Turtles all the way down

John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
This page at http://www.pitt.edu/~jdnorton/Goodies

1. Introduction

The wikipedia article gives an interesting survey of this simple, persistent idea.
The article quotes this engaging version of the story from a 1967 dissertation.

In a simple cosmology, the earth is supported by a turtle; that turtle is supported by another; and so on "all the way down." The most prudent reaction is that this is a charming piece of folklore not to be examined too closely, either historically or physically. Here I will cast prudence aside and ask whether it could make sense physically for the world to be supported by an infinite tower of turtles.

The question started as a moment of musing. Does this infinite tower work if we imagine it in the simplest physical setting: a homogeneous, Newtonian gravitational field. Will the tower stand or fall? That is, will the tower in its infinite entirely just fall downward in the gravitational field, just as would any unsupported object? Or does the infinity of the tower protect it from this fate, so that it can serve to hold up the world? We seemed to have a well-posed question in Newtonian gravitation theory. We would expect a simple and quick answer.

The moment of musing over this puzzle extended into moments as I found that things are not quite so simple. There are apparently quite cogent arguments for both possibilities, stand or fall. It took me a while to see past the problems. I found the exercise quite entertaining. This piece is written for anyone who might like a brief diversion into an innocent puzzle that proves a little messier than you may expect.

2. A Preliminary Kinematical Puzzle

The chief concern here will be whether an infinite tower stands or falls when we take proper account of the Newtonian forces at play. As a preliminary, independently of any issue of physical forces, we might expect that the tower of turtles cannot fall for a reason of kinematics:

There’s nothing for the tower to fall into!

We might reason as follows. For a body to fall, there must be empty space under it, into which it can fall. But there is no empty space under the infinite tower of turtles. So there is nowhere for it to fall. Thus it cannot fall.

Another way to put the problem is that, for the tower to fall, it can only fall into its own space. But its own space is already fully occupied by turtles.

The solution to this puzzle will be familiar to anyone who has seen the Hilbert Hotel. Having no place into which it can fall would be a problem for a body of finite volume. The turtle tower, however, has infinite volume. So it can fall in such a way that, during the fall, it occupies only a part of the volume of space initially occupied.

We can see this using a “Hilbert hotel” construction. In the first unit of time of fall:

Turtle 1 moves to where turtle 2 was.
Turtle 2 moves to where turtle 3 was.
Turtle 3 moves to where turtle 4 was.
Turtle 4 moves to where turtle 5 was.

and so on for all infinity of the turtles.

This solves the kinematical problem. An infinite tower of turtles can fall within the space it occupies.

3. The Puzzle: Arguments For and Against

If we take into account the forces acting on the turtles and the forces between them, will the tower of turtles stand or fall? There are arguments for and against; and they seem equally cogent.

3.1 The Balanced Force Argument For Standing

Consider any one turtle in the tower. The weight of the turtles above presses down on it and, in addition, a gravitational force pulls it downward. However it sits on the back of another turtle and that turtle exerts a reaction force on the first turtle that exactly cancels these downward forces.

It is like a brick in a wall. The brick carries the weight of the bricks above it, as well as the force of gravity acting on it. Those forces are balanced exactly by the reaction force from the brick beneath it.

Thus the turtle under consideration stands and does not fall. This turtle, however, can be any of the infinitely many turtles in the tower. That means that all the turtles in the tower stand. Since the behavior of the tower as a whole is just the sum of the behaviors of all the individual turtles in the tower, the tower stands.

If this argument seems too hasty, note how it fails for a finite tower of turtles. The reasoning above holds for every turtle except the lowest. If that turtle is unsupported, then there is no reaction force holding it and it must fall. It is, as a result, unable to exert a reaction force on the  turtle above it, which also must fall. We proceed up the tower to conclude that all the turtles and the world must fall.

The balanced forces argument depends essentially on their being an infinity of turtles, so that there is no, lowest unsupported turtle. Every turtle is supported by one beneath it.

