Physics - Energy and Momentum using Multivectors

Prerequisites

If you are not familiar with this subject you may like to look at the following pages first:

using Multivectors

If we specify a velocity using a 3D multivector (explained here) and choose to define this as follows:

Scalar (e) = 1
Vector (ex,ey,ez) = vx,vy,vz
BiVector (eyz,ezx,exy) = 0
TriVector (exyz) = 1

where:other definitions
symbol
description
type
units
ex,ey,ez basis vectors vector  
exyz   trivector  
vx,vy,vz components of velocity in x,y and z dimensions vector  

So the velocity multivector is:

v = 1 + ex * vx + ey * vy + ez * vz + exyz

We can new calculate the square of the velocity using multi-vector multiplication which gives:

v2 = x2 + y2 + z2 + ex * 2 * vx + ey * 2 * vy + ez * 2 * vz + eyz * 2 * vx + ezx * 2 * vy + exy * 2 * vz + exyz * 2

We can new calculate m * v2/2 which gives the following:

m * v2/2 = (x2 + y2 + z2)*m/2+ ex * m * vx + ey * m * vy + ez * m * vz + eyz * m * vx + ezx * m * vy + exy * m * vz + exyz * m

This has the following components:

Scalar (e) = (x2 + y2 + z2)*m/2 = kinetic energy
Vector (ex,ey,ez) = m * vx ,m * vy, m * vz = linear momentum
BiVector (eyz,ezx,exy) = m * vx ,m * vy, m * vz = linear momentum
TriVector (exyz) = m = mass

So it is made up of constants for linear motion, for example in a perfectly elastic collision between two objects, if we calculate this quantity for both objects before the collision and after the collision, the total of both objects before the collision should be the same as the total after the collision.

Adding rotational energy

The above only includes linear motion, it would be good if we could find a way to include rotational energy and momentum. So here we try adding a rotational velocity (w) to the velocity multivector as follows:

Scalar (e) = 1
Vector (ex,ey,ez) = vx,vy,vz
BiVector (eyz,ezx,exy) = wx,wy,wz
TriVector (exyz) = 1

where:
symbol
description
type
units
w angular velocity bivector s-1
angle angle in radians scalar none
t time scalar s
d ... /dt rate of change    

where:

So the velocity multivector is:

v = 1 + ex * vx + ey * vy + ez * vz +eyz * wx + ezx * wy + exy * wz + exyz

We can new calculate the square of this velocity which gives:

e = 1 + vx * vx + vy * vy + vz * vz - wz * wz - wy * wy - wx * wx - 1
ex = vx vx - wz * vy + wy * vz + vy * wz - vz * wy - 1 * wx - wx
ey = vy + wz * vx vy - wx * vz - vx * wz - 1 * wy + vz * wx - wy
ez = vz - wy * vx + wx * vy vz - 1 * wz + vx * wy - vy * wx - wz
exy = wz + vy * vx - vx * vy vz wz + wx * wy - wy * wx + vz
ezx = wy - vz * vx vy + vx * vz - wx * wz wy + wz * wx + vy
eyz = wx vx + vz * vy - vy * vz + wy * wz - wz * wy wx + vx
exyz = 1 + wx * vx + wy * vy + wz * vz + vz * wz + vy * wy + vx * wx + 1

Cancelling out gives:

e = vx * vx + vy * vy + vz * vz - wz * wz- wy * wy - wx * wx
ex = 2 * vx - 2 * wx
ey = 2 * vy - 2 * wy
ez = 2 * vz - 2 * wz
exy =2 * vx + 2 * wx
ezx =2 * vy + 2 * wy
eyz =2 * vz + 2 * wz
exyz = 2 + 2 * wx * vx + 2 * wy * vy + 2 * wz * vz

This is a nice symmetrical result, but I cant yet work out how to relate this to angular momentum and energy, can anyone help me?


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flag flag flag flag flag flag New Foundations for Classical Mechanics (Fundamental Theories of Physics). This is very good on the geometric interpretation of this algebra. It has lots of insights into the mechanics of solid bodies. I still cant work out if the position, velocity, etc. of solid bodies can be represented by a 3D multivector or if 4 or 5D multivectors are required to represent translation and rotation.

 

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