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(2007-10-31)   Work Done  &  Kinetic Energy
The  work done  is equal to the change in  kinetic energy.

The infinitesimal work done by the resultant  F  of all the forces applied to a point-mass of velocity  v  is equal to the  dot product  of that force by the displacement  (v dt)  of its point of application  (in the time interval  dt).  Namely:

F  v dt   =   m (dv/dt) v dt   =   m v dv   =   d ( ½ m v² )

Therefore, the work done over any finite interval of time is equal to the variation of what's called the  kinetic energy  of the point-mass, namely  ½ m v².

The modern nomenclature  (French:  travail  &  énergie cinétique)  for that  clean  result is due to  Gustave Gaspard de Coriolis  (1792-1843).

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rsg160 (2001-04-15)
I have been told that a satellite in a circular orbit that starts to enter the atmosphere actually speeds up, at first. When the satellite is outside the atmosphere there are only conservative forces (gravity) acting and, if the satellite is in a circular orbit, its [speed] is constant. When it starts to enter the atmosphere there is a small drag force, since the atmosphere is thin high up. This force always opposes the motion and I would have guessed that it would slow the satellite down. Can you help me in figuring out why the satellite actually speeds up?

You're absolutely right. As long as the drag force remains fairly small, the satellite will gain speed...  The reason for this is simple: As altitude decreases, speed increases. A drag force cannot change this unless it's large enough. The key is that a spacecraft loses altitude at a steady rate during reentry. It's fairly easy to work this out quantitatively:

Let's call M the mass of the satellite, V its speed and z its altitude. Let R be the radius of the Earth (assuming its mass distribution has perfect spherical symmetry) and let's call g the gravitational field at z = 0. 

The kinetic energy of the spacecraft is ½MV ² and its potential energy is exactly Mgz/(1+z/R)  (normalized to 0 when z = 0; it's about Mgz when z is small). The sum of these two terms is the total energy E, which is constant in the absence of drag.  In the presence of a drag force F (in a direction opposite to that of the velocity), there's a loss of total energy equal to the power FV of the drag force. In other words, E' = - FV.  We may use the above expression of E in this, neglecting the 1/(1+z/R) correction factor for the potential energy (thus making an error on the order of 1%) and obtain the relation:

M ( V V' + gz' )   =   - FV

Now, we may remark that the quantity (-z' )/ V is simply sin(a), the sine of the angle of reentry a (this is a positive quantity since z is decreasing, it would be zero for horizontal flight). Therefore, the above relation translates into:

dV/dt   =   V'   =   g sin(a) - F/M

As long as the drag force F is less than Mg sin(a), V' is therefore a positive quantity, which means that the spacecraft will indeed gain speed initially.

For proper re-entry the angle of reentry a cannot be too small, or else the spacecraft could "bounce" off the atmosphere and be back into outer space after losing just a little bit of energy. If no action is taken, an orbiting spacecraft could thus keep bouncing back until enough energy is lost and/or its reentry angle is sufficiently large --possibly dangerously so, since a large a means fast reentry and a lot of heat!

Of course, as the atmosphere becomes denser at lower altitudes, the drag force F will eventually exceed the above threshold and the spacecraft will slow down.

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(2002-01-27)   Lewis Carroll's Monkey
Consider an equilibrium realized when a "perfect" rope is passed over a frictionless and massless pulley with a ten-pound weight on one side and a ten-pound monkey on the other...  What happens when the monkey decides to climb up the rope?

This problem was popularized by the author of Alice in Wonderland, Lewis Carroll (1832-1898, né Charles Dodgson), who agonized over it.  He was a professor of mathematics at Oxford from 1855 to 1881.  The above picture once illustrated a discussion of the puzzle by the mathematical columnist Sam Loyd (1841-1911), who called the problem "Lewis Carroll's Monkey Puzzle", while stating that it was not known whether Lewis Carroll originated the question.  (Unfortunately, the solution given by Loyd happens to be erroneous.)

The answer is that the centers of inertia of the weight and the monkey will have the same vertical motion (we assume, of course, that the monkey only goes up or down but does not swing the rope). Thus, if the monkey and the weight are initially motionless at the same height, they will always face each other no matter what the monkey does.  For example, they will both be in free fall if the monkey lets go of the rope, and both falls stop when the monkey grabs the rope again.

The reason for this is simply that all the forces that are acting on either the monkey or the balancing weight are always equal.  There are only two such forces for each body; the downward weight and the upward tension of the rope.  The weights are equal because the two bodies have the same mass and the rope also exerts the same force on either body because of the numerous "ideal" assumptions made here, including the absence of swinging on the monkey's side (so that the rope exerts only a vertical force in either case).  It's also essential to assume not only the lack of any friction, but also the absence of mass for both pulley and rope (otherwise the rope's tension would not be the same on either side of an accelerating pulley and it would vary along the length of an accelerating rope).

Note also that a "perfect" rope retains its length and transmits instantly its change of tension. This is clearly unrealistic, but it's logically consistent with the axioms of classical mechanics.  Changes in tension propagate with infinite speed over the length of a "massless" rope. Such an assumption would be logically inconsistent in the context of relativistic mechanics.

When the same forces act on bodies of equal masses their speeds change in the same way, so that the speeds remain equal if they are originally so (and we're told here of an original equilibrium where both speeds are zero).  Both motions will therefore mirror each other.

From the monkey's perspective, pulling 2 feet of rope will get him only 1 foot higher from the ground, but will require as much effort (work) as would be necessary to climb 2 feet on a stationary rope. That's not surprising in view of the fact that 20 lb were lifted one foot in the process (the monkey and the weight went up one foot each), which is just as difficult a task for a 10 lb monkey as lifting his own weight up two feet...

This deceptively simple puzzle has been known to be an excellent way to start a healthy discussion about the fundamental principles of classical mechanics.

