“All the light which is radiated… will, after it has traveled a distance r, lie on the surface of a sphere whose area S is given by the first of the formulae (3). And since the practical procedure… in determining d is equivalent to assuming that all this light lies on the surface of a Euclidean sphere of radius d, it follows…4 \pi d^2 = S = 4 \pi r^2 (1 - \frac{K r^2}{3} + …);whence, to our approximation 4)d = r (1- \frac{K r^2}{6} + …), or
r = d (1 + \frac{K d^2}{6} + …).</center”

— Howard P. Robertson

Geometry as a Branch of Physics (1949)

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Howard P. Robertson 28

American mathematician and physicist 1903–1961

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“The value of the intrinsic approach is especially apparent in considering 3-dimensional congruence spaces… The intrinsic geometry of such a space of curvature K provides formulae for the surface area S and the volume V of a "small sphere" of radius r, whose leading terms are 3)S = 4 \pi r^2 (1 - \frac{Kr^2}{3} + …),
V = \frac{4}{3} \pi r^3 (1 - \frac{Kr^2}{5} + …).”

Howard P. Robertson (1903–1961) American mathematician and physicist

Geometry as a Branch of Physics (1949)

“Measurements which may be made on the surface of the earth… is an example of a 2-dimensional congruence space of positive curvature K = \frac{1}{R^2}… [C]onsider… a "small circle" of radius r (measured on the surface!)… its perimeter L and area A… are clearly less than the corresponding measures 2\pi r and \pi r^2… in the Euclidean plane. …for sufficiently small r (i. e., small compared with R) these quantities on the sphere are given by 1):L = 2 \pi r (1 - \frac{Kr^2}{6} + …),
A = \pi r^2”

Howard P. Robertson (1903–1961) American mathematician and physicist

1 - \frac{Kr^2}{12} + …
Geometry as a Branch of Physics (1949)

“[T]he astronomical data give the number N of nebulae counted out to a given inferred "distance" d, and in order to determine the curvature… we must express N, or equivalently V, to which it is assumed proportional, in terms of d. …from the second of formulae (3) and… (4)… to the approximation here adopted, 5)V = \frac{4}{3} \pi d^2 (1 + \frac{3}{10} K d^2 + …);…plotting N against… d and comparing… with the formula (5), it should be possible operationally to determine the "curvature" K.”

Howard P. Robertson (1903–1961) American mathematician and physicist

Geometry as a Branch of Physics (1949)

“The solution of (1), which represents a homogeneous manifold, may be written in the form:ds^2 = \frac{d\rho^2}{1 - \kappa^2\rho^2} - \rho^2 (d\theta^2 + sin^2 \theta \; d\phi^2) + (1 - \kappa^2 \rho^2)\; c^2 d\tau^2, \qquad (2)where \kappa = \sqrt \frac{\lambda}{3}. If we consider \rho as determining distance from the origin… and \tau as measuring the proper-time of a clock at the origin, we are led to the de Sitter spherical world…”

Howard P. Robertson (1903–1961) American mathematician and physicist

"On Relativistic Cosmology" (1928)

“The search for the curvature K indicates that, after making all known corrections, the number N seems to increase faster with d than the third power, which would be expected in a Euclidean space, hence K is positive. The space implied thereby is therefore bounded, of finite total volume, and of a present "radius of curvature" R = \frac{1}{K^\frac{1}{2}} which is found to be of the order of 500 million light years. Other observations, on the "red shift" of light from these distant objects, enable us to conclude with perhaps more assurance that this radius is increasing…”

Howard P. Robertson (1903–1961) American mathematician and physicist

Geometry as a Branch of Physics (1949)

“The Rule of Three, or Golden Rule of Arithmeticall whole Numbers. Be the three termes given 2 3 4. …To finde their fourth proporcionall Terme: that is to say, in such Reason to the third terme 4, as the second terme 3, is to the first terme 2 [Modern notation: \frac{x}{4} = \frac{3}{2}]. …Multiply the second terme 3, by the third terme 4, & giveth the product 12: which dividing by the first terme 2, giveth the Quotient 6: I say that 6 is the fourth proportional terme required.”

