Quaternion Inverse

Every nonzero Hamilton quaternion has a multiplicative inverse. If

$\displaystyle q=q_w+\mathbf q$

and

$\displaystyle q^*=q_w-\mathbf q$

is its conjugate, then

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2}, \qquad q\neq0.$

(1)

Because

$\displaystyle \lVert q\rVert^2 = qq^* = q^*q,$

equation (1) satisfies both

$\displaystyle qq^{-1}=1$

(2)

and

$\displaystyle q^{-1}q=1.$

(3)

This two sided inverse is what makes the nonzero quaternions a division algebra.

Quaternion multiplication is not commutative, however, so division cannot be written as a single ambiguous quotient. Solving

$\displaystyle ax=b$

requires

$\displaystyle x=a^{-1}b,$

while solving

$\displaystyle xa=b$

requires

$\displaystyle x=ba^{-1}.$

These two expressions are generally different. The distinction becomes important in rigid body mechanics, navigation, robotics, and any quaternion calculation in which multiplication order carries physical meaning.

For unit quaternions, the inverse simplifies to

$\displaystyle q^{-1}=q^*.$

(4)

This is the form used throughout the later PhysicsLibrary attitude articles. Under the PhysicsLibrary passive convention, inverting a unit frame quaternion reverses the direction of the coordinate map.

Definition

Let $q$ be a nonzero quaternion. A quaternion $r$ is called the inverse of $q$ if

$\displaystyle qr=rq=1.$

(5)

When such an $r$ exists, it is unique and is written $q^{-1}$ .

For Hamilton quaternions, every nonzero element has an inverse. The zero quaternion does not, because

$\displaystyle 0r=0$

for every quaternion $r$ , so no product involving zero can equal the multiplicative identity $1$ .

Derivation from conjugate and norm

The preceding PhysicsLibrary articles established

$\displaystyle q^*=q_w-\mathbf q$

(6)

and

$\displaystyle qq^*=q^*q=\lVert q\rVert^2.$

(7)

If $q\neq0$ , then

$\displaystyle \lVert q\rVert^2>0,$

so division by this real scalar is allowed. Starting from

$\displaystyle qq^*=\lVert q\rVert^2,$

divide by $\lVert q\rVert^2$ :

$\displaystyle q \left( \frac{q^*}{\lVert q\rVert^2} \right) = 1.$

Similarly,

$\displaystyle \left( \frac{q^*}{\lVert q\rVert^2} \right) q = 1.$

Therefore

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2}.$

(8)

The denominator is an ordinary positive real number, so its position in the product does not matter.

Component formula

Let

$\displaystyle q = q_w+q_x\mathbf i+q_y\mathbf j+q_z\mathbf k.$

Then

$\displaystyle q^* = q_w-q_x\mathbf i-q_y\mathbf j-q_z\mathbf k$

and

$\displaystyle \lVert q\rVert^2 = q_w^2+q_x^2+q_y^2+q_z^2.$

Therefore

$\displaystyle q^{-1} = \frac{ q_w-q_x\mathbf i-q_y\mathbf j-q_z\mathbf k }{ q_w^2+q_x^2+q_y^2+q_z^2 }.$

(9)

In scalar first component form,

$\displaystyle [q^{-1}]_{\mathrm{PL}} = \frac{1}{ q_w^2+q_x^2+q_y^2+q_z^2 } \begin{bmatrix} q_w\ -q_x\ -q_y\ -q_z \end{bmatrix}.$

(10)

The formula requires only conjugation, four squares, their sum, and a real scalar division.

Uniqueness of the inverse

Suppose both $r$ and $s$ are inverses of $q$ . Then

$\displaystyle rq=1$

and

$\displaystyle qs=1.$

Using associativity,

$\begin{equation*} \begin{aligned} r &=r1\ &=r(qs)\ &=(rq)s\ &=1s\ &=s. \end{aligned}\end{equation*}$

Therefore the inverse is unique.

The proof uses associativity but does not require commutativity.

Inverse of a product

For nonzero quaternions $p$ and $q$ ,

$\displaystyle (pq)^{-1} = q^{-1}p^{-1}.$

(11)

The factor order reverses.