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(2007-01-03)
A heavy ball is dropped on a hard floor from a height of one foot with a light ball (ping-pong ball) on top of it.  How high would the light ball bounce?

Answer :  Up to 9 feet !       Explanation : 

Shortly before impact, both balls have the same downward speed V.  An elastic bounce off the floor makes the heavy ball go upward at speed V to meet the light ball still going downward at speed V.  The speed of the ping-pong ball  relative  to the heavy ball is thus 2V.  Upon impact of the two balls that speed reverses itself (since the mass ratio of the two balls is very large).  An upward speed of 2V relative to the heavy ball translates into an upward speed of 3V...

Ideally, the  kinetic energy  of the ping-pong ball thus goes from  ½ mV²  before impact to 9 times that much after the shock !  This is enough to propel that ball, up to 9 times the height it was dropped from.  

In practice, a lesser height is reached  (because of inelastic collisions and a finite mass ratio)  but the actual demonstration remains impressive.

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(2007-10-09)   Tangential and Normal Accelerations
The tangential acceleration is  dv/dt.  The normal acceleration is  v²/R.

Consider a smooth curve in space.  Let  s  be the  curvilinear abcissa  along that curve and  T = T(s)  a  unit vector  tangent to the curve in the direction of increase of  s.  Geometrically, the  curvature  1/R  at  s  can be defined as the  length  of the vector  dT/ds = N/R  (which is perpendicular to  T, since the length of  T  remains constant).  R  is called the  radius of curvature  at that point.

When such a curve is the trajectory of a point-mass traveling at speed  v = ds/dt  its [vectorial] velocity  is  v T.  The  acceleration  of that point-mass is thus:

a   =   d(vT)/dt   =   (dv/dt) T  +  v dT/dt   =   (dv/dt) T  +  (v²/R) N

This means that the component of the acceleration which is perpendicular to the trajectory has magnitude  v²/R  (speed squared divided by radius of curvature).

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(2007-10-09)   Loop-the-Loop in a Roller Coaster   (teardrop  shaped)
Harnesses shouldn't be  needed  to keep riders on their seats.

Traditional roller-coasters rides are designed so that there's always a component of the acceleration which makes the wheels of the car push against the tracks and/or the bodies of the riders push against the seat cushions.

Loop-the-Loop  at Coney Island  (1901-1910)

At the top of a "loop-the-loop" (where riders are upside down) either condition means that the  normal acceleration  is greater than the vertical acceleration of gravity. v2 / R   >   g On the other hand, we have: m g h  >  ½ m v2 This expresses that the gravitational energy obtained by releasing a car of mass  m  from a height  h  above the top of the loop is greater than the kinetic energy it retains at that point  (the two would be equal if there was no loss of mechanical energy due to friction).  Therefore: h   >   ½ R

This is to say that a roller-coaster car should be released from a height above the top of a loop-the-loop (much) greater than half of the radius of curvature there.

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(2007-10-18)   The Conical Pendulum
A hanging bob traveling along an  horizontal  circle.

Let  L  be the length of the string and  q  the (constant) angle it makes with the vertical.  The radius  R  of the trajectory is:

R   =   L  sin q

The horizontal centripetal acceleration is  v²/R = w² R,  as the bob travels around the circle at speed  v = wR.  (This is a special case of the  normal acceleration discussed above.)  Newton's law says that this acceleration multiplied by the mass  m  of the bob is is the  vector sum  of all the applied forces, namely the weight  (mg)  and the string tension  (F).  The magnitudes of the relevant components of the acceleration are thus obtained as the sides of the right triangle pictured at right  (featuring the aforementioned angle  q ).

Thus,  v²  =  R g tan q  =  L g sin² q / cos q  =  L g (1-u² ) / u   which can be recast into the following quadratic equation, with respect to  u = cos q :

u ²  +  ( v ² / Lg )  u  -  1   =   0

Solving for u, we retain only the solution which is between 0 and 1.  We obtain:

cos q   =   u   =   ( 1 + x² )^½ - x         where   x  =  v² / 2Lg

If we put  x = sh j,  this means simply that  u = e ^j.  In other words:

q   =    arccos (exp (argsh ( v² / 2Lg )))
v ²   =   2 L g  sh (ln (cos q))

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(2008-02-13)   Ball in a Bowl
The motion of a small ball  rolling  at the bottom of a spherical bowl.

A ball released at zero speed on a spherical surface moves in a vertical plane.

Let  R  be the radius of that spherical bowl.  The radius of the ball is  r < R.  The mass of the ball is  M  and its moment of inertia  is  J.

Let  O  be the center of the spherical bowl and  C  be the center of the small ball.  q  is the (oriented) angle from the vertical to  OC  and  q'  is the derivative  dq/dt.

In a  pure roll,  the point of contact of the ball with the spherical surface has zero speed.  Since the speed of  C  is  (R-r) q',  we may deduce that the ball spins with angular velocity  (1-R/r) q'.  The potential energy of the ball is  -M g (R-r) cos q  (up to an irrelevant additive constant).  Therefore, the following expression is equal to the  total mechanical energy,  which remains constant:

½ [ M + J/r² ]  (R-r)² (q' )²  -  M g (R-r) cos q

For an homogeneous ball,  J = ²/₅ M r ²  (therefore, ²/₇ of the kinetic energy is rotational energy).  The derivative of the above boils down to:

⁷/₅ (R-r)  q''   +  g  sin q     =   0

A simple pendulum of length  40%  longer than  OC = R-r  would have that same equation of motion  (a  40%  longer length means an  18.3%  longer period).

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brentw (Brent Watts of Hickory, NC. 2001-05-04)
A 24 lb weight, attached to the end of a spring, stretches it 4 inches... What is the equation of motion if the mass is released from rest from a point 3 inches above the equilibrium position?

