# XYLENE POWER LTD.

## BASIC PHYSICAL CONCEPTS - PART D

### RIGID BODIES

#### By Charles Rhodes, P. Eng., Ph.D.

INTRODUCTION:
This web page reviews physics related to rigid bodies. A rigid body is a solid formed from a collection of molecules that remain in a fixed spacial relationship with respect to one another but the entire collection of atoms or molecules can movw linearly and/or rotate.
In a liquid the molecules can slide past one another. In a gas the molecules can fly around with minimal interaction except at the container walls.

LOSS OR QUANTIZATION:
For rigid bodies, each with a large number of atoms, the energy difference between the adjacent discrete real energy solutions is so small compared to the total system energy that from an external observers perspective the discrete changes in energy associated with changes in bulk motion are invisible. However, an important observable effect of quantization of electron energy states is the frequency dependence on temperature of random kinetic (heat) energy loss via thermal electromagnetic radiation.

TOTAL LINEAR MOMENTUM:
A real object usually consists of a cluster of spacially distributed particles.
The total linear momentum (Pc - Po) of a cluster of particles is defined as:
(Pc - Po) = [Sum of all (Pi - Poi)]
= [Sum of all {(Ei / C^2) (Vi - Vo)}]

CENTER OF MOMENTUM VELOCITY:
The point Xc, Vc is known as the Center of Momentum (CM). The CM velocity (Vc - Vo) is defined by:
(Pc - Po) = [Sum of all {(Ei / C^2) (Vi - Vo)}]
= [E / C^2][Vc - Vo]

Hence:
(Vc - Vo) = [C^2 / E][Sum of all {(Ei / C^2) (Vi - Vo)}]
= [Sum of all {(Ei / E) (Vi - Vo)}]

LAW OF CONSERVATION OF LINEAR MOMENTUM:
The law of conservaton of linear momentum is one of the most fundamental laws of physics. This law states that from the perspective of an inertial observer the linear momentum (Pc - Po) of any isolated system is constant. Hence:
(Pc - Po) = constant
or
d(Pc - Po) / dT = 0

This conservation law is actually 3 conservation equations because (Pc - Po) has 3 orthogonal components, each of which is conserved.

CENTER OF MOMENTUM OF AN ISOLATED SYSTEM IS INERTIAL:
The linear momentum of an isolated system with respect to an inertial observer can be expressed as:
(Pc - Po) = (E / C^2) (Vc - Vo)

The law of conservation of momentum requires that the term on the right hand side must be constant. The law of conservation of energy requires that E be constant. Hence for an inertial observer:
(Vc - Vo) = constant.
or
d(Vc - Vo) / dT = 0
Since the observer is inertial:
dVo / dT = 0
Hence:
dVc / dT = 0
Hence for an isolated system the Center of Momentum (CM) is inertial.

In many practical situations it is convenient to locate the reference point Xo, Vo at the CM because the CM is inertial.

CHANGE IN REFERENCE POINT:
The momentum (Pi - Poi) with respect to Xo, Vo is given by:
(Pi - Poi) = ((Ei / C^2) (Vi - Vo))

This momentum can be expressed in terms of the velocity (Vc - Vo) of CM (Xc, Vc) as:
(Pi - Poi) = ((Ei / C^2) ((Vi - Vc) + (Vc - Vo))
= (Ei / C^2) (Vi - Vc) + (Ei / C^2) (Vc - Vo)
= (Pi - Pci) + (Ei / C^2) (Vc - Vo)

Hence:
|Pi - Poi|^2 = [(Pi - Pci) + (Ei / C^2) (Vc - Vo)] * [(Pi - Pci) + (Ei / C^2) (Vc - Vo)]
= |Pi - Pci|^2 + |(Ei / C^2) (Vc - Vo)|^2 + 2(Ei / C^2) (Pi - Pci) * (Vc - Vo)

VECTOR IDENTITY:
At the web page VECTOR IDENTITIES the vector identity:
|A X B|^2 + (A * B)^2 = |A|^2 |B|^2
is proven.

