Is parallel transport exactly norm-preserving or only approximately?

[Kostik]

TL;DR

Is parallel transport exactly norm-preserving or only approximately?

Dirac ("General Theory of Relativity", pp. 12-13, see below) shows the following. Let $A_\nu (x)$ be a vector, and let
$$K_\nu (x+dx) = A_\nu (x) + dA_\nu (x)$$ be the vector after parallel transport from $x$ to $x+dx$. The formula $$dA_\nu = A_\mu \Gamma^\mu_{\nu\sigma}dx^\sigma $$ is given as equation (7.7) below.

Dirac shows (leading up to and following equation (7.8)) that $$d(A^\nu A_\nu) = 0 \,\,$$ This would seem to say that the length of $A_\nu$ remains unchanged under parallel displacement; that is, it is norm-preserving.

Yet, if you calculate:
$$K^\nu K_\nu (x+dx) = (A^\nu + dA^\nu)(A_\nu + dA_\nu) = A^\nu A_\nu + d(A^\nu A_\nu) + dA^\nu dA_\nu $$ $$ = A^\nu A_\nu + O[(dA_\nu)^2]$$ you seem to find that $||K_\nu||^2$ is equal to $||A_\nu||^2$ only up to order $O[(dx^\nu)^2]$.

This seems strange. Either parallel transport is either exactly norm-preserving, or it isn't?

Note: Dirac says at the bottom of p. 13: "It follows that, to the first order, the length of the whole vector equals that of its tangential part" (i.e., the parallel-transported vector). So, Dirac also seems to be suggesting that parallel transport is norm-preserving only to the first order in the coordinate displacement.

 

[martinbn]

It is norm preserving.

 

[Kostik]

It is norm preserving.

So the mistake in the calculation of $||K_\nu||^2$ is ..... ?

 

[Dale]

It is exact. The expression you are writing is not an equality. It is an approximation to first order.
$$K_\nu (x+dx) \approx A_\nu (x) + dA_\nu (x)$$

 

[Kostik]

It is exact. The expression you are writing is not an equality. It is an approximation to first order.

Exactly my point. Where is the error in the calculation of $||K_\nu||^2$?

 

[Dale]

The error is where you say $$K_\nu (x+dx) = A_\nu (x) + dA_\nu (x)$$

That is not true. It should be $$K_\nu (x+dx) \approx A_\nu (x) + dA_\nu (x)$$ or $$K_\nu (x+dx) = A_\nu (x) + dA_\nu (x) + O[d x^2]$$

 

[Kostik]

That is not correct. Following Dirac, $K_\nu (x+dx) = A_\nu + dA_\nu$ is the definition of $dA_\nu$ (see the equation on p. 12 just before (6.7)).

And $dA_\nu$ is given in terms of the metric as above, or in (7.7).

 

[Dale]

That is not correct. Following Dirac, $K_\nu (x+dx) = A_\nu + dA_\nu$ is the definition of $dA_\nu$ (see the equation on p. 12 just before (6.7)).

And $dA_\nu$ is given in terms of the metric as above, or in (7.7).

It is correct. Dirac said explicitly that it was only to first order:

This means in effect "in the following equations add a term $O[d x^2]$ which I am not going to bother to write"

 

[PeroK]

TL;DR Summary: Is parallel transport exactly norm-preserving or only approximately?

Dirac ("General Theory of Relativity", pp. 12-13, see below) shows the following. Let $A_\nu (x)$ be a vector, and let
$$K_\nu (x+dx) = A_\nu (x) + dA_\nu (x)$$ be the vector after parallel transport from $x$ to $x+dx$. The formula $$dA_\nu = A_\mu \Gamma^\mu_{\nu\sigma}dx^\sigma $$ is given as equation (7.7) below.

Dirac shows (leading up to and following equation (7.8)) that $$d(A^\nu A_\nu) = 0 \,\,$$ This would seem to say that the length of $A_\nu$ remains unchanged under parallel displacement; that is, it is norm-preserving.

Yet, if you calculate:
$$K^\nu K_\nu (x+dx) = (A^\nu + dA^\nu)(A_\nu + dA_\nu) = A^\nu A_\nu + d(A^\nu A_\nu) + dA^\nu dA_\nu $$ $$ = A^\nu A_\nu + O[(dA_\nu)^2]$$ you seem to find that $||K_\nu||^2$ is equal to $||A_\nu||^2$ only up to order $O[(dx^\nu)^2]$.

This seems strange. Either parallel transport is either exactly norm-preserving, or it isn't?

Note: Dirac says at the bottom of p. 13: "It follows that, to the first order, the length of the whole vector equals that of its tangential part" (i.e., the parallel-transported vector). So, Dirac also seems to be suggesting that parallel transport is norm-preserving only to the first order in the coordinate displacement.

View attachment 361010

If you look at the calculation, Dirac proves that the differential of the vector's length is zero. This is where he says that the expression in 7.8 vanishes.

His calculation does not rely on the first order approximation.

 

[PeroK]

That is not correct. Following Dirac, $K_\nu (x+dx) = A_\nu + dA_\nu$ is the definition of $dA_\nu$ (see the equation on p. 12 just before (6.7)).

