# Inertia tensor

Physical size
Surname Inertia tensor
Size type Moment of inertia
Formula symbol ${\ displaystyle \ mathbf {\ Theta}, I}$
Size and
unit system
unit dimension
SI ${\ displaystyle \ mathrm {kg \, \ cdot \, m ^ {2}}}$ ${\ displaystyle M \, \ cdot \, L ^ {2}}$
Remarks
The inertia tensor is a covariant and positively definite 2nd order tensor .

In mechanics, the inertia tensor is the property of a rigid body that describes its inertia in relation to changes in its angular momentum . Its symbol is or . It is a second order covariant tensor and positive definite for extended bodies . ${\ displaystyle \ mathbf {\ Theta}}$${\ displaystyle \ mathbf {I}}$

With the help of the inertia tensor, the relationship between the angular momentum of a body and its angular velocity can be represented in vector form as a matrix product of the inertia tensor with the angular velocity: ${\ displaystyle {\ vec {L}}}$ ${\ displaystyle {\ vec {\ omega}}}$

${\ displaystyle {\ vec {L}} = \ mathbf {\ Theta} \ cdot {\ vec {\ omega}}}$

The value of the inertia tensor depends on the choice of its reference point. This is usually set to the center of mass of the body to calculate the inertia tensor . This choice facilitates the separate calculation of the natural and orbital angular momentum . With the help of Steiner's theorem , the inertia tensor of the center of gravity can be calculated for any reference point.

In the coordinate representation of the inertia tensor with respect to an orthonormal basis with the coordinate origin at the reference point, it contains the moments of inertia and deviation for axes of rotation that are parallel to the basis vectors . The moments of inertia and deviation with respect to other axes are obtained through the reference point through coordinate transformation .

For certain axes of rotation , the angular momentum is parallel to the angular velocity. These axes are called the main axes of inertia . For every body there are at least three main axes of inertia that are perpendicular to each other. They are parallel to the eigenvectors of the inertia tensor. The corresponding eigenvalues ​​of the inertia tensor are called the main moments of inertia of the body. If the body rotates around an axis other than one of the main axes of inertia, its angular momentum and its axis of rotation are generally not parallel. Then, as a result of the conservation of angular momentum, the axis of rotation is not fixed, but also rotates: the body 'eggs'. If you hold the axis of rotation in this case by force, forces act on the bearings due to the imbalance and the angular momentum is variable.

Inertia tensors of simple bodies can be found in the list of inertia tensors .

## Analogy to mass with translatory movement

In the equations of motion of mechanics, the inertia tensor has a comparable position in relation to rotation, as does mass in relation to translation .

rotation Translation
${\ displaystyle \ underbrace {\ vec {L}} _ {\ mathrm {angular momentum}} = \ underbrace {\ mathbf {\ Theta}} _ {\ mathrm {Tr {\ ddot {a}} unit tensor}} \ cdot \ underbrace {\ vec {\ omega}} _ {\ mathrm {angular velocity}}}$ ${\ displaystyle \ underbrace {\ vec {p}} _ {\ mathrm {impulse}} = \ underbrace {m} _ {\ mathrm {mass}} \ cdot \ underbrace {\ vec {v}} _ {\ mathrm { Speed}}}$

Beyond the formally identical position as an expression of the inertia, to link the kinematic variable (angular) speed with the dynamic variable (angular) momentum , there are essential differences that distinguish the rotations from the translations:

• the mass is a scalar quantity , the inertia tensor a second order tensor.
• Momentum and velocity are always parallel, angular momentum and angular velocity are generally not.
• The mass is constant over time in all reference systems, the inertia tensor generally depends on the orientation of the body and its position in relation to the reference point. Since these can change, the components are time-dependent, while the mass is constant for translations. The components of the inertia tensor are only constant in a body-fixed reference system.

## Inertia tensor for a point mass

### Derivation and definition

The following applies to the angular momentum of a point mass with respect to the coordinate origin: ${\ displaystyle {\ vec {L}}}$

${\ displaystyle {\ vec {L}} = m \, {\ vec {r}} \ times {\ vec {v}} = m \, {\ vec {r}} \ times ({\ vec {\ omega }} \ times {\ vec {r}})}$.