APPLICATION OF VECTOR IDENTITY:
Make the substitution:
A = (Pi - Pci)

Make the substitution:
B = (Xi - Xc)

Then:
|(Pi - Pci) X (Xi - Xc)|^2 + ((Pi - Pci) * (Xi - Xc))^2
= |Pi - Pci|^2 |Xi - Xc|^2
or
(|(Pi - Pci) X (Xi - Xc)| / |Xi - Xc|)^2
+ (((Pi - Pci) * (Xi - Xc)) / |Xi - Xc|)^2
= |Pi - Pci|^2

Then:
|Pi - Poi|^2 = |Pi - Pci|^2 + |(Ei / C^2) (Vc - Vo)|^2 + 2(Ei / C^2) (Pi - Pci) * (Vc - Vo)
= (|(Pi - Pci) X (Xi - Xc)| / |Xi - Xc|)^2
+ (((Pi - Pci) * (Xi - Xc)) / |Xi - Xc|)^2
+ |(Ei / C^2) (Vc - Vo)|^2
+ 2(Ei / C^2) (Pi - Pci) * (Vc - Vo)

KINETIC ENERGY OF AN ISOLATED CLUSTER OF PARTICLES:
The law of conservation of energy gives the total energy of an isolated cluster of particles as:
E = Sum of all Ei

The kinetic energy Ek of an isolated cluster of particles is given by:
Ek = Sum of all Eki
Sum of all [Ei - Epi]
= Sum of all{Epi [1 + (C^2 / Epi^2)|Pi - Poi|^2)]^0.5 - Epi}

For |Vi - Vo| << C:
Ek ~ Sum of all {(C^2 / 2 Epi)|Pi - Poi|^2}
= Sum of all (C^2 / 2 Epi){(|(Pi - Pci) X (Xi - Xc)| / |Xi - Xc|)^2
+ (((Pi - Pci) * (Xi - Xc)) / |Xi - Xc|)^2
+ |(Ei / C^2) (Vc - Vo)|^2
+ 2(Ei / C^2) (Pi - Pci) * (Vc - Vo)}

Choose the point Xc, Vc such that the fourth term of Ek is zero. Hence:
Sum of all {(C^2 / 2 Epi) 2 (Ei / C^2) (Pi - Pci) * (Vc - Vo)} = 0
or
Sum of all {(Ei^2 / Epi C^2) (Vi - Vc) * (Vc - Vo)} = 0
or
Sum of all {(Ei^2 / Epi C^2) [(Vix - Vcx) x * (Vcx - Vox) x]
+ (Ei^2 / Epi C^2) [(Viy - Vcy) y * (Vcy - Voy) y]
+ (Ei^2 / Epi C^2) [(Viz - Vcz) z * (Vcz - Voz) z]} = 0
or
Sum of all {(Ei^2 / Epi C^2) ([(Vix - Vcx)(Vcx - Vox)]
+ [(Viy - Vcy)(Vcy - Voy)]
+ [(Viz - Vcz)(Vcz - Voz)])} = 0

The parameters Vcx, Vcy and Vcz are orthogonal. Each Vcx, Vcy, Vcz is chosen so that its term sums to zero. Hence this equation becomes three equations:
Sum of all {(Ei^2 / Epi C^2) (Vix - Vcx)} = 0
Sum of all {(Ei^2 / Epi C^2) (Viy - Vcy)} = 0
Sum of all {(Ei^2 / Epi C^2) (Viz - Vcz)} = 0