3.2 The Unbalanced Force Argument for Falling

Consider the tower as a single huge object. It sits in a gravitational field that exerts an external downward force on it. There is no compensating external force to counteract it. Hence the tower must fall.

A slightly more sophisticated version applies Newton's

Force = mass x acceleration

(1)      F = ma

There is a non-zero, net downward force acting on the tower. Through F = ma, the tower is accelerated downward.

A complication gives us a temporary respite. If the tower consists of infinitely many, equally massive turtles then the mass m in F=ma is infinite. Since the gravitational force is proportional to the mass, the gravitational force is infinite as well. Thus F=ma becomes:

Infinite force = infinite mass x a

This equation leaves the acceleration a undetermined. It could be any value: zero, finite or infinite.

That the respite is temporary follows if we consider a fanciful elaboration of the tower of turtles. What if masses m1m2, m3, ... diminish so that their sum is finite? Call that sum Mt

(2)     Mt = m1 + m2 + m3 +  ...

For example, we might have m1 = 1,  m2 =1/2,  m3 = 1/4 ..., for which Mt = 2. Then the Newtonian F=ma becomes

(3)     F = Mtg = Mta

where the constant g is the acceleration imparted by the homogeneous gravitational field. Solving for a we have that the tower accelerates downward with acceleration g. That is, a = g.

3.3 The Limit Argument for Falling

A standard method for recovering the behavior of an infinite system is to consider the behavior of finite systems and then take a limit as the number of components becomes infinitely large.

To this end, consider a tower of finitely many -- N -- turtles. For any value of N, we have a tower unsupported in the gravitational field. It will fall with acceleration g.

If we now take the limit as N goes to infinity, we arrive at an infinite tower falling with acceleration g.

3.4 The Limit Argument for Standing

There is another way to approach the this same limit of an infinite tower. Consider a tower of finitely many -- N -- turtles. This time, the lowest, Nth turtle is supported by a powerful rocket motor. Its thrust is enough to hold up the entire weight of the tower. The finite tower stands.

Now take the limit as the number of turtles N goes to infinity. We once again end up with an infinite tower, but this time the tower stands.

What of the rocket motor? It is tempting to imagine that the tower is held up by the rocket motor, which now lives on "at infinity." But the place "at infinity" is a fiction. There is no such place. The rocket motor supports the lowest turtle in the tower. In the tower of the infinite limit, there is no lowest turtle. Every turtle in the tower can be assigned a number according to its position in the tower. Pick any number N? Is that the turtle supported directly by the rocket motor? No! There is another turtle, number N+1, beneath it.

In the limit, there is no rocket motor. There is just an infinite tower of turtles that stands.

Now What?

We have a classic paradox. We have pairs of arguments, each apparently quite cogent, but arriving at conclusions that contradict. Two conclude that the tower falls. Two conclude that the tower stands.

The fun of the paradox is seeing through the logic of the arguments. Some or possibly all of the arguments are flawed. Which are they?

Here I urge the reader to take a short break and ponder. I am giving my solution below. But isn't it more fun to figure out your own solutions first?


4. Solutions

The two limit arguments can be dispatched quickly. The false presumption is that either limit process is a reliable way of inferring the behavior of an infinite tower. That is already evident from the mere fact that we have two ways of approaching the same limit of an infinite tower of turtles, but the two ways ascribe different properties to the same infinite tower.

The use of limiting procedures to infer the properties of infinite systems is delicate and sometimes fraught. Their use can  give spurious results. For some further discussion, see my "Approximation and Idealization..."

More subtlety is needed to resolve the contradiction produced by the two arguments pertaining the the balance or imbalance of forces. The resolution depends on a disanalogy between the case of a finite tower and an infinite tower.

In the case of a finite tower, we determine its future behavior by stipulating certain of its properties at present: the positions of the objects and their initial velocities. Those same properties are not sufficient to determine the future behavior of the infinite tower. Whether it stands or falls requires specification of further properties: the inter-object forces at the initial moment of time.

