This page contains some extra mathematical details to supplement the main page
on this topic. In everything that follows, we will give the three dimensions of space the coordinates *x*, *y* and *u*, where *x* and *y*
are ordinary, “space-like” dimensions, and *u* is a “time-like” dimension.
This means that we need to use a metric *g* which defines the dot product between vectors as:

v·w=g(v,w) =v_{x}w_{x}+v_{y}w_{y}–v_{u}w_{u}

and gives the squared magnitude of a vector as:

v·v=g(v,v) =v_{x}^{2}+v_{y}^{2}–v_{u}^{2}

We will choose the *y* coordinate to correspond to the up/down direction defined by gravity, and the objects
we model will be assumed to move only in the *yu*-plane.

The analysis below assumes that everything is moving slowly enough that we can use **non-relativistic** physics, so the kinetic energy
of an object of mass *m* and velocity *v* is just *K* = ½ *m* *v* · *v*. So when we talk
about forces and velocities “going to infinity”, we are not claiming that the detailed formulas we develop will continue to apply
for arbitrarily high velocities. But when the forces increase dramatically like this, most actual materials would fail long before relativistic corrections to any of the calculations would make any difference.

Consider a thin, initially vertical rod of length *L* and total mass *m*, **pivoting so that its lowest point remains
fixed**. If the hyperbolic angle of its tilt is given by λ(*t*), we can describe the
location of a point on the rod at a distance of *s* from the bottom as:

R(s,t) =s(sinh(λ(t))e_{u}+ cosh(λ(t))e_{y})

where *e*_{u} and *e*_{y} are unit vectors in the *u*- and *y*-coordinate directions,
and we have chosen to fix the bottom of the rod at the origin of our coordinate system.

The total potential energy of the rod can be found by integrating the product of the mass, the height, and the gravitational acceleration
*g* along the length of the rod, or equivalently, just treating all the mass as being concentrated at the centre of mass, *s*=½*L*. So we have:

U(t) = ½Lmgcosh(λ(t))

The velocity of any point on the rod can be found by taking the time derivative of *R*(*s*, *t*), which gives us:

v(s,t) =sλ'(t) (cosh(λ(t))e_{u}+ sinh(λ(t))e_{y})

v(s,t) ·v(s,t) = –s^{2}(λ'(t))^{2}

The total kinetic energy is obtained by integrating ½ (*m*/*L*) *v*(*s*, *t*) · *v*(*s*, *t*) d*s* along the length of the rod, yielding:

K(t) = –(1/6)L^{2}m(λ'(t))^{2}

So we have the total energy of the rod, kinetic plus potential:

E(t) = ½Lmgcosh(λ(t)) – (1/6)L^{2}m(λ'(t))^{2}

and this is equal to a constant that will depend on the initial angular velocity of the rod, λ'(0), when it is perfectly vertical, at *t*=0, λ(0)=0:

E(0) = ½Lmg– (1/6)L^{2}mλ'(0)^{2}

We can solve the conservation of energy equation, *E*(*t*) = *E*(0), to find:

λ'(t) = √[λ'(0)^{2}+ 3 (g/L) (cosh(λ(t)) – 1)]

which also gives us:

dt/dλ = 1 / √[λ'(0)^{2}+ 3 (g/L) (cosh(λ) – 1)]

This can be integrated to give a closed-form solution in terms of incomplete elliptical integrals of the first kind:

t(λ) = (–2i/ λ'(0))F(iλ/2 | 6g/[Lλ'(0)^{2}])

where we use the convention that the second argument in *F*, after the vertical bar, is the *parameter* of the elliptic integral (which is the square of the *modulus*). Here *i* is the square root of minus one. Obviously the value of *t*(λ) must be
real; the value of *F* with an imaginary argument is itself imaginary, so multiplying that by *i* gives a real result.

So, we can calculate *t*(λ) for any λ, and with numerical methods we can work backwards and calculate λ(*t*)
for any *t*.

