I Transforming to Local Inertial Coordinates

[Ibix]

I've been playing around a bit with the Kerr orbit program I wrote, and have been thinking about ways to set the initial conditions. One thing I'd like to be able to do is specify a launch from some event in terms that would be convenient for an observer at that event with some given four-velocity - but the orbit program works with global coordinates (Boyer-Lindquist in this case, although I don't think that's important). I think I've worked out (in painful detail below) how to transform between local coordinates and global, but would appreciate a sanity check (apologies if this better belongs in the homework section).

The metric at the launch point is $g_{ab}$ specified in some global coordinate system, and the launching observer has four-velocity $U^a$, specified in that same system. What I'm looking for is the transform from this system to local Minkowski coordinates with time axis parallel to $U^a$. I'm going to work with the metric with upper indices so I can apply the results to four-velocities without further mucking around.

Diagonalising the inverse metric with a matrix of eigenvectors, $\mathbf E$ is textbook. The result, $D^{a'b'}=E^{a'}{}_{a}E^{b'}{}_bg^{ab}$, has the eigenvalues of $g^{ab}$ on the diagonal ($D^{ii}=\lambda^{(i)}$ where no summation is assumed). Presumably one will have the opposite sign to the other three. I construct another matrix, $S^{a''}{}_{a'}$, which is diagonal with $S^i{}_i=|\lambda^{(i)}|^{-1/2}$ (where, again, no summation is assumed), and then $S^{a''}{}_{a'}E^{a'}{}_{a}S^{b''}{}_{b'}E^{b'}{}_bg^{ab}$ should be a Minkowski metric.

Now I can apply this transform to $U^a$ to get the launching observer's four velocity in this Minkowski frame. I can then find a Lorentz transform, $\Lambda^{a'''}{}_{a''}$, so that $\Lambda^{a'''}{}_{a''}S^{a''}{}_{a'}E^{a'}{}_{a}U^a$ has no spatial components. I may also wish to apply this composite transform to vectors parallel to the global spatial coordinate axes and append a rotation, $R^{a''''}{}_{a'''}$, to enforce a simple relationship between the global and local spatial coordinates - that their spatial projections are parallel, for example.

Then I have a Jacobean $J^{a''''}{}_{a}=R^{a''''}{}_{a'''}\Lambda^{a'''}{}_{a''}S^{a''}{}_{a'}E^{a'}{}_{a}$ that transforms global coordinates to local ones, and whose inverse takes local coordinates to global ones.

Right? Or am I over-complicating things?

 

[pervect]

I've been playing around a bit with the Kerr orbit program I wrote, and have been thinking about ways to set the initial conditions. One thing I'd like to be able to do is specify a launch from some event in terms that would be convenient for an observer at that event with some given four-velocity - but the orbit program works with global coordinates (Boyer-Lindquist in this case, although I don't think that's important). I think I've worked out (in painful detail below) how to transform between local coordinates and global, but would appreciate a sanity check (apologies if this better belongs in the homework section).

The metric at the launch point is $g_{ab}$ specified in some global coordinate system, and the launching observer has four-velocity $U^a$, specified in that same system. What I'm looking for is the transform from this system to local Minkowski coordinates with time axis parallel to $U^a$. I'm going to work with the metric with upper indices so I can apply the results to four-velocities without further mucking around.

Diagonalising the inverse metric with a matrix of eigenvectors, $\mathbf E$ is textbook. The result, $D^{a'b'}=E^{a'}{}_{a}E^{b'}{}_bg^{ab}$,

With the right choice of a transfomration matrix $L^{a}{}_{b}$ one can make the metric $g_{ab}$ Minkoskian at a point. Following the conventions in my textbook, this would be $g_{a'b'} = L^{a}{}_{a'}L^b{}_{b'}g_{ab}$, which is different than what you wrote. (My textbook is MTW). Apparently the conventions in my text aren't universal from another thread, but it's what I'm used to. The set of coordinates defined by the transfomation would serve as the convenient local coordiantes you metnioned. It's not a unique transformation - you can go further and make the derivatives of $g_{a'b'}$ zero at a point as well as making it Minkowskian. Around a point, this is called Riemann normal coordinates, there's a version around worldlines that is called Fermi normal coordinates.

See https://en.wikipedia.org/w/index.php?title=Normal_coordinates&oldid=774138316 for normal coordinates, Riemann normal coordinates are normal coordinates on a Riemanian manifold (at least that's my understanding).

There are other ways to do this using frame fields, perhaps I'll write those up, that are also convenient.

