I recently came across an example of spontaneous symmetry breaking in a context that was new to me and a priori, unexpected. It is the situation of a loop of a perfectly flexible, frictionless, narrow, and inextensible string/cable of uniform weight density looped around and across two horizontally separated narrow pegs/nails/pins, such that the string/cable can freely slide over and around each peg without any friction or binding. In order for the loop to fit over and around both of the pegs the circumference of the cable loop must be at least twice the separation distance between the pegs or else it couldn't reach between them. Since the cable is inextensible we are not allowed to stretch it over the pegs if it is too short. The limiting case of the loop's circumference being exactly just twice the peg separation means there will be no sagging of the loop when it is installed in that limiting case. But in that limiting case the tension on the loop will be unrealizably infinite because it is so tight so as to prevent any sagging of the cable in between the pegs.
When the loop is longer than this minimal limiting case both segments of the loop will sag somewhat between the pegs. One of these segments will come off of the underside of the pegs and the other one will come off of the upper sides of the pegs. But since we are neglecting the thickness/diameter of the pegs we consider both cable segments as having concurrent common endpoints at the pegs. Now it is a standard statics problem to determine that the shape of each of those segments between the pegs will be a catenary (i.e. a hyperbolic cosine function shape) when hanging in equilibrium. *If* the peg-to-cable contacts at the ends were not frictionless, but sufficiently sticky or binding, we could adjust the lengths of the two segments any way we wished by making one segment longer and the other one shorter, as desired, as long as the sum of the lengths of both catenary segments adds up to the circumference of the loop, and as long as the shortest segment is at least as long as the inter-peg separation. We could partition the total excess slack in the loop between the two segments any way we choose, and each separate segment will have its own catenary shape depending on how much slack is given to each segment. The segment with the longer length will droop and sag more than the one with less of the slack. For both segments the tension is maximal at the endpoints and is minimal in the center of each one's sag half way between the pegs where the cable is locally horizontal. The shorter segment has a greater central minimal tension. Locally along each catenary segment the tension on an infinitesimal piece of the cable is proportional to the secant of the tilt angle of the infinitesimal piece. Thus as the cable segment dips into its sag its local tension there also dips because the tilt angle is zero in the middle and becomes greater as the slope increases as the ends are approached in either direction from the middle.
Now imagine removing all frictional and binding forces at the pegs. This means when the system is in equilibrium the tension at each segment catenary's endpoints must be the same as for the other segment's catenary. Otherwise, the higher tension segment will pull on the lower tension segment at the pegs and steal some of other one's extra slack for itself, thus relieving its excess tension, and tightening up the other one. Once everything equilibrates both segments will have then have the same tension at their endpoints. The behavior of each segment's catenary is related to the other segment's by the fixed total loop circumference constraint *and* by the equal-tension-at-the-ends constraint.
Clearly the symmetric situation of both segments having equal length must satisfy both constraints wherein both segments have exactly half of the total extra length, and both catenaries have the same shape, and lie everywhere against each other between the pegs. (Remember they don't exert any frictional forces on each other, even though they are always touching everywhere). Let D be the distance between the two pegs, and let L be the total loop circumference of the cable. This symmetric situation requires that each segment be L/2 long (remember L/2 > D). This symmetric situation is the correct and only stable equilibrium situation if
1 < L/(2*D) < 1.257736456168094 .
But the remarkable and unexpected thing is that when
L/(2*D) > 1.257736456168094
a pair of *asymmetric* different length segments is *also* an equilibrium solution, and this asymmetric solution pair is the stable solution, whereas the symmetric solution is now an *unstable* equilibrium, and the system spontaneously goes *away* from the symmetric case, and settles in on the asymmetric situation with one segment being longer than the other one. But which of the two segments is the longer one and which is the shorter one doesn't matter as both of these unequal outcomes are allowed and the system picks one of them to be the short segment and the other one to be the long one. In this case the equilibrium has one catenary segment being long and droopy while the other one is short and hardly sagging at all. The special case of
L/(2*D) = 1.257736456168094
is a *critical point* for the system and the potential energy for it has a very flat bottomed locally quartic minimum at the symmetric equilibrium point. In the first situation when
1 < L/(2*D) < 1.257736456168094
the potential energy function has a single locally quadratic U-shaped minimum at the symmetric equi-partition of excess length. In the case of asymmetric equilibrium when
L/(2*D) > 1.257736456168094
the potential energy function has a rounded w (or lower case omega) shape with the central maximum being the symmetric equi-partition of the excess length, and each minimum on either side of it being the pair of asymmetric partitions of the excess length. The symmetric potential energy maximum is a frowning locally quadratic maximum, and each of the two asymmetric partition minima are smiling locally quadratic minima of equal depth (and symmetrically placed on either side of the symmetric maximum).