If we take the time derivative of *E*(*t*) = *E*(0), we obtain the simple result (assuming λ'(*t*)≠0):

λ''(t) = (3/2) (g/L) sinh(λ(t))

We can then compute the acceleration of the centre of mass of the rod, by taking the time derivative of *v*(½*L*, *t*)
and substituting the formulas we have for λ'(*t*) and λ''(*t*). Multiplying that acceleration by the mass of the
rod, and subtracting the fixed weight vector, we obtain the force that must be provided by the ground at the pivot point in order for the
centre of mass to accelerate this way:

Ground force= (m/4) ( sinh(λ(t)) (g(9 cosh(λ(t)) – 6) + 2Lλ'(0)^{2})e_{u}+ (g(1 – 3 cosh(λ(t)))^{2}+ 2Lcosh(λ(t)) λ'(0)^{2})e_{y})

As the rod topples, λ(*t*) will increase without bound, and both components of the ground force will also increase without bound.

Suppose that we again have a thin, initially vertical rod of length *L* and total mass *m*, but now **let the bottom of the rod slide freely over the ground**. If the ground is entirely frictionless, the force it exerts on the rod will be directed entirely
vertically, and the weight of the rod will also act vertically. So if the centre of mass of the rod is initially at rest, its horizontal
*u* coordinate will remain fixed, and we can choose coordinates so it lies at *u*=0. We can give the rod an initial angular velocity
that sets it toppling, without giving its centre of mass any horizontal velocity, by applying equal and opposite horizontal pushes to the top and
bottom of the rod.

If the hyperbolic angle of the tilt is given by λ(*t*), we can describe the
location of a point on the rod at a distance of *s* from the bottom as:

R(s,t) = (s– ½L) sinh(λ(t))e_{u}+scosh(λ(t))e_{y}

The total potential energy of the rod is exactly the same as before, as a function of λ(*t*):

U(t) = ½Lmgcosh(λ(t))

The velocity of any point on the rod is:

v(s,t) = λ'(t) ((s– ½L) cosh(λ(t))e_{u}+ssinh(λ(t))e_{y})

v(s,t) ·v(s,t) = –(s^{2}+ ¼L(L– 4s) cosh(λ(t))^{2}) (λ'(t))^{2}

The kinetic energy is obtained by integrating ½ (*m*/*L*) *v*(*s*, *t*) · *v*(*s*, *t*) d*s* along the length of the rod, yielding:

K(t) = –(1/24)L^{2}m(λ'(t))^{2}(4 – 3 cosh(λ(t))^{2})

The last factor in the kinetic energy will be positive for low values of λ(*t*), but will cross zero when λ(*t*) reaches arccosh(2/√3) ≈ 0.549306.
What this means is that, as λ(*t*) approaches that critical angle, the kinetic energy becomes less and less sensitive to
λ'(*t*), and the angular velocity will need to increase without bound just to maintain a constant value of the total energy.

The total energy of the rod, kinetic plus potential, is:

E(t) = ½Lmgcosh(λ(t)) – (1/24)L^{2}m(λ'(t))^{2}(4 – 3 cosh(λ(t))^{2})

and this is equal to a constant that will depend on the initial angular velocity of the rod, λ'(0), when it is perfectly vertical, at *t*=0, λ(0)=0:

E(0) = ½Lmg– (1/24)L^{2}mλ'(0)^{2}

We can solve the conservation of energy equation, *E*(*t*) = *E*(0), to find:

λ'(t) = √[λ'(0)^{2}+ 12 (g/L) (cosh(λ(t)) – 1)] / √[4 – 3 cosh(λ(t))^{2}]

This gives us d*t*/dλ, but unlike the result for the pivoting rod, this doesn’t seem to give a closed-form
solution for *t*(λ). Nevertheless, we can use numerical methods to find both *t*(λ) and λ(*t*).

