User:WillowW/Affine connection

This is a sandbox for random snippets that will assembled later at affine connection and maybe a few other articles.

Context of the affine connection

Vectors are often represented in Cartesian coordinates by a sum of components multiplied by unit vectors e_i

$\mathbf {V} =\sum _{i=1}^{D}V^{i}\mathbf {e} _{i}$

where D is the dimension of the space. For example, a three-dimensional vector may be written as

$\mathbf {V} =V^{x}\mathbf {i} +V^{y}\mathbf {j} +V^{z}\mathbf {k} =V^{1}\mathbf {e} _{1}+V^{2}\mathbf {e} _{2}+V^{3}\mathbf {e} _{3}=\sum _{i=1}^{3}V^{i}\mathbf {e} _{i}$

This case is particularly simple, because the vectors e_i have unit length (e_i·e_i = 1), are perpendicular to one another (e_i·e_j = 0), and do not vary from position to position.

Things become a little more complicated when we consider orthogonal coordinates, such as spherical coordinates. To be sure, the basis vectors e_i still have unit length and are perpendicular to one another (also called orthogonal); yet they vary from position to position. For example, the unit vector e_r in spherical coordinates points radially outwards from any point on a sphere; thus, seen from outer space, it points upwards at the North Pole and downwards at the South Pole (add Figure here). A vector V at any point r can always be expressed as a sum in components as before

$\mathbf {V} =V^{r}\mathbf {e} _{r}+V^{\theta }\mathbf {e} _{\theta }+V^{\phi }\mathbf {e} _{\phi }$

but the components (V^r, V^θ, V^φ) may vary with the position r, even if the vector V itself does not change.

The ultimate limit is to remove all restrictions and express vectors in terms of basis vectors ε_i that may not have unit length, may not be orthogonal and may vary from place to place, albeit smoothly.

$\mathbf {V} =\sum _{i=1}^{D}V^{i}{\boldsymbol {\epsilon }}_{i}$

Such a situation may arise when an evil tyrant decrees that different measurement systems should be applied in different locations and even in different directions. Less ominously, it may also arise if the chains used to survey a property change length with temperature (which may vary with location) or with magnetic field (different directions). In such cases, the affine connection may be used to correct for such changes in the basis vectors. :)

Perhaps talk about a much-beloved sister who lives in a faraway town; consider that the vectors "North", "South", "East" and "West" are not the same for her as for you, when seen from absolute space. However, they may be connected by parallel transport to one another. Introduce direct angelic transport in an n dimensional embedding space? secant vs. tangent transport

Introduce qⁱ here and sundry formulae such as the symmetric metric tensor

$g_{ij}={\boldsymbol {\epsilon }}_{i}\cdot {\boldsymbol {\epsilon }}_{j}=g_{ji}$

Define contravariant metric tensor by the equation

$g^{ik}g_{kj}=\delta _{j}^{i}$

where δⁱ_j is the Kronecker delta and thence to the dot product

${\boldsymbol {\epsilon }}^{i}\cdot {\boldsymbol {\epsilon }}_{j}=\delta _{j}^{i}$

Raising and lowering properties

Derivation of the simplest affine connection

Need to mention the summation convention and dummy indices above

Need to introduce covariant vectors above; maybe not?

Show example of spherical coordinates and g_ij above, or hold off until the Schwarzschild solution?

The affine connection arises when one takes the derivative of a vector with respect to a coordinate q^j. The product rule shows that the variation in V is partly due to the intrinsic variation in its components and partly due to the changes in the basis vectors

${\frac {\partial \mathbf {V} }{\partial q^{j}}}={\frac {\partial }{\partial q^{j}}}V^{i}{\boldsymbol {\epsilon }}_{i}={\frac {\partial V^{i}}{\partial q^{j}}}{\boldsymbol {\epsilon }}_{i}+V^{i}{\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}$

The variations in the basis vectors should be a linear combination of the basis vectors themselves; the weighting terms are the definition of the affine connection

${\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}\equiv \Gamma _{ij}^{k}{\boldsymbol {\epsilon }}_{k}$

Substitution of this result and changing the dummy summation index yields the formula