In all these cases we imagine the potential energy as being plotted against the 'order parameter' for the broken symmetry. Although a better name for this parameter, which we call σ here, would be the 'asymmetry parameter'. This parameter, σ, is 0 when both segments have equal length, is 1 when one catenary segment has all of the excess slack, and is -1 when the other segment has all of the excess slack. When the symmetry is broken this asymmetry parameter has a spontaneous nonzero value whose absolute value locates how far the position of the potential energy minimum is on either side of the symmetric maximum. In general, σ has an equilibrium value somewhere in the range -1 < σ < 1. Because the potential energy, and indeed the entire problem, is even in the value of σ, we can, without any loss of generality, just consider σ to be its absolute value, and consider its range to be 0 ≤ σ < 1. We below consider some special cases of the symmetric, critical point, and broken symmetry situations. But before considering them we first define some convenient dimensionless parameters, variables and functions for this problem.
Let Δ ≡ L/(2*D) - 1. Here Δ is the dimensionless *average* excess length of each catenary segment over its minimum possible value in units of the inter-peg separation distance, D. Note the total loop circumference, and L is therefore given by L = 2*D*(Δ + 1).
Let λ_s be the dimensionless excess length of the *shorter* segment catenary in units of the inter-peg distance, D. This means D*(λ_s + 1) is the total length of the shorter catenary segment.
Let λ_l be the dimensionless excess length of the *longer* segment catenary in units of the inter-peg distance, D. This means D*(λ_l + 1) is the total length of the longer catenary segment. Note, these definitions mean Δ = (λ_l + λ_s)/2 because the total lengths of both segments is the loop circumference, L. This is why we call Δ the average (dimensionless) excess length.
Let σ ≡ (λ_l - λ_s)/(λ_l + λ_s) be the asymmetry parameter (i.e. 'order parameter') for the system. With this definition we see λ_l = Δ*(1 + σ) and λ_s = Δ*(1 - σ).
(Only in the symmetric situation do we have Δ = λ_l = λ_s, i.e. when σ = 0.)
Let w ≡ linear weight density of the cable, i.e. the cable's weight per unit length.
Let T_m ≡ common maximum cable tension at the endpoint pegs for both catenary segments.
Let T_s ≡ minimal cable tension at the horizontal center of the *shorter* catenary segment.
Let T_l ≡ minimal cable tension at the horizontal center of the *longer* catenary segment.
Let τ_m ≡ 2*T_m/(w*D) be the dimensionless common maximum endpoint tension in units of the weight (w*D/2) of a piece of cable half as long as the inter-peg distance..
Let τ_s ≡ 2*T_l/(w*D) be the dimensionless minimum center tension of the *shorter* catenary segment in units of the weight (w*D/2) of a piece of cable half as long as the inter-peg distance.
Let τ_l ≡ 2*T_s/(w*D) be the dimensionless minimum center tension of the *longer* catenary segment in units of the weight (w*D/2) of a piece of cable half as long as the inter-peg distance.
Let H_s ≡ central sag depth in the middle of the shorter catenary segment below the level of the pegs.
Let H_l ≡ central sag depth in the middle of the longer catenary segment below the level of the pegs.
Let η_s ≡ 2*H_s/D be the dimensionless central sag of the shorter segment in umits of half-distance between the pegs, D/2.
Let η_l ≡ 2*H_l/D be the dimensionless central sag of the longer segment in umits of half-distance between the pegs, D/2.
Let V ≡ total gravitational potential energy of the system relative to the zero level being the common height of the pegs.
Let p ≡ 4*V/(w*D^2) be the dimensionless potential energy in units of the work (w*(D/2)^2) required to lift a piece of cable whose length is D/2 upward a distance of D/2.
Let α_s ≡ 1/τ_s. Here α_s is a convenient variable pertaining to and parameterizing the shorter segment.