Here are:

• ${\ displaystyle m}$: the mass of the point mass
• ${\ displaystyle {\ vec {r}}}$: the position vector of the point mass
• ${\ displaystyle {\ vec {v}} = {\ dot {\ vec {r}}}}$: the speed of the point mass
• ${\ displaystyle {\ vec {\ omega}}}$: the angular velocity of the point mass relative to the coordinate origin

Using the BAC-CAB formula , the unit tensor and the operator for the dyadic product, this can be transformed into: ${\ displaystyle \ mathbf {1}}$${\ displaystyle \ otimes}$

${\ displaystyle {\ vec {L}} = m \, [({\ vec {r}} \ cdot {\ vec {r}}) \, \ mathbf {1} - {\ vec {r}} \ otimes {\ vec {r}}] \ cdot {\ vec {\ omega}}}$

With the definition of the inertia tensor : ${\ displaystyle \ mathbf {\ Theta}}$

${\ displaystyle \ mathbf {\ Theta}: = m \, [({\ vec {r}} \ cdot {\ vec {r}}) \, \ mathbf {1} - {\ vec {r}} \ otimes {\ vec {r}}]}$

the above-mentioned relationship between angular momentum and angular velocity results . ${\ displaystyle \ textstyle {\ vec {L}} = \ mathbf {\ Theta} \ cdot {\ vec {\ omega}}}$

### calculation

The matrix representation of the inertia tensor with respect to the orthonormal basis with the unit vectors is obtained from the bilinear form , with the indices numbering the coordinates: ${\ displaystyle \ mathbf {\ Theta}}$${\ displaystyle {\ hat {e}} _ {1,2,3}}$ ${\ displaystyle \ Theta _ {ij} = {\ hat {e}} _ {i} \ cdot \ mathbf {\ Theta} \ cdot {\ hat {e}} _ {j}}$${\ displaystyle i, j}$

{\ displaystyle {\ begin {aligned} \ Theta _ {ij} = & {\ hat {e}} _ {i} \ cdot \ mathbf {\ Theta} \ cdot {\ hat {e}} _ {j} \ \ = & {\ hat {e}} _ {i} \ cdot m \, [({\ vec {r}} \ cdot {\ vec {r}}) \, \ mathbf {1} - {\ vec { r}} \ otimes {\ vec {r}}] \ cdot {\ hat {e}} _ {j} \\ = & m \, {\ hat {e}} _ {i} \ cdot [({\ vec {r}} \ cdot {\ vec {r}}) {\ hat {e}} _ {j} - ({\ vec {r}} \ cdot {\ hat {e}} _ {j}) {\ vec {r}}] \\ = & m \, [({\ vec {r}} \ cdot {\ vec {r}}) ({\ hat {e}} _ {i} \ cdot {\ hat {e }} _ {j}) - r_ {j} ({\ hat {e}} _ {i} \ cdot {\ vec {r}})] \\ = & m \, [({\ vec {r}} \ cdot {\ vec {r}}) \ delta _ {ij} -r_ {i} \, r_ {j}] \\\ Rightarrow \ Theta = & m \, {\ begin {pmatrix} (r_ {2} ^ {2} + r_ {3} ^ {2}) & - r_ {1} r_ {2} & - r_ {1} r_ {3} \\ - r_ {1} r_ {2} & (r_ {1} ^ {2} + r_ {3} ^ {2}) & - r_ {2} r_ {3} \\ - r_ {1} r_ {3} & - r_ {2} r_ {3} & (r_ {1 } ^ {2} + r_ {2} ^ {2}) \ end {pmatrix}} \\\ end {aligned}}}

Here are also:

• ${\ displaystyle {\ vec {r}} = (r_ {1}, \, r_ {2}, \, r_ {3})}$the coordinates of the position vector
• ${\ displaystyle \ delta _ {ij} = {\ begin {cases} 1 & {\ text {for}} \; i = j \\ 0 & {\ text {otherwise}} \ end {cases}}}$the Kronecker Delta

The inertia tensor is a symmetric tensor because it always holds . ${\ displaystyle \ Theta _ {ij} = \ Theta _ {ji}}$

## Structure of the inertia tensor

The elements of the inertia tensor in a coordinate representation have direct physical meaning:

### Moment of inertia with respect to any axis

The three elements of the main diagonal are the body's moments of inertia when rotating around the respective axis of the coordinate system. The moment of inertia about an axis in the direction of any unit vector is given by ${\ displaystyle \ Theta _ {ee}}$${\ displaystyle {\ hat {e}}}$

${\ displaystyle \ Theta _ {ee} = {\ hat {e}} \ cdot \ mathbf {\ Theta} \ cdot {\ hat {e}}}$.

