Interpreting Path Ordered Exponentials with Non-Sensical Integration Variables

[etotheipi]

I see a term like$$\int_0^{t''} \left[ \int_{t'}^t \omega(t'') \omega(t') dt' \right] dt''$$here $\omega$ is a matrix. How to interpret this integral - the integration variables are in the limits, in places where they don't make sense. Is that a mistake? It's given that the range of integration is $0 < t' < t'' < t$, i.e. a triangle in the $t'$-$t''$ plane. Reference is top of page 50 here. Thanks!

 

[fresh_42]

I think $\omega\, : \,[0,1] \longrightarrow \operatorname{SO}(3)$ is simply a path on the manifold. The point is to demonstrate why general, i.e. non-commuting matrices cannot be handled the same way as (commuting) scalars. The "let's pretend as if" part is a bit confusing, since he uses the same variables.

 

[wrobel]

Hope it will be of some use

formula (3.11) at page 49 means that $\omega(t)\in T_e(SO(3))$

formula (3.16) follows from the standard successive procedure of the contraction mapping principle for the equation $R=E+\int_0^t\omega Rd\xi$.

 

[etotheipi]

That's interesting! Maybe a bit above my level of maths. Let me see what I understand; by the contraction mapping principle, the successive approximations to $R(t)$ are $$R_n(t) = 1 + \int_0^t \omega(\xi) R_{n-1}(\xi) d\xi$$So for instance$$R_1(t) = 1 + \int_0^t \omega(\xi) d\xi$$and the next approximation is$$\begin{align*}

R_2(t) = 1 + \int_0^t \omega(\xi) R_1(\xi) d\xi &= 1+ \int_0^t \omega(\xi) \left[ 1 + \int_0^t \omega(\eta) d\eta \right] d\xi \\

&= 1 + \int_0^t \omega(\xi) d\xi + \int_0^t \int_0^t \omega(\xi) \omega(\eta) d\eta d\xi

\end{align*}$$and then continue on and take $R(t) = \lim_{n \rightarrow \infty}R_n(t)$. Except, I have some different limits in the second integral here to Tong...

 

[wrobel]

You write $R_1(\xi)$ it is correct ; but then you must substitute $R_1(\xi)=1+\int_0^\xi\omega(s)ds$

 

[etotheipi]

You write $R_1(\xi)$ it is correct ; but then you must substitute $R_1(\xi)=1+\int_0^\xi\omega(s)ds$

Ah, yes that's it, thanks ☺