These equations can be rewritten as:
Sum of all {(Ei^2 / Epi C^2) ((Vix - Vox) - (Vcx - Vox))} = 0
Sum of all {(Ei^2 / Epi C^2) ((Viy - Voy) - (Vcy - Voy))} = 0
Sum of all {(Ei^2 / Epi C^2) ((Viz - Voz) - (Vcx - Voz))} = 0
or
(Vcx - Vox) = [Sum of all {(Ei^2 / Epi C^2) ((Vix - Vox)}] / [Sum of all (Ei^2 / Epi C^2)]
(Vcy - Voy) = [Sum of all {(Ei^2 / Epi C^2) ((Viy - Voy)}] / [Sum of all (Ei^2 / Epi C^2)]
(Vcz - Voz) = [Sum of all {(Ei^2 / Epi C^2) ((Viz - Voz)}] / [Sum of all (Ei^2 / Epi C^2)]

These equations fully define:
Vc - Vo = (Vcx - Vox) x + (Vcy - Voy) y + (Vcz - Voz) z

In vector notation these equations can be summarized as:
[Sum of all (Ei^2 / Epi C^2)][Vc - Vo]
= [Sum of all {(Ei^2 / Epi C^2) (Vi - Vo)}]

If |Vi - Vo| << C
then:
Epi ~ Ei
giving:
[Sum of all (Ei^2 / Epi C^2)][Vc - Vo]
~ [Sum of all (Ei / C^2)][Vc - Vo]
= [E / C^2][Vc - Vo]
= (Pc - Po)
and
[Sum of all {(Ei^2 / Epi C^2) (Vi - Vo)}]
~ [Sum of all {(Ei / C^2) (Vi - Vo)}]
= (Pc - Po)

Thus:
[Sum of all {(Ei^2 / Epi C^2) (Vi - Vo)}]
= [Sum of all (Ei^2 / Epi C^2)][Vc - Vo]

Hence for |Vi - Vo| << C:
Sum of all (C^2 / 2 Epi){2(Ei / C^2) (Pi - Pci) * (Vc - Vo)} ~ 0

Hence for a cluster of particles in which |Vi - Vo| << C Ek simplifies to:
Ek = Sum of all (C^2 / 2 Epi){(|(Pi - Pci) X (Xi - Xc)| / |Xi - Xc|)^2
+ (((Pi - Pci) * (Xi - Xc)) / |Xi - Xc|)^2
+ |(Ei / C^2) (Vc - Vo)|^2}
= rotation kinetic energy
+ CM kinetic energy

With a gas or a vapor generally only the CM kinetic energy can do useful work. The rotation kinetic energy and radial kinetic energy generally manifest themselves as heat.

ANGULAR MOMENTUM:
The concept of conservation of angular momentum is only true at speeds much less than the speed of light. However, since rigid bodies generally move slowly compared to the speeed of light the concept of angular momentum is convenient for analysing the motion of rigid bodies.

ANGULAR MOMENTUM OF A PARTICLE:
The term "angular momentum" is confusing because conservation of angular momentum is really conservation of rotational kinetic energy. In Newtonian mechanics, where V << C, the angular momentum Lci of particle i within a cluster of particles is defined by:
Lci = ((Pi - Pci) X (Xi - Xc))
where:
Pci is the vector component of linear momentum Pi through the cluster CM and Xc is the position of the cluster CM.

RIGID BODY:
For a rigid body rotating around an axis through its CM for every particle i:
|Xi - Xc| = constant
or
d[|Xi - Xc|^2] / dT = 0
or
d[(Xi - Xc) * (Xi - Xc)] / dT = 0
or
(Vi - Vc) * (Xi - Xc) = 0
or
Ei (Vi - Vc) * (Xi - Xc) = 0
or
(Pi - Pci) * (Xi - Xc) = 0

Hence for a rigid body rotating around an axis through its CM the radial kinetic energy is zero.