An oscillation where the acceleration is proportional to the distance from a fixed point is a sinusoidal motion about that point known as an  harmonic motion.

Hooke's law states that the force exerted by a spring is proportional to its elongation. Under standard gravity (g=9.80665 m/s²), the weight (force) Mg of a mass M of 24 lb (10.88621688 kg) is 24 lbf (106.757318766 N). If this corresponds to an elongation of magnitude L=4 inches (0.1016 m), the force corresponding to an elongation y is therefore -y(Mg/L).

Now, if we decide to count upward elongations positively, the acceleration y" is therefore such that My"=-Mg-y(Mg/L) (Newton's Law). If we consider instead the position x=y-L above the equilibrium position, we have x"+(g/L)x=0 (in the absence of friction or damping). Introducing the quantity w such that w²=g/L (in other words w is Ö(9.80665/0.1016) or about 9.82457 rad/s), x is therefore equal to A cos(wt) + B sin(wt), for some constants A and B. Since we are told that, at t=0, x is 3 inches and the speed x' is zero (the mass is just "released from rest"), we know that B is zero and A is 3 inches (76.2 mm). Therefore, the equation of motion is simply x = A cos(wt), with A=3"=76.2 mm and w=9.82456847...rad/s (corresponding to a period of oscillation T=2p/w of about 0.64 seconds).

Note that the value of the mass M turns out to be irrelevant here (the only thing that matters is the magnitude L of the elongation at rest), but the value of the gravitational field g does matter: The value of the period of oscillation T would be different under a gravitational field other than the "standard" one.

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gatman (Central Florida. 2001-03-31)
How fast (rpm and mph) are
electrons going around a nucleus?
What factors affect it and how?

For small objects like electrons, quantum mechanics states that the very notion of trajectory breaks down.  You can't measure both the momentum and the position of a particle:  The product of the uncertainties in the measurements of such conjugate quantities cannot be less than the so-called Heisenberg limit.  Thus, the electron does not have a precise speed in the classical sense. However, you can work out what the expected value of the momentum would be if you were to measure it with infinite precision (which would mean, then, that you would not know at all where the electron is).  From that momentum, you may derive some kind of expected speed...

This being said (and it had to be said), you may work out things numerically using something as crude as the old-fashioned semi-classical Bohr model of the atom (circular "trajectories" --oh, well-- with an angular momentum which is only allowed to be n times the rationalized Planck constant "h-bar"  = h/2p).  The number n is the principal quantum number you may find listed in chemistry books; it's normally equal to 1 for the electron around an hydrogen nucleus.

All told, you'll find that the binding energy of the electron (a negative quantity) in the Bohr model is E(n) = -chR/n², where c = 299792458 m/s is the speed of light (Einstein's constant), h = 6.626 10^-34 J.s is the (unrationalized) Planck constant, and R=10973731 m^-1 is Rydberg's constant.  (I'll neglect the so-called normal mass shift correcting factor of 1/(1+m/M), where m/M is the ratio of the mass of the electron to that of the nucleus).  In other words, E(n) = (-2.18 10^-18 J) / n².

As is the case for planets around the Sun, it turns out that the kinetic energy  ½ mv²  is equal to -E(n).  The mass of the electron is m = 0.911 10^-30 kg.  So, what you get is a speed of the electron which is inversely proportional to n, namely:  v = (2190 km/s) / n.  The largest speed is for n = 1 and corresponds to about 0.73% of the speed of light, which means it was OK to neglect relativistic effects at our casual level of precision.  If you insist on having the speed expressed in mph, the formula is v = (4900000 mph)/n.

To obtain the period of rotation around the nucleus, you need to know the radius of the "orbit" (again, this is not to be seriously taken as a real trajectory), which is a_on², where a_o= 0.53 10^-10 m is Bohr's radius.  The period is equal to 2p/v times this, which means it is proportional to n³:  T = (1.52 10^-16 s) n³.  The frequency is thus inversely proportional to the cube of n:  f = (6.58 10¹⁵ Hz)/n³, or if you insist on using rpm's:  f = (395 000 000 000 000 000 rpm)/n³, since 60 rpm = 1 Hz.

All this pertains to the hydrogen atom (Z = 1).  For a lone electron around a nucleus with Z protons (a so-called hydrogenoid ion), the energy gets multiplied by Z² and the speed is therefore multiplied by Z.  Since the size of the "orbit" is divided by Z, the frequency is multiplied by Z².

To summarize, a lone electron would be expected to go around a nucleus with Z protons at a speed  (Z/n) V_o  and a frequency  (Z²/n³ ) f_o , where :

• V_o is about 0.0073 c, 2190 km/s, 7900 000 km/h, or 4900 000 mph.
• f_o is about 6.58 10¹⁵ Hz or 3.95 10¹⁷ rpm.

It's worth noting that a single electron  (of charge  q = 1.6 10^-19 C )  going around a circuit at frequency  f₀  is equivalent to a current  q f₀  of about  1 mA.

The so-called  Bohr magneton  (q/2m  =  9.27400915(23) 10^-24 A.m²)  is an important quantity which serves as a unit for whatever  magnetic moment  may be associated with the orbiting electron.  Classically, the magnetic moment associated with a current  I  circulating in a loop of vectorial area  S  is  m = IS.  The Bohr magneton would be the moment classically associated with an electron actually moving at the above speed along a circular path with a radius equal to the aforementioned  Bohr radius.

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(2008-03-16)   Thermal Expansion Coefficients
The thermal expansivity is 3 times the coefficient of linear expansion.

