Three-Body Inertia Tensor

We derive a general formula for the inertia tensor of a three-body system. By employing three independent Lagrange undetermined multipliers to express the vectors corresponding to the sides in terms of the position vectors of the vertices, we present the general covariant expression for the inertia tensor of the three particles of different masses. If $m_a/a=m_b/b=m_c/c=\rho$, then the center of mass coincides with the incenter of the triangle and the moment of inertia about the normal axis passing the center of mass is $I=\rho abc$, where $m_a$, $m_b$, and $m_c$ are the masses of the particles at $A$, $B$, and $C$, respectively, and $a$, $b$, and $c$ are the lengths of the line segments $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. The derivation and the corresponding results are closely related to the famous Heron's formula for the area of a triangle.


I. INTRODUCTION
The center of mass (CM) and inertia tensor are fundamental physical quantities for a system of particles. In usual textbooks like Refs. 1 on classical mechanics, two-body systems and specific rigid bodies like rods, plates, discs, cubes, and spheres are considered. However, students are not frequently exposed to a three-body system that constructs a triangle to investigate the fundamental mechanical properties of the CM and the inertia tensor.
Normally undergraduate students are well aware of various formulas involving trigonometric identities that project the fundamental nature of a three-body system. In this work, we are particularly interested in an arbitrary three-body system of pointlike masses m a , m b , and m c placed at points A, B, and C, respectively. It is straightforward to find the CM of this three-body system. The computation of the inertia tensor of the system involves the computation of m a A 2 + m b B 2 + m c C 2 , where A , B , and C are the lever-arm vectors from the CM to the vertices A, B, and C, respectively. We can express the lever-arm vectors as linear combinations of vectors a, b, and c that represent the sides opposite to the vertices.
Because A , B , and C are coplanar, those linear combinations are not unique. We employ Lagrange's method of undetermined multipliers to find the inverse transformation 2-6 . The complicated dependence on the multipliers eventually cancels when we compute a physically measurable quantities like the scalar products of any two vectors.
By making use of the bra-ket notation and completely covariant approach 7 , we have computed the inertia tensor in an arbitrary frame of reference. As a special example, we consider the case in which the CM coincides with the incenter of the triangle. In that case, the perpendicular distance from the CM to every side is the same. We have demonstrated that the condition is satisfied if m a : m b : m c = a : b : c. The solutions that we discarded involves unphysical mass contributions. The moment of inertia about the normal axis passing the center of mass is found to be I = ρabc when ρ = m a /a = m b /b = m c /c. The derivation and the corresponding results are closely related to the famous Heron's formula for the area of a triangle. This paper is organized as follows: In Sec. II, we list the definitions of various physical variables that we frequently use in the remainder of this paper. Section III contains the computation of the scalar products of the lever-arm vectors by employing Lagrange's method of undetermined multipliers. The features of the special case that the CM coincides with the incenter are considered in Sec. IV. The relationship among our mechanical variables and Heron's formula for the area of a triangle is presented in Sec. V. The computation of the inertia tensor of the three-body system is given in Sec. VI that is followed by the discussion in Sec. VII.

II. DEFINITIONS
As shown in Fig. 1, there are three particles of mass m a , m b , and m c placed at A, B, and C, respectively. We denote the total mass by M = m a + m b + m c . The CM 1 is at We employ the bra-ket notation 8-11 U |V ≡ U · V for any Euclidean vectors U and V .
The inertia tensor operator can then be defined by Here, 1 is the identity operator whose matrix representation is 1 = (δ ij ). In any Cartesian coordinates with the unit basis vectorsê i , the inertia tensor is the matrix element of I = (I ij ) that can be projected out from (2) as where ê i |ê j = δ ij , and V i = ê i |V = V |ê i for any Euclidean vector V .
We attach a superscript to represent a vector in the CM frame such that Thus in the CM frame, the three lever-arm vectors A , B , and C from the CM are coplanar: Parallel-axis theorem can be employed to express the inertia tensor as We define the displacements − − → BC, −→ CA, and −→ AB that are independent of the frame of reference as

III. LAGRANGE MULTIPLIERS AND INVARIANTS
We want to express the lever-arm vectors A , B , and C in terms of the side vectors a, b, and c. According to (5), only two of the three vectors A , B , and C are linearly independent. Thus the inverse transformation is not unique. We introduce three Lagrange undetermined multipliers λ 1 , λ 2 , and λ 3 with the dimensions of mass inverse. By adding , which is actually vanishing according to (5), to the constraint equations in Eq. (7), we can make the transformation invertible: where the matrix Λ(λ 1 , λ 2 , λ 3 ) is defined by where the determinant D of Λ(λ 1 , λ 2 , λ 3 ) is given by It is manifest that the determinant D vanishes when λ 1 + λ 2 + λ 3 = 0. Except for that case, we are free to choose any values λ i 's. The introduction of the multipliers brings in parameter dependence because the coefficient of each parameter that is actually vanishing is shared by A , B , and C nonuniformly. The explicit dependence on the parameters λ i 's is present only at the vector level. We shall find that the parameter dependence disappears once we compute the invariant quantities like scalar products.
According to (7) and Fig. 1(b), the angle between a and b is π − ∠C and so on. Here, we denote the internal angle of the triangle at vertex C by ∠C, and similarly for ∠A and ∠B.
Therefore, the scalar products of unit vectorsâ,b, andĉ parallel to a, b, and c, respectively, While each of the lever-arm vectors A , B , and C are dependent on the parameters λ i , the scalar products are free of the parameter dependence: where is the reduced mass of a two-body system.

