed, a lot “harder” than inverse-Y in (6.4).

   ,     ,   ,     (6.5b)

This is called an adiabatic or “fast” force law. Collisions are so fast that an isothermal-seeking “Robin Hood” in the ceiling hasn’t time to steal m₂’s energy when it’s judged too energy-rich or give energy back when m₂ becomes energy-poor. So m₂ can get hotter and hit m₁ harder and more often as gap Y shrinks.

Conservative forces and potential energy functions

Either force law (5.9) and (6.5) actually conserves the energy of the big-m₁ ball in the long run. By that we mean that m₁ will come out with practically the same energy that it had when it went in.

The adiabatic case is easier to see. Each bang conserves energy as demanded by the kinetic energy (KE) conservation relation (3.5a). Little-ball velocity v₂=const./Y from (6.5b) is used here.

                        =const.                           (6.6)

The first term is m₁’s kinetic energy KE₁. The second term, which is really m₂’s kinetic energy, is called m₁’s potential energy PE₁ or just plain PE, and it is labeled U(Y) since it varies according to height Y of m₁ only.

                           (6.7)

The PE is energy that m₁ lends to m₂ each time m₁ moves a distance ÄY closer so m₁ does a little bit of work ÄW on m₂. Work is defined as force times distance. (ÄW=F·ÄY) Power, the rate of work done, is defined as force times velocity. Here distance is a small ÄY and the force F in (6.5b) is m₂ const.²/Y³. But “work” force might be plus-or-minus (±)m₂ const.²/Y³. Which sign? (+) or (−)? Conflicting sign conventions make force-physics confusing. The sign depends on how force and direction are defined. (It’s all relative!)

Is it +or-? Physicist vs. mathematician and the 3^rd law

A physicist’s force F^phys is what is felt by a free object (Here that’s m₁.) whose motion is driven by force field F=F^phys. A mathematician’s force F^math is what is needed to hold back the object in the force field. (How apropos! A physicist lets it go but a constipated mathematician holds it back!) They differ by (±) sign only, that is, F^math =-F^phys, and F^math is the equal-but-opposite force by an object (m₁ here) on its field or force agent(s) (m₂ here). (This is essentially Newton’s 3^rd law. (NEWTON-THREE) )

Force is momentum flow. Momentum is stuff that’s conserved, so the flow rate F^phys of this stuff into an object m₁ must be balanced by an equal-but-opposite negative flow, F^math =-F^phys, out of the forcing agent(s) (m₂ here), and, vice versa, whatever flows out of m₁ flows into m₂. Momentum p=mv and force F are both vector quantities and a ±sign gives direction to-or-fro, another confusing (±) sign to bother us. But, whatever the flow rate F^phys seen by m₁, then m₂ sees the opposite rate F^math =-F^phys.

Let’s define positive Y and F direction to be away from the wall in Fig. 6.1. So incoming m₁ has negative velocity v₁=-ÄY/Ä t , but after m₁ reverses V=ÄY/Ä t is positive. Positive V=-v₁ (increasing Y) and positive F^phys means both momentum and energy of m₁ are being increased by force F^phys. Each bit of energy or work ÄW=F^physÄY gained by m₁ is energy lost by the force-field’s potential “bank” that is m₂. (ÄU=- ÄW)

                                                       (6.8)

In other words, power Ð =F^phys.V into m₁ is power (- ÄU/Ä t ) out of the field. (V=ÄY/Ä t is velocity of m₁.)

                                (6.9)

But is this consistent? Does force F^phys in (6.8) really equal minus the slope of potential (6.7)?

   (6.10)

It checks!! Note that F=- ÄU/Ä Y needs that ¹/₂ on kinetic energy ¹/₂ m₂v₂². (Recall discussion of (3.5).)

Isothermal “Robin Hood”and “Fed rules”

The isothermal case is a weird one. The little “force-field agent” m₂ maintains it kinetic energy at around the same initial value ¹/₂ m₂v₂² no matter how much the big mass m₁ loses or gains kinetic energy.

It’s as though a “Robin-Hood” in the ceiling acts like a big Federal Reserve Bank. (“The Fed.”) Whatever energy m₂ earns from m₁ is taken and stored away if its over initial deposit (m₂v₂²)=T, but if m₂ deposits falls below that value, the Fed makes up the difference. This energy or deposit limit is determined by a prevailing allowed “temperature” of the ceiling or the current money supply. (I’m not making this up. It’s what happens in nature and very roughly what happens in our economy. It becomes a problem if the Fed stops being a Robin Hood and becomes a robbing hood!)

Under ideal conditions, force agent m₂ makes a much “softer” 1/Y force field F=m₂v₂²/Y given by (5.9). Definition (6.9) of force F as negative-U-slope -ÄU/Ä Y then gives a log_eY=lnY potential.

                      (6.11)

It may seem weird that we can define a useful potential while energy-funds are being siphoned in and out. Nevertheless, the ceiling “Robin Hood” is true to his word. (Analogy with “The Fed” ends here!) He puts back all the energy that m₁ gave up to m₂ (the potential U) on the way in, so that, except for small-change or “tips” left with m₂ after the final parting collision, m₁ recovers the energy it originally had. Such a force field, if determined by such a reliable potential, is also a conservative one. We discuss later the details of what is needed for general multi-dimensional fields to be labeled “conservative.”