You can see this simply in the matrix representation above, if you expand the selected unit vector by two further unit vectors to form an orthogonal basis. Because the diagonal elements are the moments of inertia around the directions of the basis vectors. ${\ displaystyle {\ hat {e}}}$

### Moments of deviation

The off-diagonal elements are called moments of deviation . They indicate (after multiplying by ) the torques that must be exerted by the bearings so that the axis of rotation maintains its direction. ${\ displaystyle \ omega ^ {2}}$

### Principal axes of inertia and principal moments of inertia

In general . From the positive definiteness of the tensor it follows that in three spatial dimensions there are also three positive eigenvalues and associated eigenvectors for which applies . ${\ displaystyle {\ vec {L}} = \ mathbf {\ Theta} \ cdot {\ vec {\ omega}}}$${\ displaystyle \ mathbf {\ Theta}}$ ${\ displaystyle \ Theta _ {k}}$ ${\ displaystyle {\ vec {\ omega}} _ {k}}$${\ displaystyle {\ vec {L}} = \ Theta _ {k} {\ vec {\ omega}} _ {k}}$

The eigenvectors of the inertia tensor are called the principal axes of inertia and its eigenvalues ​​are the principal moments of inertia .

With the main moments of inertia and their main axes of inertia, the inertia tensor has a particularly simple diagonal form :

${\ displaystyle \ mathbf {\ Theta} = {\ begin {pmatrix} \ Theta _ {1} && \\ & \ Theta _ {2} & \\ && \ Theta _ {3} \ end {pmatrix}}}$

### Symmetry considerations

Each axis of symmetry is a main axis of inertia. The following applies:

• In the case of straight prismatic bodies with a base in the form of a circle or a regular polygon, two of the three main moments of inertia are equal to one another. Their main axes of inertia are parallel to the base, the third main axis of inertia is perpendicular to it.
• In surface- symmetrical bodies, one main axis of inertia is perpendicular to the plane of symmetry, the other two in the plane of symmetry.
• If the body has two mutually perpendicular planes of symmetry, then their normals and their straight lines of intersection are the main axes of inertia.
• In the case of a tetrahedron , a cube , the other three regular bodies and the sphere , each spatial direction is the main axis of inertia.
• If , and are different from one another in pairs, there is no rotational symmetry with respect to an axis through the reference point, e.g. B. because the reference point is not in the center of mass or the body is not rotationally symmetrical with respect to any axis.${\ displaystyle \ Theta _ {1}}$${\ displaystyle \ Theta _ {2}}$${\ displaystyle \ Theta _ {3}}$

## Angular momentum and rotational energy in the main axis system fixed to the body

In the coordinate system, the three base vectors of which are defined by the main axes of inertia, the angular velocity is expressed as follows: ${\ displaystyle {\ hat {e}} _ {k}}$

${\ displaystyle {\ vec {\ omega}} = \ omega _ {1} {\ hat {e}} _ {1} + \ omega _ {2} {\ hat {e}} _ {2} + \ omega _ {3} {\ hat {e}} _ {3}}$

Then applies to the angular momentum

${\ displaystyle {\ vec {L}} = \ mathbf {\ Theta} \ cdot {\ vec {\ omega}} = \ Theta _ {1} \ omega _ {1} {\ hat {e}} _ {1 } + \ Theta _ {2} \ omega _ {2} {\ hat {e}} _ {2} + \ Theta _ {3} \ omega _ {3} {\ hat {e}} _ {3}}$.

and for the rotational energy

${\ displaystyle E _ {\ text {red}} = {\ frac {1} {2}} {\ vec {\ omega}} \ cdot \ mathbf {\ Theta} \ cdot {\ vec {\ omega}} = { \ frac {1} {2}} (\ Theta _ {1} \ omega _ {1} ^ {2} + \ Theta _ {2} \ omega _ {2} ^ {2} + \ Theta _ {3} \ omega _ {3} ^ {2})}$.