IV. WHEN CM IS IDENTICAL TO THE INCENTER
We consider a special case in which the CM becomes identical to the incenter of the triangle. We shall find that the CM is identical to the incenter of the triangle at which the following ratios are all equal: Let us introduce the position vector A for the foot of the normal from the X CM to a.
B and C are defined in a similar manner as shown in Fig. 2. Then the vectors A , B , and C bisect the angles ∠A, ∠B, and ∠C, respectively. We define Then the normal vectors A , B , and C are expressed as Hence, We define the perpendicular distance from the CM to a, b, and c by By making use of the trigonometric identity sin θ 2 = (1 − cos θ)/2, we find r 2 a , r 2 b , and r 2 c from Eqs. (12) and (13) as follows: where If the position of the CM coincides with the incenter of the triangle, then r a = r b = r c .
The first solution is given by The other solutions are where Källén function 12-15 λ(a 2 , b 2 , c 2 ) is defined by According to the first line of (28), λ(a 2 , b 2 , c 2 ) must be negative definite because a+b+c > 0, a + b > c, b + c > a, and c + a > b. Thus we discard the solution (27a) which gives imaginary mass. We also discard the solution of Eq. (27b) that involves a negative mass.
Listing the sides in order as c ≥ b ≥ a, we can show that By imposing the conditions in (29) into Eq. (27b), we find that at least one particle must be massless or of negative mass. Thus, we discard this unphysical solution, too. As a result, (26) is the unique physical solution for r 2 a = r 2 b = r 2 c and is consistent with (14). We substitute the constraint (14) into (25) to find that r a = r b = r c = r, where According to (24), we find that where we used the trigonometric identity tan θ 2 = (1 − cos θ)/(1 + cos θ) and define s as We substitute the constraint (14) into the scalar products in Eq. (13) to determine the scalar products of the lever-arm vectors: By making use of the identities in Eq. (12), we obtain where Källén function λ(a 2 , b 2 , c 2 ) is defined in Eq. (28). Heron's formula for the area of a triangle is reproduced: where we have made use of the identities (30) and (31).

VI. INERTIA TENSOR
The inertia tensor for arbitrary masses and length parameters can be computed by making use of (6). In this section, we compute that tensor. One of the principal axis is alongn perpendicular to the triangle passing the CM. The moment of inertia about that axis is Because A , B , and C are coplanar andn is perpendicular to the triangle, the coefficient of |n n| must be the same as that for 1: where 1 ⊥ is the two-dimensional identity operator on the triangle plane. We choose the cyclic orthonormal basis vectors asê Then 1 ⊥ = |ê 1 ê 1 | + |ê 2 ê 2 | and where Here, X and X ⊥ are the components of X that are parallel and perpendicular to A , respectively, as shown in Fig. 3. We can replace the components of C with those of A and B by making use of the CM condition The inertia tensor operator becomes most compact if we choose the principal axes to be the basis vectorsq ± where The principal axes {q + ,q − } can be obtained by rotating {ê 1 ,ê 2 } as |q + = cos θ|ê 1 + sin θ|ê 2 , where the angle θ is The rotationally invariant forms of A , B , and C are Substituting Eq. (44) into Eq. (43) and applying Eq. (13), we express the principal moments In, I + , and I − in terms of only the masses and lengths: where the Källén function λ(a 2 , b 2 , c 2 ) is defined in Eq. (28). It is manifest that the above expressions satisfy the perpendicular-axis theorem, I + + I − = In.
If the incenter of the triangle coincides with the CM, then the condition Eq. (14) requires Our derivation for the inertia tensor involved a full exploitation of Dirac's bra-ket notation. The inertia tensor operator was expressed as the sum of operators before the projection onto a specific coordinate system. Thus during the full procedure of the derivation, we were free of complicated vector indices. Extension to the fully covariant derivation 7 can be directly obtained from the operator representation. Explicit coordinate dependence can then be projected out from the covariant expression. Ordinarily, undergraduate physics major students are exposed to the bra-ket notation for the first time when they take a course on quantum mechanics. We argue that a lot of experiences can be made regarding the bra-ket notation when they study classical mechanics.
As a specific example, we have considered the case in which the CM coincides with the incenter of the triangle. In that case, the perpendicular distance from the CM to every side