Oscillator force field and potential

            Consider a mass m₁ between two walls and two little speeding m₂ masses as in Fig. 5.5. m₁ feels a force like that of an oscillator. As m₁ moves distance x off center the left wall space expands to Y+x and the right wall space shrinks to Y-x. Two opposing forces (6.11) then are unbalanced. (Only x², x⁴,… terms cancel.)

                         

Here we let Y=1 be a unit interval and assume an isothermal kinetic constant  for each side. For small x (x<<1) the force F^total has a linear or Hooke’s law form, and the potential U^total is quadratic.        

                                                             (6.12)

            Harmonic oscillator (HO) linear forces and quadratic potentials are, perhaps, the most useful ones in AMO physics because they approximate any stable system. Normally, they are analogized by a mass on a spring, rubber band, or pendulum, only rarely (if ever) in a context like Fig. 6.3. HO motion is sinusoidal  with angular frequency and period independent of the oscillator amplitude A or phase . The calculation of period for Fig. 6.3c is left as an exercise.

            The 2^nd most useful field is probably the Coulomb potential U=-k/r and force F=k/r². (See Ch. 7 for electrostatics and Earth gravity, which also have 2D HO potentials at their cores.) After that, the 2D Coulomb U=k·ln(r) and F=k/r is an important field shown in Unit 10. (The latter is like (6.11). A pair of them underlies Fig. 6.3 for the isothermal case.)

You should be warned that an oscillator like Fig. 6.3 is not as simple as it might appear, and as we will see, neither are springs, rubber bands, or pendulums. Also, balls bouncing against moving objects are particularly dicey devices. A simple model with one ball and one oscillating wall is called a Fermi oscillator, and is quite chaotic.  The thing in Fig. 6.3 can be even more devilish if m₂ is not very small. Caveat emptor!

Fig. 6.3 Oscillator force and potential (a) Off center with (-)force (b) On center at equilibrium. (c) Quasi-harmonic oscillation of M=50 in adiabatic force of two m=0.1 masses of speed v₀=20 and range Y₀=3.

The simplest force field F=const.

We have mentioned power-law forces F_adiab=k/y³=ky^-3 (6.5), F_Coul=k/y²=ky^-2, F_isoT=k/y=ky^-1 (6.4), and lastly F_osc=-ky (6.12), but have forgotten the simplest, namely zero power law F_const=k =ky⁰. This last one is like a constant near-Earth-surface gravity force ==mg =-m|g| on a mass m. ( (-) sign  for downward.) Acceleration of gravity near Earth’s surface is nearly -10 meters per second per second and very nearly –9.8. (g=-9.7997m/s²) Terrestrial objects experience this whether they are bundled together or not.

All power-law forces F=ky^p have power-law potentials U=-∫F·dy=-ky^p/(p+1), except for p=-1 where F_isoT=k/y has a logarithmic U_isoT=-k ln(y). (6.11) Earth-surface potential is linear in height y=h. This we use to compute height of a superball toss by equating its floor level KE=1/2mV² to maximum PE=mgh.

              (6.13a)                                                             (6.13b)

Ejection height goes as the square of ejection velocity. A 3-fold velocity gain means 3²=9-fold height gain.

Introducing Action. It’s conserved (sort of)

It is remarkable that a bouncing mass has a physical property called action that is more or less constant even if its position x momentum P and kinetic energy KE are driven crazy. Action is defined by the area of a one-cycle loop swept out in a momentum vs position phase-plot (P vs x). That is analogous to an energy or power-plot of force vs position (F vs x) whose loop area is work per cycle.

Conservation of momentum and conservation of energy are each a rigorously obeyed axiom or theorem for an isolated classical system. However, conservation of action is “more or less” or “sort of” and “it depends” for a driven system. The concept of action is both subtle and deep and it lies at the heart of quantum theory and accounts for a lot of how we affect and are affected by the world around us.

Here we use a geometric construction of a bouncing ball trajectory to quantify action conservation or lack thereof. We suppose the little mass m₂ is caught as before in Fig. 5.1 and Fig. 6.1 between a rock and a hard place, that is, bouncing between a big mass m₁ (moving in at a constant velocity v₁= 1 from the left) and a hard elastic wall. The big ball path is indicated in Fig. 6.4 by a line of slope=1= v₁ that hits an initially fixed m₂ following a vertical line (slope=0=v₂) that then gets knocked up to a line of slope=2=v₂ (after Bang(1)). Throughout the imagined collision sequence we suppose the big ball is so much more massive that its change in velocity is not noticeable. This is in spite of the fact that it is absorbing more and more momentum from the little ball with each bang. (Surely, something in it is going to break eventually!)

Each time the small ball is banged elastically by the big one it picks up two more units of velocity v₁ that it maintains, apart from change in sign, through its subsequent bang with the elastic wall. Each time it returns for more, is banged again, and increases its speed by two units. (Recall Fig. 6.2.)

The horizontal dashed lines in Fig. 6.4 indicate the range Äx available to the small ball at each instant of its bang with the wall. Note that the product of the range Äx and the speed v₂ is a constant three units even as spatial range Äx rapidly decreases and the velocity range Äv=2|v₂| increases just as rapidly.

