Feynman, French Curves, and Fragility

 

Commend me to them,

And tell them that, to ease them of their griefs,

Their fears of hostile strokes, their aches, losses,

Their pangs of love, with other incident throes

That nature's fragile vessel doth sustain

In life's uncertain voyage, I will some kindness do them.

                                  Shakespeare

 

In Richard Feynman’s popular book “Surely You’re Joking, Mr. Feynman!” he tells how during his college years he “often liked to play tricks on people”. Most of these tricks were designed to show how dumb people are. For example, in a mechanical drawing class at MIT where the students were taught to use a drawing instrument called a French curve (a curly piece of plastic for drawing smooth curves), Feynman informed the other students that “the French curve is made so that at the lowest point on each curve, no matter how you turn it, the tangent is horizontal”. He reports that the other students in the class were excited by this discovery, and began holding up their French curves and turning them in various ways, trying to verify that the curve was always horizontal at the lowest point. Feynman found this funny, because they had already taken calculus and supposedly learned that the derivative of the minimum of any curve is zero. (Of course, it’s also intuitively obvious: If a curve at a given point is not flat, the point is obviously not the minimum.) Feynman says “I don’t know what’s the matter with people: they don’t learn by understanding; they learn by some other way – by rote or something. Their knowledge is so fragile!”

 

A few years later, when he was at Princeton, Feynman played a similar trick, this time on an assistant of Einstein and hence presumably well-versed in Einstein’s theory of gravitation. Feynman posed him a problem:

 

You blast off in a rocket which has a clock on board, and there’s a clock on the ground. The idea is that you have to be back when the clock on the ground says one hour has passed… Exactly what program of time and speed should you make so that you get the maximum time on your clock?

 

According to Feynman, the assistant of Einstein worked on the problem for quite a bit before realizing that the answer is simply the natural free ballistic trajectory of an object, because the fundamental principle of general relativity is that free-fall trajectories (i.e., geodesic paths) maximize the elapsed proper time along the path. Feynman comments that this should have been immediately obvious to Einstein’s assistant

 

But when I put it to him, about a rocket with a clock, he didn’t recognize it. It was just like the guys in the mechanical drawing class, but this time it wasn’t dumb freshmen. So this kind of fragility is, in fact, fairly common, even with more learned people.

 

Of course, the question was designed to be misleading. As in the case of Sandford’s elevator, the misdirection is introduced by referring to the ground, which is actually irrelevant. If the irrelevant introductory statements had been omitted, and the question simply posed as “What program of time and speed should you make so that you maximize the elapsed time on your clock?”, the assistant of Einstein would presumably have recognized the geodesic condition immediately.

 

Feynman’s obvious enjoyment of this kind of trick was not his most loveable quality, although there is undeniably some validity to his complaint about the fragility of the understanding of many supposedly intelligent and well educated people. Indeed, someone recently quoted Feynman’s rocket question (without the punch line) on an internet physics forum, and participants in the forum began to propose various ways of trying to solve the problem (unsuccessfully), without realizing that the answer was simply the geodesic path. My first thought when reading the question was that it was obviously just describing free trajectories, and I couldn’t believe that Feynman hadn’t realized this, so I checked the book and found that (sure enough) he had posed it as a trick question.

 

Incidentally, the internet poster insisted that it hadn’t been his intent to repeat Feynman’s trick (perish the thought). Instead, he claimed to be genuinely interested in the details of how one would go about actually computing the path of the object that maximizes the elapsed proper time. The posted responses were all unsatisfactory for numerous reasons. For one thing, the responses failed to actually connect the metric line element to the expression to be extremized. For another, the responses took the Euler-Lagrange equation as given, whereas one would think the derivation of this equation is precisely what was being sought. Also, the responses tended to assume the strength of gravity was constant over the trajectory, even though clearly a one-hour ballistic up-down trajectory would carry an object to a significant height, where gravity has only a small fraction of the strength relative to the strength at the surface of the Earth.

 

The exact solution for radial trajectories is a standard result in general relativity (see the note on Radial Paths, building on the variational results from Relatively Straight) but, just for fun, suppose we were trying to give a simple but complete and self-contained explanation of how to derive the (approximate) path of maximal proper time to someone who knows just basic calculus and is willing to accept the Schwarzschild metric line element for proper time

 

 

where m is the mass of the gravitating body (the Earth in our case), and we are using geometrical units such that G = c = 1. Thus the differential of proper time along a radial path can be written as


In the weak and slow limit the quantities m/r and (dr/dt)2 are both extremely small numbers (orders of magnitude smaller than 1), so we can neglect all products of those two quantities and all higher order terms.  Therefore the spatial curvature part of the metric can be neglected (which would not be the case if we were considering something involving high speed like the deflection of a light pulse), and we can take the lowest-order binomial expansion (1+x)1/2 ≈ 1 + x/2 of the square roots to give the approximate expression


The elapsed proper time Δτ along a path r(t) from t1 to t2 is the integral of this, i.e.,


We want to find the function r(t) that maximizes this elapsed proper time. To understand conceptually how we do this, it may be helpful to recall how we find the maximum of an ordinary function f(x). We find x such that df/dx = 0. Strictly speaking, this just locates a point where the function is stationary, i.e., flat (like the bottom of Feynman’s French curve), so it can represent either a maximum, a minimum, or an inflection point. But the context usually makes it clear which of these applies.