To verify this, multiply on the right:

$\begin{equation*} \begin{aligned} (pq)(q^{-1}p^{-1}) &= p(qq^{-1})p^{-1}\ &= p1p^{-1}\ &= 1. \end{aligned}\end{equation*}$

Multiplying in the opposite order gives

$\begin{equation*} \begin{aligned} (q^{-1}p^{-1})(pq) &= q^{-1}(p^{-1}p)q\ &= q^{-1}1q\ &= 1. \end{aligned}\end{equation*}$

Thus $q^{-1}p^{-1}$ is the two sided inverse of $pq$ .

For a longer product,

$\displaystyle (q_1q_2\cdots q_n)^{-1} = q_n^{-1}\cdots q_2^{-1}q_1^{-1}.$

(12)

This reversal is the inverse analogue of the conjugation rule

$\displaystyle (pq)^*=q^*p^*.$

Inverse and conjugation

Starting from

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2},$

conjugate both sides:

$\displaystyle (q^{-1})^* = \frac{q}{\lVert q\rVert^2}.$

But

$\displaystyle (q^*)^{-1} = \frac{(q^*)^*}{\lVert q^*\rVert^2}.$

Since

$\displaystyle (q^*)^*=q$

and

$\displaystyle \lVert q^*\rVert=\lVert q\rVert,$

we obtain

$\displaystyle (q^{-1})^* = (q^*)^{-1}.$

(13)

Also,

$\displaystyle (q^{-1})^{-1}=q.$

(14)

Norm of the inverse

Using multiplicativity of the norm,

$\displaystyle 1 = \lVert qq^{-1}\rVert = \lVert q\rVert \lVert q^{-1}\rVert.$

Therefore

$\displaystyle \lVert q^{-1}\rVert = \frac{1}{\lVert q\rVert}.$

(15)

This also follows directly from equation (8).

Unit quaternions

If $q$ is unit,

$\displaystyle \lVert q\rVert=1.$

Then equation (8) reduces to

$\displaystyle q^{-1}=q^*.$

(16)

Thus inversion of a unit quaternion requires only reversing the sign of its vector part.

$\displaystyle q=q_w+\mathbf q,$

then

$\displaystyle q^{-1}=q_w-\mathbf q.$

This simple inverse is one of the main computational advantages of using unit quaternions for orientation.

Pure quaternions

Let

$\displaystyle v=\mathbf v$

be a nonzero pure quaternion. Its conjugate is

$\displaystyle v^*=-v.$

Its squared norm is

$\displaystyle \lVert v\rVert^2 = \mathbf v\cdot\mathbf v.$

Therefore

$\displaystyle v^{-1} = -\frac{v}{\lVert v\rVert^2}.$

(17)

If $u$ is a unit pure quaternion,

$\displaystyle \lVert u\rVert=1,$

then

$\displaystyle u^{-1}=-u.$

(18)

Because a unit pure quaternion also satisfies

$\displaystyle u^2=-1,$

the result $u^{-1}=-u$ follows immediately.

Real quaternions

If $q=a$ is a nonzero real quaternion, then

$\displaystyle q^*=a$

and

$\displaystyle \lVert q\rVert^2=a^2.$

Hence

$\displaystyle q^{-1} = \frac{a}{a^2} = \frac{1}{a}.$

(19)

Quaternion inversion therefore reduces to ordinary reciprocal division on the real subalgebra.

Left and right division

Because quaternion multiplication is noncommutative, an expression resembling

$\displaystyle \frac{b}{a}$

is ambiguous unless multiplication order is stated.

There are two distinct equations to consider.

Solving

Suppose

$\displaystyle ax=b$

with $a\neq0$ .

Multiply on the left by $a^{-1}$ :

$\displaystyle a^{-1}(ax)=a^{-1}b.$

Associativity gives

$\displaystyle (a^{-1}a)x=a^{-1}b,$

$\displaystyle x=a^{-1}b.$

(20)

Solving

Now suppose

$\displaystyle xa=b.$

Multiply on the right by $a^{-1}$ :

$\displaystyle (xa)a^{-1}=ba^{-1}.$

Associativity gives

$\displaystyle x(aa^{-1})=ba^{-1},$

$\displaystyle x=ba^{-1}.$

(21)

In general,

$\displaystyle a^{-1}b\neq ba^{-1}.$

Therefore one must never cancel a quaternion factor without preserving the side on which the inverse acts.