Specifying these forces one way gives a tower that falls, as the unbalanced forces argument predicts. Specifying them another way gives a tower that stands, as the balanced forces argument predicts.

In short, both balanced and unbalanced forces arguments succeed, but only because they apply to different cases. Further specification of the initial conditions allow to decide which case we have at hand.

Here is a more sustained development of this resolution.

4.1 Initial conditions for a finite tower

The initial conditions required for a finite tower are

(i) the positions of all the component turtles and world;
(ii) their initial velocities (which are all zero).

In one case, the components include an immovable base at rest. What results is a finite tower of turtles, resting on the base. It remains so and does not fall.

In another case, the components do not include the immovable base. The lowest turtle is unsupported and falls. With it, all the remaining turtles fall.

4.2 Initial conditions for an infinite tower

The initial conditions required for an infinite tower are

(i) the positions of all the component turtles and world;
(ii) their initial velocities (which are all zero);
(iii) the initial inter-turtle and turtle-world forces.

The extra condition (iii) requires specification of the reaction forces exerted between neighboring turtles and on the world.

Their specification is not needed in the finite case. Whether reaction forces are there or not follows, for the finite tower, from the initial conditions (i) and (ii).

If there is an immovable base, gravity cannot accelerate the lowest turtle through the base, so the lowest turtle exerts a force on it. Because of its immovability the base exerts an equal but opposite reaction force on the lowest turtle. This lowest turtle is immobile. This reasoning propagates up the tower. Each turtle exerts a loading force on the one beneath. That turtle exerts an equal but opposite reaction force on the one above. All turtles of the tower are immobile.

If there is no immovable base, the lowest turtle enters into free fall and exerts no reaction force on the turtle above it. This next turtle enters into free fall; and so on up the tower.

However, for the case of an infinite tower, initial condition (i) and (ii) alone are insufficient to determine the future behavior of the tower. We must posit in (iii) whether there are loading and reaction forces between the turtles. If we posit their existence, then the initial conditions lead to a tower that stands. If we posit none, we have a tower that falls.

5. Reaction forces made explicit

Since the reaction forces can grow arbitrarily large, it is simplest to conceive of springs that do not obey Hooke's law, but instead a non-linear law, such that the force grows rapidly with the compression of the spring.

A further complication is that systems consisting of an infinite chain of mass-spring-mass-spring-... are generically indeterministic. The system can spontaneously excite through disturbances that, loosely speaking, propagate in from infinity. See Approximation and Idealization... (Appendix). We can preclude them with a fourth condition:

(iv) The positions of the turtles relative to each other remain fixed over time.

A way see how this last conclusion works, is to make the presence or absence of loading and reaction forces in the infinite tower visible by connecting the turtles by springs.

A pair of loading and reaction forces between one turtle and the one above will manifest as a compression of the spring connecting them. If there is no force, then the spring is unextended and uncompressed. It is at its equilibrium length in which in exerts no force. If there is a  force, then the spring is compressed.

How we set the initial condition in (iii) will then determine whether the infinite tower of turtles will stand or fall:

• If we initialize the infinite tower so that each turtle is supported by a reaction force from the turtle below it, the tower will stand. (Shown on the right.) Those supporting reaction forces will persist unchanged over time.

• If we initialize the infinite tower so that there are no reactions forces supporting each turtle, the tower will fall. (Shown on the left.) While it falls, the connecting springs will remain in their zero-force uncompressed state. The tower will continue its free fall.

5.1 The governing equations

We can give a more substantive analysis in Newtonian physics of these two cases.

Sign conventions: the x axis increases downward in the direction of the field. Positive forces are directed downward.

