Why ##A_{\nu:\sigma}=0## in flat space?

[Kostik]

TL;DR Summary

Dirac asserts that in flat space, $A_{\nu:\sigma}=0$ in flat space?

In Dirac's GTR. Sec. 12 (p. 22), he wants to show the equivalence of:
(a) Vanishing of the curvature tensor $R^\beta_{\sigma\nu\rho}=0$; or equivalently, the equality of mixed second covariant derivatives $A_{\nu:\sigma:\rho}=A_{\nu:\rho:\sigma}$.
(b) Path independence of parallel transport
(c) The existence of rectilinear coordinates (hence a constant metric).

The equivalence (a) $\leftrightarrow$ (b) is easy to show. Then Dirac simply says that if you take a vector $A_{\nu}$ at a point $x$ and parallel transport it to every point in the space, you have a vector field $A_{\nu}(x)$ with $A_{\nu:\sigma}=0$. That's where I am stuck.

He is saying that if the parallel transport of $A_\nu$ from $x$ to $y$ is path-independent, then $A_{\nu:\sigma}=0.$

Now, $A_{\nu}(y)=A_{\nu}(x)+\int_{C}A_{\nu:\sigma}dx^\sigma$ for any path $C$ from $x$ to $y.$ Equivalently, $\oint_{C}A_{\nu:\sigma}dx^\sigma=0.$ How does this imply $A_{\nu:\sigma}=0$?

It seems like quite a strong statement to assert that $A_{\nu:\sigma:\rho}=A_{\nu:\rho:\sigma}$ implies $A_{\nu:\sigma}=0$.

Is there a simple explanation in the simple (calculus-component based) language of Dirac's book?

Edit: Dirac's assertion cannot be correct. The contrapositive would be $A_{\nu:\sigma}\neq 0$ implies $A_{\nu:\sigma:\rho}\neq A_{\nu:\rho:\sigma}$, which is plainly false: take $A_{\nu:\sigma}(x) = \text{constant}.$

 

[Ibix]

Clearly the integral is zero because your two $A_\nu$ are equal. If the integral is zero it's either because the integrand is zero everywhere or because it's positive in some places and negative elsewhere. If it's positive in some places I can pick a part of your curve $C$ that excludes some of the places where the integrand is positive without excluding any of the places where it's negative and get a non-zero value of the integral.

In short, the integral must be zero for all paths between all pairs of points, and the only way for that to hold is for the integrand to be everywhere zero.

 

[Kostik]

Clearly the integral is zero because your two $A_\nu$ are equal. If the integral is zero it's either because the integrand is zero everywhere or because it's positive in some places and negative elsewhere. If it's positive in some places I can pick a part of your curve $C$ that excludes some of the places where the integrand is positive without excluding any of the places where it's negative and get a non-zero value of the integral.

In short, the integral must be zero for all paths between all pairs of points, and the only way for that to hold is for the integrand to be everywhere zero.

The integral $\oint_{C}A_{\nu:\sigma}dx^\sigma=\oint_{C}A_{\nu:\sigma} \cdot d\vec{l}=0$ is a line integral (actually one integral for each $\nu$). By Stokes' theorem, $\int_{S} (\nabla\times A_{\nu:\sigma}) \cdot d\vec{S}=0$, where $C$ is the oriented boundary of $S$. This seems to suggest that $\nabla\times A_{\nu:\sigma}=0$, which then implies that $A_{\nu:\sigma}$ is itself some kind of gradient ... but this can't be what Dirac was getting at.

 

[renormalize]

$A_{\nu}(y)=A_{\nu}(x)+\int_{C}A_{\nu:\sigma}dx^\sigma$ for any path $C$ from $x$ to $y.$

This equation is incorrect. Where did you find it?

$\oint_{C}A_{\nu:\sigma}dx^\sigma=\oint_{C}A_{\nu:\sigma} \cdot d\vec{l}=0$
$\int_{S} (\nabla\times A_{\nu:\sigma}) \cdot d\vec{S}=0$

These equations are also wrong. They nonsensically mix 4D tensor quantities with 3D vector manipulations.
Look at, e.g., https://scipp.ucsc.edu/~haber/ph171/parallel_transport15.pdf for a proper discussion of integrating parallel-transport around an infinitesimal closed loop.

 

[Kostik]

This equation is incorrect. Where did you find it?

These equations are also wrong. They nonsensically mix 4D tensor quantities with 3D vector manipulations.
Look at, e.g., https://scipp.ucsc.edu/~haber/ph171/parallel_transport15.pdf for a proper discussion of integrating parallel-transport around an infinitesimal closed loop.

