Re: [Phys-L] Fictitious Mathalon.

[David Bowman <David_Bowman@georgetowncollege.edu>]

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