Searching for a Solid Proof of Chasles/Mozzi/Cauchy's Theorem

  • Thread starter Thread starter ObsessiveMathsFreak
  • Start date Start date
  • Tags Tags
    Proof Solid Theorem
AI Thread Summary
Chasles' Theorem states that any rigid body displacement can be represented as a screw motion, involving a circular helical path about a common axis. The original poster seeks a solid, linear algebraic proof, expressing dissatisfaction with existing proofs that lack rigor. They discovered a highly recommended paper by Dunham Jackson that thoroughly explains rigid body motions and their decomposition into screw motions, although it lacks diagrams. The poster has developed a semi-proof for the 2D case and is interested in exploring the 3D case further. The discussion highlights the importance of finding clear, convincing proofs for complex mathematical theorems.
ObsessiveMathsFreak
Messages
404
Reaction score
8
I've unsuccessfully been looking for a decent proof of Chasles' Theorem which states that any rigid body displacement whatsoever can be decomposed into a screw motion. In other words, no matter what the displacement is, you can consider it the result of the partiles having moved to their positions by following a circular helixical path about a common axis, with a common angular speed.

I suppose I'm mostly looking for a (linear) algebraic type proof. Most proofs I've encountered have been fairly loose and unconvincing. Does anyone know of a solid proof of this theorem?

Awkwardly, this theorem is also variously known as Mozzi's or Cauchy's screw theorem.
 
Physics news on Phys.org
In answer to my own question:

Shortly after making this post I stumbled across one of the best papers I've ever read.
http://www.jstor.org/view/00029890/di991259/99p1550p/0 by Dunham Jackson. It's old, but great, and completely and totally explains rigid body motions and their decompositions into rotations and translations and finally into screw motions. He refers to the theorem as Mozzi's theorem.

It's a great paper. I'd seriously recommend anyone to give it an hour of their time. It's an easy read, and the proofs are just brilliant. The only thing I could fault it on is a lack of diagrams.
 
Last edited by a moderator:
Thanks for the link, this is a problem I've been annoyed with for a while too. I've worked out a semi-proof for the 2d case using similar triangles, but I would like to see the 3d case as well. Do you have a link for a free copy of the paper?
 
EFuzzy said:
Thanks for the link, this is a problem I've been annoyed with for a while too. I've worked out a semi-proof for the 2d case using similar triangles, but I would like to see the 3d case as well. Do you have a link for a free copy of the paper?
Attached is a copy of the paper.
 

Attachments

Thanks!
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...

Similar threads

Replies
12
Views
5K
Replies
7
Views
3K
Replies
9
Views
4K
Replies
6
Views
4K
Replies
1
Views
2K
Back
Top