Worked example 1: numerical inverse

Let

$\displaystyle q=1+2\mathbf i-\mathbf j+2\mathbf k.$

Its conjugate is

$\displaystyle q^* = 1-2\mathbf i+\mathbf j-2\mathbf k.$

Its squared norm is

$\displaystyle \lVert q\rVert^2 = 1^2+2^2+(-1)^2+2^2 = 10.$

Therefore

$\displaystyle q^{-1} = \frac{1}{10} \left( 1-2\mathbf i+\mathbf j-2\mathbf k \right).$

(22)

A direct multiplication gives

$\displaystyle qq^{-1}=q^{-1}q=1.$

Worked example 2: left and right division

Let

$\displaystyle a=1+\mathbf i, \qquad b=\mathbf j.$

First compute

$\displaystyle a^*=1-\mathbf i$

and

$\displaystyle \lVert a\rVert^2=2.$

Thus

$\displaystyle a^{-1} = \frac12(1-\mathbf i).$

Consider first

$\displaystyle ax=b.$

The solution is

$\displaystyle x=a^{-1}b.$

Hence

$\begin{equation*} \begin{aligned} x &= \frac12(1-\mathbf i)\mathbf j\ &= \frac... ...hbf i\mathbf j)\ &= \frac12(\mathbf j-\mathbf k). \end{aligned}\end{equation*}$

Now consider instead

$\displaystyle xa=b.$

The solution is

$\displaystyle x=ba^{-1}.$

Thus

$\begin{equation*} \begin{aligned} x &= \mathbf j\frac12(1-\mathbf i)\ &= \frac... ...hbf j\mathbf i)\ &= \frac12(\mathbf j+\mathbf k). \end{aligned}\end{equation*}$

Therefore

$\displaystyle a^{-1}b = \frac12(\mathbf j-\mathbf k)$

(23)

while

$\displaystyle ba^{-1} = \frac12(\mathbf j+\mathbf k).$

(24)

The two division problems have different answers because the inverse acts on different sides.

Worked example 3: inverse of a product

Let

$\displaystyle p=1+\mathbf i, \qquad q=1+\mathbf j.$

Their inverses are

$\displaystyle p^{-1} = \frac12(1-\mathbf i)$

and

$\displaystyle q^{-1} = \frac12(1-\mathbf j).$

The inverse of the product is

$\displaystyle (pq)^{-1} = q^{-1}p^{-1}.$

Therefore

$\begin{equation*} \begin{aligned} (pq)^{-1} &= \frac14(1-\mathbf j)(1-\mathbf i)... ...c14 \left( 1-\mathbf i-\mathbf j-\mathbf k \right). \end{aligned}\end{equation*}$

Since

$\displaystyle pq = 1+\mathbf i+\mathbf j+\mathbf k,$

its conjugate is

$\displaystyle (pq)^* = 1-\mathbf i-\mathbf j-\mathbf k$

and its squared norm is

$\displaystyle \lVert pq\rVert^2=4.$

The direct inverse formula gives the same result.

Passive frame interpretation

The algebraic inverse formula is independent of active or passive interpretation. Its frame meaning depends on the convention used for orientation.

PhysicsLibrary uses a unit quaternion

$\displaystyle {}^{B}q_A$

to map coordinates from frame $A$ into frame $B$ :

$\displaystyle {}^{B}v = {}^{B}q_A\, {}^{A}v\, ({}^{B}q_A)^*.$

(25)

Since a frame quaternion is unit,

$\displaystyle ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*.$

The inverse quaternion therefore represents the reverse coordinate map:

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*.$

(26)

Consequently,

$\displaystyle {}^{A}v = {}^{A}q_B\, {}^{B}v\, ({}^{A}q_B)^*.$

(27)

The frame labels make the direction of the inverse explicit. Inverting a frame quaternion does not merely change a sign; it reverses which coordinate system is mapped into which.