Simon Stevin (1548–1620) Flemish scientist, mathematician and military engineer

Disme: the Art of Tenths, Or, Decimall Arithmetike (1608)

“In the work of Vieta the analytic methods replaced the geometric, and his solutions of the quadratic equation were therefore a distinct advance upon those of his predecessors. For example, to solve the equation x^2 + ax + b = 0 he placed u + z for x. He then hadu^2 + (2z + a)u +(z^2 + az + b) = 0.He now let 2z + a = 0, whence z = -\frac{1}{2}a,and this gaveu^2 - \frac{1}{4}(a^2 - 4b) = 0.
u = \pm \frac{1}{2} \sqrt{a^2 - 4b}.andx = u + z = -\frac{1}{2}a \pm \sqrt{a^2 - 4b}.</center”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.449

“A new point is determined in Euclidean Geometry exclusively in one of the three following ways:
Having given four points A, B, C, D, not all incident on the same straight line, then
(1) Whenever a point P exists which is incident both on (A, B) and on (C, D), that point is regarded as determinate.
(2) Whenever a point P exists which is incident both on the straight line (A, B) and on the circle C(D), that point is regarded as determinate.
(3) Whenever a point P exists which is incident on both the circles A(B), C(D), that point is regarded as determinate.
The cardinal points of any figure determined by a Euclidean construction are always found by means of a finite number of successive applications of some or all of these rules (1), (2) and (3). Whenever one of these rules is applied it must be shown that it does not fail to determine the point. Euclid's own treatment is sometimes defective as regards this requisite.
In order to make the practical constructions which correspond to these three Euclidean modes of determination, correponding to (1) the ruler is required, corresponding to (2) both ruler and compass, and corresponding to (3) the compass only.
…it is possible to develop Euclidean Geometry with a more restricted set of postulations. For example it can be shewn that all Euclidean constructions can be carried out by means of (3) alone…”

E. W. Hobson (1856–1933) British mathematician

Source: Squaring the Circle (1913), pp. 7-8

“These seven stages we shall name as follows:
1. Mixture
2. Gestation
3. Expansion
4. Age of Conflict
5. Universal Empire
6. Decay
7. Invasion”

Carroll Quigley (1910–1977) American historian

Source: The Evolution of Civilizations (1961) (Second Edition 1979), Chapter 5, Historical Change in Civilizations, p. 146

“The discovery of Hippocrates amounted to the discovery of the fact that from the relation
(1)\frac{a}{x} = \frac{x}{y} = \frac{y}{b}it follows that(\frac{a}{x})^3 = [\frac{a}{x} \cdot \frac{x}{y} \cdot \frac{y}{b} =] \frac{a}{b}and if a = 2b, [then (\frac{a}{x})^3 = 2, and]a^3 = 2x^3.The equations (1) are equivalent [by reducing to common denominators or cross multiplication] to the three equations
(2)x^2 = ay, y^2 = bx, xy = ab[or equivalently…y = \frac{x^2}{a}, x = \frac{y^2}{b}, y = \frac{ab}{x} ]Doubling the Cube
the 2 solutions of Menaechmusand the solutions of Menaechmus described by Eutocius amount to the determination of a point as the intersection of the curves represented in a rectangular system of Cartesian coordinates by any two of the equations (2).
Let AO, BO be straight lines placed so as to form a right angle at O, and of length a, b respectively. Produce BO to x and AO to y.
The first solution now consists in drawing a parabola, with vertex O and axis Ox, such that its parameter is equal to BO or b, and a hyperbola with Ox, Oy as asymptotes such that the rectangle under the distances of any point on the curve from Ox, Oy respectively is equal to the rectangle under AO, BO i. e. to ab. If P be the point of intersection of the parabola and hyperbola, and PN, PM be drawn perpendicular to Ox, Oy, i. e. if PN, PM be denoted by y, x, the coordinates of the point P, we shall have

\begin{cases}y^2 = b. ON = b. PM = bx\\ and\\ xy = PM. PN = ab\end{cases}whence\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.
In the second solution of Menaechmus we are to draw the parabola described in the first solution and also the parabola whose vertex is O, axis Oy and parameter equal to a.”

Thomas Little Heath (1861–1940) British civil servant and academic

The point P where the two parabolas intersect is given by<center><math>\begin{cases}y^2 = bx\\x^2 = ay\end{cases}</math></center>whence, as before,<center><math>\frac{a}{x} = \frac{x}{y} = \frac{y}{b}.</math></center>
Apollonius of Perga (1896)

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