I know the Stokes Theorem comment was somewhat fanciful, I was just trying to emphasize that the integral in question is an oriented line integral.

Back to the original question: path independence of parallel transport is easily seen to be equivalent to the vanishing of the curvature tensor (or the commutativity of covariant derivatives). I do not see how to show that $A_{\nu:\sigma}=0$ and I provided a counterexample above.

I suspect all we can say is that $A_{\nu:\sigma}=\phi_{\nu,\sigma}$ for some scalar field $\phi_\nu$ (considering $\nu$ as fixed), i.e., $A_{\nu:\sigma}$ is a gradient.

Edit: if it is true that $\oint A_{\nu:\sigma}=0$ implies $A_{\nu:\sigma}=\phi_{\nu,\sigma}$ for some scalar field $\phi_\nu$, considering $\nu$ as fixed, (using Stokes plus Poincaré’s theorems), then I can show $A_{\nu:\sigma}=0$, but this cannot be what Dirac had in mind.

 

[martinbn]

Then Dirac simply says that if you take a vector $A_{\nu}$ at a point $x$ and parallel transport it to every point in the space, you have a vector field $A_{\nu}(x)$ with $A_{\nu:\sigma}=0$. That's where I am stuck.

He is saying that if the parallel transport of $A_\nu$ from $x$ to $y$ is path-independent, then $A_{\nu:\sigma}=0.$

No, he is not saying that. He says that if you start with a tangent vector at a point $x$, you can parallel transport it to all points (this is well defined because transport is path independent) and obtain a vector field whose covariant derivative is zero.

 

[Kostik]

Yes I know. How do you show it?

 

[martinbn]

Yes I know. How do you show it?

How do you show what?

 

[Kostik]

How do you show that the covariant derivative is zero. Given that the curvature tensor vanishes (ie second covariant derivatives commute).

Please see original post.

 

[martinbn]

How do you show that the covariant derivative is zero. Given that the curvature tensor vanishes (ie second covariant derivatives commute).

Please see original post.

They statement he makes is that the covariant derivative of the field he constructs is zero. And it is zero by constructions. Parallelly transported means that the derivative along the curve is zero.

 

[Kostik]

Parallel transport of $A_\nu(x)$ to $x+dx$ is a vector $K_\nu(x+dx)=A_\nu(x+dx)-A_{\nu:\sigma}dx^\sigma$.

Path independence means the curvature tensor vanishes. Equivalently, $A_{\nu:\sigma:\rho}=A_{\nu:\rho:\sigma}$?

How does this imply $A_{\nu:\sigma}=0$?

 

[martinbn]

To parallel transport a vector along a curve means to find a vector field along the curve whose derivative along the curve is zero.

 

[Kostik]

Define “derivative along the curve”. Do you mean $A_{\nu:\sigma}(x)$?

 

[martinbn]

Define “derivative along the curve”. Do you mean $A_{\nu:\sigma}(x)$?

It is the covariant derivative of the field in the direction of the tangent vector of the curve. In the usual notation if the field is $V$ and the curve is $\gamma(t)$, then the derivative is $\nabla_{\dot{\gamma}(t)}V$.

 

[martinbn]

https://en.wikipedia.org/wiki/Parallel_transport

 

[Kostik]

Dirac is transporting an arbitrary vector to every point in the space. The vector need not be tangent to the path of transport.

 

[martinbn]

Dirac is transporting an arbitrary vector to every point in the space. The vector need not be tangent to the path of transport.

Of course not. Its derivative along the path needs to be zero (by definition).

 

[anuttarasammyak]

TL;DR Summary: Dirac asserts that in flat space, $A_{\nu:\sigma}=0$ in flat space?

He is saying that if the parallel transport of Aν from x to y is path-independent, then Aν:σ=0.

He says let A be a vector field so that A in original point is parallel transported to all the places. He introduced it to make rectilinear coordinates. I am afraid that you take it holds for general vectors.

 

[Kostik]

Yes, I see now that $A_\nu$ is not arbitrary. But I’m not seeing how $A_{\nu:\sigma}=0$. I’m only seeing that $A_{\nu:\sigma:\rho} = A_{\nu:\rho:\sigma}$.

 

[Kostik]

Sorry - the whole issue just became clear to me. By definition $A_\nu(x+dx)=K_\nu(x+dx)$ ($=$ the parallel transported vector $A_\nu(x)$ from $x$ to $x+dx$), hence $A_{\nu:\sigma}=0$.