Let α_l ≡ 1/τ_l. Here α_l is a convenient variable pertaining to and parameterizing the longer segment.
We also define a couple of convenient functions, δ() & arcδ(). In particular, we define
δ(α)≡ sinh(α)/α - 1 &
arcδ(λ) ≡ inverse of δ(), i.e. α = arcδ(λ) when λ = δ(α). Thus λ ≡ δ(arcδ(λ)) & α ≡ arcδ(δ(α)).
Using these definitions we have the following relationships between these quantities:
Here are some special cases for various Δ values below, at and above the critical point:
An example of a stable symmetric case below the critical point is when Δ = 0.15, then σ = 0, λ_s = λ_l = 0.15, α_s = α_l = 0.928477184420735, η_s = η_l = 0.498562345387235, τ_s = τ_l = 1.077032388926053,τ_m = 1.575594734313288, and p = -0.734901555342283.
At the critical point Δ = λ_s = λ_l = 0.257736456168094, σ = 0, α_s = α_l = 1.199678640257734, η_s = η_l = 0.675323001937355, τ_s = τ_l = 0.833556559600965, τ_m = 1.508879561538320, and p = -1.064216727291271.
An example of a stable *asymmetric* case above the critical point is when Δ = 0.36, then σ = 0.725352324580598, λ_s = 0.0988731631509846, λ_l = 0.621126836849015, α_s = 0.759208447922104, α_l = 1.783486473962996, η_s = 0.398191755342614, η_l = 1.154653559010278, τ_s = 1.317161318129329, τ_l = 0.560699514461666, τ_m = 1.715353073471943 , and p = -1.393949763626345. If in this case the symmetry wasn't broken then the (dimensionless local maximum) potential energy would have been p_symmetric = -1.374827663970676 which is above the actual stable asymmetric minimum potential energy.
Above the critical point as Δ increases the asymmetry becomes ever greater. If Δ is increased to Δ = 0.5, then σ = 0.881677615975199, λ_s = 0.0591611920124004, λ_l = 0.940838807987600, α_s = 0.590620165207311, α_l = 2.12393632129417, η_s = 0.303994996445853, η_l = 1.526306640985646, τ_s = 1.693135552947120, τ_l = 0.470823908407326, τ_m = 1.997130549392973, and p = -1.913716093412236. If in this case the symmetry wasn't broken then the unstable (dimensionless local maximum) potential energy would have been p_symmetric = -1.816136254680986 which is above the actual stable asymmetric minimum potential energy.
Finally, suppose the loop's circumference is precisely twice as long as the minimal possible limit, i.e. suppose L=4*D. This case has Δ = 1 and this is well above the critical point and the equilibrium solution is quite asymmetric. The various parameters are: σ = 0.978920553942392, λ_s = 0.0210794460576082, λ_l = 1.978205539423918, α_s = 0.354520068887770, α_l = 2.82766411450992, η_s = 0.1791243995845190, η_l = 2.64619039228320, τ_s = 2.82071478530760, τ_l = 0.353648792608918, τ_m = 2.99983918489212, and p = -4.41249658082597 If in this case the symmetry wasn't broken then the unstable (dimensionless local maximum) potential energy would have been p_symmetric = -3.64483385359642 which is above the actual stable asymmetric minimum potential energy.
Another mundane situation involving a possibly un expected spontaneous symmetry breaking is the case of a uniform rectangular plank or block straddling and balanced on a sideways circular cylinder. What I have in mind here is a cylinderical drum laid on its side and wedged so it cannot roll on the floor and which has a uniform rectangular block or plank balanced on top of it such that a side view of the system is of a rectangle sitting symmetrically on top of a circle so that the center of mass of the plank is directly above the contact point between the plank and the cylinder. We also suppose that the static friction coefficient between the plank and the cylinder is sufficient to prevent any slipping of the plank if the plank is tipped somewhat, so that the contact point rolls over the cylinder with the tipping plank so no slipping occurs. Like the looped double catenary, this system has a symmetric equilibrium case, a critical point case, and an instability of the symmetric equilibrium having a spontaneous breaking of the symmetry. But the similarity is, at best, superficial, in that the behavior is in some ways the precise opposite of the looped double catenary. Consider the gravitational potential energy function of the system as a function of the tipping angle of the plank from horizontal.