                                                             Äx v₂ =3.0 = Äx Äv/2 

This is an example of conservation of action mentioned before. If we define the small ball’s “range of velocity” by Äv=2|v₂| then this relation takes the form of a weird kind of uncertainty relation, that is, it looks like Heisenberg’s famous minimum uncertainty relation Äx Äp =h=(constant) for position and momentum. It happens that the two are related even though the constant used by Heisenberg is an unimaginably tiny Planck constant (h~10^-34Js) compared to a constant 3.0 appearing above. (Ours has gadzillions of wave quanta!)

The geometry behind this relation is exposed in Fig. 6.4 (b). It is obtained by considering intersections between lines of integral speeds or slopes v₂ =±1, ±2, ±3, ±4, ±5, ±6, ±7,… that are relevant to the bang sequence. They are also relevant to quantum theory where the speeds of a particle in a box are indeed quantized to integers times a tiny number. (This is where that tiny h comes in.) That is simply a reflection (pun intended) of the fact that mutually reflecting waves require that an integral (or half-integral) number of the wavelengths fit perfectly between mirroring containment walls or cavities.

Now we might ask if the action area Äx Äv in Fig. 6.4c-e stays the same if the big-ball speed v₁ varies. Action variance was argued hotly by Einstein and the “quantum gang” at the1920 Solvay Conference. They imagined a hotel chandelier being dragged up or down by a clerk holding its support cable upstairs. They concluded that if the clerk could not detect the swinging pendulum phase or frequency, then he would seldom be able to change its action. However, if he could synchronize his oscillations then he could drive the chandelier exponentially to destruction. We shall review this important and explosive process known as parametric resonance in later units. It is fundamental to mechanics and particularly quantum wave mechanics. Action and its wiggly antics deserve our attention.

Monster mass M₁ and Galilean symmetry (It’s deja vu all over, again.)

 “Monster mass” M₁ bongs hapless m₂-atoms in Fig. 6.4 using Galilean symmetry. To show symmetry we imagine two head-on monster M₁‘s going at ±V₁=±1 in Fig. 6.5. A mirror image of Fig. 6.4 lies in extended m₂-path lines. The red paths of even integral velocity v₂=0, ±2, ±4,… are copies of Fig. 6.4 paths. Odd integral velocity v₂=±1, ±3,… paths mesh with even ones to make a full grid. Any initial v₂ between ±V₁ has a path on the grid. A blue path is drawn thru a series of bongs with v₂=-0.2,+2.2,-4.2,+6.2,...in Fig. 6.5.

 

 

Fig. 6.4 Bang sequence for small ball between big ball and wall. (a) Spacetime paths. (b-c) Geometry of constant product Y·V_Y of velocity and coordinate ranges.

Fig. 6.5 Symmetric pair of head-on V₁=±1 monster-m₁-masses pong tiny-m₂-atoms to higher speeds.

 

            Monster M₁/m₂-ratios have simple V₁-v₂-plots shown in Fig. 6.6a. (Recall Fig. 6.2.) Wall M₁ simply adds twice its speed (2V₁) to incoming speed v₂ of atom m₂ as M₁ bounces m₂ out at that speed v^FIN₂=v^IN₂+2V₁. Monster M₁ is the COM so its path bisects in-and-out paths as it balances v^IN and v^FIN paths of atom m₂. (In its COM frame each bong is simply a change of sign for velocity. Recall balance in Fig. 2.6.)

            The geometry of adding slope 2V₁ to speed v₂ is shown if Fig. 6.6a. It is based on the unit square and unit velocity V₁=1. Incoming -v^IN₂ is an altitude of a right triangle with vertical base V₁=1, and it is reflected thru the square diagonal to +v^IN₂ then added to 2V₁ to give sum v^FIN₂=v^IN₂+2V₁ as long side of the triangle  with right side vertical base V₁=1 in Fig. 6.6a. The hypotenuse is the final path with final slope v^FIN₂. Each m₂-path and slope originates or terminates at base pt-B₋ or else pt-B₊ . These are ends of the double-unit square bisected by unit slope path of M₁ terminating at B₀. Fig. 6.6.c shows quadrilateral B₋B₊A₊A₋ bisected by M₁ path B₀CA₀. Similar triangles explain multiple coincidences.

Fig. 6.6 Bisection geometry of Fig. 6.5.

 

Fig. 6.7 contains time plots for paths in different Galilean reference frames. An excerpt plot in Fig. 6.7a shows how Fig. 6.4 (copied in Fig. 6.7b) appears to a frame traveling at V=1 with each velocity in Fig. 6.7b reduced by V=1 in Fig. 6.7a. Also shown in Fig. 6.7a is the extension of lines connecting the two plots and this highlights this remarkable symmetry. All collision times in Fig. 6.7a match perfectly with ones in Fig. 6.7b though all velocities are shifted. Galileo’s symmetry wouldn’t have it any other way.

Fig. 6.7 (a) Galilean frame shift by frame velocity V=1 of collision sequence in Fig. 6.4 (shown in (b)).

