Since Newton's laws are defined for point particles, I'd like to derive some laws of motions for rigid bodies only by considering a rigid body as a system of particles such that the distances from every particle to every other particle doesn't change with time. I think I derived that the force applied on one particle of a rigid body must be the same for every other particle of the rigid body in one dimension by the following:

Consider two particles on a line $P_1$ and $P_2$ both with masses $dm$ and positions $x_1$ and $x_2$. Let's say that a force $F_1$ acts on the particle $P_1$. By Newton's second law we get: $$F_1 = dm\frac{d^2x_1}{dt^2}$$ By the definition of a rigid body, the distance between $P_1$ and $P_2$ doesn't change with time. Define $r$ as this distance ie. $r = x_1 - x_2$. Therefore: $$\frac{dr}{dt} = 0$$ Taking the derivative of both sides we further get that $$\frac{d^2r}{dt^2} = 0$$ $$\frac{d^2(x_1 - x_2)}{dt^2} = 0$$ $$\frac{d^2x_1}{dt^2} = \frac{d^2x_2}{dt^2}$$ By Newton's second law this is the same as: $$\frac{F_1}{dm} = \frac{F_2}{dm}$$ (where $F_2$ is the force acting on $P_2$), and since $dm \ne 0$ finally: $$F_1 = F_2$$

These steps can be done for arbitrary amount of particles, and so we get that in one dimension, if a force is applied on one of the particle of a rigid body, every other particle of the rigid body experiences the same force.

The problem is that I cannot do a similar proof for two dimensions by defining the distance $r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, but I'm sure that it can be done, and that if done torque, moment of inertia and center of mass would arise. Can someone do a similar proof for two dimensions, if it can be done like this at all?

  • $\begingroup$ Is the goal to see how the force on one particle in the rigid body ends up producing some combination of torque and center-of-mass motion, starting with Newton's laws for point particles? $\endgroup$ Commented Apr 7, 2019 at 16:07
  • $\begingroup$ @ChiralAnomaly Yes, that would be the main goal. Even if it doesn't get produced by this method. $\endgroup$
    – pavle
    Commented Apr 7, 2019 at 16:58
  • $\begingroup$ @Steeven If I did that, would I get the same result, that the force on one particle must be the same on any other, because clearly that isn't right, since there is rotation in 2D rigid body dynamics, but this can only produce translation. I'm not really sure what you mean. $\endgroup$
    – pavle
    Commented Apr 7, 2019 at 17:02
  • $\begingroup$ @AyyLmao Aha, I see what you mean. I will remove the comment, as it is a bit misleading. $\endgroup$
    – Steeven
    Commented Apr 7, 2019 at 17:05
  • $\begingroup$ In two or three dimensions the forces on every particle are NOT the same if the rigid body motion involves rotation, because different particles have different accelerations. The equations are derived in any university-level textbook on mechanics, unless you really want to re-invent them by yourself (and that might be hard, if you are not as smart as Newton was!) $\endgroup$
    – alephzero
    Commented Apr 7, 2019 at 18:12

2 Answers 2


I'd like to derive some laws of motions for rigid bodies only by considering a rigid body as a system of particles such that the distances from every particle to every other particle doesn't change with time.

The beautiful answer by ja72 was already posted before I wrote this one, but this one uses a different approach, so I decided to post it anyway. A distinguishing feature of this approach is that it works for $D$-dimensional space with arbitrary $D\geq 2$.


Notation: A rotation about the origin is described by a $D\times D$ matrix $R$ whose transpose is equal to its inverse and whose determinant is equal to $1$. A vector $\mathbf{x}$ can be represented by a matrix with $D$ components in a single column. With this notation, the result of applying a rotation $R$ to a vector $\mathbf{x}$ is the vector $R\mathbf{x}$, using ordinary matrix multiplication.

