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I apologize for the goofy commutator $\left[\left[\_,\_\right]\right]$ notation. MathJax doesn't like my \llbracket \rrbracket notation. And I religiously use $\left[\dots\right]$ for function arguments.


Edit to add for future reference: \left[\![\_,\_\right]\!] $\to\left[\![ \_,\_ \right]\!]$
This is from Chapter 9, Box 9.2 of Gravitation, by Charles W. Misner, Kip S. Thorne & John Archibald Wheeler. It seems incorrect to me. Is it?

MTW Box 9.2 A

A. Pictoral representation of flat spacetime

  1. For ease of visualization, consider flat spacetime, so the two vector fields $\mathfrak{u}\left[\mathscr{P}\right]$ and $\mathfrak{v}\left[\mathscr{P}\right]$ can be laid out in spacetime itself.

  2. Choose an event $\mathscr{P}_{0}$ where the commutator $\left[\left[ \mathfrak{u},\mathfrak{v}\right]\right] $ is to be calculated.

  3. Give names $\mathscr{P}_{1},\mathscr{P}_{2},\mathscr{P}_{3},\mathscr{P}_{4}$ to the events pictured in the diagram.

  4. Then the vector $\mathscr{P}_{4}-\mathscr{P}_{3},$ which measures how much the four-legged curve fails to close, can be expressed in the coordinate basis

$$ \mathscr{P}_{4}-\mathscr{P}_{3}=\left(\mathfrak{u}\left[\mathscr{P}_{0}\right]+\mathfrak{v}\left[\mathscr{P}_{1}\right]\right)-\left(\mathfrak{u}\left[\mathscr{P}_{2}\right]+\mathfrak{v}\left[\mathscr{P}_{0}\right]\right)$$

$$ =\left(\mathfrak{v}\left[\mathscr{P}_{1}\right]-\mathfrak{v}\left[\mathscr{P}_{0}\right]\right)-\left(\mathfrak{u}\left[\mathscr{P}_{2}\right]-\mathfrak{u}\left[\mathscr{P}_{0}\right]\right) $$

$$ =\left(v^{\beta}{}_{,\alpha}u^{\alpha}\mathfrak{e}_{\beta}\right)_{\mathscr{P}_{0}}-\left(u^{\beta}{}_{,\alpha}v^{\alpha}\mathfrak{e}_{\beta}\right)_{\mathscr{P}_{0}}+\text{errors} $$

$$ =\left[\left[\mathfrak{u},\mathfrak{v}\right]\right]_{\mathscr{P}_{0}}+\text{errors}, $$ where $\text{errors}$ consits of terms such as $v^{\beta}{}_{,\mu\nu}u^{\mu}u^{\nu}\mathfrak{e}_{\beta}.$

  1. Notice that if $\mathfrak{u}$ and $\mathfrak{v}$ are halved everywhere, then $\left[\left[ \mathfrak{u},\mathfrak{v}\right]\right] $ is cut down by a factor of 4, while the error terms in the above go down by a factor of 8.

From what is given, $v^{\beta}{}_{,\alpha}$ and $u^{\beta}{}_{,\alpha}$ are evaluated at $\mathscr{P}_{0},$ so they remain constant as $\mathfrak{u}$ and $\mathfrak{v}$ are reduced in magnitude. Call $\left\{ u^{\beta}{}_{,\alpha}\right\} _{\mathscr{P}_{0}}=\left\{ a^{\beta}{}_{\alpha}\right\} $ and $\left\{ v^{\beta}{}_{,\alpha}\right\} _{\mathscr{P}_{0}}=\left\{ b^{\beta}{}_{\alpha}\right\} $, which are constants in the limit as $\mathfrak{u}$ and $\mathfrak{v}$ go to $\mathfrak{0}$; as are $\left\{ \mathfrak{e}_{\beta}\right\} _{\mathscr{P}_{o}}$. So

$$ \left(v^{\beta}{}_{,\alpha}u^{\alpha}\mathfrak{e}_{\beta}\right)_{\mathscr{P}_{0}}-\left(u^{\beta}{}_{,\alpha}v^{\alpha}\mathfrak{e}_{\beta}\right)_{\mathscr{P}_{0}} $$

$$ =\left(b^{\beta}{}_{\alpha}u^{\alpha}-a^{\beta}{}_{\alpha}v^{\alpha}\right)\mathfrak{e}_{\beta} $$

$$ =\left[\left[ \mathfrak{u},\mathfrak{v}\right]\right] _{\mathscr{P}_{0}}, $$

and

$$ \left[\left[ \frac{\mathfrak{u}}{2},\frac{\mathfrak{v}}{2}\right]\right] _{\mathscr{P}_{0}}=\frac{1}{2}\left[\left[ \mathfrak{u},\mathfrak{v}\right]\right] _{\mathscr{P}_{0}}. $$