${\frac {\partial \mathbf {V} }{\partial q^{j}}}=\left({\frac {\partial V^{i}}{\partial q^{j}}}+V^{k}\Gamma _{kj}^{i}\right){\boldsymbol {\epsilon }}_{i}=V_{;j}^{i}{\boldsymbol {\epsilon }}_{i}$

where the covariant derivative is defined by

$V_{;j}^{i}\equiv {\frac {\partial V^{i}}{\partial q^{j}}}+V^{k}\Gamma _{kj}^{i}$

For a covariant vector, the covariant derivative differs by a sign

$V_{i;j}\equiv {\frac {\partial V_{i}}{\partial q^{j}}}-V_{k}\Gamma _{ij}^{k}$

Properties of the affine connection (may be sacrificed for brevity?)

Taking the dot product with the contravariant basis vector ε^m yields the affine connection by itself

$\Gamma _{ij}^{m}={\boldsymbol {\epsilon }}^{m}\cdot {\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}$

since, by construction, ε^m · ε_k equals the Kronecker delta δ^m_k. It is relatively easy to show that the affine connection is symmetric in its lower two indices

$\Gamma _{ij}^{k}=\Gamma _{ji}^{k}$

since the order of two partial differentiations can be swapped

$\Gamma _{ij}^{k}{\boldsymbol {\epsilon }}_{k}={\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}={\frac {\partial ^{2}\mathbf {r} }{\partial q^{i}\partial q^{j}}}={\frac {\partial ^{2}\mathbf {r} }{\partial q^{j}\partial q^{i}}}={\frac {\partial {\boldsymbol {\epsilon }}_{j}}{\partial q^{i}}}=\Gamma _{ji}^{k}{\boldsymbol {\epsilon }}_{k}$

Another version of the affine connection is obtained by lowering the index

$[ij,k]\equiv g_{mk}\Gamma _{ij}^{m}=g_{mk}{\boldsymbol {\epsilon }}^{m}\cdot {\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}={\boldsymbol {\epsilon }}_{k}\cdot {\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{j}}}$

Differentiating the metric tensor g_ij = ε_i · ε_j yields the formula

${\frac {\partial g_{ij}}{\partial q^{k}}}={\frac {\partial {\boldsymbol {\epsilon }}_{i}}{\partial q^{k}}}\cdot {\boldsymbol {\epsilon }}_{j}+{\boldsymbol {\epsilon }}_{i}\cdot {\frac {\partial {\boldsymbol {\epsilon }}_{j}}{\partial q^{k}}}=[ik,j]+[jk,i]$

It is relatively simple to show that

$[ij,k]={\frac {1}{2}}\left\{{\frac {\partial g_{ik}}{\partial q^{j}}}+{\frac {\partial g_{jk}}{\partial q^{i}}}-{\frac {\partial g_{ij}}{\partial q^{k}}}\right\}$

and thus

$\Gamma _{ij}^{m}=g^{km}[ij,k]={\frac {1}{2}}g^{km}\left\{{\frac {\partial g_{ik}}{\partial q^{j}}}+{\frac {\partial g_{jk}}{\partial q^{i}}}-{\frac {\partial g_{ij}}{\partial q^{k}}}\right\}$

This formula helps in calculating the divergence of a vector field, which is defined by

${\boldsymbol {\nabla }}\cdot \mathbf {V} =V_{;i}^{i}={\frac {\partial V^{i}}{\partial q^{i}}}+V^{k}\Gamma _{ik}^{i}$

since

$\Gamma _{ik}^{i}={\frac {1}{2}}g^{im}{\frac {\partial g_{im}}{\partial q^{k}}}={\frac {1}{2g}}{\frac {\partial g}{\partial q^{k}}}={\frac {1}{\sqrt {g}}}{\frac {\partial {\sqrt {g}}}{\partial q^{k}}}={\frac {1}{2}}{\frac {\partial \log g}{\partial q^{k}}}$

since the theory of determinants gives

${\frac {\partial g}{\partial q^{k}}}=gg^{im}{\frac {\partial g_{im}}{\partial q^{k}}}$

Taken together, this yields the divergence formula

${\boldsymbol {\nabla }}\cdot \mathbf {V} =V_{;i}^{i}={\frac {1}{\sqrt {g}}}{\frac {\partial }{\partial q^{k}}}\left(V^{k}{\sqrt {g}}\right)$

Compare to orthogonal coordinates result for divergence, connect to Laplacian for swift GR derivation in Newtonian approximation.