Consider a rigid body made of point particles. Let $m_n$ be the mass of the $n$th particle, and let $\mathbf{b}_n$ be its position in a coordinate system attached to the body, so $\mathbf{b}_n$ doesn't change as the body shifts or rotates. Choose the origin of the coordinate system so that $\sum_n m_n\mathbf{b}_n=0$, where $0$ denotes the zero vector.

Let $\mathbf{x}_n$ be time-dependent position of the $n$th particle in some inertial coordinate system. The assumption that the body is rigid means $$ \mathbf{x}_n = R\mathbf{b}_n+\mathbf{x} \tag{1} $$ where $R$ is a time-dependent rotation matrix and $\mathbf{x}$ is the time-dependent position of the center of mass: $$ \mathbf{x} := \frac{\sum_n m_n\mathbf{x}_n}{\sum_n m_n}. \tag{2} $$ The goal is to derive equations for the time-dependence of $R$ and $\mathbf{x}$ in terms of the forces applied to the particles.

The motion of the center of mass

Let $\mathbf{f}_n$ be the force applied to the $n$th particle, not including the inter-particle forces that keep the body rigid. This section derives the result $$ \sum_n \mathbf{f}_n = m\ddot{\mathbf{x}} \hskip1cm \text{with } \hskip1cm m:=\sum_n m_n \tag{3} $$ where each overhead dot denotes a derivative with respect to time. Equation (3) says that the net applied force $\sum_n \mathbf{f}_n$ is equal to the total mass $m$ times the acceleration $\ddot{\mathbf{x}}$ of the center of mass.

To derive (3), use the fact that the total force on the $n$th particle is $\mathbf{F}_n := \mathbf{f}_n+\sum_k\mathbf{f}_{nk}$ where $\mathbf{f}_{nk}$ is the force that the $k$th particle must exert on the $n$th particle to keep the body rigid. The total force on the whole body is then $$ \sum_n\mathbf{F}_n=\sum_n\mathbf{f}_n +\sum_{n,k}\mathbf{f}_{nk}. \tag{4} $$ Since the forces that the $k$th and $n$th particles exert on each other must be equal and opposite, the last term in equation (4) is zero, so $$ \sum_n\mathbf{F}_n=\sum_n\mathbf{f}_n. \tag{5} $$ Newton's law for each individual particle says $\mathbf{F}_n=m_n\ddot{\mathbf{x}}_n$, and using this on the left-hand side of equation (5) gives equation (3), remembering the definition (2) of $\mathbf{x}$. This completes the derivation of (3).

Rotational motion

The usual way of describing rotational motion in $3$-dimensional space involves some special conventions that only make sense in $3$-dimensional space. The generalization to $D$-dimensional space is easier if we use the different set of conventions that I'll introduce here. The only difficult part is relating these more-natural conventions to the more-traditional conventions that only work in $3$-dimensional space. I'll omit that difficult part, but a hint is given near the end of this post.

For the $n$th particle, we can combine the vectors $\mathbf{f}_n$ and $\mathbf{x}_n$ into a square ($D\times D$) matrix $\mathbf{f}_n\mathbf{x}_n^T$, where the superscript $T$ denotes transpose. We will be interested in the antisymmetric part of this matrix. Geometrically, the antisymmetric part of this matrix corresponds to the plane spanned by the vectors $\mathbf{f}_n$ and $\mathbf{x}_n$, with an associated magnitude that goes to zero whenever the two vectors are proportional to each other (because then they don't define a plane). This is the $D$-dimensional replacement for the traditional "cross product." More generally, for any square matrix $B$, we can construct the antisymmetric matrix $$ \Delta(B) := B-B^T. \tag{6} $$ Using this notation, the net torque applied to the body is defined to be the antisymmetric matrix $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right). \tag{7} $$ This is the $D$-dimensional generalization of the torque "vector," but one of the things we learn from this generalization is that torque is not a vector! Torque is a bivector, represented here by an antisymmetric matrix. In $3$-d space, we can get away with using a vector-like notation only because there is a unique line through the origin orthogonal to any given plane through the origin.

