Fermat’s Principle of Least Time

Pierre de Fermat was born in Beaumont-de-Lomagne, France on August 17, 1601. Written in an article by Boyer, Fermat was “called the founder of the modern theory of numbers” (Fermat, 2016). 

He was considered one of the most profound mathematicians during the 17th century. It is said he studied law, foreign languages, literature, science, and even mathematics (2016).

​From his studies, we know that he was primarily a lawyer and did “mathematics in his spare time” according to wolfram.

Fermat was a very influential man, as you may can tell. In an article on the story of mathematics site it is said that he was able to “discover several new patterns in numbers which had defeated mathematicians for centuries” (Article).

It also goes on to say that he is credited “for early developments that led to modern calculus, and for early progress in probability theory” (Article). Fermat was actually so amazing at everything he did that he was fluent in four different languages like Italian and even Greek.

Over time Fermat discovered many different theorems within mathematics. For example, the Two Square Theorem which shows any prime number that is “divided by 4 leaves a remainder of 1 [and] can always be rewritten as the sum of two squares” (Article).

The one theorem I like best is called Fermat’s Principle of Least Time. He used the methods that were influenced by Heron. The principle says that a path between two points by a ray or a light is a path that can be traversed in the least possible time.

In an article on MAA, it says that “Fermat became interested in finding a theoretical derivation of Snell’s Law” (Sanchis, 2014).

Snell’s Law was discovered by Willebrord Snell, it dealt with “the change in direction that occurs when a beam of light crosses a boundary from one medium into another” (2014). The proof of this principle could be solved using algebraic optimization.

If we were looking at a coordinate plane, we are going to find C(x,0) to minimize the travel time of a light beam alone path A to C to B.

We will assume the velocity above the x-axis is \(v_{1}\) and below is \(v_{2}\). Since \( Time=\dfrac{Distance}{Velocity} \), the time of travel is \( T(x) = \dfrac{\sqrt{(x-a)^{2}+b^{2}}}{v_1} + \dfrac{\sqrt{(x-c)^{2} + d^{2}}}{v_2} \). 

Now we will need to derive this and set it equal to 0, \( \dfrac{\dfrac{1}{2}[(x-a)^{2}+b^{2}]^{\dfrac{-1}{2}}*2(x-a)}{ v_{1}}+\dfrac{\dfrac{1}{2}[(x-c)^{2}+d^{2}]^{\dfrac{-1}{2}}*2(x-c)}{v_{2}}=0 \). 

This simplifies to \( \dfrac{x-a}{v_{1}\sqrt{(x-a)^{2}+b^{2}}}+\dfrac{c-x}{ v_{2}\sqrt{(x-c)^{2}+d^{2}}}=0 \).

Since, \( \sin{\theta_{1}}=\dfrac{x-a}{\sqrt{(x-a)^{2}+b^{2}}}\) and \(\sin{\theta_{2}}=\dfrac{c-x}{ \sqrt{(x-c)^{2}+d^{2}}} \). 

We now have, \( \dfrac{\sin{\theta_{1}}}{v_{1}}-\dfrac{\sin{\theta_{2}}}{v_{2}}=0 \).

This simplifies to \( \dfrac{\sin{\theta_{1}}}{\sin{\theta_{2}}}=\dfrac{ v_{1}}{v_{2}} \) which is Snell’s Law.

​Thus giving us the answer of, \( v_{2} \sin{\theta_{1}} = v_{1}\sin{\theta_{2}} \) (2014).