Hence, for |Vi - Vo| << C, the kinetic energy of a rigid body simplifies to:
Ek = Sum of all [(C^2 / 2 Epi){|(Pi - Pci) X (Xi - Xc)|^2 / |Xi - Xc|)^2
+ |(Ei / C^2) |Vc - Vo|^2}]
= Sum of all [(C^2 / 2 Epi){|(Pi - Pci) X (Xi - Xc)|^2 / |Xi - Xc|^2}
+ (Ei^2 / 2 Epi C^2) |(Vc - Vo)|^2]
~ Sum of all [(C^2 / 2 Epi) |(Pi - Pci) X (Xi - Xc)|^2 / |Xi - Xc|^2]
+ (E / 2 C^2) |Vc - Vo|^2

SUMMARY:
The velocity (Vi - Vc) of a particle at Xi relative to Xc can be resolved into a radial velocity component along (Xi - Xc) and a tangential velocity component at Xi that is perpendicular to (Xi - Xc).

With proper choice of Xc, Vc for a rigid body rotating around an axis through its CM the sum of the radial momentum components equals zero leaving only rotational momentum and linear momentum components.

For an observer that cannot see the rotation the rotational kinetic energy becomes part of the apparent CM rest energy Ecp. If the invisible rotational kinetic energy is quantized then the apparent rest energy is also quantized.

TOTAL KINETIC ENERGY OF A RIGID BODY:
The total kinetic energy of a rigid body becomes:
Ek = Sum of all [(C^2 / 2 Epi) {|(Pi - Pci) X (Xi - Xc)|^2 / |Xi - Xc|^2]
+ {(E / 2 C^2) |Vc - Vo|^2}
= Sum of all [(C^2 / 2 Epi) (|Lci|^2 / |Xi - Xc|^2)]
+ {(E / 2 C^2) |Vc - Vo|^2}

In English what this equation states is that for a rigid body with |Vi - Vo| << C:
Total kinetic energy = kinetic energy of rotation about the CM + kinetic energy of linear motion of energy E as if it were located at the CM at Xc, Vc.

ANGULAR MOMENTUM OF A RIGID BODY:
Recall that:
Ek = Sum of all [(C^2 / 2 Epi) (|Lci|^2 / |Xi - Xc|^2)]
+ {(E / 2 C^2) |Vc - Vo|^2}
The total angular momentum Lc caused by revolution of a rigid body about Xc, Vc is defined by:
Ek ~ [(C^2 / 2 E) (|Lc|^2 / |Xe - Xc|^2)]
+ {(E / 2 C^2) |Vc - Vo|^2}
where |Xe - Xc| is the radius of an equivalent ring containing the same rest energy Ep and the same kinetic energy Ek as the rotating cluster of particles forming the rigid body.

The total momentum of a rigid body is:
(P - Po)
= (Pc - Po) + (P - Pc)
= (E / C^2) (Vc - Vo) + Lc / |Xe - Xc|

Note that the linear and angular momentum terms are orthogonal to each other. Hence the products of linear and angular momentum terms are zero, giving:
|(P - Po)|^2
= |(E / C^2) (Vc - Vo)|^2 + |Lc|^2 / |Xe - Xc|^2
+ 2(E / C^2) (Vc - Vo) * Lc / |Xe - Xc|
= |(E / C^2) (Vc - Vo)|^2 + |Lc|^2 / |Xe - Xc|^2

Hence the generalized form of the Einstein equation for a rigid body is:
E^2 = Ep^2 + C^2 |P - Po|^2
= Ep^2 + [C^2 |P - Pc|^2] + [C^2 |Pc - Po|^2]
where:
|Pc - Po|^2 = |(E / C^2) (Vc - Vo)|^2
= (linear momentum of body}^2
and
|P - Pc|^2 = |Lc|^2 / |Xe - Xc|^2
= (rotational momentum of body)^2,
where:
Lc = angular momentum of body.

If the rotation is invisible in order to keep E constant Ep must increase to absorb the rotational kinetic energy.

ROTATION INCREASES APPARENT REST ENERGY Ep:
The rest energy is determined by measurement of rest mass.

From the perspective of only linear motion the rotation of a charged particle such as an electron or proton about its own CM cannot be seen, so such rotational kinetic energy is included in the particles apparent rest energy.

From the perspective of only linear motion the revolution of the electrons and protons of an atom around their common CM cannot be seen, so the kinetic energy of such revolution is included in the atom's apparent rest energy.

