# Terminologies for moment of inertia

Perhaps someone can suggest the right terms for the following mathematical objects related to moment of inertia?

1. A inertia tensor $I$. $$I \equiv \begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\\end{bmatrix}$$
2. A product of inertia is an off-diagonal entry in the tensor: $I_{1,2} = I_{2,1}$, $I_{1,3} = I_{3,1}$, or $I_{2,3} = I_{3,2}$.
3. A principal moment of inertia is a diagonal entry in the tensor: $I_{1,1}$, $I_{2,2}$, or $I_{3,3}$. This is the semantic of moment of inertia discussed in elementary treatment of Physics.
4. What is the term for $I_2$ and $I_3$ in the last line below? \begin{align*} I_{1,1} &= \sum_{j} m_j\;\left(r^2_{j,2} + r^2_{j,3}\right) \\ &= \sum_{j} m_j\,r^2_{j,2} + \sum_{j} m_j\,r^2_{j,3} \\ I_{1,1} &= I_2 + I_3 \end{align*}
• I have a hard time to see what the question could possibly be. You seem to be asking about the right terminology but the question itself seems to contain the answers. Commented Jun 23, 2014 at 17:59
• I don't know the term for $I_2$ and $I_3$ in number 4. Commented Jun 23, 2014 at 18:33
• The semantic of $I_2$ (or $I_3$) is the perpendicular distance of a mass to $x_2$-axis (or $x_3$-axis) after the position $r$ of the mass has been projected onto the $x_2$$x_3-plane. Commented Jun 23, 2014 at 18:55 • Mathematically, I_2 \neq I_{2,2} and I_3 \neq I_{3,3} because by definition, I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right), I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right), and I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right), and therefore, I_{1,1} \neq I_{2,2} + I_{3,3}. Commented Jun 23, 2014 at 19:09 • @TadeusPrastowo You said "The semanatic ... plane". This definition is not conistent because I_2 would have dimensions of length. Are you reading any book in particular? Commented Jun 23, 2014 at 19:43 ## 2 Answers A inertia tensor I.$$I \equiv \begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\\end{bmatrix}$$A product of inertia is an off-diagonal entry in the tensor: I_{1,2} = I_{2,1}, I_{1,3} = I_{3,1}, or I_{2,3} = I_{3,2}. True. A principal moment of inertia is a diagonal entry in the tensor: I_{1,1}, I_{2,2}, or I_{3,3}. This is the semantic of moment of inertia discussed in elementary treatment of Physics. This is not true, the principal moments of inertia are the diagonal elements (eigenvalues), only if you have diagonalized the inertia matrix (which you can always do). What is the term for I_2 and I_3 in the last line below?$$I_{1,1} = I_2 + I_3 $$If I_{2}=I_{2,2} and I_{3}=I_{3,3}, once you have diagonalized I. In general, we have (dropping the indices):$$I = m \begin{bmatrix}y^2+z^2 & I_{1,2} & I_{1,3} \\ I_{2,1} & x^2+z^2 & I_{2,3} \\ I_{3,1} & I_{3,2} & x^2+y^2\\\end{bmatrix}$$But in a planar body one of the coordinates is zero, in the example below (continuos body but same concept), y=0. Then you can see that:$$I = \int d\mathbf{m} \begin{bmatrix}z^2 & I_{1,2} & I_{1,3} \\ I_{2,1} & x^2+z^2 & I_{2,3} \\ I_{3,1} & I_{3,2} & x^2\\\end{bmatrix}$$So the moment of inertia associated to the z axis is the sum of the other two. • Is there any term for a diagonal element in an inertia tensor when the matrix is not yet diagonalized? Commented Jun 23, 2014 at 19:02 • @TadeusPrastowo By term do you mean expression? Do you know how to calculate eigenvalues? Commented Jun 23, 2014 at 19:03 • No, I mean the terminology. When I communicate with someone else and I need to say the noun for a diagonal element in the moment of inertia tensor matrix that is not yet diagonalized, what is the noun? Commented Jun 23, 2014 at 19:05 • I_2 \neq I_{2,2} and I_3 \neq I_{3,3} because by definition, I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right), I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right), and I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right), and therefore, I_{1,1} \neq I_{2,2} + I_{3,3}. Commented Jun 23, 2014 at 19:08 • That is why I ask whether there is any name for I_2 and I_3 because they are not equal to I_{2,2} and I_{3,3}, respectively. I_{2,2} and I_{3,3} are called moments of inertia about an axis. Commented Jun 23, 2014 at 19:11 They are just the equation above rewritten to a shorter format using the definition of the moment of inertia about an axis:$$ I = \sum_{i=1}^N{m_ir_i^2}. $$The equation above uses r_{j,2} and r_{j,3}, which correspond to I_2 and I_3. • But, I_2 (or I_3) is not a perpendicular distance to x_2-axis (or x_3-axis) while the Wikipedia article defines r in the "moment of inertia about an axis" to be the perpendicular distance of a mass to an axis. The semantic of I_2 is the perpendicular distance of a mass to x_2-axis after the position of the mass has been projected onto the x_2$$x_3$-plane. Commented Jun 23, 2014 at 18:53
• Mathematically, $I_2 \neq I_{2,2}$ and $I_3 \neq I_{3,3}$ because by definition, $I_{1,1} = \sum_j m_j\,\left(r^2_{j,2} + r^2_{j,3}\right)$, $I_{2,2} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,3}\right)$, and $I_{3,3} = \sum_j m_j\,\left(r^2_{j,1} + r^2_{j,2}\right)$, and therefore, $I_{1,1} \neq I_{2,2} + I_{3,3}$. Commented Jun 23, 2014 at 19:09