Usually, the size of a body increases with temperature.  The  relative  increase is a characteristic of the material  (under given conditions of temperature and pressure)  known as the thermal expansion coefficient.  Unfortunately, there are  two  flavors of this coefficient depending on whether a length (L) or a volume (V) is used to gauge the aforementioned  size.  Because volume is proportional to the cube of length, the volumetric (or cubical) coefficient  b  is exactly  3 times  the linear coefficient  a  (HINT:  dV/V = 3 dL/L).  The coefficient  b  is also known as the  thermal expansivity  of the material:

a  =

1

æ è

¶L

ö ø

p

b  =

1

æ è

¶V

ö ø

p

=   3 a

L

¶T

V

¶T

[ This notation is dominant, although some authors swap the rôles of a and b. ]  A traditional source of confusion is that the  linear coefficient  (a)  is usually tabulated for  solids  whereas the  volumetric coefficient  (b)  is commonly tabulated for fluids.  (What's  really  confusing is that some such tables call "a" the cubical coefficient.)  Here is an  unambiguous  table:

Coefficients of expansion ( in  ppm/K, at 20°C)	Linear ( a = b/3 )	Cubical ( b = 3a )
Ideal Gas  (b=1/T)		3411
Ethanol		750
Water		207
Mercury (Hg)		182
Aluminum (Al)	24	72
Brass	19	57
Silver  (Ag)	18	54
Stainless Steel  (18% Cr, 8% Ni)	17.3	51.9
Copper  (Cu)	17	51
Gold  (Au)	14	42
Steel  (1% C)	11.8	35.4
Iron  (Fe)	11.1	33.3
Ordinary glass	9	27
Tungsten (W)	4.3	13
Pyrex	3.3	9.9
Invar  (64% Fe, 36% Ni)	0.62	1.9
Fused quartz	0.59	1.8

Invar and Elinvar: Nobel lecture by Charles-Edouard Guillaume (1920)

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(2002-06-02)   What's the speed of sound in a solid?

The waves that propagate in the midst of a solid's bulk are called body waves. For an homogeneous isotropic material, there are (only) two types of these.  Although both could possibly qualify as "sound", the term "speed of sound" is best reserved for Vp, the celerity of P-waves (which are always faster than any other mechanical waves for a given solid).  This is also called longitudinal velocity (VL = Vp) in contradistinction with the slower transverse velocity (VT = Vs) of S-waves.

• P waves of speed Vp (primary waves or pressure waves) are longitudinal compression waves, for which material moves back and forth along the direction of propagation.  They're also called push-pull waves. 
• S waves of speed Vs (secondary waves or shear waves) are transverse waves for which material moves side-to-side, in a direction perpendicular to the direction of propagation.  The "S" may also be remembered as standing for "slow", "shake" or "shock" (since S-waves often cause the most damage in actual earthquakes because of amplitudes that are often much larger than P-waves).

At the surface of a solid, different propagation conditions prevail, corresponding to surface waves (known to seismologists as Rayleigh waves or Love waves).

In the Lab, it's most convenient to study a given material in the form of a thin rod.  Two types of waves leave invariant the axis of such a rod:

• Extensional waves have the following speed (Ve): 

  Ve  =  Vs	Ö	(3Vp 2 -4Vs 2 ) / (Vp 2 -Vs 2 )

  They are triggered in a thin rod by some longitudinal stress (for example, a hammer hitting one end in the direction of the axis). However, any material with a nonzero Poisson's ratio [that's to say almost any material] will respond to an axial force [or stress] with both an axial and a lateral elongation [or strain], which means that the cross section of the rod varies accordingly and extensional waves thus involve some radial motion.  In materials with a very small Poisson's ratio (like beryllium or cork), these are virtually identical to P-waves. 
• Torsional waves propagate a change in torsion.  They have the same speed (Vs) as shear waves.

Besides the solid's density ( r), the following dynamic quantities are relevant:

• Poisson's Ratio (n) is the ratio of the lateral shrinking to the longitudinal elongation which occurs in the direction of a pull.

  In a few rare so-called auxetic materials, this may be negative, which indicates the expansion is both longitudinal and lateral.

• Elasticity (E) is the increase in tension per unit of cross-sectional area or a small relative increase in the elongation of a wire.  E  is often called Young's modulus, in contradistinction with other elasticity coefficients.

  A relative increase in length is a dimensionless quantity called strain, and E is thus a "stress to strain ratio".  So are the other elasticity coefficients G, K and l, described below.  All of these are in units of pressure (or stress): pascals (Pa, N/m² ) or GPa.

• Rigidity (G), also called modulus of rigidity, shear modulus, or torsional modulus: ... 
• Stiffness or Bulk modulus (K): The increase in pressure for a small relative decrease in volume.  K is the reciprocal of compressibility (k_s ).

  K   =   -V	æ è	¶p	ö ø	S	[Subscript "S" is for isentropic, see below.]
  		¶V

• Lamé modulus (l):   l = K-2G/3   is one of the two Lamé constants sometimes used in basic elasticity theory (the other one is   m = G ).
  Since l = 3K n / (1+n),  it's negative for auxetic materials (-1 < n < 0 ).

The following relations hold between the above quantities [do multiply by 10⁹ the values tabulated in GPa for the elasticity coefficients E, G and K]:

9/E   =   3/G + 1/K

E  =  2(1+n)G  =  3(1-2n)K  =  9 KG / (3K+G)
(Vp)² = 3K/r  (1-n) / (1+n)   =  (K + 4G/3) / r
(Vs)² = G/r
(Ve)² = E/r      [Note:   Ve > Vs   if   n > -½ ]

Ve  =  Vs	Ö	(3Vp 2 -4Vs 2 ) / (Vp 2 -Vs 2 )

x + n  =  x n + ½       [where   x = (Vs/Vp)² ]
1/x  =  4/3 + K/G  =  1 + 3K/E
x  =  (n-½) / (n-1)

Thermodynamics of Elasticity :