Commutativity of covariant differentiation and curvature

For a scalar field, covariant differentiation commutes

$\phi _{;\mu \nu }=\left(\phi _{\mu }\right)_{;\nu }=\phi _{,\mu \nu }-\phi _{,\sigma }\Gamma _{\mu \nu }^{\sigma }=\left(\phi _{\nu }\right)_{;\mu }$

For a vector field, however, this is not true

$V_{\sigma ;\mu \nu }-V_{\sigma ;\nu \mu }=V_{\rho }R_{\sigma \mu \nu }^{\rho }$

where the curvature tensor is defined by

$R_{\sigma \mu \nu }^{\rho }\equiv \Gamma _{\sigma \nu ,\mu }^{\rho }-\Gamma _{\sigma \mu ,\nu }^{\rho }+\Gamma _{\sigma \nu }^{\alpha }\Gamma _{\alpha \mu }^{\rho }-\Gamma _{\sigma \mu }^{\alpha }\Gamma _{\alpha \nu }^{\rho }$

As an aside that is peculiar to four dimensions, owing to the many symmetries of this tensor, only 20 of its 256 components (4 values of 4 indices = 4⁴ components) are independent.

It is straightforward to derive the Bianchi relations for the curvature tensor

$R_{\sigma \mu \nu ;\alpha }^{\rho }+R_{\sigma \alpha \mu ;\nu }^{\rho }+R_{\sigma \nu \alpha ;\mu }^{\rho }=0$

which are so useful in showing that the physical field equations conserve energy and momentum.

Contraction of indices leads to the symmetric Ricci tensor

$R_{\sigma \mu }=R_{\sigma \mu \nu }^{\nu }=\Gamma _{\sigma \nu ,\mu }^{\nu }-\Gamma _{\sigma \mu ,\nu }^{\nu }+\Gamma _{\sigma \nu }^{\alpha }\Gamma _{\alpha \mu }^{\nu }-\Gamma _{\sigma \mu }^{\alpha }\Gamma _{\alpha \nu }^{\nu }$

which may be further contracted to obtain the curvature scalar

$R=g^{\sigma \mu }R_{\sigma \mu }$

Parallel transport (maybe right after covariant derivative?)

If a vector V is constant at every point, then its components much change to compensate for the changes in the basis vectors; in other words, the covariant derivative must be zero.

Lead into geodesics, show stationary property

Take the same vector by parallel transport along two paths, show that they disagree at their destination, if curvature is present; use sphere surface example

Discussion of curvature; the difficulties of surveying the U.S. states into square county grids; the surveyors weren't always drunk or corrupt! ;)

Mention equivalence of "curved space" and "funky rulers" in a magical world, rulers that change dimensions at different places and in different directions

Parallel transport occurs when the covariant derivative is zero

${\frac {\partial \mathbf {V} }{\partial q^{k}}}=0$

which occurs when the changes in components exactly compensate the changes in the local basis vectors

${\frac {\partial V^{i}}{\partial q^{k}}}=-V^{m}\Gamma _{mk}^{i}$

or, expressed in covariant coordinates,

${\frac {\partial V_{i}}{\partial q^{k}}}=V_{m}\Gamma _{ik}^{m}$

Let there be a path q^k(λ) parametrized by a variable λ, and let the vector V be transported parallelly along this path. Then we have

${\frac {dV_{i}}{d\lambda }}=V_{m}\Gamma _{ik}^{m}{\frac {dq^{k}}{d\lambda }}$

which we may integrate

$V_{i}(\lambda _{2})-V_{i}(\lambda _{1})=\int _{\lambda _{1}}^{\lambda _{2}}{\frac {dV_{i}}{d\lambda }}d\lambda =\int _{\lambda _{1}}^{\lambda _{2}}V_{m}\Gamma _{ik}^{m}{\frac {dq^{k}}{d\lambda }}d\lambda$

Let the path be a closed path, so that we end up at our starting position

$\oint _{\lambda _{1}}^{\lambda _{2}}{\frac {dq^{k}}{d\lambda }}d\lambda =0$

The affine connection Γ may be expanded in a Taylor series about the starting position Q