To derive an equation that describes the rotational motion of the rigid body, start with the definition of torque, equation (7). Using equation (4) with $\mathbf{f}_{kn}+\mathbf{f}_{nk}=0$ and with the fact that $\mathbf{f}_{kn}$ is directed along the line between the $k$th and $n$th particles, we can derive $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right)= \Delta\left(\sum_n\mathbf{F}_n \mathbf{x}_n^T\right). \tag{8} $$ This says that the applied torque (the left-hand side) is equal to the total torque (the right-hand side), where the total torque includes that due to the internal forces that keep the body rigid. Now use $\mathbf{F}_n=m_n\ddot{\mathbf{x}}_n$ in the right-hand side of (8), and then use equation (1). Some terms cancel due to the antisymmetry, leaving the result $$ \Delta\left(\sum_n\mathbf{f}_n \mathbf{x}_n^T\right)= m\Delta\big(\ddot{\mathbf{x}}\mathbf{x}^T\big) +\Delta\big(\ddot RM_b R^T\big) \tag{9} $$ with $$ M_b := \sum_n m_n\mathbf{b}_n\mathbf{b}_n^T. \tag{10} $$ Equation (9) relates the applied torque to the time-dependence of the body's rotation $R$ and of the body's center-of-mass $\mathbf{x}$.

The first term on the right-hand side of equation (9) has an easy interpretation: the matrix $\Delta\big(\ddot{\mathbf{x}}\mathbf{x}^T\big)$ is the angular acceleration of the body's center of mass. Geometrically, this has an orientation represented by the plane spanned by $\mathbf{x}$ and $\ddot{\mathbf{x}}$.

The second term on the right-hand side of equation (9) describes the rotation of the body about its center of mass. The matrix (10) is the special combination of particle-masses and their body-fixed positions that controls how the body responds to torque. The subscript $b$ on $M_b$ means "body-fixed." The second term on the right-hand side of equation (9) may be rewritten using the identity $$ \Delta\big(\ddot RM_b R^T\big) =\frac{d}{dt}L \tag{11} $$ where $L$ is the angular momentum bivector $$ L := \Delta\big(WM\big) \tag{12} $$ where $M := RM_b R^T$ is just $M_b$ expressed in the inertial coordinate system, and $W$ is the angular velocity bivector $$ W := \dot R R^T. \tag{13} $$ This matrix is already antisymmetric, so we don't need to use $\Delta$ here. (Proof: Take the time-derivative of both sides of the identity $R R^T=1$.)

Relationship to traditional notation when $D=3$

This is the tricky part, and this post is already long, so I'll omit the details. Here's a hint: For $D=3$, a bivector is represented by a $3\times 3$ antisymmetric matrix. Such a matrix has only $3$ independent components, because the components below the diagnal are the negatives of those above the diagonal, and the diagonal components are zero. By arranging these $3$ components into a "vector" and re-writing the preceding equations in that vector-like notation, we can recover the traditional formulation for $3$-dimensional space.

The messiest part is showing how the matrix (10) is related to the thing that is traditionally called the "moment of intertia tensor." The matrix (10) is actually simpler than the traditional moment of inertia tensor, and it conveys the same information, so we could have called (10) the "momentum of inertia tensor" instead... but history has already taken its course, and I won't try to change it here.

The easist case: $D=2$

In this case, a bivector (antisymmetric matrix) has only one indepenent component, which makes things relatively easy: all of the equations involving $\Delta$ reduce to essentially scalar equations. Actually, "pseudoscalar" is a better name for it: a pseudoscalar changes sign when any one direction is reflected, but a scalar in the strict sense does not. In $D=2$, a bivector is the same thing as a pseudoscalar. (In $D=3$, a bivector is the same thing as a "pseudovector.")