The above elasticity coefficients involved in wave propagation are the isentropic (or adiabatic) ones.  Static measurements of these coefficients may also be done fairly slowly under isothermal conditions.  A correction is to be applied to translate such an isothermal coefficient into an adiabatic one.  For example, the adiabatic bulk modulus (K = KS) is slightly larger than its isothermal counterpart (KT).  At some absolute temperature T, we may call Cp the specific heat capacity at constant stress (in J/K per kg) and b the expansivity (the relative change in volume per kelvin, also known as  thermal coefficient of cubical expansion ).  If we introduce the quantity W = r Cp / Tb²  [which is homogeneous to a pressure], we have:   1/KT  =  1/KS + 1/W   and:

KS   =   KT / (1 - KT / W )
KT   =   KS / (1 + KS / W )

b  =	1		æ è	¶V	ö ø	p	is  3 times  the relative change in length per kelvin  (a).
	V			¶T

Usually, the quantity W is much larger than K and the relative difference between the adiabatic and the (smaller) isothermal coefficient is thus very close to K/W. For example, in the case of iron,  b  is listed at 35.4 10^-6/K, whereas Cp is about 448 J/kg K and r is 7874 g/L.  At 20°C (T = 293.15 K), this makes W roughly 9600 GPa.  As K itself is about 56 times smaller than that (about 170 Gpa), we obtain a relative difference of 1.8 %, which is of borderline relevance here.

For the record, the isothermal and adiabatic versions of rigidity are identical (G = GS = GT). On the other hand, the two flavors of Young's modulus obey the same type of relation we just described above for the bulk modulus, except that the linear coefficient of expansion  (a = b/3)  takes the place of the cubical one (b) which ends up involving 9W instead of the quantity W defined above, so that we have: 1/ET  =  1/ES + 1/9W.  (You may notice that these relations make the identity  9/E = 3/G + 1/K  hold in both the adiabatic and the isothermal case.)  We are proposing to call the pressure W the (thermal) wring of the material.

Further Reading, References & Online Data...

• David R. Lide (Editor-in-Chief)
  CRC Handbook of Chemistry and Physics
  76th Edition (1995-1996)  [page 14-34; data from 1950 to 1960]
• Geophysical Data for the Geologist by Kristoffer T. Walker.

  About auxetic materials (negative Poisson's ratio)

• Negative Poisson's Ratio by Rod Lakes (University of Wisconsin). 
• Cristobalite and Molecular Dynamics (MD) simulation. 
• Amir Yeganeh-Haeri, Donald J. Weidner, John B. Parise (1992).
  Elasticity of a-cristobalite: A silicon dioxide with a negative Poisson's ratio.
  Science 257: 650-652.

  About Utrahard Fullerite :

• V.D. Blank, V.M. Levin, V.M. Prokhorov, S.G. Buga, G.A. Dubitskii, N.R. Serebryanaya
  Elastic properties of ultrahard fullerites
  Journal of Experimental and Theoretical Physics.
  October 1998, Volume 87, Issue 4, pp. 741-746. 
• V.D. Blank, A.A. Nuzhdin, V.M. Prokhorov, R.Kh. Bagramov
  [...] thermodynamic characteristics of fullerite and various covalent elements
  Physics of the Solid State.  July 1998, Volume 40, Issue 7, pp. 1261-1263.

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(2002-06-09)   [ SAW  =  Surface Acoustic Waves ]
What's the speed of a Rayleigh wave?

Rayleigh waves were first described mathematically in 1885 by Lord Rayleigh  (born John William Strutt, 1842-1919) before being actually observed in earthquakes. 

A Rayleigh wave is a surface acoustic wave (SAW) which propagates in an homogeneous and isotropic solid at the planar boundary with some medium having little or no inertia (air or vacuum).

In the midst of such a solid, body waves propagate with two different celerities.  The so-called P-waves travel at the longitudinal celerity Vp whereas S-waves travel at the transverse celerity Vs.  We may call x the ratio Vs/Vp  [which is at most ½Ö3 (86.6%) but only exceeds ½Ö2 (70.7%) in the rare case of auxetic substances].  This parameter x is tied to the solid's Poisson's ratio (n):

n + x²  =  n x² + ½ ;     n  =  (1-2x² ) / (2-2x² ) ;     x²  =  (1-2n) / (2-2n)

If we call VR the celerity of Rayleigh waves, we may call h the ratio VR/VS and the following relation holds:

h⁶ - 8 h⁴ + 8 (3-2x² ) h² - 16 (1-x² )   =   0         [ V_R = h V_S ]

As a cubic polynomial of h², this equation may have other positive roots with a physical meaning (see article by Edouard G. Nesvijski) but there's only one which is less than unity and is relevant to pure Rayleigh propagation.  More precisely, h decreases from about 0.95531250 for rubbery materials (x » 0) to 0.87403205 for Ideal Cork (x = ½Ö2) or, possibly, even down to 0.68889218 for the most auxetic substances we could dream of (x » ½Ö3).  An approximative formula for ordinary materials (positive n) was proposed by L. Bergmann (in 1954):

h   »   (0.87 + 1.12 n) / (1+n)     =     (2.86 - 3.98 x² ) / (3 - 4 x² )

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(2002-06-10)
Is there anything harder than diamond?

Several substances are harder than diamond :

• Ultrahard Fullerite (C₆₀ );  positively!
• Carbon Nitride ( b-C₃N₄ );  probably...
• Rhenium Diboride ( ReB₂ );  in certain directions only  (2007-04-19).
• Aggregated Diamond Nanorods (ADNR).   [ ESRF ]

Ultrahard Fullerite

Ultrahard fullerite is a polymerized phase of fullerene discovered in 1995, which is the object of several patents awarded to Dr. Michael Yu. Popov.  It is currently used in the NanoScan (NS) scanning force microscope (SFM).  Some experimental studies indicate that ultrahard fullerite is about twice as hard as diamond (approximately 300 GPa, vs. 150 GPa for diamond).