 

Exercise 1.6.1 Suppose Fig. 6.3 shows a mass m₁=1kg ball trapped between two smaller mass m₂=1gm balls of high speed (v₂(0)=1000m/s for x=0) that provide m₁ with an effective force law F(x) based on isothermal approximation (6.11) while assuming m₁ moves only moderately far or fast from equilibrium at x=0.

 

(a) A further approximation is the one-Dimensional Harmonic Oscillator (1D-HO) force and PE in (6.12). If each mass m₂ start in an interval Y₀=1m, derive approximate 1D-HO frequency and period for mass m₁.

 

(b) What if the adiabatic approximation is used instead? Does the frequency decrease, increase, or just become anharmonic? Compare isothermal and adiabatic quantitative results for m₁=1kg ball being hit by two m₂=1gm balls each having speed of v₂(0)=1000m/s as each starts bouncing in a space of Y₀=1m on either side of the equilibrium point x=0 for the 1kg ball.

 

(c) How does the frequency decrease or increase in isothermal case versus the adiabatic case if we shorten the run interval Y₀=1m to one-quarter meter?…What if we reduce the mass ratio m₁/ m₂ by one-quarter?

 

(d) Derive the adiabatic frequency for the case M=50kg in adiabatic force of two m=0.1kg masses of initial speed v₀=20m/s and range Y₀=3m. Compare with Fig. 1.6.3c.

 

Exercise 1.6.2 The moving ballwall-trapped-ball constructions in Fig. 6.4 involves a plot of a ballwall coming in with unit slope (velocity). Consider a construction where it has a velocity of 1/2 and intercepts a trapped ball of velocity –1 at space-time point (x=-2, t=4) that is 2 units from the fixed wall. Construct five or more back-and-forth collisions and comment on what, if any, differences exist with Fig. 6.4. If you can, also construct one or two prior collisions (before t=4).

Evaluate approximate or average action values as described in class or after Fig. 6.4 in Unit 1.

 

Chapter 7 Interaction Forces and Potentials in Collisions

            Derivation of force field potentials in Ch. 6 used elementary bangs by tiny m₂’s on a big M₁. (Ch.5) We predicted elementary bangs between a ball and floor, ceiling, or another ball without knowing potentials. However, three (or more) objects having a ménage a trois are not so easy to predict, and outcomes of 3-body interactions depend sensitively on whatever interaction potential or force law couples the participants.

Geometry of superball force law

            When a superball or any elastic sphere hits the floor or ceiling it dents itself and, maybe it dents the surface it’s hitting a little bit, too. But, if the floor, wall, or ceiling is much harder than the ball, we might assume only the ball develops a “flat-tire” as shown in the Figure 7.1a below.

Fig. 7.1 Superball collides with solid wall. (a) “flat” (b) Saggital (“Bow”) mean geometry

 

            The radius r of the ball’s “flat” is indicated by an altitude in Fig. 7.1b and is the geometric mean of the depression distance x and the remainder 2R-x of the ball diameter. (Recall Thales geometry in Fig. 1.9a.)

                                                                   (7.1a)

Solving approximately for depression x gives the Saggital (“bow”) formula. (It’s used for thin arc lenses.)

                                                                                            (7.1b)

How much force F(x) is needed to depress the ball by distance x?

            The answer is, “It depends.” A hollow rubber ball or balloon with pressure P pushes back with force equal to product P·A of pressure and area of contact A=πr². It is a linear (Hooke) force law of a spring.

                          F_balloon(x) = P·A = P πr² Å 2πPRx                                                     (7.2)

(Recall (6.12) and Fig. 6.3.) Another example is gravity inside the Earth. (See (9.4) or Fig. 9.6 in Ch. 9.)

            However, the pressure and force in a solid ball varies non-linearly with x. Even if force varies only linearly with volume of the x-dent in Fig. 7.1b, it’s still non-linear in x. As seen in (7.4) below, sector volume varies roughly as quadratic x² function. Superballs involve even higher power laws. (Superpower!)

 

                  (7.4)

(Here we check that our integral gives the whole ball volume 4πr³/₃ for x=2R. That’s the equivalent of crushing the superball into a black hole (or black spot). It’s likely to complain before we get that far!)

Dynamics of superball force: The Project-Ball story

            One of the interesting things to come out of Project Ball was the superball’s peculiar force law behavior. The USC mechanical engineering department took an interest in this crazy project when it showed up on NBC News “Ray Duncan Reports.” They offered to measure the superball force curve on a precise tension meter. But, that curve never worked. It didn’t predict the bounces the students were observing. Nothing was making any sense even though we had a big analog computer working it all out.

            That was a low point in the project. Even with all this fancy experiment, computers, and theory, I looked like I didn’t know what the heck I was doing. So, what’s new? That’s science most of the time! But, to make things worse we got kicked out of the Project Ballroom, the old basement Lab 69 that we’d squatted in. It was up to be repainted so we had to drag all our stuff out of there and store it down the hall.

            Well,  after that I had to do something with the students so I arranged for a visit to Whammo Mfg. Co. in San Gabriel, California, where superballs and other goofy stuff was made. The Whammo man said maybe we could talk business about selling our super-elastic toy. So, a day or so later, with $$-signs in our eyes, we piled into our cars and drove down to the plant.