  • 1
    $\begingroup$ Between equations (4) and (5) - Not because the body is rigid, but because of Newton's 3rd law. $\endgroup$
    – mr_e_man
    Commented Jun 10, 2022 at 2:37
  • $\begingroup$ @mr_e_man You're right! I must have been typing faster than I was thinking... I edited the answer to delete the incorrect phrase. I also used the opportunity to fix a mathjax formatting issue (too much vertical space below the double-dots). Thank you! $\endgroup$ Commented Jun 10, 2022 at 16:44

You are in the correct path. Let's see if you can follow along:

  1. Consider the kinematics. That is all the possible motions that keep the particle distances constant. This is resolved by Chasle's Theorem, which states that relative to an arbitrary point A the velocity of another point B is given by $$ \vec{v}_B = \vec{v}_A + \vec{\omega} \times ( \vec{r}_B - \vec{r}_A )$$ where $\vec{r}_A$ and $\vec{r}_B$ are the instantaneous position vectors of A and B respectively and the vector $\vec{\omega}$ is the angular velocity vector which is shared among all the points 1.

  2. Describe the center of mass of a system of particles. Consider $n$ particles each with mass $m_i$, velocity $\vec{v}_i$. At each time frame, a point C can be described such as $$\sum_i^n m_i \vec{r}_i = \left(\sum_i^n m_i \right) \vec{r}_C = m\, \vec{r}_C$$ The point C is called the center of mass. Now the position of each particle $\vec{r}_i$ is decoposed into two parts. The position of the center of mass, and the relative position $\vec{d}_i$ of the particle to the center of mass $$\vec{r}_i = \vec{r}_C + \vec{d}_i$$ NOTE: This implies that $\sum \limits_i^n m_i \vec{d}_i = \vec{0}$ from the equation above it.

    Using the kinematics from above, we can pick the center of mass as the reference point and describe the kinematics of each particle (the derivative of the above) as $$ \vec{v}_i = \vec{v}_C + \dot{\vec{d}}_i = \vec{v}_C + \vec{\omega} \times \vec{d}_i $$

  3. Describe momentum of a system of particles. The total momentum of the system is $$\require{cancel} \begin{aligned} \vec{p} & = \sum_i^n (m_i \vec{v}_i) = \sum_i^n m_i (\vec{v}_C + \vec{\omega} \times \vec{d}_i) \\ & = \left( \sum_i^n m_i \right) \vec{v}_C + \vec{\omega} \times \left(\cancel{ \sum_i^n m_i \vec{d}_i }\right) \\ & = m \, \vec{v}_C \end{aligned}$$

  4. Newton's 2nd Law applies here to each particle, where a small force $\vec{F}_i$ is applied on each particle resulting on the total load on the body $\vec{F} = \sum \limits_i^n \vec{F}_i$. Newton's second law states for each particle that $\vec{F}_i = \frac{\rm d}{{\rm d}t} \vec{p}_i = \frac{\rm d}{{\rm d}t}( m_i \vec{v}_i)$. When summed together the total force is $$ \begin{aligned} \vec{F} & = \frac{\rm d}{{\rm d}t} \vec{p} = \frac{\rm d}{{\rm d}t} (m\,\vec{v}_C) \\ \vec{F} & = m \vec{a}_C \end{aligned} $$

    where $\vec{a}_C = \frac{\rm d}{{\rm d}t} \vec{v}_C$ is the acceleration of the center of mass.

  5. Describe angular momentum of a system of particles. By taking the moment of momentum $\vec{L}_i = \vec{r}_i \times (m_i \vec{v}_i)$ and summing for all particles we describe the total angular momentum