Normally, fullerene crystallizes as a fairly soft yellow solid of low density (1680 g/L) where buckyballs are held together by  van der Waals  forces, similar to what holds together the hexagonal carbon planes underlying the structure of graphite  (the density of graphite is  2267 g/L).

However, the polymerization of fullerene which occurs at high temperature under gigapascals of pressure yields an ultrahard phase whose density  (3170 g/L)  compares to that of diamond  (3516 g/L).  Under normal conditions, the resulting stable structure leaves little room for compression...  The material is more than three times stiffer than either diamond or osmium, according to the above table  (itself based on published  acoustic  properties of ultrahard fullerite).

From borazon (1957) to carbon nitride and beyond...

In 1989, Marvin Cohen and his graduate student Amy Liu (then at UC Berkeley) devised a theoretical model to predict a crystal's stiffness [its bulk modulus], which was thought until recently to be a good indicator of the more elusive quality called hardness.  Noteworthy candidates which did not live up to expectations, according to this model, included cubic boron nitride.  (CBN is the hard form of "BN".  It was first synthesized in 1957 by Dr. Robert H. Wentorf of the General Electric Company.  It's known in the trade as borazon.)

On the other hand, Cohen's model clearly indicated that a carbon nitride crystal should be stiffer (and possibly harder) than diamond.  The race was on to obtain the stuff in crystalline form and measure its properties.  Some early efforts by the team of Yip-Wah Chung (at Northwestern University) resulted in a layered composite of titanium nitride and carbon nitride (a so-called superlattice) which was, surprisingly, almost as hard as diamond.

Crystals of carbon nitride were apparently first synthesized at Lawrence Berkeley Lab (LBL) by Eugene Haller and William Hansen, using an approach similar to the synthesis of industrial diamonds.  A 1993 patent for this new superhard material was subsequently awarded to Cohen, Haller and Hansen.  However, the jury seems still out... [See March 1998  APS conference, and May 2000  Nature article.]

Because ultrahard fullerite is so much harder than diamond, we may venture the guess that it's harder than carbon nitride as well, but we won't know for sure until someone manages to make a carbon nitride crystal big enough to test...

Recently (February 2004) a new kind of synthetic diamond was found to be at least 50% harder than natural diamond.  This was obtained at the Carnegie Institution's Geophysical Laboratory  (Washington, DC) by submitting to extreme temperatures and pressures (2000°C,  5-7 GPa) crystals synthesized [much faster than before] by a new chemical vapor deposition (CVD) process.

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(2002-06-10)   How is Hardness Defined ?
Hardness is the resistance to permanent surface damage...

It's totally different from elastic resilience, in spite of a loose correlation.

Unlike stiffness, which is an elastic property of a solid [quantitatively, stiffness is simply the solid's bulk modulus], hardness actually indicates the resistance of a solid's surface to permanent deformation (scratching or indentation) by another solid.  Hardness is somewhat correlated with elastic moduli (the larger the moduli, the harder the material is expected to be).  This correlation is far from perfect, as was spectacularly demonstrated in 2002 by a measurement proving osmium stiffer than diamond, although osmium is not nearly as hard as diamond.  (The term indentability is a less ambiguous alternative for hardness.)

Some news reports have wrongly described osmium as a "soft" metal, whereas it is one of the hardest metals, with a Mohs hardness of 7...

Hardness does not have a theoretical definition.  Instead, it is evaluated using a number of practical scales, which are only roughly compatible with each other.  Hardness values obtained by conversion between such scales are notoriously fuzzy and/or unreliable.

The oldest scale of hardness is the Mohs' scale, which can barely be called quantitative.  It was first devised in 1812 by the German mineralogist Friedrich Mohs (1773-1839) who published it in 1822.  This scale is based on comparisons with the materials below, which are assigned the values listed.  If a material scratches another, it's said to be of equal or greater hardness.  Mohs' scale is popular with geologists in the field, who can use it by carrying a kit containing the standard Mohs minerals and/or other substances of intermediate hardness:  Lead is 1½, a fingernail is 2½ (so are galena and gold), a knife blade is 5½ (so is window glass), a steel file is 6½, tungsten is 7½ (tungsten carbide is almost 9).  It's not necessary to have an expensive diamond in such a kit, except to "identify" other diamonds, because all other minerals are much softer (as of 2002, only one or two synthetic substances are known to be harder than diamond, as discussed above).

MH 1	MH 2	MH 3	MH 4	MH 5
Talc Mg3H2(SiO3)4	Gypsum  CaSO4	Calcite   CaCO3	Fluorite CaF2	Apatite Ca5PO4(OH)
Basic Mohs Hardness (MH) Scale
MH 6	MH 7	MH 8	MH 9	MH 10
Orthoclase KAlSi3O8	Quartz SiO2	Topaz Al2SiO4 X2	Corundum Al2O3	Diamond C

Some english speakers have been known to memorize this sequence using the dubious sentence: Those Girls Can Flirt And Other Queer Things Can Do.

The rungs in this hardness "ladder" are uneven:  According to some plastometric hardness scales (described below), diamond (MH 10) is about 3½ times harder than corundum (MH 9), whereas fluorite or fluorspar (MH 4; sometimes misspelled  "flourite" or "flourspar") is only 20% harder than calcite (MH 3). 

Pure corundum is colorless.  Colored corundum gemstones are called either ruby if red (because of the presence of chromium) or sapphire otherwise (mostly blue, in the presence of titanium).  Widely used for grinding and polishing, the emery abrasive contains » 60% corundum (Al₂O₃ ), mixed with magnetite (Fe₃O₄ ) and spinel (MgAl₂O₄ ; MH 8)

This problem with Mohs' scale has been somewhat corrected in a so-called extended scale, which departs from the above beyond MH 6 and assigns a hardness of 15 to diamond (instead of 10).  This extended scale remains much less popular with geologists than the above original one...