Inverse of a passive frame chain

Suppose

$\displaystyle {}^{B}q_A$

maps coordinates from $A$ into $B$ , and

$\displaystyle {}^{C}q_B$

maps coordinates from $B$ into $C$ .

The direct passive chain is

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A.$

(28)

Invert both sides:

$\displaystyle ({}^{C}q_A)^{-1} = ({}^{B}q_A)^{-1} ({}^{C}q_B)^{-1}.$

Using the reversed frame labels gives

$\displaystyle {}^{A}q_C = {}^{A}q_B\,{}^{B}q_C.$

(29)

The order reversal in the algebra is exactly what is needed to traverse the frame chain in the opposite direction.

Passive axis angle interpretation

For a positive frame rotation through angle $\theta$ about unit axis $\widehat{\mathbf u}$ , PhysicsLibrary uses

$\displaystyle {}^{B}q_A = \cos\frac{\theta}{2} - \widehat{\mathbf u}\sin\frac{\theta}{2}.$

(30)

Because this quaternion is unit,

$\displaystyle ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*.$

Therefore

$\displaystyle {}^{A}q_B = \cos\frac{\theta}{2} + \widehat{\mathbf u}\sin\frac{\theta}{2}.$

(31)

The inverse corresponds to traversing the same frame relation in the opposite direction.

Relationship to direction cosine matrices

Let

$\displaystyle {}^{B}C_A = C({}^{B}q_A)$

be the passive direction cosine matrix associated with a unit frame quaternion.

The inverse coordinate transformation is

$\displaystyle ({}^{B}C_A)^{-1}.$

Because the matrix is orthogonal,

$\displaystyle ({}^{B}C_A)^{-1} = ({}^{B}C_A)^T.$

The quaternion inverse produces the same map:

$\displaystyle C(({}^{B}q_A)^{-1}) = C({}^{A}q_B) = ({}^{B}C_A)^T.$

(32)

Thus quaternion inversion corresponds to matrix transposition for unit orientation transformations.

Numerical considerations

The inverse formula

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2}$

is mathematically valid for every nonzero quaternion, but numerical implementations should consider several practical issues.

If $\lVert q\rVert$ is extremely small, division by $\lVert q\rVert^2$ can amplify floating point error.
If is intended to be an orientation quaternion, it should normally be maintained close to unit norm. In that case the inverse should be close to the conjugate.
Do not silently replace $q^{-1}$ with unless unit norm is known or enforced.
In software, preserve multiplication order when solving quaternion equations. A left inverse operation and a right inverse operation are not interchangeable.
Frame labels should be propagated with the inverse. If ${}^{B}q_A$ maps into , its inverse is ${}^{A}q_B$ .

Common pitfalls

Using the conjugate as the inverse for a nonunit quaternion.
In general,

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2}.$

Only a unit quaternion satisfies $q^{-1}=q^*$ .
Trying to invert the zero quaternion.
The zero quaternion has no multiplicative inverse.
Forgetting inverse order reversal.
The correct identity is

$\displaystyle (pq)^{-1}=q^{-1}p^{-1}.$
Writing quaternion division as an ambiguous fraction.
The expressions $a^{-1}b$ and $ba^{-1}$ solve different equations and are generally unequal.
Cancelling on the wrong side.
For , multiply by $a^{-1}$ on the left. For , multiply by $a^{-1}$ on the right.
Changing Hamilton algebra because the attitude interpretation is passive.
The inverse formula is algebraic and does not change when passive frame maps are adopted.
Inverting a frame quaternion without reversing its labels.
Under the PhysicsLibrary convention,

$\displaystyle ({}^{B}q_A)^{-1} = {}^{A}q_B.$
Assuming inverse order reversal is merely notation.
It follows from noncommutative multiplication and is required for both algebraic cancellation and correct reversal of frame chains.

Exercises

The exercises are stated first so the article can be used for self study. Complete solutions follow afterward.