## Ellipsoid of inertia

Define the length of the position vector in each direction by the equation ${\ displaystyle {\ vec {r}}}$

${\ displaystyle 1 = {\ vec {r}} \ cdot \ mathbf {\ Theta} \ cdot {\ vec {r}}}$,

then the end points of these vectors lie on a closed surface in the form of an ellipsoid ( proof ). In each direction, the distance between the surface and the origin is equal to the reciprocal of the square root of the moment of inertia for the axis in that direction:

${\ displaystyle r = {\ frac {1} {\ sqrt {\ Theta _ {rr}}}}}$

The three axes of the ellipsoid are the main axes of inertia. The longest has the direction of the axis of rotation with the smallest possible moment of inertia for the given arrangement of the masses, the shortest semiaxis the direction with the greatest possible moment of inertia. These axes have fixed directions in the body's own reference system, because their spatial position is determined by the position of the body.

## Calculation of the inertia tensor

### For a system of mass points

The angular momentum of a composite system is the sum of the angular momentum of the components of the system . ${\ displaystyle {\ vec {L}}}$${\ displaystyle {\ vec {L}} _ {n}}$

${\ displaystyle {\ vec {L}} = \ sum _ {n} {\ vec {L}} _ {n} = \ sum _ {n} \ mathbf {\ Theta} _ {n} \ cdot {\ vec {\ omega}} _ {n}}$

If the angular velocities of the components are all identical and identical , then: ${\ displaystyle {\ vec {\ omega}} _ {n}}$${\ displaystyle {\ vec {\ omega}}}$

${\ displaystyle {\ vec {L}} = \ sum _ {n} \ mathbf {\ Theta} _ {n} \ cdot {\ vec {\ omega}}}$

And so the following applies to the inertia tensor of the system: ${\ displaystyle \ mathbf {\ Theta}}$

{\ displaystyle {\ begin {aligned} \ mathbf {\ Theta} & = \ sum _ {n} \ mathbf {\ Theta} _ {n} \\ & = \ sum _ {n} m_ {n} \, [ ({\ vec {r}} _ {n} \ cdot {\ vec {r}} _ {n}) \, \ mathbf {1} - {\ vec {r}} _ {n} \ otimes {\ vec {r}} _ {n}] \\ & = \ sum _ {n} m_ {n} {\ begin {pmatrix} (y_ {n} ^ {2} + z_ {n} ^ {2}) & - x_ {n} y_ {n} & - x_ {n} z_ {n} \\ - y_ {n} x_ {n} & (x_ {n} ^ {2} + z_ {n} ^ {2}) & -y_ {n} z_ {n} \\ - z_ {n} x_ {n} & - z_ {n} y_ {n} & (x_ {n} ^ {2} + y_ {n} ^ {2}) \ end {pmatrix}} \\ & = {\ begin {pmatrix} \ sum m_ {n} (y_ {n} ^ {2} + z_ {n} ^ {2}) & - \ sum m_ {n} \ , x_ {n} y_ {n} & - \ sum m_ {n} \, x_ {n} z_ {n} \\ - \ sum m_ {n} y_ {n} x_ {n} & \ sum m_ {n } (x_ {n} ^ {2} + z_ {n} ^ {2}) & - \ sum m_ {n} \, y_ {n} z_ {n} \\ - \ sum m_ {n} z_ {n } x_ {n} & - \ sum m_ {n} \, z_ {n} y_ {n} & \ sum m_ {n} \, (x_ {n} ^ {2} + y_ {n} ^ {2} ) \ end {pmatrix}} \ end {aligned}}}

Here are also:

• ${\ displaystyle m_ {1 \ ldots N}}$ the masses of the mass points from which the system is composed,
• ${\ displaystyle {\ vec {r}} _ {1 \ ldots N} = (x_ {1 \ ldots N}, \, y_ {1 \ ldots N}, \, z_ {1 \ ldots N})}$the coordinates of their position vectors

### With continuous mass distribution

In the transition to a continuous mass distribution of the mass density, an integral replaces the sums : ${\ displaystyle \ rho ({\ vec {r}})}$