The trip to Whammo

            By the time we got there, the inventors were on an all-day “alpha-wave break.” That’s a 60’s fad where you try to increase your creativity by looking at your brain waves. I said, “Maybe, I could use some of that stuff!” But, the company lawyer wanted to show us around. After awhile, he said he thought our invention was cool, but its product liability potential looked too high to make a commercial toy.

            We all must have looked pretty sad after hearing that. So he went in a back room and dragged out a big collection of superballs that had been rejected for one reason or another. “Here, take as many as you want!” We thanked him and loaded the balls into some boxes and headed back to USC.

            When we got back to Rm 69, the painters were done but the paint wasn’t quite dry. So I said, ”Let’s drop off our new balls so we’re ready for tomorrow.” The students took “drop” to mean literally and dumped them out of the boxes into the empty room. Right away the balls bounced into the wet paint and made lots of little polka-dot spots all over the floor and wall. What fun! What a mess.

Eureka! Polka-dots save Project Ball

            But, suddenly, it occurred to me what was wrong with our force analysis and how we might fix it. The engineers had carefully and slowly produced a static or isothermal force curve, but what we really needed was a fast-response or adiabatic force curve. I thought, “Maybe that force law can be told by the polka-dots!”

            From a polka-dot radius r made by a superball of mass M and radius R dropped from a height h we could relate gravitational potential energy Mgh to an adiabatic superball potential energy U,  and then find a U(x) curve for each value of x=r²/2R in formula (7.1b) by plotting height h against x given by dot radius r. Then the adiabatic force curve F(x) can be found from the slope dU(x)/dx of a U(x) curve.

            Just as the adiabatic F=1/Y³ in (6.5) force curve is steeper and curvier than the isothermal F=1/Y in (6.4) so was the polka-dot bounce curve steeper than what we had been using. We stuck our new F(x) on the analog computer’s diode function generator and started getting good predictions. Now we could work out the deadly Model-X3, a 3-ball super tower! (This is described later in Chapter 8.)

The “polka-dot” potential

            First, let’s look carefully at this “polka-dot” potential theory. What we did, like most of physics, was an approximation. Using gravitational potential to estimate superball U(x) is a neat trick only if the superball forces are large and quick compared to the gravitational force or weight mg of the ball.

            Fig. 7.2a shows a massive (Bowling-ball sized) superball at its (V=0) drop point h, where potential energy is mgh. Kinetic energy rises from zero as the ball falls and flattens on the floor until it passes a point where the upward floor force cancels the ball’s downward weight mg. That point-x_static of static equilibrium is at the bottom of the total potential energy curve in Fig. 7.2b. The ball would sit still if put gently at x_static with no kinetic energy. It’s a point of zero slope since total force F(x_static) is zero there.

            After passing x_static the ball slows down due to upward force. (That’s positive F(x) for x<x_static.) Finally it stops at its maximum penetration point x_max where the total energy line intersects the total potential line in Fig. 7.2c. Now the ball’s initial gravity potential mgh₀ has been converted completely into potential energy U(x_max)  due to compressing rubber a distance x_max. (We’re ignoring tiny frictional heat.)

            In the example, the ball’s weight is almost as large as the inertial bang-force driving the ball into the floor. An indication of this is how flat the ball is in Fig. 7.2 b when its weight and compressive force are equal. A standard superball sits stiffly on a table with no noticeable depression, and mg is a tiny part of the total force. It’s so stiff that its bang force is several times its weight and lasts only a few hundredths of a second. Very stiff rebounding potentials are shown in the later Fig. 7.3 and Fig. 7.4 b in which gravity is a negligible force and stiff rebound forces dominate during the collision.

By comparison, the ball in Fig. 7.2 is heavy and its potential is not so stiff. Instead it is so soft it has a big “flat” if sits still with zero KE at x_static just as it does when passing that point in Fig. 7.2 b. The collision shown in Fig. 7.2 a-c is less like a bang and more like a lingering smooch! Similarly soft collision energy for a linear rebound force and quadratic potential is shown in parts (d) and (e) of Fig. 7.4.

 

Fig. 7.2 Geometry of ball hitting floor (a) Ball is dropped. (b) Ball at max speed. (c) Ball at low point.

 

 

Force geometry: Work and impulse vs. energy and momentum

            TV daredevils jump off 30-meter towers and belly-flop into kiddy-pools that are less than 1 meter deep. What a way to earn a buck! And, how do they ever survive such stunts?

            Two important physical quantities tell about survival chances. The first is the product F_˙x of force-times-distance, or, more precisely, the integral ∫Fdx of force over distance. The second is the product F_˙t of force-times-time, or, more precisely, the integral ∫Fdt of force over time. (Recall the fundamental Galileo-Newton relations (3.16) and (6.0).)

            The first quantity ∫Fdx is work done or energy -U(x) acquired. U(x) is area under an -F vs. x plot.

                                    (7.5a)

If energy is stored as potential energy U(x), then force -F(x) is the slope of a U(x) plot at point x.

                                                                                                                          (7.5b)

(Recall the discussion of force and potential leading up to (6.10).)

            A second quantity ∫Fdt is impulse done or momentum P(t) acquired and area under an F vs.t plot.

