Let's begin by taking an example of a block sliding across a completely level, frictionless table, as shown in the strobe picture below. Based on our definition of average speed as total distance divided by total time we can see that the straight line path followed by the dark block images would be the shortest path. Any deviation from this path, for example, the block's checkered position at t = 0.3 seconds, would require a faster speed to still reach the end in the same amount of time. This observation lays the foundation for our discussion of using energy to determine the shortest path between two points. To continue our analysis, let's change our focus from the tabletop picture above to a graph of the block's **Position vs. Time**. On this graph, x_{2}, will be the optional position through which the block might pass at t = 0.3 seconds. The method we will use is to determine nature's preferred path is called the Principal of Least Action. The action, S, is a scalar quantity defined by the integral where L(x, v) is called the Lagrangian. We can restate this algebraically as where DS an incremental value of the action for a time segment Dt while KE_{bar} and PE_{bar} represent the average values of the object's kinetic and potential energy during the same interval. Mathematicians and scientists have discovered that Nature's preferred path will be one having a minimal action - that is, the path that has the smallest value of when all of the DS for each time interval are added together. In our example, all of the time intervals, Dt, represent the same amount of time and the only differences in the paths are the position of x_{2}, our deviated position at 0.3 seconds. So each DS equals Where the expressions represent the slopes of the line segments between A and x_{2} as well as between x_{2} and C. Remember that the slope of a position-time graph represents the object's velocity. The total action between A and C is found by adding DS_{1} and DS_{2}. Note that Dt has been distributed throughout the terms in the equation. To find the minimal action, we now need to take the derivative of S with respect to the only value that can change in our equation, x_{2}. Since our table's surface is level, PE is constant, that is, never-changing, throughout the 0.5 seconds the block is sliding. Setting this derivative equal to zero will give us our optimal value for x_{2}. Notice in the last step that we have shown that our original two slopes must be equal to each other to minimize the action. This means that x_{2} must lie exactly along the line between A and C, no deviation is allowed. The fact that the slopes of the segments Ax_{2} and x_{2}C must be equal to satisfy the Principal of Least Action is equivalent to Newton's Law of Inertia: an object will remain in a state of constant velocity unless acted upon by an outside, unbalanced force. |