${\ displaystyle \ Theta = \ int _ {V} \ rho ({\ vec {r}}) \, [({\ vec {r}} \ cdot {\ vec {r}}) \ mathbf {1} \ , - \, {\ vec {r}} \ otimes {\ vec {r}}] \ mathrm {d} V}$

with the individual moments of inertia

${\ displaystyle \ Theta _ {ij} = \ int _ {V} \ rho ({\ vec {r}}) \, [({\ vec {r}} \ cdot {\ vec {r}}) \ delta _ {ij} \, - \, r_ {i} r_ {j}] \ mathrm {d} V}$

### Example: inertia tensor of a homogeneous cube

In the center of mass of a cube with edge length , a Cartesian coordinate system is placed so that the coordinate axes are parallel to the cube edges. Because of the homogeneity, the density is constant and can be put in front of the integral: ${\ displaystyle d = 2a}$

${\ displaystyle \ Theta _ {ij} = \ Theta _ {\ beta \ alpha} = \ varrho \, \ int _ {V} (r ^ {2} \ delta _ {ij} -r_ {i} r_ {j }) \ mathrm {d} V}$

Now the six independent tensor components can be determined: These are three moments of inertia and three moments of deviation, because the tensor is symmetrical . For the cube with edge length , from to is integrated in all three spatial directions to calculate the inertia tensor with respect to the origin . For the cube we get: ${\ displaystyle \ Theta _ {ij} = \ Theta _ {ji}}$ ${\ displaystyle 2a}$${\ displaystyle -a}$${\ displaystyle + a}$

{\ displaystyle {\ begin {aligned} \ Theta _ {xx} = & \ varrho \ int _ {V} (y ^ {2} + z ^ {2}) \ mathrm {d} V = \ varrho {\ frac {16} {3}} a ^ {5} \\\ Theta _ {yy} = & \ varrho \ int _ {V} (x ^ {2} + z ^ {2}) \ mathrm {d} V = \ varrho {\ frac {16} {3}} a ^ {5} \\\ Theta _ {zz} = & \ varrho \ int _ {V} (y ^ {2} + x ^ {2}) \ mathrm {d} V = \ varrho {\ frac {16} {3}} a ^ {5} \\\ Theta _ {xy} = & \ Theta _ {yx} = - \ varrho \ int _ {V} yx \ mathrm {d} V = 0 \\\ Theta _ {yz} = & \ Theta _ {zy} = - \ varrho \ int _ {V} zy \ mathrm {d} V = 0 \\\ Theta _ {zx} = & \ Theta _ {xz} = - \ varrho \ int _ {V} xz \ mathrm {d} V = 0 \ end {aligned}}}

It was

{\ displaystyle {\ begin {aligned} \ int _ {- a} ^ {a} \ mathrm {d} x = & \ left [x \ right] _ {- a} ^ {a} = 2a \\\ int _ {- a} ^ {a} x \ mathrm {d} x = & \ left [{\ frac {x ^ {2}} {2}} \ right] _ {- a} ^ {a} = 0 \ \\ int _ {V} x ^ {2} \ mathrm {d} V = & \ int _ {x = -a} ^ {x = a} \ int _ {y = -a} ^ {y = a} \ int _ {z = -a} ^ {z = a} x ^ {2} \, \ mathrm {d} z \ mathrm {d} y \ mathrm {d} x = \ int _ {x = -a} ^ {x = a} x ^ {2} \, \ mathrm {d} x \ int _ {y = -a} ^ {y = a} \ mathrm {d} y \ int _ {z = -a} ^ {z = a} \ mathrm {d} z = \ left [{\ frac {x ^ {3}} {3}} \ right] _ {- a} ^ {a} (2a) ^ {2} = { \ frac {8} {3}} a ^ {5} \ end {aligned}}}

used, the same applies in - and - direction. With these results, the edge length and the mass of the cube, the tensor gets the shape ${\ displaystyle y}$${\ displaystyle z}$${\ displaystyle d = 2a}$${\ displaystyle m = \ varrho d ^ {3}}$

${\ displaystyle \ mathbf {\ Theta} = \ varrho {\ frac {16} {3}} a ^ {5} {\ begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \ end {pmatrix}} = {\ frac {\ varrho} {6}} d ^ {5} \ mathbf {1} = {\ frac {m} {6}} d ^ {2} \ mathbf {1}}$.