Let the mass of the world be m0 and the masses of the turtles m1m2, m3, ... as before. The world and turtles are located at positions x0, x1, x2, x3, ... The net force acting on the ith object is

(4)      f0 = m0a0 = m0g + f01
= miai = mig + fi,i-1 + fi,i+1 for i>0

where fik is the force exerted on the ith object by the kth object. (The only cases we will consider are those in which i and k are one number separated, so that k=i-1 or k=i+1.) The acceleration of each object is ai.

The weight of all the objects above the ith object is transmitted to it as  fi,i-1 which is the force with which the i-1th object bears down on the ith object. The ith object exerts a reaction force back on the i-1th object of fi-1,i

Newton's third law requires that the force with which the ith object acts on the kth object is equal but opposite in sign to the force with which the kth object acts on the ith object. That is:

(5)     fik = -fki

5.2 The tower stands

The case of the tower standing arises when all the forces vanish. Then the accelerations must also vanish:

(6)     0 = f0 = f1 = f2 = f3 = ...      0 = a0 = a1 = a2 = a3 = ...

We can find the forces fik that realize this case by solving equations (4), (5) and (6) iteratively:

(7)     -f01 = f10 = m0g
         -f12 = f21 = m1g + f10 = m1g + m0g
         -f23 = f32 = m2g + f21 = m2g + m1g + m0g
         -fi-1,i, = fi,i-1 = mi-1 + ... + m2g + m1g + m0g

If the tower is initialized with these inter-object forces, the springs will all be compressed by just the amount needed to sustain these forces. Since the net force on each object is zero, the objects do not accelerate. They remain separated by the same distances, so that the springs retain their compression and the forces of (7) remain unchanged through time.

The tower stands, supported by the forces in (7).

5.3 The tower falls

The case of the tower falling arises when there are no inter-object forces. That is, the springs connecting the objects are in their uncompressed states. Then we have

(8)     fik = -fki = 0

Equations (4) are then easily solved to give

(9)     ai = g

All the objects fall with the same acceleration g. It follows that the distance between them remains the same, so no inter-object forces arise during the fall.

6. A center of mass theorem

The unbalanced force argument of Section 3.2 above assumed that what determines the overall motion of the tower of turtles is the net external force acting on the tower. A non-zero net external force, as is the force of gravity, must accelerate it.

This assumption cannot be made without recalling the conditions that lead to it. It is derived as a theorem in elementary mechanics from the summation of all the forces acting on the component objects of the system. The theorem provides a summary description of the combination of all those forces. Under ordinary conditions, the summation leads to the familiar result that we can ignore all the inter-object forces, when we consider the system as a whole. The effect of all external forces is to impart an acceleration to the system's center of mass that is proportional to its total mass.

The theorem is easy to derive. Here is the derivation for the special case of the tower of objects, that is, the world plus its supporting turtles.

6.1 A center of mass theorem for a finite tower

For a finite tower with N turtles, equations of motion (4) must be replaced by

(10)      f0 = m0a0 = m0g + f01
= miai = mig + fi,i-1 + fi,i+1 for 0 < i < N
            fN = mNaN = mNg + fN,N-1

These equations for the motion of each individual object then determine an equation governing the center of mass of the whole tower. To see it, sum the equations of (10) to find the total, net force F acting:

          F = f0 + f1 + f2 +... + fN = m0a0 + m1a1 + m2a2 + ... + mNaN
= (m0 + m1 + m2 +... + mN)g
+ f01 + (f10 + f12) + (f21 + f23) + (f32 + f34) +
                                                               ... + (fN-1,N-2 + fN-1,N) + fN,N-1

All that remains is to simplify this expression. The masses sum to M:

          M = m0 + m1 + m2 +... + mN

The acceleration is written as a second derivative

          ai = (d2/dt2) xi

in time t. Then we can rewrite

         m0a0 + m1a1 + m2a2 + ... + mNaN
(d2/dt2) (m0x0 + m1x1 + m2x2 + ... + mNxN)
               = M (d2/dt2) (m0x0 + m1x1 + m2x2 + ... + mNxN)/M
               = M (d2/dt2) X

where the center of mass of the tower is defined as

         X = (m0x0 + m1x1 + m2x2 + ... + mNxN)/M

If we redistribute the parentheses in the term that sums inter-object forces and apply (5), we find that this term vanishes