 

[Ibix]

Thanks, pervect. The Riemann Normal prompt (which I shouldn't have needed!) led me here: http://users.monash.edu.au/~leo/research/papers/files/lcb96-01.pdf, which says (equation 3.7.2) that RNCs $x^\mu$ can be expressed in terms of generic coordinates $z^\eta$ with $$x^\mu=\Lambda^\mu{}_\eta\left(\Delta z^\eta+\frac 12\Gamma^\eta_{\alpha\beta}\Delta z^\alpha\Delta z^\beta+\frac 16\left(\Gamma^\eta_{\alpha\tau}\Gamma^\tau_{\beta\rho}+\Gamma^\eta_{\alpha\beta,\rho}\right)\Delta z^\alpha\Delta z^\beta\Delta z^\rho\right)$$(plus higher order corrections) where $\Delta z^\eta=z^\eta-z^\eta_0$, and $z^\eta_0$ is the origin of the RNCs, and $\Lambda^\mu{}_\eta$ is an arbitrary Lorentz transform to align the axes as you like. I think this makes the transformation matrix$$
\begin{eqnarray*}\frac{\partial x^\mu}{\partial z^{\mu'}}=\Lambda^\mu{}_\eta\bigg(&&\delta^\eta_{\mu'}+\Gamma^\eta_{\alpha\mu'}\Delta z^\alpha\\
&&+\frac 16\left(\Gamma^\eta_{\alpha\tau}\Gamma^\tau_{\beta\rho}+\Gamma^\eta_{\alpha\beta,\rho}\right)\left(\delta^\alpha_{\mu'}\Delta z^\beta\Delta z^\rho+\Delta z^\alpha\delta^\beta_{\mu'}\Delta z^\rho+\Delta z^\alpha\Delta z^\beta\delta^\rho_{\mu'}\right)\bigg)\end{eqnarray*}$$The question is whether that's the same as what I was proposing. I will investigate.

 

[PAllen]

I'm wondering whether you need coordinate transform at all for your stated purpose. The initial condition for a geodesic is a point (event) plus a 4 velocity. If you want to be able to specify various 4 velocities in reference to a given one, it seems to me you don't need a coordinate transform. Given a 4 velocity at an event expressed in your global coordinates, pick any coordinate tangent not a multiple of this 4 velocity. Then take this tangent vector minus itself dot the starting 4-velocity time the starting 4-velocity. Normalize this. You now have a unit vector normal to the starting 4-velocity. Then pick any other coordinate direction vector, and subtract its dot product with each of these two orthonormal vectors, times each vector. You now have a third orthonormal vector. Similarly for the 4th. Then, any direction expressed as combination of these orthonormal vectors (as expressed in the original coordinate basis) is an appropriate 4-velocity for a geodesic.

In effect, if all you want is a direction expressed in some orthonormal basis at a point, you hardly need a coordinate transform for a finite region.

 

[Ibix]

Good point. I am over-complicating it. I can easily pick a set of four orthonormal vectors to use at a point, starting with one specified time-like one to guarantee that it's relevant to my observer and adding some sensible relationship to global coordinates. Then I specify my launch velocity in terms of those basis vectors and just multiply the ith component by the ith local basis vector (expressed in hlobal coordinates) and sum over i.

Thanks.

 

[pervect]

Good point. I am over-complicating it. I can easily pick a set of four orthonormal vectors to use at a point, starting with one specified time-like one to guarantee that it's relevant to my observer and adding some sensible relationship to global coordinates. Then I specify my launch velocity in terms of those basis vectors and just multiply the ith component by the ith local basis vector (expressed in hlobal coordinates) and sum over i.

Thanks.

Yes, picking four orthonormal vectors - ore co-vectors - defines a frame field at a point. Wiki has an article on frame-fields, which may or may not be readable. Picking four orthonormal basis vectors can also be interpreted (locally!) as a linear coordinate transformation. (add). The orthonormal basis vectors span the tangent space, the coordinates span the manifold. Locally, the tangent space is a good description of the manifold, this can be made more formal.

When viewed as a coordinate transformation, the linear transformation can make the metric Minkowskian at a point, though it can't make the first derivatives vanish as Riemann normal coordinates do. One needs a nonlinear coordinate transformation to make the derivatives of the metric vanish at a point. . But if one doesn't need to make the first derivatives vanish, a linear transformation is easier, and is equivalent to a frame-field defined by four orthonormal basis vectors (or co-vectors).

If one chooses four orthonormal co-vectors, the whole process of finding the orthonormal co-vectors can be done with algebra. A co-vector is a map from a vector to a scalar. The process of algebraically manipulating the line element into the Minkowskian form $-dT^2 + dX^2 + dY^2 + dZ^2$ by linear algebraic substitutions i.e. $X= \alpha_{00}\,t + \alpha_{01}\,x + \alpha_{02}\,y + \alpha_{03}\,z$ is the same algebraic process as manipulating the co-vectors into the Minkowskian form, ##-dT^2 + dX^2 + dY^2 + dZ^2##. The only difference in the manipulations is that dX is a scalar, which isn't a function of anything, while dX is a map from a vector to a scalar, so it has a scalar value but also takes an vector argument, i.e. it's a functional.

If co-vectors aren't familiar, though, the process can be viewed without them, co-vectors merely allows one to interpret the manipulations as algebraic.