“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…”

"On Relativistic Cosmology" (1928)

Adopted from Wikiquote. Last update June 3, 2021. History

Help us to complete the source, original and additional information

Do you have more details about the quote "The solution of (1), which represents a homogeneous manifold, may be written in the form:ds^2 = \frac{d\rho^2}{1 - \kap…" by Howard P. Robertson?

Howard P. Robertson28

American mathematician and physicist 1903–1961

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

Geometry as a Branch of Physics (1949)

“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

“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)

“Consider an event, for example the outburst if a nova… Suppose this event is observed from two stars in line with the nova, and suppose further that the two stars are moving uniformly with respect to each other in this line. Let the epoch at which these stars passed by each other be taken as the zero of time measurement, and let an observer A on one of the stars estimate the distance and epoch of the nova outburst to be x units of length and t units of time, respectively. Suppose the other star is moving toward the nova with velocity v relative to A. Let an observer B on the star estimate the distance and epoch of the nova outburst to be x' units of length and t' units of time, respectively. Then the Lorentz formulae, relating x' to t', arex' = \frac {x-vt}{\sqrt{1-\frac{v^2}{c^2}}}; \qquad t' = \frac {t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}
These formulae are… quite general, applying to any event in line with two uniformly moving observers. If we let c become infinite then the ratio of v to c tends to zero and the formulae becomex' = x - vt; \qquad t' = t.”

Gerald James Whitrow (1912–2000) British mathematician

p, 125
The Structure of the Universe: An Introduction to Cosmology (1949)

“[Zuanne de Tonini] da Coi… impuned Tartaglia to publish his method, but the latter declined to do so. In 1539 Cardan wrote to Tartaglia, and a meeting was arranged at which, Tartaglia says, having pledged Cardan to secrecy, he revealed the method in cryptic verse and later with a full explanation. Cardan admits that he received the solution from Tartaglia, but… without any explanation. At any rate, the two cubics x^3 + ax^2 = c and x^3 + bx = c could now be solved. The reduction of the general cubic x^3 + ax^2 + bx = c to the second of these forms does not seem to have been considered by Tartaglia at the time of the controversy. When Cardan published his Ars Magna however, he transformed the types x^3 = ax^2 + c and x^3 + ax^2 = c by substituting x = y + \frac{1}{3}a and x = y - \frac{1}{3}a respectively, and transformed the type x^3 + c = ax^2 by the substitution x = \sqrt[3]{c^2/y}, thus freeing the equations of the term x^2. This completed the general solution, and he applied the method to the complete cubic in his later problems.”

David Eugene Smith (1860–1944) American mathematician

Source: History of Mathematics (1925) Vol.2, p.461

“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)

“Vieta was the first algebraist after Ferrari to make any noteworthy advance in the solution of the biquadratic. He began with the type x^4 + 2gx^2 + bx = c, wrote it as x^4 + 2gx^2 = c - bx, added gx^2 + \frac{1}{4}y^2 + yx^2 + gy to both sides, and then made the right side a square after the manner of Ferrari. This method… requires the solution of a cubic resolvent.
Descartes (1637) next took up the question and succeeded in effecting a simple solution… a method considerably improved (1649) by his commentator Van Schooten. The method was brought to its final form by Simpson”

David Eugene Smith (1860–1944) American mathematician

1745
Source: History of Mathematics (1925) Vol.2, p.469

“[T]he formalist school, of whom the most eminent representative is Hilbert, have concentrated on the propositions of mathematics, such as '2 + 2 = 4'. They have pronounced these to be meaningless formulae to be manipulated according to arbitrary rules, and they hold that mathematical knowledge consists in knowing what formulae can be derived from what others consistently with the rules…. for example…'2' is a meaningless mark occurring in these meaningless formulae. But… '2' occurs not only in '2 + 2 = 4', but also in 'It is 2 miles to the station', which is not a meaningless formulae, but a significant proposition, in which '2' cannot conceivably be a meaningless mark.”

Frank P. Ramsey (1903–1930) British mathematician, philosopher

The Foundations of Mathematics (1925)

“U don't have 2 be rich
2 be my girl
U don't have 2 be cool
2 rule my world
Ain't no particular sign I'm more compatible with
I just want your extra time and your
Kiss.”

Prince (1958–2016) American pop, songwriter, musician and actor

Kiss
Song lyrics, Parade Under the Cherry Moon (1986)

Help us to complete the source, original and additional information

Howard P. Robertson28

Related quotes

Related topics