So I get a factor of 2, not 4. The graphic actually uses differentiable vector fields. When I reduce the vectors by half, the commutator is reduced by half. Apparently the polygon representing the open quadrilateral retains its shape as the vectors are uniformly reduced in magnitude.


Edit to add: I believe I figured this out. The displacements along $\mathfrak{u}\left[\mathscr{P}\right]$ and $\mathfrak{v}\left[\mathscr{P}\right]$ represent unit changes in coordinate values, so reducing the vectors by half reduces the coordinate mesh, and therefore the scale by which we are differentiating.
Edit to retract the previous suggestion. It is exactly wrong. The curves to which $\mathfrak{u}$ and $\mathfrak{v}$ are tangent are not in general coordinate curves.

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  • $\begingroup$ It should just be as simple as (omitting indices) $[u,v]=u\partial v-v\partial u$ so if both $u$ and $v$ are reduced by half then the products $u\partial v$ and $v\partial u$ are reduced by 1/4 since constants aren't affected by derivatives. $\endgroup$ – enumaris Jul 20 '18 at 18:37
  • $\begingroup$ It doesn't work to use the result of a proof to prove the result. $\endgroup$ – Steven Thomas Hatton Jul 20 '18 at 18:46
  • $\begingroup$ Ah, sorry, I'm not able to actually view the image. I thought you were just trying to visualize the commutator $[u,v]$ and got stuck on a specific aspect of the visualization. $\endgroup$ – enumaris Jul 20 '18 at 18:54
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    $\begingroup$ Why do you say that derivatives remain constant if the vectors are halved? That's not the case. $\endgroup$ – Javier Jul 21 '18 at 14:57
  • $\begingroup$ With the amount of information given, it's hard to understand what is intended. I assume there are two continuous families of parameterized curves which intersect, at least in some cases. The tangent vectors form the vector fields. These curves are somehow prior to the introduction of their parameterization. I assume I can compare proper time and proper distance along these curves using identical clocks and measuring rods, but cannot measure "diagonally". Since there are partial derivatives, there must be a coordinate system. I assume no affine parameterization. $\endgroup$ – Steven Thomas Hatton Jul 21 '18 at 17:01
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This is far from rigorous, but it explains why the partial derivatives $u^{\beta}{}_{,\alpha}$ and $v^{\beta}{}_{,\alpha}$ scale proportionally with a uniform scaling of the parameters.

I shall assume that the families of curves to which the vector fields $\mathfrak{u}$ and $\mathfrak{v}$ are parallel remain fixed as parameters are uniformly scaled. The included figure shows a very simply example representing only one family of tangent curves. These are the horizontal blue $v$ "curves". The vertical black line at the left side represents the set of points on adjacent $v$ curves with a parameter value of $0$. It is assumed that the parameterization changes continuously as "adjacent" curves are encountered when "cutting across the grain" of the family of $v$ curves.

The various curves which begin vertically at the bottom represent points of equal parameter value along the $v$ curves. The black arrows lying under the red arrows depict the tangent vectors to the originally parameterized $v$ curves. The read arrows represent the tangent vectors when the parameter units are uniformly scaled by .5. The black and red arrows share common tails, intentionally placed at a common integer x value.

The rectangular Cartesian coordinate system is shown as a light green grid.

Between the tips of the black arrows is a line segment approximating the rate of change in the originally parameterized tangent vectors, separated in the y direction by a unit change in coordinate value. The line segment joining the tips of the red arrows approximates the rate of change in the tangent vectors after scaling the parameters.

The "slopes" $\frac{\Delta x}{\Delta y}$ of the joining line segments approximate the partial derivatives $v^{x}{}_{,y}$ which I had mistakenly taken to be constant under a uniform change of parameter.

The red arrows are approximately half as long as the corresponding black arrows, and become exactly half in the limit as the parameters are reduced to zero.

enter image description here

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