Now, to find the function r(t) that maximizes the integrated Δτ, we proceed in a similar way, by finding a path for which Δτ is stationary for small variations in the path. For any small constant u and any arbitrary function q(t) defined from t1 to t2 with q(t1) = q(t2) = 0, let Δτ(u) denote the integrated proper time for the disturbed path r(t) + uq(t). Thus we have


The disturbed path represented by r(t) + uq(t) has the same starting and ending points as r(t), but deviates from r(t) between those points. Our objective is to find a function r(t) such that d(Δτ)/du = 0 at u = 0 for any q(t).  This condition signifies that Δτ is stationary under any small variations of the path.

 

To differentiate the integral for Δτ(u) with respect to u we can bring the differentiation inside the integral, so we have

 

where accent symbols signify derivatives with respect to t. At u = 0 this reduces to

 


Now, recalling from basic calculus that (r′q)′ = r′q′ + r″q, we can write this as

 


The integral of the middle term from t1 to t2 is simply r′(t2)q(t2) – r′(t1)q(t1), which equals zero because q(t) is zero at both t1 and t2. Hence that term drops out, and we have


To make r(t) a stationary path, this derivative must equal zero for all possible deviation functions q(t), which implies that the expression in the square brackets must vanish. Hence we have the condition for the path length to be stationary


This is the geodesic equation for radial trajectories (to the lowest order of approximation), and of course it is just Newton's inverse square law of gravitation combined with the first law of motion, confirming that general relativity does indeed reduce to Newtonian gravity in the weak slow limit. (Coincidentally, if we replace the coordinate time t with the proper time τ, this formula is also exact in general relativity for a momentarily stationary free test particle, as explained in Radial Paths.)

Our next task is to solve this equation for the actual path. This too is a standard result in classical mechanics. (See for example the note on Free-Fall Equations.) Integrating both sides of the geodesic equation over dr from an arbitrary initial separation r(0) to the separation r(t) at some other time t gives

 

 

The left hand integral can be rewritten as

 

 

Therefore, the previous equation can easily be integrated to give

 

 

Solving the equation for r′(t), we have

 

 

Rearranging this gives

 

 

To simplify the expressions, we put r0 = r(0), v0 = r′(0), r = r(t), and ρ = r/r0. In these terms, the preceding expression can be written

 

 

If K is positive the trajectory is bounded, and there is some point on the trajectory (the apogee) at which v = 0. Choosing this point as our time origin t = 0, we have K=1, and the standard integral gives

 

 

It’s useful to define a parameter θ such that

 

 

Inserting this expression for r into the preceding equation, we find that t can be expressed in terms of the parameter θ by

 

 

These two parametric equations describe a cycloid, which is the curve traced by a point fixed on the perimeter of a wheel rolling along a flat surface, as illustrated in the figure below.

 

 

Our final task is to insert the specific conditions of the problem, i.e., a one hour radial round-trip beginning and ending at the surface of the Earth at radius re, to determine the actual the height r0 of the apogee where θ = 0 and t = 0, and the cycloid parameter θ* where r = re. We know θ = 0 and t = 0 at the apogee, where r = r0, and we also know r equals the radius of the Earth re at the times t = ±1800 seconds. (It takes half an hour to reach the apogee, and another half hour to fall back to Earth). We need to determine the value of θ corresponding to the beginning and ending of the trip.

 

From the radial parametric equation we have

 

 

Solving this for r0 and substituting into the temporal parametric equation gives

 

 

The mass of the Earth in geometric units is m = 0.00443 meters, the radius of the Earth is re = 6378000 meters, and the value of t* in geometric units is (5.4)1011 meters (which is half an hour times the speed of light). Inserting these values, we have

 

 

which can be solved numerically to give θ* = 1.494. The radial parametric equation then gives the radial position of the apogee r0 = 11850000 meters. (Since this is nearly twice the radius of the Earth, the acceleration of gravity at that height is only 1/4 as great as at the Earth’s surface, which confirms that we cannot assume constant gravity.) A plot of the relevant portion of the cycloid is shown below.

 

 

Also from the ratio of dr/dθ and dt/dθ the velocity is given by

 

 

and so the velocity of the projectile at the Earth’s surface is 7596 meters/second, which is about 68% of escape velocity. These parametric equations represent the “program of time and speed so that you get the maximum time on your clock”. To give the speed as an explicit function of radial position, we can solve the radial parametric equation for θ and substitute into the speed equation, giving the familiar result

 

 

As expected, this gives dr/dt = 0 at the apogee r = r0, and it gives dr/dt = 7596 meters/second at the Earth’s surface r = re.

 

Since we’ve been working only to the lowest order of approximation, it was permissible to evaluate the trajectory for a one-hour lapse of proper time for the ballistic object, because the difference between that and the proper time for a stationary clock on the Earth’s surface is extremely small. To evaluate this difference, we can use the exact expression for the elapsed coordinate time derived in the note on Radial Paths, and then multiply this by dτ/dt (also derived in that note) for the Earth’s surface to give the trip time as measured by the ground clock. For a round trip that takes 3600 seconds (1 hour) on the ground clock, the elapsed proper time on the ballistic clock exceeds this by 1.772 μsec, and the elapsed coordinate time exceeds the ground clock by 3.655 μsec.

 

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