(11)     f01 + (f10 + f12) + (f21 + f23) + (f32 + f34)+ ... + (fN-1,N-2 + fN-1,N) + fN,N-1
           = (f01 + f10) + (f12 + f21) + (f23 + f32) + ... + (fN-1,N + fN,N-1)
           = 0 + 0 + 0 + ... + 0 = 0

This vanishing is as expected. It tells us that the inter-obect forces contribute nothing to the net force acting on the tower as a whole.

Combining, we recover the final expression

(12)     F = M (d2/dt2) X = Mg

That is, the net force F acting on the tower is just the external gravitational force Mg and its effect is an acceleration of the center of mass A = (d2/dt2) X that conforms with F = MA.

6.2 A center of mass theorem for an infinite tower

Equation (12) shows us that we can ignore the inter-object forces in a finite tower. We can treat the tower as a single object that is acted on only by external gravitational forces and whose center of mass is accelerated accordingly.

The unbalanced forces argument of Section 3.2 above assumed that we can treat the infinite tower in the same way. For that to be so, we must have a center of mass theorem for an infinite tower analogous to the one just demonstrated for a finite tower.

As we shall now see, it turns out that we do not have a corresponding center of mass theorem that vindicates the unbalanced forces argument. We will see that the applicable theorem can sustain a tower that falls or one that stands, according to the assumptions we make about the inter-object forces.

To find the corresponding theorem, we proceed as in Section 6.1 by summing all the forces in (4) acting on the individual objects in the tower. We find:

          F = f0 + f1 + f2 +...  = m0a0 + m1a1 + m2a2 + ...
= (m0 + m1 + m2 +... )g
+ f01 + (f10 + f12) + (f21 + f23) + (f32 + f34) + ...

To continue we would want to introduce a center of mass

         X = (m0x0 + m1x1 + m2x2 + ... )/M

where the total mass M is

           M = m0 + m1 + m2 + ...

However if this last mass M is infinite and the summation of terms in mixi diverges, then the center of mass X will be undefined. In this generic case, a center of mass theorem cannot be recovered. The reasoning of the unbalanced forces argument of Section 3.2 is blocked.

However, we can take the expedient of Section 3.2 and assume that the total mass M is finite. In addition we can assume that the positions xi do not grow too fast with i so that the sum m0x0 + m1x1 + m2x2 + ... converges. Then the center of mass X is defined and a center of mass theorem is recoverable. (See here for computation of specific cases.)

With these assumptions, the analysis proceeds as in Section 6.1 and we arrive at

(13)     F = M (d2/dt2) X
               = Mg  + f01 + (f10 + f12) + (f21 + f23) + (f32 + f34) + ...

The simplest case is that of no inter-object forces in which all fik = 0. Then (13) reduces to

          (d2/dt2) X = g

and the center of mass of the tower falls with acceleration g.

The more difficult case is that of non-vanishing inter-object forces. For then the partial sums of the infinite sum of inter-object forces in (13)

          f01 + f10 + f12 + f21 + f23 + f32 + f34 + ...

are each a sum of alternative positive and negative values, such that the infinite sum does not converge uniformly.

We can escape this lack of uniform convergence by noting that the physics of the problem prescribes a particular grouping of terms in the summation that does lead to convergence.