From the perspective of only linear motion the random thermal vibration of molecules in a solid cannot be seen, so such vibration kinetic energy is included in the solid's apparent rest energy.

From the perspective of only linear motion the rotation of a multi-atom gas molecule around its CM cannot be seen, so such rotation is included in the molecule's apparent rest energy.

From the perspective of an external observer confined random linear motion of gas molecules cannot be seen, so the kinetic energy due to such confined random motion is included in the molecule's apparent rest energy. The kinetic energy due to such confined random molecular motion is known as heat.

From the perspective of only linear motion the rotation of a fly wheel around its CM cannot be seen, so such rotation adds to the fly wheel's apparent rest energy.

ANGULAR MOMENTUM:
For any real object angular momentum about its CM at Xc, Vc contributes to the apparent potential (rest) energy Ecp at the CM as seen by an external inertial observer that observes only linear motion of the CM.

APPROXIMATE CONSERVATION OF ANGULAR MOMENTUM OF AN ISOLATED SYSTEM DURING A CHANGE IN ROTATIONAL KINETIC ENERGY:
For an isolated cluster of singularities and for |Vi - Vo| << C angular momentum is also approximately conserved. However, that apparent conservation arises from application of the law of conservation of energy at non-relativistic speeds. Conservation of angular momentum is not a true conservation law.

Recall that:
Eki = [(C^2 / 2 Epi) (|Lci|^2 / |Xi - Xc|^2)]
+ {(Ei / 2 C^2) |Vc - Vo|^2}
where:
|Lci| = |(Ei / C^2) (Vi - Vc) X (Xi - Xc)|
= (Ei / C^2) Wi Ri^2
Where:
Wi = angular velocity of singularity i around the axis of rotation
Ri = radius from Xi to the axis of rotation.

Hence the rotational part of Eki is:
Ekir = [(C^2 / 2 Epi) (|Lci|^2 / |Xi - Xc|^2)]
= [(C^2 / 2 Epi) (|(Ei / C^2) (Wi Ri^2)|^2 / |Ri|^2)]

For |Vi| << C:
Ekir ~ (Ei / 2 C^2)(Wi Ri)^2

Hence:
dEkir = (Ei / C^2) (Wi Ri) d[Wi Ri]
= (Ei / C^2) (Wi Ri) [dWi Ri + Wi dRi]

Centrifugal force = (Ei / C^2) Wi^2 Ri
Hence:
dEkir = - (centrifugal force) dRi
= - (Ei / C^2) Wi^2 Ri dRi

Use the law of conservation of energy to equate the two expressions for Ekir to get:
(Ei / C^2)(Wi Ri) [dWi Ri + Wi dRi] = - (Ei / C^2) Wi^2 Ri dRi
or
[dWi Ri + Wi dRi] = - Wi dRi
or
dWi Ri = - 2 Wi dRi
or
dWi / Wi = - 2 dRi / Ri
or
Wib / Wia = (Ria / Rib )^2
or
Wib Rib^2 = Wia Ria^2

Hence:
Wi Ri^2 = constant

Recall that:
|Lci| = (Ei / C^2) Wi Ri^2
Since for an isolated system in which |Vi - Vo| << C both Ei and (Wi Ri^2) are constant:
|Lci| = constant

Total angular momentum is given by:
Lc = Sum of all Lci
Hence, for |Vi - Vo| << C, for an isolated cluster of singularities total angular momentum Lc is approximately constant.

An element of momentum dP can transfer from one object to another object within an isolated cluster of interacting objects but the Center of Momentum (CM) Xc, Vc for the cluster remains inertial and subject to |Vi - Vo| << C the total angular momentum remains approximately constant. This issue has major implications in terms of planetary motion and attempting to deflect from planet Earth potentially impacting comets and other projectiles. Even if these projectiles are blown up their CM paths and total angular momenta remain unchanged.

This web page last updated December 27, 2013.