                                 (7.5c)

If momentum is stored in kinetic velocity V(t)= P(t)/M then force F(t) is slope of the P(t) plot at time t.

                                                                                                                              (7.5d)

The time equation (7.5c-d) is just Newton’s 2^nd law given by (6.0). The space force law (7.5a-b) is just the slope rule first stated (with the physicist’s minus-sign) in (6.9). Both laws deal with conserved stuff. If you, a daredevil, acquire x of this stuff (energy or momentum) sooner or later you are going to have to find something or someone help you get rid of x. Or else!

            A daredevil falling 30 meters acquires energy equal to gravity force (body weight Mg) times thirty meters. Fig. 7.3a-b plots a constant F=-Mg and a linear potential U(y)=Mg y from y=30 to y=0. The 1m kiddy-pool must get rid of the 30Mg (Newton meters) of energy in one meter, by applying a force of 30Mg (Newtons) steadily over the entire meter from y=0 to y=-1. (That’s a 30g~300ms^-2 deceleration. Human survivability is somewhere around 50g.) An alternative is to get rid of that energy in the concrete below the pool in about 1millimeter, a 30 thousand g deceleration. (That is not survivable!)

Kiddy-pool versus trampoline

            Suppose the daredevil falls onto a special trampoline that applies exactly the same constant force as the kiddy-pool, but stores the energy as potential instead of dissipating it all by dousing the audience with a huge splash. (Recall Ka-Bong! versus Ka-Runch!  in Ch. 1.) The trampoline could then toss the daredevil back up to the 30 m tower to do the fall over again. (My gosh! What a daredevil has to do to satisfy a sated TV audience these days!) Such a potential is plotted by a steep-slope line U(y)=-30y in Fig. 7.3b.

Fig. 7.3 Force and potential plots. (a-b) Strong (30g) deceleration. (c-d) Medium (6g) deceleration.

 

            Suppose the Americans for Humane Daredevilry (AHD) demand that the deceleration distance be increased from 1 meter to 5 meters. (That’s what Olympic divers get for a 10 m fall.) As shown in Fig. 7.3c this reduces the deceleration by a factor of 5 from 30g to only 6g. (A walk in the park!)  The sloping U(x) lines are tallying the area-accumulation under the F(x) lines. Starting on the right hand side, U(x) drops by 30 units in 30 meters in Fig. 7.3b to correspond to the –30 units of area under the gravitational F=-1 unit line for the same distance in Fig. 7.3a. The daredevil’s kinetic energy must increase by 30 units to conserve total energy. So trampoline or pool is hit at 24 meters per sec. or 55 mph. (Recall (6.13).)

                                      ¹/₂ M V²=30 Mg                        or:  V=√(60g) = √588=24.2m/sec.

Getting rid of this 30 J potential deficit means climbing a steep 30 J high slope between y=0 and -1 in Fig. 7.3b or a medium slope of the same height between y=0 and –5 in Fig. 7.3d. Both cases have the same +30 J area under a force line, but having 5 meters instead of just one reduces the force to ³⁰/₅=6.

            Time functions F(t) and MV(t)=P(t) relate to F(x) and U(x) using Newton II: F=M^dV/_dt in (7.5d).

      or:                   (7.6a)

                                      or:                   (7.6b)

The first relation is total energy conservation (KE+PE=const.) first stated in (6.6) and (6.7).

Linear force law, again (But, with constant gravity, too)

Let’s imagine the AHD demands further protection of daredevils from themselves by outlawing constant-force targets that turn on a full force suddenly upon entry. Claiming that “high-jerk” is bad, the AHD requires linear-force targets, instead. Physicists comply happily since a harmonic-oscillator linear-force-quadratic-potential (6.12) is the favorite force law. It also describes inside-Earth oscillation in Chapter 9.

            Plots of linear-force-quadratic-potentials are shown in Fig. 7.4. Just like the preceding Fig. 7.3, a constant gravitational force F_grav=-Mg is present both in and out of the (y<0)-region where the linear F=-ky force and the U(y)=¹/₂ky² potential exist as a sum of constant and linear forces for (y<0).

                  

                                                            (7.7a)                                                               (7.7b)

If a linear potential b·y is added to a quadratic a·y² potential we get the same parabolic curve U=a·y², but that curve is shifted to the left by y_shift=-b/2a and down by U_shift=-b²/4a as follows.

                                            (7.8a)

                                                       (7.8b)

The nose or tip of the parabola, which is the equilibrium resting point, follows an upside-down copy of the U-parabola itself! This important geometric fact is shown in Fig. 7.4. The geometry does not reveal itself until we look in Fig. 7.4e at a “soft ball” that is soft enough to clearly show its gravitational shifts. A hard superball is more like Fig. 7.4b that barely shows such a small shift.