To see it, let FN be the sum of the forces acting on the first N objects

          FN = f0 + f1 + f2 +... + fN

Correspondingly, MN is the sum of the first N masses

          MN = m0 + m1 + m2 +... + mN

and XN is the resulting center of mass

           XN = (m0x0 + m1x1 + m2x2 + ... + mNxN)/MN

Replicating the inferences above we arrive at an expression for FN that is almost exactly the same as that of the finite case:

(14)     FN = MN (d2/dt2) XN
               = MN + f01 + (f10 + f12) + (f21 + f23) + (f32 + f34) +
                                               ... + (fN-1,N-2 + fN-1,N) + (fN,N-1 + fN,N+1)

The crucial different is the presence of one additional term in the summation of inter-object forces, which is indicated in enlarged text: + fN,N+1). This is the reaction force acting on the Nth object from the (N+1)th object. This term does not appear in the analysis of the finite tower, since there is no N+1th turtle.

If we proceed as before and regroup the terms in the summation of intermolecular forces, we recover for this sum:

       f01 + (f10 + f12) + (f21 + f23) + (f32 + f34) +
                              ... + (fN-1,N-2 + fN-1,N) + (fN,N-1 + fN,N+1)
      = (f01 + f10) + (f12 + f21) + (f23 + f32) + (f34 +
                              ... + fN-1,N-2) + (fN-1,N + fN,N-1) + fN,N+1
      = 0 + 0 + 0 + ... + 0 + fN,N+1 = fN,N+1

We have from (7) that

       fN,N+1 = -(m0 + m1 + m2 +... + mN)g = -MNg

Equation (14) reduces to the simpler:

(15)     FN = MN (d2/dt2) XN = MNg - MNg = 0

Might one object to the use of this infinite limit, recalling how limits can lead to unwanted problems? If one rejects this limit, then the net force on the entirety of the infinite tower is undefined. That means that we cannot set up a center of mass theorem in the first place and we must set aside concerns associated with it.

To conclude, we define the total force F on the tower to be the limit:

            F = Lim (N → ∞) FN = 0  

Since it vanishes then so must

           (d2/dt2) X = 0.

The center of mass X is unaccelerated.

That is, when the inter-object forces are given by (7) for a standing tower, the applicable version of the center of mass theorem tells us that there is no net force on the total system and its center of mass X remains unaccelerated.

This result comes about because of a failure of a familiar intuition concerning inter-object forces. We naturally assume that they all cancel out and leave no residual force to act on the tower as a whole. That assumption fails because of the infinity of the number of objects. For any finite part of the tower, there is always a residual force, the reaction force exerted on the Nth object by the N+1th object,  fN,N+1 = -MNg . This residual force survives in the limit as the number of turtles becomes infinite. It provides exactly the force -Mg needed to counteract the external forces +Mg of gravity.

7. Conclusion

What this analysis concludes is that the usual presentation of the problem of the infinite tower of turtles leaves the state of the tower incompletely described. This incompleteness is what enables there to be apparently cogent arguments for contradictory conclusions. If we want to know whether the tower will stand or fall, we need more information than the initial positions of the world and each turtle and their initial velocities. We need also to know what forces are acting pairwise between the world and all the turtles. One specification leads to a tower that falls. Another specification leads to a tower that stands.

The longest and technically messiest part of the analysis is in Section 6. It deals with the argument that the tower must fall since, as a totality, it is a body acted on by a gravitational force, but without any external force to counteract it.

This unbalanced forces argument depends on the assumption that one can neglect all the inter-object forces when one analyses the tower as a whole. That assumption is supported by a center of mass theorem formulated to apply to finite towers only.

We can develop a version of the center of mass theorem that applies to infinite towers. Its scope is limited since, in generic cases, its quantities diverge so no result is derivable. If we contrive circumstances in which these divergences are avoided, then the infinity of turtles leads to a curious result. The totality of forces acting between the towers do not cancel. There is a residual that balances the external gravitational forces exactly and allows the tower to stand.

One should not become too comfortable with this last result. If we allow the sizes of the object to shrink so that the entirety fits into a finite volume of space, this same mechanism will allow an infinity of objects to sustain themselves in mid-air by (metaphorically) "pulling on their own bootstraps." It also provides the architectural design for the foundations of a castle that can float in the air.

John D. Norton, April 18, 2018