The author of the above article[1] used Newton’s polygonal approximation method to derive the centrifugal forces of uniform rotational motion around circles on a hyperbolic plane of constant curvature, −1, and a circle in the elliptic plane of curvature, +1.
In a previous publication[2], Lamphere found that the Euclidean result,
−m \frac{v^2}{ r }, (1)
where m is the mass of the particle, v is the constant, uniform velocity and r is the radius of the circle, should be replaced by
−m \frac{v^2}{ tanh(r)}
in the hyperbolic plane of constant curvature. It would then follow that the centrifugal acceleration would be
−m \frac{v^2}{ tan(r)} (2)
for a sphere of unit radius. This is because hyperbolic and trigonometric functions are related by tan(r) = −itanh(ir), it suffices to treat either case. However, both results are wrong, and it is the purpose of this note to correct them. To motivate our discussion, let us consider the Newtonian gravitational potential in the hyperbolic plane of constant curvature. A Lobachevskian straight line is tanh(r), so that Newton’s potential,
V = \frac{µ}{ tanh(r) },
where µ = GM, the gravitational parameter, gives rise to a force
\frac{dV}{ dr} = − \frac{µ}{ sinh2 (r) }.
The corresponding metric is given by
ds^2 = dr^2 + sinh^2 (r)dφ^2 ,
for polar coordinates (r, φ). This replaces the Euclidean metric,
ds^2 = dr^2 + r ^2 dφ^2 ,
for which the Newtonian law
\frac{dV}{ dr} = − \frac{µ}{ r^2} ,
in the hyperbolic plane of constant curvature. In fact, the hyperbolic distance between points (r, φ) and (r, φ + dφ) in the hyperbolic plane is
ds = sinh(r)dφ,
instead of the Euclidean arc length, ds = rdφ. And instead of the circumference, 2πr, the hyperbolic circumference of a unit circle is 2π sinh(r), which was first derived by Gauss. 2
Turning to equation (3) of Lamphere’s paper, the impulse is
f∆t = −2mv sin θ.
And referring to Figure 1(a) we apply the hyperbolic law of sines to the isosceles triangle SCB we get
\frac{1}{ sinh(r)} =\frac{ 2 cos φ}{ sinh(∆s)} ,
where φ is the complement of the angle θ, and we used sin(π − 2φ) = 2 sin φ cos φ. Introducing this expression into the impulse gives
f∆t = −m\frac{ sinh(∆s)/∆s}{ sinh(r)} v∆s.
And if there are n sides to the polygon, then the total impulse is
fn∆t = −m \frac{sinh(∆s)/∆s}{ sinh(r)} vn∆s.
Letting n → ∞, and both ∆t → 0, and ∆s → 0 such that n∆t = T, and n∆s = L with L = vT, we come out with the hyperbolic centrifugal force
f = −m \frac{v^2}{ sinh(r)} ,
in the hyperbolic plane of constant curvature. Analogously, we have
f = m\frac{ v^2}{ sin(r)} ,
for the sphere of unit radius in elliptic geometry. In this case, as r → π/2, the magnitude of the centrifugal force approaches its liming value of mv^2 , instead of vanishing as it would if (2) were valid, or if the Euclidean result, (1), applied in the limit of r → ∞.
Regarding paper[2], consider Lamphere’s modified Fig. 2 Let the secant RL be further inclined so that
Figure 1: Lamphere’s modified Fig.2
point L on the circumference of the circle intersect with the extension of line SP through the origin. Call the angle α. Now if L, Q, and P are three distinct points on a circle, the line LP is a diameter, then the ∠LQP is a right triangle. This is known as Thales’ theorem, 3 and is proven as part of the 31st proposition in the IIIrd book of Euclid’s Elements[3]. Thus, △QPR is also a right triangle. The angles
∠LP Q = ∠QRP = π /2 − α,
since ∠QP Q = α. The sides
RP = vδt (3)
QR = \frac{1}{3}\frac{f}{m} (δt)^2. (4)
Applying the law of sines to △LRP,
\frac{sin α}{sin(RP)}= \frac{1}{ sin(LR)} =\frac{ cos α}{ sin(LP).}
The last equality gives
cos α = \frac{sin(LP)}{ sin(RP) }. (5)
Now, applying the law of sines to the second right triangle △RQP, we get
\frac{sin α}{sin(QR)}= \frac{1}{(RP)} .
Eliminating sin α between the two sets of expressions gives
sin(QR) × sin(RL) = sin^2 (RP),
which is none other than the elliptic form of the Euclidean formula
RQ × LQ = RP^2 ,
that Newton used to derive his expression for the centrifugal force. Introducing (3) and (4) into the elliptic form of the tangent-secant law, and letting α → 1 in (5) gives
sin (\frac{1/2} \frac{f}{ m} (δt)^2= \frac{sin^2(vδt)}{ 2 sin(r)} , (6)
where we introduced the radius r = SP of an elliptic circle of circumference 2π sin(r). The same limit
\frac{ lim}{ x → 0} \frac{sin(x)}{x} = 1,
will also work for tan(x) as Lamphere used. Thus, he should have obtained f = mv^2 sin(r) , for elliptic geometry, and not with tan(r) [2]. Analogous for hyperbolic geometry of constant curvature the radius becomes imaginary resulting in the centrifugal force law,
f = m\frac{v^2}{ sinh(r) },
and not with the Lobachevskian ‘straight’ line tanh(r).
References [1] R. L. Lamphere, “Math. mag.” pp. 366–369, 2010. [2] ——, “Amer. math. monthly,” pp. 650–655, 2002. [3] T. L. Heath, “The thirteen books of euclid’s elements,” 1908. 6
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