Direct inverse.
Find the inverse of

$\displaystyle q=2+\mathbf i-2\mathbf j+\mathbf k.$
Verify both sides.
For the quaternion in Exercise 1, verify explicitly that

$\displaystyle qq^{-1}=1$

and

$\displaystyle q^{-1}q=1.$
Unit quaternion.
Let

$\displaystyle q=\frac12 \left( 1+\mathbf i+\mathbf j+\mathbf k \right).$

Show that is unit and find $q^{-1}$ .
Pure quaternion.
Find the inverse of

$\displaystyle v=2\mathbf i-\mathbf j+2\mathbf k.$
Inverse of a product.
Let

$\displaystyle p=1+\mathbf i, \qquad q=1+\mathbf k.$

Compute $p^{-1}$ and $q^{-1}$ , then verify

$\displaystyle (pq)^{-1}=q^{-1}p^{-1}.$
Left division.
For

$\displaystyle a=1+\mathbf i, \qquad b=\mathbf k,$

solve

$\displaystyle ax=b.$
Right division.
Using the same and , solve

$\displaystyle xa=b.$

Compare your answer with Exercise 6.
Norm of the inverse.
Prove that

$\displaystyle \lVert q^{-1}\rVert = \frac{1}{\lVert q\rVert}$

for every nonzero quaternion.
Passive frame inverse.
If ${}^{B}q_A$ maps coordinates from into , write the quaternion that maps coordinates from into .
Frame chain reversal.
Starting from

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A,$

derive the inverse chain from frame back to frame .

Solutions

1. Direct inverse

Let

$\displaystyle q=2+\mathbf i-2\mathbf j+\mathbf k.$

The conjugate is

$\displaystyle q^* = 2-\mathbf i+2\mathbf j-\mathbf k.$

The squared norm is

$\displaystyle \lVert q\rVert^2 = 2^2+1^2+(-2)^2+1^2 = 10.$

Therefore

$\displaystyle q^{-1} = \frac{1}{10} \left( 2-\mathbf i+2\mathbf j-\mathbf k \right).$

(33)

2. Verify both sides

By construction,

$\displaystyle qq^* = q^*q = 10.$

Therefore

$\displaystyle qq^{-1} = q \frac{q^*}{10} = \frac{10}{10} = 1.$

Similarly,

$\displaystyle q^{-1}q = \frac{q^*q}{10} = 1.$

3. Unit quaternion

For

$\displaystyle q=\frac12 \left( 1+\mathbf i+\mathbf j+\mathbf k \right),$

the squared norm is

$\displaystyle \lVert q\rVert^2 = 4\left(\frac12\right)^2 = 1.$

Thus $q$ is unit and

$\displaystyle q^{-1}=q^*.$

Therefore

$\displaystyle q^{-1} = \frac12 \left( 1-\mathbf i-\mathbf j-\mathbf k \right).$

4. Pure quaternion

For

$\displaystyle v=2\mathbf i-\mathbf j+2\mathbf k,$

the squared norm is

$\displaystyle \lVert v\rVert^2 = 4+1+4 = 9.$

Since $v^*=-v$ ,

$\displaystyle v^{-1} = -\frac{v}{9}.$

Hence

$\displaystyle v^{-1} = -\frac{2}{9}\mathbf i +\frac{1}{9}\mathbf j -\frac{2}{9}\mathbf k.$

5. Inverse of a product

For

$\displaystyle p=1+\mathbf i,$

we have

$\displaystyle p^{-1} = \frac12(1-\mathbf i).$

For

$\displaystyle q=1+\mathbf k,$

we have

$\displaystyle q^{-1} = \frac12(1-\mathbf k).$

Therefore

$\begin{equation*} \begin{aligned} q^{-1}p^{-1} &= \frac14(1-\mathbf k)(1-\mathbf... ...c14 \left( 1-\mathbf i+\mathbf j-\mathbf k \right). \end{aligned}\end{equation*}$

Now

$\displaystyle pq = (1+\mathbf i)(1+\mathbf k) = 1+\mathbf i-\mathbf j+\mathbf k.$