Better quantitative ratings of hardness are mostly obtained with two very different kinds of instruments, which may well measure different characteristics of the material under test.  One of these is known as a durometer, the other is called a plastometer (of various types, named after the specific indenter used).

A durometer ["dur" is French for "hard"] is simply a diamond-pointed hammer which slides under its own weight in a glass tube and rebounds off the surface of the material under test.  The height of the rebound is measured and compared with what would be obtained for some reference material.  If a conventional rating of 100 is assigned to high carbon steel, this principle defines the so-called Shore scale, which is divided into overlapping subscales (A, B, C, D, O, OO) covering progressively softer materials with different measurement specifics.

Other hardness scales are based on the size of the indentation left by a plastometer after pressing [usually for 30 seconds] an object (indenter) of known geometry with a calibrated force against a planar surface of the material under test.

The above picture shows an oblique view of the imprint obtained on a silicon nitride surface by applying a force of 10 kgf (about 98 N) to a so-called "Vickers indenter" (a diamond square pyramid with an angle of 136° between opposing faces).  Such a picture offers clues that the observed remnant is smaller than what the indentation used to be when the indenter was still in it.  The elastic recovery which took place is properly disregarded in the evaluation of hardness, which is supposed to be a measure of how difficult it is to inflict permanent indentations (or scratches) on the surface of a solid.

A plastometric hardness is then defined (in units of pressure) as the ratio of the calibrated force to the total surface area of that part of the indenter at rest which has the same cross-sectional area as the observed indentation.

In those cases where the indenter is not much harder than the material under test a theoretical correction may be needed to estimate the size of an indentation that would have been left by an infinitely hard indenter.  This is also the correct way to extend to diamond (and/or substances like ultrahard fullerite, the hardest stuff known to Man) a scale like the Vickers scale, which is normally based on direct readings from instruments with diamond indenters (valid for ordinary materials, compared to which diamond may be considered "infinitely hard").

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Kelly (Bakersfield, CA. 2001-08-29; e-mail)
I am from Guam, but now live in Bakersfield, California. [...] I get a sunburn quicker in Guam than in Bakersfield.
I say that it's because being in Guam [latitude 13.5°N] puts me closer to the Sun than being in Bakersfield [35.4°N].  Am I correct?   [...]

Not quite so.  What you want to compare is what happens in the two locations at the same local solar time, say noon.  You may as well compare the situations of two points B and G on the half meridian directly facing the Sun (noon local time) and located at the respective latitudes of Bakersfield and Guam (the longitudes of Bakersfield and Guam are irrelevant).

Let's do a rough calculation first (always a good idea).  The Earth is almost a perfect sphere of radius R = 6371000 m.  Because the sun is so far away, the difference in the solar distances of two points A and B on the Earth is accurately estimated as the distance between the two parallel planes containing A and B that are perpendicular to the rays from the Sun.  Therefore, you may observe that the difference between the solar distances of two illuminated points on Earth may not exceed the Earth radius (R).  This overestimate is good enough for our next point...

The difference in solar distances at noon is thus [much] less than 6371 km.  Since the Sun is about 150 000 000 km away, this amounts to less than 0.0000425 of the distance to the Sun.  Now, the energy received from the Sun per unit of area (physicists call it the radiant illumination) is inversely proportional to the square of the distance to the Sun.  The difference in radiant illumination due to the difference in solar distance is thus no more than 0.0085%.  Obviously, such a minuscule difference could not possibly account for the observation concerning sunburns.  There's another explanation...

What's important is the angle of the Sun's rays, not the distance to the Sun! This matter of angles also explain why summers are warmer than winters [in the Northern Hemisphere], in spite of the fact that the Earth happens to be closer to the Sun in Winter than in Summer. [Believe it or not.]  The basic reason why it's colder in Winter is that each square mile of the Earth's surface "sees" the Sun at a more oblique angle and thus receives a narrower beam of sunlight.  Also, there's less time available between sunrise and sunset to receive the Sun's energy.

Your observation about sunburns, however, involves yet another angular aspect. Sunburns are directly related to UV exposure in the middle of the day.  It's important to realize that the atmosphere is a natural UV filter.  The lower you see the Sun on the horizon, the thicker the filter.  At noon, the Sun is higher in Guam and will therefore burn you faster.  At sunrise or at sunset, you can't possibly get sunburned.  In Guam or in Bakersfield...

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master_at_games_not   (Yahoo! 2007-07-22)   Kelvin's Thunderstorm
A nice suggestion for an experimental project in physics.   [ 12^th grade ]

In 1867, Lord Kelvin (1824-1907) devised a simple way to generate high voltages by harnessing the power of... falling drops of water.

To the best of my knowledge, this experiment has not yet been performed in a high-school context, but it should make a great student project...

You may want to watch a video demonstration by Walter Lewin at MIT.

Kelvin Water Dropper (Wikipedia)   |   Kelvin's Thunderstorm by Bill Beaty (1995)
Lord Kelvin's Electrostatic Water Drop Experiment   |   Spark-Museum

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(2007-07-24)   The Coriolis Effect:  A Simple Case
A dropped object always falls to the east of a plumb line.

The Coriolis force (or Coriolis acceleration) is observed only for something that moves with respect to a rotating frame of reference.

