Re: [Phys-L] Fictitious Mathalon.

["Richard Tarara" <rtarara@saintmarys.edu>]

James Burke's "The Day the Universe Changed" episode "Infinitely Reasonable" depicts Kepler doing just what David suggests below.

Richard W. Tarara
Professor of Physics, emeritus
Saint Mary’s College
Notre Dame, Indiana

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-----Original Message-----
From: Phys-l <phys-l-bounces@mail.phys-l.org> On Behalf Of David Bowman
Sent: Wednesday, July 6, 2022 7:11 AM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] Fictitious Mathalon.

Regarding Don Polvani's question:

> Does anyone know how Kepler (died 1630) computed the area and 
> perimeter of an ellipse?  Newton was born in 1642, so Kepler didn't 
> have the advantage of calculus.

I don't know for certain how he did it since I've never actually read that much about J Kepler's precise calculational techniques.  But based on famous drawings involving Mars' orbit and Kepler's 2nd law of planetary motion I suspect he may have used brute force by breaking up an ellipse into a large number of nearly triangular wedges with a common interior apex point and then separately added up the areas and  outer edge lengths of those triangles.  He probably did a sequence of an increasing number of triangles with an ever finer angular splay and may have attempted to project the results to the infinitely fine limit, i.e. use a numerical method of doing the needed calculus.

But regardless of Kepler's methods one can easily find the area of an ellipse without calculus by a simple application of one dimensional scale factors.  Suppose one has any closed planar figure whose finite area is A_0.  Next imagine uniformly scaling (just) one of 2 orthogonal dimensions of the figure by a scale factor, S, so that one dimension of the figure is now S times its previous value while the other orthogonal dimension is left alone.  Then the new area, A_1 of the figure after scaling/stretching one of its dimension's by S is given by A_1 = S*A_0.

Now suppose we start with an circle of radius a.  Since the time of the ancient Greeks it has been known that the area of that circle is π*a^2.  Now imagine scaling one dimension of that circle by a factor of b/a so that dimension's radius turns into a semi-axis of (b/a)*a = b.  The other orthogonal dimension is left alone and thus its semi-axis remains the original radius a.  Then the new area of the scaled circle (which by virtue of the scaling has now turned into an ellipse with a x b semi-axes) is (b/a)*(π*a^2) = π*a*b.

The trick also works for finding volumes of one-dimensionally scaled solids in 3 dimensions as.  The new volume of the scaled solid is the old volume times the 1-dimensional scale factor.

When I used to teach an intro astronomy course, and was discussing black-body radiation and the Planck formula for the radiation intensity vs wavelength, I used the above scaling property trick and the Planck function's scaling properties to infer the T^4 form of behavior for the integrated bolometric black-body radiation intensity, without using any calculus.  From the functional form of the wavelength-dependent Planck distribution function, I(λ,T), one can see that it can be written in the suggestive form: I(λ,T)= (T^5)*F(λ*T) where F(x) is a function of a single variable argument.  This means a graph of I vs λ for a fixed T has a width that inversely proportional to T since the argument, λ, appears only in the product λ*T in the function.  But the vertical height of the graph is proportional to T^5 because of the T^5 factor out front.  Now the graph of I(λ,T) vs λ and the area between it and the horizontal λ-axis forms a 2-dimensional figure on the plane of the graph with a finite area.  Suppose we start with such a graph with a given initial (absolute) temperature, T_0, and it has an integrated area below the curve of A_0.  Next, we change the temperature to T_1.  This change of temperature scales the width of the graph by a factor of (T_1/T_0)^(-1), and scales the height of the graph by a factor of (T_1/T_0)^5.  This means the new area under the curve for the new temperature, T_1, is A_1 = A_0*[(T_1/T_0)^(-1)]*[(T_1/T_0)^5] = A_0*(T_1/T_0)^4 after rescaling both orthogonal dimensions.  So we thus have the T^4 proportionality for the total integrated intensity, A, for all wavelengths without ever actually integrating the area under the Planck function.

David Bowman
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