Its conjugate is

$\displaystyle (pq)^* = 1-\mathbf i+\mathbf j-\mathbf k,$

and

$\displaystyle \lVert pq\rVert^2 = 4.$

Hence

$\displaystyle (pq)^{-1} = \frac14 \left( 1-\mathbf i+\mathbf j-\mathbf k \right) = q^{-1}p^{-1}.$

6. Left division

Given

$\displaystyle a=1+\mathbf i, \qquad b=\mathbf k,$

we have

$\displaystyle a^{-1} = \frac12(1-\mathbf i).$

For

$\displaystyle ax=b,$

the solution is

$\displaystyle x=a^{-1}b.$

Thus

$\begin{equation*} \begin{aligned} x &= \frac12(1-\mathbf i)\mathbf k\ &= \frac... ...hbf i\mathbf k)\ &= \frac12(\mathbf k+\mathbf j). \end{aligned}\end{equation*}$

Therefore

$\displaystyle x = \frac12(\mathbf j+\mathbf k).$

7. Right division

For

$\displaystyle xa=b,$

the solution is

$\displaystyle x=ba^{-1}.$

Hence

$\begin{equation*} \begin{aligned} x &= \mathbf k\frac12(1-\mathbf i)\ &= \frac... ...hbf k\mathbf i)\ &= \frac12(\mathbf k-\mathbf j). \end{aligned}\end{equation*}$

Therefore

$\displaystyle x = \frac12(-\mathbf j+\mathbf k).$

The solutions to Exercises 6 and 7 differ because quaternion multiplication is not commutative.

8. Norm of the inverse

Because

$\displaystyle qq^{-1}=1,$

norm multiplicativity gives

$\displaystyle \lVert q\rVert \lVert q^{-1}\rVert = 1.$

Since $q\neq0$ ,

$\displaystyle \lVert q\rVert>0.$

Therefore

$\displaystyle \lVert q^{-1}\rVert = \frac{1}{\lVert q\rVert}.$

9. Passive frame inverse

The reverse coordinate map is

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^{-1}.$

Since frame quaternions are unit,

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^*.$

10. Frame chain reversal

Start from

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A.$

Invert:

$\displaystyle ({}^{C}q_A)^{-1} = ({}^{B}q_A)^{-1} ({}^{C}q_B)^{-1}.$

Relabeling the reverse maps gives

$\displaystyle {}^{A}q_C = {}^{A}q_B\,{}^{B}q_C.$

The reversed multiplication order agrees with the reversed sequence of frames.

What comes next

The first five algebra articles have now supplied the main operations needed for quaternion orientation:

$\displaystyle q^*, \qquad \lVert q\rVert, \qquad q^{-1}.$

For a unit quaternion these collapse to the particularly simple relation

$\displaystyle q^{-1}=q^*.$

The next PhysicsLibrary article develops the axis angle representation and shows how a physical frame rotation through angle $\theta$ about a unit axis $\widehat{\mathbf u}$ produces the passive quaternion

$\displaystyle {}^{B}q_A = \cos\frac{\theta}{2} - \widehat{\mathbf u}\sin\frac{\theta}{2}.$

From Q06 onward, the quaternion series can build rotation geometry directly on the algebra developed in Q01 through Q05.

Sources and historical notes

Hamilton's quaternions form a division algebra: every nonzero quaternion has a two sided multiplicative inverse. Classical quaternion texts often express division through reciprocal quaternions, conjugation, and the historical tensor or norm notation.

Modern notation makes the inverse especially compact:

$\displaystyle q^{-1} = \frac{q^*}{\lVert q\rVert^2}.$

The distinction between left and right division is a direct consequence of noncommutative multiplication and should be preserved explicitly in engineering applications.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., edited by C. J. Joly, Longmans, Green, and Co., 1899. Public domain historical source. Internet Archive scan
2: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public domain historical source. Internet Archive scan
3: A. S. Hathaway, A Primer of Quaternions, 1896. Public domain historical source. Project Gutenberg edition
4: Wikibooks contributors, “Abstract Algebra/Quaternions.” Openly licensed instructional source. Wikibooks quaternion article

License

Unless otherwise noted, this PhysicsLibrary entry is intended for release under the Creative Commons Attribution ShareAlike 4.0 International license.

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