And $dA_\nu$ is given in terms of the metric as above, or in (7.7).

Dirac nay be mixing notations here. If $dx$ is a differential or an infinitesimal, then the equation is exact- not to first order. It only makes sense to say to first order if $dx$ is a small finite change.

 

[PeroK]

For example, if $y = \sin x$, then:
$$dy = \cos x \ dx$$But:$$\Delta y \approx \cos x \ \Delta x$$

 

[Dale]

For example, if $y = \sin x$, then:
$$dy = \cos x dx$$But:$$\Delta y \approx \cos x \Delta x$$

And the explicit “to first order” means “don’t worry about the distinction between the two”

 

[Kostik]

View attachment 361011

Dang, you're right: of course, the key step here was the Taylor approximation which you pasted. Many thanks - this is much appreciated.

 

[Kostik]

If you look at the calculation, Dirac proves that the differential of the vector's length is zero. This is where he says that the expression in 7.8 vanishes.

His calculation does not rely on the first order approximation.

Well, that's not exactly true. As @Dale pointed out, if we write $K_\nu (x+dx) = A_\nu + dA_\nu$ (exactly), then we have, instead of Dirac's (7.7), the more complete equation $$dA_\nu = A^\mu \Gamma_{\mu\nu\sigma}dx^\sigma + O(dx^2) \,\,.$$ Dirac drops the $O(dx^2)$ in his calculation of $d(A^\nu A_\nu)$. Thus, more correctly Dirac should get $$d(A^\nu A_\nu)=O(dx^2) \,\,.$$ So, the vector $A_\nu$ parallel transported by $dx$ has the same norm, with an error of $O(dx)$ (after taking square roots).

 

[Orodruin]

So, the vector Aν parallel transported by dx has the same norm, with an error of O(dx) (after taking square roots).

No. The norm is exactly preserved under parallel transport with a metric compatible affine connection.

For any vector $X$ parallell transported along a curve $\gamma$ with curve parameter $s$ holds that
$$
\frac{d(g(X,X))}{ds}
= \nabla_{\dot\gamma} g(X,X)
= [\nabla_{\dot\gamma} g](X,X) + 2 g(X,\nabla_{\dot\gamma}X)
= 0
$$
The first term vanishes due to the connection being metric compatible and the second by the assumption that $X$ is parallel transported along $\gamma$.

 

[PeroK]

Well, that's not exactly true. As @Dale pointed out, if we write $K_\nu (x+dx) = A_\nu + dA_\nu$ (exactly), then we have, instead of Dirac's (7.7), the more complete equation $$dA_\nu = A^\mu \Gamma_{\mu\nu\sigma}dx^\sigma + O(dx^2) \,\,.$$ Dirac drops the $O(dx^2)$ in his calculation of $d(A^\nu A_\nu)$. Thus, more correctly Dirac should get $$d(A^\nu A_\nu)=O(dx^2) \,\,.$$ So, the vector $A_\nu$ parallel transported by $dx$ has the same norm, with an error of $O(dx)$ (after taking square roots).

That's not valid. You are mixing differentials with small finite changes. Alternatively, if $dx$ is an infinitesimal then $dx^2 =0$ by definition.

 

[Kostik]

That's not valid. You are mixing differentials with small finite changes. Alternatively, if $dx$ is an infinitesimal then $dx^2 =0$ by definition.

$dx$ is what it is, $dx$. No need to call it "an infinitesimal". "$dx^2 =0$ by definition" makes no sense.

 

[Kostik]

No. The norm is exactly preserved under parallel transport with a metric compatible affine connection.

For any vector $X$ parallell transported along a curve $\gamma$ with curve parameter $s$ holds that
$$
\frac{d(g(X,X))}{ds}
= \nabla_{\dot\gamma} g(X,X)
= [\nabla_{\dot\gamma} g](X,X) + 2 g(X,\nabla_{\dot\gamma}X)
= 0
$$
The first term vanishes due to the connection being metric compatible and the second by the assumption that $X$ is parallel transported along $\gamma$.

Would be helpful if you could write this in the language Dirac uses. Only need to consider the two points $x$ and $x+dx$.

 

[PeroK]

$dx$ is what it is, $dx$. No need to call it "an infinitesimal". "$dx^2 =0$ by definition" makes no sense.

Physicists often treat differentials as infinitesimals. That predates the rigorous formulation of infinitesimals.

In any case, the differential form is exact. It is not a Taylor linear approximation.

Dirac was mixing notations when he talked about "to first order".

 

[PeroK]

See the discussion here, for example, starting with the answer "there is an old tradition":

https://physics.stackexchange.com/questions/92925/how-to-treat-differentials-and-infinitesimals

For more on infinitesimals:

https://www.sjsu.edu/faculty/watkins/infincalc.htm

 

[Dale]

Alternatively, if $dx$ is an infinitesimal then $dx^2 =0$ by definition.

For dual numbers this is true, but for hyperreals it is not. I know there are other classes of numbers that have infinitesimals, but I only know this for the hyperreals and dual numbers

 

[Dale]

No need to call it "an infinitesimal". "dx2=0 by definition" makes no sense

It makes sense in the dual numbers. Just like $i^2=-1$ makes sense in the complex numbers.