            Hardball total potential is u(y)=8y²+y with a total force function f(y)=-16y-1 in graph units of Fig. 7.4(a-b). A medium total potential is u(y)=y²+y with a total force function f(y)=-2y-1 is plotted in Fig. 7.4(c-d). The latter clearly shows the equilibrium or lowest “sag” point of zero force. The softball total potential is u(y)=(1/4)y²+y with a total force function f(y)=-(1/2)y-1 in Fig. 7.4e. The hardball potential requires about 6 meters (Y=-6 or y=-0.6) to cancel the energy from the 30 meter fall (from Y=30 or y=3) and maximum force of about F=10. This is much more than the constant F=6 that stopped the same daredevil in 5 meters in Fig. 7.3c because a linear force has only the area under a triangle which has a factor of ¹/₂. Here ¹/₂(F=10)(Y=-6) gives the necessary energy of 30 Joules. So the AHD ruling has actually increased the maximum force on the daredevil! (But, only during the final milliseconds is F large.)

            Note that the focus of the U(y) parabola is on the y-axis because we plot gravity with slope=1. Can you find a geometrical a way to locate that focus given some allowed stopping distance?

            Parabolic geometry of an oscillator potential subject to a uniform (or nearly uniform) force field is an important one in physics. Electronic charges pinned to an atomic potential well behave like oscillators in an electric field of a passing light wave. Generally the light wavelength of 0.5 micron (0.5E-6m) is several thousand times as long as the atomic radius of a few Angstrom (1E-10m). So the effective potential is a rigid parabola like Fig. 7.4e shifting to and fro and up and down at some frequency.

 

Fig. 7.4 Linear deceleration force after constant falling force. (a-b) Hard (c-d) Medium (e)Soft

 

            As we mentioned, superball force function is non-linear; approximately F_ball(y)~y⁴ plotted in Fig. 7.2 and Fig. 7.5 below. Compare this to the linear balloon-like force curve F_balloon(y)~y¹ in Fig. 7.4e above. (Recall (7.2).) F_balloon(y) is a pair of straight lines bent at contact point y=0, while F_ball(y) has a long flat region below y=0. For either case, the force integrals ∫F^total(y)dy and the areas they represent cancel between any two points y=h and y=y_max that have the same potential energy U(h)=E=U(y_max). If that energy is the total energy E then these points y=h and y=y_max are classical turning points. The mass M stops with zero KE (no speed) to turn around and fall backward or forward, respectively, into the potential valley lying between h and y_max. PE curve U^total(y) near bottom (y_static) in Fig. 7.2-5 is nearly parabolic as is U(x) in Fig. 6.3. The difference for Fig. 7.4 is that all of the U^total(y) curves are perfectly parabolic for y<0. (See exercise 1.7.1.)           

            Fig. 7.5 Force and potential for soft nonlinear (F=ky⁴) superball dropped from height h

           

Why super-elastic bounce?

            Super-elastic bounce involving two balls was introduced way back in Fig. 4.5 and “explained” by the 2-Bang model sketched there. Is that the only explanation? Certainly not! Is it even right? Well, yes and no. Here is a chance to discuss how science works or doesn’t work. It is, after all, a human endeavor. (To err is…)

RumpCo versus Crap Corp

            Let’s imagine a big scientific fight between two research groups something like real ones I’ve seen. We’ll imagine it’s about superball dynamics. On one side is a small but creative group working for the Rumpany Company® that first discovers the effect and explains it with the 2-Bang model. But their small budget limits them to things you can do cheaply with a ruler and compass.

On the other side is the huge Crap Corporation®. With unlimited military contracts, CrapCorp can afford any kind of computer or lab equipment. They hear about RumpCo’s discovery and decide to develop and sell it to the Army as a bomb detonation system.

            I hope you’ll excuse a scatological nomenclature and contempt for shortsighted and mindless goals often associated with post-modern cash-flow-science. My allegorical objective is to encourage curiosity-driven-science that is now becoming regarded as quaint. I do believe that humans are capable of creating much more than fertilizer and should be strongly encouraged to do better. If earning gets in the way of learning, then humans do poorly. I have watched big labs in government, industry, and university die of a pernicious groupthink fueled by the acquisitive rather than the inquisitive human drives. People lose their ability to reflect and become happy to merely genuflect. A novel Radiance by Carter Scholz (Picador 2002) is a “Star Wars” romaine a’clef exposing foibles of scientists at Livermore and Los Alamos.

            On one side of our allegory is poor but resourceful little RumpCo full of ideas but nowhere to go. Their 2-Bang model of super-elastic bounce is simple, elegant, but appears wrong. The powerful CrapCorp, on the other hand, knows where it’s going and what’s right. It has every resource imaginable. Except wisdom.

            CrapCorp’s first move is to discredit RumpCo’s work. They set up a computer that uses lab observed potential functions to fully analyze a 2-ball bounce. Let’s compare two competing vu-graphs side-by-side.

 

 

Fig. 7.6 RumpCo theory versus CrapCorp’s simulation. (RumpCo) Finite initial gap (CrapCorp) NO gap

 

            One thing is clear. CrapCorp does fancy-schmancy vu-graphs! They resemble wedding invitations.

And, while CrapCorp’s 10-figure precision is dubious, we note their V₁=0.62  and V₂=2.29  disagree with RumpCo’s predictions (Recall Fig. 4.4.) of final V₁=0.5 and V₂=2.5 by a little. Furthermore, RumpCo uses an independent 2-ball bang model. They assume or idealize an initial gap separating mass m₁ from m₂ so Bang-1 of m₁ with the floor is independent of Bang-2 between m₁ and m₂. So V₁ and V₂ result from 2-body energy-momentum conservation. RumpCo’s results are not sensitive to force functions.