The beautiful mathematics involved does repay study, but it's nice to demonstrate the essentials by analyzing a simple case, using only  elementary methods :

You're on a beach near the Equator, on top of a tower (or a palm tree) of height h.  First, you draw a plumb line to make a small mark in the sand below.  Then, you drop a steel marble.  If there's absolutely no wind, you might expect that marble to fall  exactly  on the mark you just made.  Well, it does not.  Instead, it falls a few millimeters  east  of the mark.  Why?  Let's be quantitative:

Let h be the height of your tower and R be the equatorial radius of the Earth  (it's nominally equal to 6378137 m, but we won't need the exact value).  Let g be the normal acceleration of gravity at the Equator  (g = 9.780327 m/s² ).  Finally, let w denote the sidereal angular rate of rotation of the Earth, namely:

w   =   2p / 86164.09   =   7.2921159 ´ 10^-5 rad/s

Looking from the south in a nonrotating frame of reference (momentarily) at rest with respect to the center of the Earth,  the key observation is that the top of your tower and the beach have  different horizontal velocities,  namely:

w (R+h)     and     w R

What matters is not the huge value of those speeds  (both are about  465.1 m/s,  1674.4 km/h  or  1040.4 mph)  but their tiny  difference.

Another way to put it is that, in a nonrotating frame of reference (momentarily) at rest with respect to the beach, the top of the tower is seen to have a velocity  w h.  A marble dropped with zero velocity with respect to the top of the tower is really dropped with an horizontal velocity  w h  with respect to the beach and it will maintain that horizontal velocity throughout the time  (t)  it takes to hit the ground.  Thus, the marble will land at a distance   Dx  =  w h t   to the  east  of the point marked by the plumb line.  Since  h = ½ gt², we have:

Dx   =   w h t   =   w Ö(2/g)  h^3/2   =   0.00003297554  h^3/2

The numerical coefficient applies when  Dx  and  h  are both expressed in meters.

Coriolis effect at the Equator  (without air drag)

Dropping Height	Deviation to the East
10 m	1.04 mm
20 m	2.95 mm
50 m	11.66 mm
100 m	32.98 mm
200 m	93.27 mm
300 m	171.35 mm

If the experiment is carried out at a nonzero  geodetic latitude  q,  then the above remains applicable, except that the horizontal velocity of the top of the tower with respect to the ground  (in an inertial frame "tangent to" the rotating one)  is now  w h cos q.  Also, the value of  g  to be used (which determines the duration of the fall)  is clearly the local value, which is greater than the above equatorial value.

Dx   =   cos(q) w Ö(2/g)  h^3/2

The  geodetic latitude  q  of a location is the angle from the equatorial plane to the local vertical.  This is the only type of latitude used in geography.  It differs from the so-called geocentric latitude  (the angle between the equatorial plane and a line going through the center of the Eartn)  since the Earth is shaped like an oblate ellipsoid.  The "local vertical" is perpendicular to the "plane of the horizon" (which is tangent to the Earth surface) and it does not go through the center of the Earth, except for points located on the Equator or at the geographic poles.

Eiffel Tower  (from below)

The  Eiffel Tower  is at latitude  48° 51' 32''.  Its top  floor  is  309.63 m  above surrounding grounds.  The local value of  g  in Paris  (as measured before 1901)  is  9.80991 m/s2. Neglecting air drag, a marble dropped from the top floor of the Eiffel Tower would hit the ground  118 mm  east  of the dropping vertical,  after a fall lasting about 7.9452 s.

Air resistance (no wind!) would only slow down the fall, thus  increasing  the deviation to the east in  direct proportion  to the increase in duration.

Dx   =   (14.8548 mm/s)  Dt

The Eiffel Tower has a long history of being linked with the study of air resistance, starting in 1903  with the experiments of  Gustave Eiffel (1832-1923) himself:  By dropping objects along a vertical cable hanging from the second floor  (149.23 m  above the ground)  Eiffel obtained the most accurate aerodynamical data of that era.  He improved on this setup by building a wind tunnel next to the Tower in 1909.  The larger of the two wind tunnels he would build at Auteuil (in 1912) is still in use today  (making Auteuil the oldest aeronautical laboratory in the World).

Similarly, the  Leaning Tower of Pisa  is located at a latitude of  43° 43' 23'' N  (and a longitude of  10° 23' 47'' E).  At that latitude  q,  the rotation of the Earth makes the top of the Tower  (h = 55 m above the ground)  moves at a speed  wh cos q  relative to the ground.  This results in a Coriolis deviation  Dx  to the east, which is related to the time of fall  Dt  by the equation:

Dx   =   (w h cos q)  Dt   =   (2.89846 mm/s)  Dt

Neglecting air resistance, the time of fall  Dt  from a height of  55 m  is about  3.35 s.  This yields a Coriolis deviation of about  9.7 mm.  Air resistance (no horizontal wind) would only increase that Coriolis deviation a bit.  The effect is roughly 12 times less than for the Eiffel tower because, at a similar latitude, it is proportional to the height  (h)  raised to the power of 1.5...

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(2007-07-26)   Terminal Velocity   (Settling Velocity)
In the air, the velocity of a falling object has an upper limit.

In the main, the dissipative forces which oppose the motion of a nonrotating smooth  sphere  in a fluid are forces which are opposite to the sphere's velocity (relative to the fluid).  They are essentially of two distinct types:

• Viscous resistance, proportional to the speed  v  (mostly for liquids).
• Quadratic drag, proportional to  v²  (mostly for gases).

If the resistive forces are (somewhat artificially) limited to the sum of those two terms, we obtain a differential equation which can be solved analytically.

Vertical fall against air resistance   |   Optimal gear ratio for a car

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(2007-09-22)   Angular Momentum & Torque
Spin and  orbital  angular momentum are conceptually distinct.

A body of mass  M,  and center of mass  C  has a a total (linear) momentum  p  which is obtained by adding the elementary contributions  v dm  of its massive elements  (v is the vectorial velocity of the infinitesimal element of mass  dm):

p   =   ò  v dm

In classical mechanics, the  angular momentum  L of an object about a fixed origin  O  is the sum of two terms:  Its  spin  (or intrinsic angular momentum about its center of mass)  and its orbital angular momentum about the arbitrary point O.

L   =   S  +  r ´ p   =   S  +  OC ´ p