## H4    Principle of Maximum Proper Time

No matter which path a clock travels from O to P through space-time - the time elapsed on the clock will always be OQ as measured by any inertial observer. The red and blue path, however, differ in that our 'black' inertial observer measures different elapsed times on his own clocks for the blue and the red path. For the blue path, it is simply the length of the segment OP = OA. But since everything always travels the same distance through space-time, we need only stretch the red path to find the corresponding elapsed time OB for black.

The proper time OQ that elapses on the clock is thus independent of the chosen path, but not, however, the quotient formed from this proper time and the time elapsed for an observer from any other inertial frame. For the straight blue path from O to P the qotient OQ / OP = OQ / OA has its maximum value, compared to all other paths from O to P as seen from the arbitrary chosen black system. The quotient is maximal because the denominator OA is minimal!

The name “principle of maximum proper time” is a bit misleading in this context since what is actually maximized for the blue path is the ratio Δτ / Δt, when we, like Epstein, denote proper time with τ and coordinate time with t. Δτ is measured on a single clock moving from O to P and influenced by gravitation and velocity, while Δt is measured on a clock of the distant observer in OFF, who in GTR really has a special position. While in STR we have given preference to symmetric presentations, we are now forced by gravitation to use the point of view of the distant observer in the OFF to compare other measurements.

Principle of maximum proper time: If no force is acting on a body, it moves from X to Y through space-time along a path for which the ratio Δτ / Δt is maximized.

In the STR, these paths are straight lines. Any other path through space-time is longer and thus increases the denominator Δt. Before we look at an example of the behavior in GTR, I would like to emphasize that here we do not consider the path that light takes from one location A to another location B through space - we will later. Here we consider “free fall lines” through space-time.

Now for an example:

What is the force-free path from a point A on the surface of the earth which leads 2 seconds later to the same place? It is a vertical throw with an initial speed of 10 m/s. During the first second the object rises, ever more slowly, to a height of 5 m and in the second second it falls freely from its height back to the starting point. We know the equations of motion and the path in space from Newtonian physics:

h(t) = v0• t – 0.5 • g • t2 ≈ 10 • t – 5 • t2 mmmm and mmm v(t) = v0 – g • t ≈ 10 – 10 • t

Why does this movement satisfy the principle of maximum proper time, why is it not more advantageous simply to remain on the ground or to rise to a height of 20 meters? First consider the related Epstein diagram:

The elapsed time Δt for an observer in the OFF is simply the length of the arc AB since the starting point traverses this path through space-time. Now the denominator in the term Δτ / Δt is constant and it is really a question of maximizing the elapsed proper time in the numerator. Why is the blue path more advantageous? Because higher up the clock is running faster! Why does it not then go even further? Because then it must go up and down so fast that the time dilation due to the high velocity would negate the advantage of the greater height! GTR advocates height, STR advocates small speeds - and the optimal compromise is our vertical throw!

What time increase does the vertical throw bring? We already deduced in G4 how the time difference depends on the height difference:

We must sum the increase per unit of time for the entire flight time of 2 seconds. The corresponding integral is harmless
This is the additional elapsed proper time in seconds due to the “hop”. We must balance this with the loss of proper time due to the velocity. According to the STR
We know that v(t) ≈ 10 – 10 • t and can therefore again integrate from 0 to 2 over time t. This integral is also quite harmless:

The loss from STR is thus only half as large as the gain from GTR and the overall positive balance is  +100 / (3 • c2) seconds. In problem 4, you will be asked to show that the increase of proper time is greatest for this parabola; that for higher paths the loss due to the STR increases faster than the gain due to GTR - and vice versa. The solution of Galileo and Newton corresponds exactly to the parabola which fulfills the principle of maximal proper time!