Energy, momentum and linear algebra | Wild Linear Algebra B 36 | NJ Wildberger

We introduce some elementary mechanics relating to conservation of momentum and energy using linear algebra. For simplicity we work in a one dimensional situation, but introduce a two dimensional space-time to interpret (elastic) collisions geometrically.

We obtain a pleasant butterfly collision diagram which explains what happens when two particles of different masses and speeds collide.

************************

Here are the Insights into Mathematics Playlists:

Here are the Wild Egg Maths Playlists (some available only to Members!)

************************

[pickeyberry]

Video Content

00:00 Introduction
03:04 Energy
06:34 Momentum
11:10 Conservation of energy
14:15 Elastic collision
23:20 Final velocity formula
28:37 Butterfly collision diagram

[lund]

Contains a few surprises which I enjoyed very much! Thanks.

[arthurbaraov]

Three dimensional situation is also amenable to linear algebra treatment. Furthermore, working out three dimensional case is not more complicated than one dimensional one.

The trick is to introduce a vector of momentum exchange ΔP that takes place in the elastic collision of material points m1 and m2. Let V1 and V2 stand for the velocities before the collision, and U1 and U2 – after the collision, accordingly; V1, V2, U1, U2 are three dimensional vectors here. As the result of elastic collision one object (say m1) is losing momentum in the amount of ΔP, while the other is gaining the same amount of momentum ΔP; so using dot product notation conservation of kinetic energy can be presented as follows: 
 
m1(V1.V1)/2 +  m2(V2.V2)/2 = m1[(V1–ΔP/m1).(V1–ΔP/m1)]/2 + m2[(V2+ΔP/m2).(V2+ΔP/m2)]/2

After little algebra we get

ΔP = 2m1m2(V1–V2)/(m1+m2).

Therefore
U1 = V1 – ΔP/m1 = V1 – 2m2(V1–V2)/(m1+m2),
U2 = V2 + ΔP/m2 = V2 + 2m1(V1–V2)/(m1+m2).

As always, Norman Wildberger presents excitingly new ways of seeing old things in mathematics. But, perhaps,  it is the physics that needs most a new way of seeing old things. For example,  isn't it odd that the classical formula for kinetic energy contains one mass only, while the gravitational potential energy involves two masses. Or, did it ever occur to you to ask: where does kinetic energy reside?  
 
See a discussion of similar questions here: 

[jehovajah]

At 17:47 you do not use a signed speed for the initial condition! So or a collision to occur you must distinguish cases.

[davidkeirsey]

This is great, I think I can explain different "kinds" of "time", "mass", "energy" and "spaces" (Lie and "thin" groups).  Thanks heaps.  (That's partial ordered Lie heaps ;-)  )  You explained "what mass is" in the limited context, but from here one can generalize without ambiguity.

[shawnheneghan]

I'm surprised that you use v (the length of the velocity vector) squared in energy equation instead of the correct inner product of the vector v with itself.

[hyperduality]

Symmetry is dual to conservation -- the duality of Noether's theorem.
Space is dual to time -- Einstein.
Action is dual to reaction -- Sir Isaac Newton (the duality of force).
Potential energy is dual to kinetic energy.
"Always two there are" -- Yoda.

[pauluk]

The world-line of a particle (in slide 7, beginning 23:20) is very interesting to think about. A horizontal line seems out of the question because it implies such a 'point' (our concept for the object) must be 'everywhere' at once. Physically impossible but mathematically allowed I suppose. The point-object must utilize a second dimension (time) in order to take any different value of position to make its way through the space-time, which implies only finite velocities.

A point with 'zero' velocity however traces a vertical world-line, which _is_ allowed in our physics, but for 'how long' can we sit and observe it always in exactly the same position? We can only look at a given object's world-line 'locally' (for a certain period) in practice.

[TheLetterW]

Is this formulation possible for a system of any number of particles?  My naive intuition would be to simply add more column entries to the transformation matrix and adding more row entries to the final and initial velocities in the final velocity formula.  Could the column entries be calculated using similar methods used here, or does the fact that two particles are used in this example somehow make the problem simple?

[jehovajah]

At 33:00 signs of points are confusing! Why does v2 have a negative sign or why does v1 have a minus sign?they are on different sides of the x collision point at t0!

[robkim]

Does Noether's theorem extend to all coordinates [ i.e in String Theory]??? where there are 10 to 11 coordinates.

[jehovajah]

There are physical difficulties with this derivation if it is moved outside of a line! The lineal algebra approach is interesting in this case, but too special fir general application. While it is in keeping to use the term vector, it is confusing to do so. 3 different kinds of" vector" are used in this de, Ritation!

[relikep]

Incidentally, is Noether the same person that Noetherian rings is named after?

[shawnheneghan]

You kind of assumed that v1 .NE. u1 and v2 .ne. u2 otherwise you could not divide equation 4 by eq 3.