                  CrapCorp can compute the difficult 3-body collision between m₂ , m₁, and m₀ (the Earth) all together just like what’s really happening on the floor. CrapCorp ‘s curvy V₁ vs. V₂ plot in Fig. 7.6 is very sensitive to each force function F(y) between each pair of colliding bodies. When (and if) CrapCorp values check out with experiment, they’ll happily sneer at the primitive pair of straight lines in the RumpCo velocity plot.

Does RumpCo have nearly the right (V₁,V₂) for wrong reasons? Not entirely. The reason a 2-Bang model works at all is that the force function for these balls is highly non-linear. A quartic function F(y)=y⁴ has a flat bottom as noted before Fig. 7.5. That allows the floor-m₁ collision to nearly finish before the m₁-m₂ bang really gets going even though the balls are in contact all during the collision.

            Realizing this, the RumpCo researchers suggest that CrapCorp try a linear force F(y)=y¹ simulation to see if super-elastic bounce disappears. They do, it does, and the rest is history. As seen in Fig. 7.7, m₁ and m₂ bounce up in unison. It’s a pax de deux. Super-elastic bounce goes away!

Fig. 7.7 Linear force kills super-elastic bounce. (Collaborative effort.)

 

The two groups decide to stop feuding and join forces. A corporate merger results in a multi-national conglomerate Carumpany Ltd. based in the Caymans. They lived happily ever after. (Well, sort of.)

Seatbelts and buckboards

Another important physics lesson from this section is, “Fasten your seatbelts…tightly!” To avoid great and damaging force you need to avoid non-linear force functions and fasten yourself with linear ones that can start working off your kinetic energy and momentum most immediately after a collision. The non-linear force with its “flat” region applies little or no force at first but then has to make up for its procrastination with deadly high force after it’s too late. Note how nonlinear force in Fig. 7.5 finishes much higher than the linear force in Fig. 7.4. Even worse is having no seatbelt at all. That’s like a very non-linear force of, say, F(x)=kx¹⁰⁰. It’s a flat gap with a practically vertical wall waiting to crush you!

            One of the most dangerous vehicles in the Wild West of the early US was the buckboard, a wagon with no suspension except for a set of springs right under the rider’s seat. When the buckboard hit a bump it generally lived up to its name. Unfortunate riders ended up like a little m₁ superball knocked skyward by a big m₂ wagon. A safer and more comfortable ride is had in a car with a body as much heavier than the wheels and suspension as possible. So-called “Monster trucks” have the worst kind of ratio possible for stability.

Friction and all that “dirty” stuff

            Slowly we have put back some of the “real-world” features of the superball collisions that our idealized “Bang-Bang” models of Ch.4 ignored in order to make the problems more easily solvable. The effects of gravity during collision have been introduced and applied to interacting zero-gap superballs.

More such effects will be studied in what follows since interacting linear forces are very common in nature and there are ways to make them easily solvable, too. The oscillating neutron star in Ch. 9 provides a taste of what is to come in the study of waves and oscillation in Unit 4 and orbits in Unit 5.

            But even the neutron star model neglects what is the bane of the purist physicist, the dreaded frictional forces. These are among the most neglected and poorly treated physical effects in physics. If anything goes wrong with a theory, we just blame it on friction! Often we have little choice in this matter.

            Friction is a result of having more particles than we’d like to admit. Consider one m₁=72 gram superball. That’s about a mole of Carbon C₆ rings and a mole has 6.02E23 (That’s Avogodro’s number.) of these C₆ molecules. So we’re dealing with not one mass m₁ particle but an enormous heap with an unimaginably huge number 60,200,000,000,0000,000,000,000 of particles that individually are (mostly) friction-free and well behaved, but their mob-behavior is just plain abominable!

You’ve got to get down to at least the individual molecular level before “internal-friction” is pretty much a non-existent phenomena and pure quantum wave mechanics rules. So what we call “frictional loss” is simply the best accounting we can do of 60.2 gazillion chiseling thieves stealing bits of energy that turn up later as “heat.” In conservative economics the effect is known as “supply side” or “trickle-down.” Let’s see if we can account for energy chiseled by just three thieves. (And, then we’ll hire more thieves until we bankrupt the whole operation!)

 

 

 

Important atomic and molecular force geometry

1.1.2 A most important mechanics problems is that of atomic oscillators affected by electric fields since it is basic to all spectroscopy. A useful approximate model is potential V^atom(x)=k x²/2 function of center x of charge Q where k is a spring constant of atomic polarizability. A uniform electric field E is assumed to apply a force F=Q·E to the charge by adding a potential V^E(x) to V^atom(x). (Give V^E(x)=______ and F^E(x)=_____)

Consider the resulting potential V^total(x) for an atom for unit constants k=1 and Q=1. Derive and plot the new values for equilibrium position x^equil(E), energy V^equil(E), dipole moment p^equil(E). Plot V^total(x) for field values of E=-2,-1, 0, 1, and 2.

Does charge oscillation frequency ù^equil(E) change? If so express in terms of ù^equil(0) and E?