Kinetic Energy Paradox - kepardox.txt [1/1]

[John T. Lowry]

The Kinetic Energy "Paradox"

Several recent rec.aviation.xx threads have featured or involved the
apparent paradox that an aircraft which turns downwind thereby
undergoes a large gain of kinetic energy (KE) w.r.t. (with respect
to) the earth but no gain of kinetic energy w.r.t. the air mass. Initial
(incorrect) physical reasoning is often along these lines: there is
only a constant velocity (that of the wind) difference between CSs
(coordinate systems) fixed w.r.t. the earth and w.r.t. the air. Hence
time rates of change of velocities (accelerations) are the same in
the two systems (the derivative of a constant being zero); hence so
are the forces the same in both (the mass never varies). Since it
takes a difference in forces to produce a difference in kinetic
energies, there SHOULD be no such KE difference. But there is.

The error is in the seemingly innocuous statement that two
identical forces F(t), t the time, produce the same KE difference.
The solution is to see that even though the forces, as seen in the
two CSs, are identical, the DISTANCES moved under those
forces, and the parallelism between force and displacement, are
NOT the same. Delta_KE = Int(F dot dr) = Int(F dot v)dt=<F dot
v>delta_t; Int' stands for integral, F and dr and v are vectors and
dot' is that of the dot or scalar product, the product of the two
flankers times the cosine of the angle between them. Angular
brackets, as <X >, stand for the average of quantity X. If one keeps
track of the displacements, and their orientation with respect to the
forces, which is what vector analysis does, then the kinetic energy
difference computes properly and, as always, there is NO
PARADOX. The truth is that the forces ARE the same, but the
kinetic energy changes w.r.t. the two systems are NOT the same.

Here's a simple example. A jet is heading North, unit vector i, at
speed V w.r.t. the earth. The stewardess, Linda, of mass m, is
walking South at velocity -si w.r.t. the jet. V might be 600 mph, s
might be 4 mph. She hands a copy of Mountain Pilot magazine to a
passenger, reverses course, and walks North at velocity +si.

W.r.t. the jet, Linda has undergone a momentum change delta_p =
2msi. And, also w.r.t. the jet, a KE change delta_KE = 0.

W.r.t. the earth, Linda has undergone a momentum change delta_p
= 2msi (the same as w.r.t. the jet). And, also w.r.t. the earth, a KE
change delta_KE = (m/2)*(V+s)^2 - (m/2)*(V-s)^2 = 2mVs. Not at
all the same as w.r.t. the jet.

The force of the deck carpeting on Linda's pumps, F(t), is in the +i
direction during her turnabout maneuver; some sort of Gaussian
peak over a relatively small time interval delta_t. Ignoring the
small walking speed difference, her speed w.r.t. the earth
throughout that maneuver was essentially V. (A relatively tiny bit
less before she got stopped, a relatively tiny bit more afterwards.)
So <F dot V> = <F>V (they're in the same direction). Now her
momentum change is Int(F dt) = <F> delta_t = 2msi. We can use
this fact to evaluate <F>delta_t, plug it in the average of the
integral formulation in the second paragraph, and find that indeed
her delta_KE = 2msV, just as calculated by taking simple
differences.

The short time intervals, integrals approximated as averages, etc.,
are only incidental features which have no bearing on the essential
argument. (Not any more than accelerations at the end and
beginning of one twin's journey at 0.9c has any real bearing on the
twin "paradox" in special relativity.) The only trouble with treating
the original airplane case, which I've done, is that you get two and
a half pages of fairly dense vector calculus and the Internet is not
yet sophisticated enough to let us put that across uniformly in this
kind of forum.

The result: though the forces are the same in the two coordinate
systems (air mass and earth), the greater distance travelled in the
earth-based system, and the fact that those displacements in space
are not perpendicular to the forces (as they always are in the air
mass based system), means that indeed kinetic energy IS gained in
a downwind turn w.r.t. the earth. Here's the final result:
Delta_KE(w.r.t. earth) = -m*omega*R*Vw*(cos(omega*tf)-1),
where omega is the angular speed (yaw rate) in radians/sec, R is
the radius of the turn w.r.t. the air mass, Vw is the wind speed, and
tf is the time of flight. When you do a 180-degree turn, tf = T/2,
where T is the time needed to make a circle w.r.t. the air mass.
Then delta_KE = 2m*omega*R*Vw. If you continue around to
make a 360, you find, as intuition suggests, that delta_KE(w.r.t.
earth) = 0.

NO paradox. There never is. There's only (hopefully, temporary)
CONFUSION. Physics is wonderful.