If the thickness of the block/plank is greater than the diameter of the drum then the potential energy is shaped like an *inverted* U. This means the symmetric balanced equilibrium position of the plank is an *unstable* equilibrium where the potential energy has a locally quadratic maximum. In this case there is no asymmetric equilibrium at a lower potential energy. What happens to the block in this situation is the block just spontaneously tips over at ever greater angles until eventually the static friction gives way and the greatly tipped block then just slides off the drum the rest of the way and crashes on the floor. As the tipping process gets started the tipping torque is initially proportional to the tipping angle and the tipping process accelerates in an initially exponential manner before some other complications set in complicating the tipping process.
When the block/plank's thickness exactly matches the diameter of the drum then the symmetric equilibrium is at a *critical point*. In this case the potential energy function still has a maximum at the horizontally balance orientation, but in this case the maximum is locally a quartic maximum. This means, although the equilibrium is still technically unstable, it has an anomalously long time to tip over as the tipping torque on the block is proportional to the cube of that angle. So when the tipping angle is infinitesimal the tipping torque is so low that it takes a long time for the tipping process to get started, but once it is well underway the tipping angular velocity quickly accelerates just before the block falls off the drum.
But when the plank's thickness is strictly less than the drum's diameter then the symmetrically balanced equilibrium is now stable, and the plank can rest on the drum indefinitely. In fact, if the plank is bumped by a small amount the plank will execute an oscillatory rocking motion about the equilibrium orientation until dissipative effects cause it to settle down back into the symmetric equilibrium. In this situation the potential energy function is shaped like the letter m (as a function of the tilt angle) and there are now 3 equilibrium angles & configurations. The symmetric one is now the stable potential energy minimum. But there are also a pair of tipped (asymmetric) unstable equilibrium configurations each tilted by opposite angles corresponding to a pair of potential energy maxima (i.e. the humps of the letter m). If the system is precisely tipped up to one of these two angles and carefully put at rest and released the plank will be precariously balanced for a short while before spontaneously *either* tilting further and falling off the drum, *or* possibly reduce its tilt and swing back through the horizontal orientation at maximal angular velocity toward the potential energy maximum with the opposite tilt angle. Because of dissipation top of the other potential energy barrier will not quite be reached, and the plank will end up oscillating with a decreasing amplitude until the symmetric stable equilibrium is again reached.
Let L be the plank/block's length, let H be the plank/block's thickness, let D be the drum's diameter, and let g be the acceleration of gravity. Then when the symmetric equilibrium is stable the period, T of very small rocking oscillations about the symmetric equilibrium is given by
T = 2*π*√((H/g)*(4 + (L/H)^2)/(6*(D/H - 1))) .
Note the argument of the square root is positive only if D/H > 1. And the period diverges to infinity as the critical point is approached. On the other hand suppose the symmetric equilibrium is an unstable potential energy maximum. In that case a very tiny displacement from the unstable equilibrium initially grows exponentially with time with a doubling time, T_2 given by
In this case the initial doubling time for the unstable exponential growth in the tilt angle also diverges as the critical point is approached from the unstable side.
Note the *qualitative* shape of the potential energy function for the plank-on-a-drum system is an *upside down version* of the potential energy function of the looped double catenary system. The letters m and m are inverted versions of each other as are the letter U and its inverted version. Also the quartic critical point minimum & maximum are mutually inverted versions of each other as well. The mutually upside down versions of the corresponding potential energy functions means that the two systems have their corresponding stable and unstable equilibria interchanged with each other. But they both have a single local quadratic extremum continuously deform under the action of some system parameter changing the local quadratic extremum into a local quartic extremum at the critical point, and with further deformation the single extremum becomes 3 locally quadratic extrema with the symmetric one situated between the 2 asymmetric ones (which are symmetrically positioned with respect to each other). In the case of the looped double catenary the parameter that continuously deforms the potential energy function (of the asymmetry parameter, σ) is the average relative excess length, Δ of the loop relative to the peg separation. In the case of the plank-on-a-drum system the parameter that deforms the potential energy function (of the tilt angle) is the ratio of the plank thickness to the drum diameter.
I'm curious if anyone out in phys-l land has their own favorite examples of a mundane simple physical system (i.e. not advanced stat mech, quantum field theory, or GR examples) with a spontaneous symmetry breaking that shows up in an a priori unexpected way.