    $$ \require{cancel} \begin{aligned} \vec{L} & = \sum_i^n \vec{r}_i \times (m_i \vec{v}_i) = \sum_i^n m_i (\vec{r}_C + \vec{d}_i) \times (\vec{v}_C + \vec{\omega} \times \vec{d}_i) \\ & = \sum_i^n m_i \left( \vec{r}_C \times \vec{v}_C + \vec{r}_C \times (\vec{\omega} \times \vec{d}_i) + \vec{d}_i \times \vec{v}_C + \vec{d}_i \times (\vec{\omega} \times \vec{d}_i) \right) \\ &= \vec{r}_C \times (m \vec{v}_C) + \vec{r}_C \times ( \vec{\omega} \times \cancel{ \sum_i^n m_i \vec{d}_i} ) + (\cancel{ \sum_i^n m_i \vec{d}_i}) \times \vec{v}_C + \sum_i^n m_i \vec{d}_i \times (\vec{\omega} \times \vec{d}_i) \\ \vec{L} &= \vec{r}_C \times \vec{p} + \mathrm{I}_C \vec{\omega} \end{aligned}$$

    where $\mathrm{I}_C \vec{\omega} = \sum_i^n m_i \vec{d}_i \times (\vec{\omega} \times \vec{d}_i)$ if $\mathrm{I}_C$ is the 3×3 symmetric mass momemnt of inertia matrix. It is commonly defined as $$ \mathrm{I}_C = \sum_i^n m_i [\vec{d}_i \times][\vec{d}_i \times] = \sum_i^n m_i \begin{vmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{vmatrix} $$ with $\vec{d}_i = \pmatrix{x & y & z}$, and $[\vec{d}_i \times]$ the 3×3 skew symmetric cross product operator.

  6. Euler's rotation law for the system of rigid bodies. The angular mometum about the center of mass is $\vec{L}_C = \vec{L} - \vec{r}_C \times \vec{p} = \mathrm{I}_C \vec{\omega}$ (due to the parallel axis theorem). Eulers rotation law states that the net torque about the center of mass equals the rate of change of agular momentum about the center of mass

    $$ \begin{aligned} \vec{\tau} & = \frac{\rm d}{{\rm d}t} \vec{L}_C =\frac{\rm d}{{\rm d}t} ( \mathrm{I}_C \vec{\omega} ) \\ & = \mathrm{I}_C \vec{\alpha} + \vec{\omega} \times \vec{L}_C \end{aligned} $$

    The derivation of the above requires the rules of differentiating a vector on a rotating reference frame to complete.


The equations of motion of a rigid body consisting of many particles of constant distance are as follows

$$ \begin{array}{r|ll} & \text{linear} & \text{angular} \\ \hline \text{momentum} & \vec{p} = m \vec{v}_C & \vec{L}_C = \mathrm{I}_C \vec{\omega} \\ \text{force} & \vec{F} = m \vec{a}_C & \vec{\tau}_C = \mathrm{I}_C \vec{\alpha} + \vec{\omega} \times \mathrm{I}_C \vec{\omega} \end{array} $$


  • 1 Proof that distances are kept constant under common rotation. Distance between A and B is $\text{(distance)}=\sqrt{(\vec{r}_B-\vec{r}_A) \cdot (\vec{r}_B-\vec{r}_A)}$, (where $\cdot$ is the vector inner product) $$\begin{aligned} (\vec{r}_B-\vec{r}_A) \cdot (\vec{r}_B-\vec{r}_A) & = \text{(const.)} & & \text{square both sides} \\ (\vec{v}_B-\vec{v}_A) \cdot (\vec{r}_B-\vec{r}_A)+(\vec{r}_B-\vec{r}_A) \cdot (\vec{v}_B-\vec{v}_A) & = 0 & & \text{derivative with time} \\ 2 (\vec{r}_B-\vec{r}_A) \cdot (\vec{v}_B-\vec{v}_A) & = 0 & & \text{commutative dot product} \\ (\vec{r}_B-\vec{r}_A) \cdot (\vec{v}_A + \vec{\omega} \times (\vec{r}_B-\vec{r}_A)-\vec{v}_A) & = 0 & & \text{substitute kinematics} \\ (\vec{r}_B-\vec{r}_A) \cdot (\vec{\omega} \times (\vec{r}_B-\vec{r}_A)) & \equiv 0 & & \text{property of cross product $\times$}\end{aligned}$$

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