$\Gamma _{ik}^{m}(q^{n})=\Gamma _{ik}^{m}(Q^{n})+(q^{n}-Q^{n})\Gamma _{ik,n}^{m}+\cdots$

as can the vector components themselves

$V_{i}(q^{n})=V_{i}(Q^{n})+(q^{n}-Q^{n}){\frac {\partial V_{i}}{\partial q^{n}}}+\cdots =V_{i}(Q^{n})+(q^{n}-Q^{n})\Gamma _{in}^{m}V_{m}(Q^{n})+\cdots$

Putting these Taylor expansions together yields the formula

$V_{i}(\lambda _{2})-V_{i}(\lambda _{1})=\left(\Gamma _{ik,l}^{m}+\Gamma _{ik}^{s}\Gamma _{sl}^{m}\right)V_{m}(Q^{n})\oint _{\lambda _{1}}^{\lambda _{2}}q^{l}{\frac {dq^{k}}{d\lambda }}d\lambda$

Swapping the kl indices and using the antisymmetry

$\oint _{\lambda _{1}}^{\lambda _{2}}q^{l}{\frac {dq^{k}}{d\lambda }}d\lambda =\oint _{\lambda _{1}}^{\lambda _{2}}{\frac {d}{d\lambda }}\left(q^{k}q^{l}\right)d\lambda -\oint _{\lambda _{1}}^{\lambda _{2}}q^{k}{\frac {dq^{l}}{d\lambda }}d\lambda =-\oint _{\lambda _{1}}^{\lambda _{2}}q^{k}{\frac {dq^{l}}{d\lambda }}d\lambda$

(which results from integration by parts) yields the curvature tensor

$V_{i}(\lambda _{2})-V_{i}(\lambda _{1})={\frac {1}{2}}R_{ikl}^{m}V_{m}(Q^{n})\oint _{\lambda _{1}}^{\lambda _{2}}q^{l}{\frac {dq^{k}}{d\lambda }}d\lambda$

Thus, if there's no curvature, parallel transport around a closed curve always results in the original vector.

Angelic transport

Let the curved human space be embedded in an uncurved, D-dimensional angelic space. Let its angelic coordinates be R, an D-dimensional vector, and the d coordinates within the manifold be qⁱ. Let R be a function of q, and consider two neighboring points separated by δq

$\delta \mathbf {R} ={\frac {\partial \mathbf {R} }{\partial \mathbf {q} }}\delta \mathbf {q}$

which may be written in components as

$\delta R^{I}={\frac {\partial R^{I}}{\partial q^{k}}}\delta q^{k}$

The uppercase index I indicates that the index runs from 1 to D, whereas the lowercase indices such as k run from 1 to d.

The distance between the two neighboring points in the angelic space is given by Pythagoras' law

$\delta s^{2}=\eta _{IJ}\delta R^{I}\delta R^{J}=\eta _{IJ}{\frac {\partial R^{I}}{\partial q^{k}}}\delta q^{k}{\frac {\partial R^{J}}{\partial q^{l}}}\delta q^{l}=g_{kl}\delta q^{k}\delta q^{l}$

which gives a relation between the angelic metric tensor η and the human metric tensor g

$g_{kl}=\eta _{IJ}{\frac {\partial R^{I}}{\partial q^{k}}}{\frac {\partial R^{J}}{\partial q^{l}}}$

Let there be a vector V that lies in the manifold at the first point, and let the angels transport it unchanged to the second point. Because of the curvature of the space, V might not lie in the manifold at the new point. Let's assume that there are constraint forces that require it to lie in the manifold, constraints that immediately cancel out any part of V that lies outside the manifold. (Is this necessary? why not keep them, if they're not hurting anything and are invisible to people within the manifold? No, the normal component must be destroyed; otherwise, you won't have non-commutativity of covariant derivatives. This would make a good Figure comparing the two ideas on the unit sphere.) Seen from the angels' perspective, the transported vector has tangent and normal components

$V^{I}=V_{\mathrm {tan} }^{I}+V_{\mathrm {norm} }^{I}$

To isolate the tangential component, we take the dot product with each of d basis vectors Ε_j, each of which are D-dimensional and lie in the manifold at the second point.

