Feynman's lost lecture The motion of the planets around the Sun

Page: 0 1 2 3 4 5 6

Start with a given length AB and two points F and F'. Construct G' so that FG' = AB. Construct the line FG' and the perpendicular bisector of F'G'. The point where these two lines cross is P. The importance of this construction is in that if G' moves freely on a circle centred at F then P appears to move in an ellipse, as shown here:

To prove it is an ellipse we need to prove that the perpendicular bisector (shown in green) is indeed a tangent to the curve. Feynman's proof involves selecting another point Q on this perpendicular bisector and joining Q to F' and G'. Clearly F'Q and G'Q are equal. Join Q to F.

It should be easy to see that the total distance from FQ+QF' is equal to the total distance FQ+QG'. Now compare FQ+QG' with FP+PG'. Clearly FP+PG' is shorter. So, if we wanted to reach Q with a length of string from tacks at F and F', the string would have to be longer than the one needed to reach the unique point P. Thus all points like Q are outside the ellipse and so the line is indeed a tangent to the curve. Feynman has shown that the shortest path from F' to the tangent line and back to F is the path that reflects light at point P. This is a special case of Fermat's principle that light always takes the shortest path between two points and it is closely related to Feynman's approach to quantum electrodynamics, which won him his Nobel Prize.