As before, we can compute the acceleration of the centre of mass of the rod, by taking the time derivative of *v*(½*L*, *t*). Multiplying that acceleration by the mass of the
rod, and subtracting the fixed weight vector, we obtain the force that must be provided by the ground in order for the
centre of mass to accelerate this way:

Ground force=

m[2g(4 + 3 cosh(λ(t)) (cosh(λ(t)) – 2)) +Lcosh(λ(t)) λ'(0)^{2}] /

[2 (4 – 3 cosh(λ(t))^{2})^{2}]e_{y}

As the rod approaches λ(*t*) = arccosh(2/√3) ≈ 0.549306, this will increase without bound.

The **image on the right** shows the rod at a sequence of positions separated by equal time intervals, with the last one placing the rod at the critical
angle when the forces become infinite. The force vectors are drawn for the previous location in the sequence, when they would still be finite.

If we replace our idealised thin rod with a rectangular prism, we can model it as a number of distinct segments, and compute the shear forces that arise between these segments. Eventually, these forces will become great enough to break the material of the rod.

The calculations for a pivoting rectangular prism proceed much as they did for a pivoting rod, with the total energy given by:

E(t) = ½mg(Lcosh(λ(t)) +wsinh(λ(t))) – (1/6) (L^{2}–w^{2})m(λ'(t))^{2}

where *w* is the width of the prism in the *u* direction. We then end up with a slightly more complex expression for the angular velocity:

λ'(t) = √[λ'(0)^{2}+ 3 (g/(L^{2}–w^{2})) (L(cosh(λ(t)) – 1) +wsinh(λ(t)))]

Integrating the reciprocal of this to
find time as a function of λ does still yield a closed-form expression, but it is substantially
more complicated than the one for the rod, and involves Appell hypergeometric functions. In any case,
we can express all the accelerations and forces as functions of λ alone, independently of
the precise time progression given by knowing λ as a function of *t*.

The acceleration of an arbitrary point in the prism can be determined in the usual way, making use of the formula above and its derivative, so we can find the acceleration of the centre of mass of each segment, and the contact forces needed to produce that part of the acceleration not accounted for by gravity alone.

These contact forces can be resolved into normal forces and shear forces; the **image on the right**
illustrates the size of the shear forces, which in this case peak at about 2/5 of the way up the prism.
As the prism tilts,
it will eventually shear apart, leaving the separate pieces to repeat the process until they end up with a small
enough height that they are stable against further toppling.

In more detail: if we calculate the magnitude of the shear force across a planar boundary that lies a fraction μ along the way from the bottom of the prism to the top, this turns out to be a quadratic in μ. The value of μ where the shear is at a maximum is given by:

μ_{max}= [(L^{2}–w^{2}) (5gsinh(λ) – λ'(0)^{2}w) + 3gLw] / [3gL(Lsinh(λ) +wcosh(λ))] – 1

= [(1 – α^{2}) (5 sinh(λ) – λ'(0)^{2}w/g) + 3 α] / [3 (sinh(λ) + α cosh(λ))] – 1

where we define α=*w*/*L*. For λ=0 this is actually negative (so the maximum shear will be at μ=0 for small λ), while for
large λ there is a simple asymptotic formula:

lim_{λ→∞}(μ_{max}) = (2 – 5 α) / 3

If the prism is initially tall and skinny, when it first shears in two this will
happen at a value of μ that is not too low; for example, for α = *w*/*L* = 1/5
and a large value of λ, it will be at μ≈1/3. But as the pieces grow shorter and α increases, the maximum shear will occur closer to
the bottom, and eventually the ground will be scraping off thin layers of material. This will
continue until the remainder of the prism is short enough that the process ceases.

The **plot on the left** gives μ_{max} as a function of α, for a
selection of values of λ, in the limit where λ'(0) is small and so we can discard the
term λ'(0)^{2} *w*/*g*.