A fresh start on angelic transport

Use uppercase letters to discern vectors in embedding space, and lowercase for vectors within the manifold.

Consider a curved manifold of dimension d with coordinates q. Let this curved space be embedded in an uncurved space of higher dimension D; vectors in this space are designated with capital letters. We poor creatures are confined to the manifold and cannot see out of it; by contrast, the angels can move freely in the D-dimensional space, carrying vectors from one point to another without altering their direction (add nice Figure here).

At any given point on the manifold, the derivative of the D-dimensional position vector R with respect to the manifold coordinates q form a set of d basis vectors

$\mathbf {E} _{i}={\frac {\partial \mathbf {R} }{\partial q^{i}}}$

Let there be a vector V that lies completely in the manifold at that given point r, which may be expressed with this set of basis vectors

$\mathbf {V} =V^{i}\mathbf {E} _{i}$

Now the angels transport the vector V unchanged to a new position, separated from the first position by a small vector of components δq^k. In general, the vector will not lie within the manifold, due to its curvature. Therefore, we have to eliminate the normal component and keep only the tangential part, the part that can be expressed in terms of the new basis vectors Eˡ_i at the new point. The new basis vectors may be expanded in a Taylor series in the coordinate variations

$\mathbf {E} _{i}^{\prime }=\mathbf {E} _{i}+{\frac {\partial \mathbf {E} _{i}}{\partial q^{k}}}\delta q^{k}+\cdots =\mathbf {E} _{i}+\mathbf {E} _{m}\Gamma _{ik}^{m}\delta q^{k}+\cdots$

To derive the new components at the second point, we take the dot product

$V_{i}^{\prime }=\mathbf {V} \cdot \mathbf {E} _{i}^{\prime }=\mathbf {V} \cdot \mathbf {E} _{i}+\mathbf {V} \cdot \mathbf {E} _{m}\Gamma _{ik}^{m}\delta q^{k}+\cdots =V_{i}+V_{m}\Gamma _{ik}^{m}\delta q^{k}+\cdots$

Therefore, angelic transport — which is the same as parallel transport to first order — is described by the equation

$\delta V_{i}=V_{i}^{\prime }-V_{i}=V_{m}\Gamma _{ik}^{m}\delta q^{k}$

which is consistent with the covariant derivative above.

Schwarzschild solution and black holes (neat!)

Let's assume a spherically symmetric black hole that is utterly static, so that the metric tensor has no time dependence. Adopting spherical coordinates (t, r, θ, φ) (note to self:note that the units don't agree; conversion factors go into the metric), the most general spherically symmetric metric is

$ds^{2}=-A(r)dt^{2}+B(r)dr^{2}+C(r)r^{2}\left(d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right)$

We redefine r to make C(r)=1, and re-write A and B in terms of exponentials

$ds^{2}=-e^{2\nu }dt^{2}+e^{2\lambda }dr^{2}+r^{2}\left(d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right)$

from which we can extract the 16 components of the metric tensor. Taking the derivatives of these components yields the affine connection components Γ^k_ij, which in turn yield the Ricci tensor R_μν. Using Einstein's law that the Ricci tensor should be zero in empty space, we can show that the first derivatives of λ and ν with respect to r sum to zero, using the equations R₀₀ = R₁₁ = 0. Similarly, the space must flatten out at infinite radii, so both λ and ν should go to zero in this limit; taken together with the derivative equation, we get that

$\lambda +\nu =0$

at all radii r. Finally, using the equation R₂₂=0, we get that

$1+2r{\frac {d\nu }{dr}}=e^{-2\nu }$

an ordinary differential equation with the solution (NB! we're using G = c = ħ = 1)

$g_{00}={\frac {1}{g_{11}}}=e^{2\nu }=1-{\frac {2m}{r}}$

The equation of motion for a particle falling radially inwards very close to the event horizon r = 2m is given by

${\frac {dt}{dr}}=-2m\log \left(r-2m\right)$

which goes to infinity as r goes to 2m. Thus, it takes an infinitely long time for a particle to fall into a black hole, when observed from without.

However, a finite proper time passes for a person falling into black hole; neglecting the awesome tidal forces, such a cosmonaut would continue to fall inwards even past the event horizon. This may be shown by transforming coordinates.