Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I'm trying to construct some coordinates on Minkowski spacetime based on a world line, $C$, ($\dot{C}\cdot\dot{C}=-1$) and forward light cone. I want the "time" coordinate of a point, $p$, to be the "retarded time", i.e the time $t(p)$ at the (unique when it exists) point $C(t(p))$ on $C$ joining $p$ by a null geodesic $\gamma$ (on the forward light cone of $C(t(p))$: $\gamma'\cdot\dot{C}<0$). I want the "distance" coordinate to then be the `light distance': $r(p)=-(p-C(t(p)))\cdot\dot{C}(t(p))$ and some angular coordinates defined by the "direction" (possibly via a projection onto an instantaneous 3-space) of $\gamma$ at $C$.

The problem I'm having is constructing the metric. Say I start with the vector field $\dot{C}$ and a triad ("forward") of linearly independent null vectors $N_i$ (so that $g(N_i,N_j)\neq 0$). If I Fermi-Walker transport this triad along $C$: \begin{equation} \nabla_{\dot{C}}N_i=(N_i\cdot \ddot{C})\dot{C}-(N_i\cdot \dot{C})\ddot{C}\quad (\ddot{C}=\nabla_{\dot{C}}\dot{C}) \end{equation} then they remain a null triad and do not rotate (I think). I could then choose to parallel translate $\dot{C}$ and $N_j$ along the cone to obtain a local basis field.

But seeing as the $N_i$ are null I'm not quite sure how to construct the metric (that is, assuming my previous steps are ethical).

The reason I would like these coordinates is to study solutions to Maxwell's equations \begin{equation} \mbox{d}F=0\qquad \mbox{d}\ast F = j \end{equation} and hopefully obtain something like the Liénard-Wiechert potential in the language of exterior calculus. So any references would also be nice (I've tried the standard texts, i.e. Jackson and Rohrlich).

Cheers,

Mat L

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

Let me first rephrase your construction of the coordinate system. Given your curve $C$, foliate the space-time by future null-cones emanating from points on $C$. Call each of the cones $\mathcal{N}_t$, indexed by the time-coordinate on the base point. This gives the $t$ coordinate of your points. Let $p$ be a point and $t(p)$ its time coordinate. There exists a unique null vector $v(p)\in T_{C(t(p))}M$ with the following properties:

  • $v(p) \cdot \dot{C}(t(p)) = -1$
  • $v(p)$ generates the null geodesic connecting $C(t(p))$ to $p$.

The on-light-cone coordinate $u$ can be taken to the affine parameter of the point $p$ relative the the generator $v(p)$. That is, $u(p) = \exp^{-1}_{v(p)} p$.

The angular coordinates can be obtained by the following: take the projection of $v(p)$ to the orthogonal subspace to $\dot{C}(t(p))$. This is a norm 1 space-like vector at $C(t(p))$ orthogonal to the world line. Fermi-transport this to $C(0)$, and it will be a norm 1 space-like vector orthogonal to $\dot{C}(0)$, in other words, it would be an element of the unit sphere in the orthogonal complement of $\dot{C}(0)$. And there we have the angular variable $\omega$.


This is a perfectly fine coordinate system, though personally I would prefer using an advanced-retarded coordinate (aka double null foliation).

The power of the Fermi-Walker transport is in the following. Consider a vector $V$ that is FW transported. Now look at its inner product against itself. You have

$$ \nabla_{\dot{C}} g(V,V) = 2 g(V, \nabla_{\dot{C}}V) = 2 g(V,0) = 0 $$

so the FW transport preserves inner-product. Furthermore, however, consider the inner product against $\dot{C}$ along $C$:

$$ \nabla_{\dot{C}} g(\dot{C},V) = 2 g(\ddot{C}, V) + g(\dot{C},\dot{C})g(\ddot{C},V) - g(\dot{C},\ddot{C})g(\dot{C},V) $$

where using $g(\dot{C},\dot{C}) = -1$ you get that the inner-product against $\dot{C}$ is also constant. This means that you can choose your null triad such that along $C$ they are always null and their inner-products against $\dot{C}$ is fixed to be $-1$.

Lastly, take $V,W$ two vectors that are FW-transported. A similar computation as above will show you that $g(V,W)$ is also preserved along $C$. So, this tells you that along $C$, the metric in the tetrad given by $(t,N_1,N_2,N_3)$ is the constant matrix

$$ \begin{pmatrix} -1 & -1 & -1 & -1 \\ -1 & 0 & -a & -b \\ -1 & -a & 0 &-c \\ -1 & -b & -c & 0 \end{pmatrix} $$

where $a,b,c$ are positive constants (using that the three null triad vectors are all future pointing).

Parallel transport also preserves inner products, so you have that your metric tensor can be globally written in a tetrad frame as above. Even the rotation coefficients are not that difficult to compute: everthing is based on the values you get along $C$, and you can easily check that the curvature is 0.

The difficulty in this construction, as I see it, is that your choice of tetrad is initiall chosen to be divorced from your coordinate system. So it is not exactly clear to me the complexity of trying to covert between coordinate vectors and your tetrad, and associating to each coordinate point its tetrad vectors.

share|improve this answer
    
Hi Willie, thanks. I guess the natural thing to do is to use a Frenet-frame based on $C$, that is $X_1$=$\ddot{C}/|\ddot{C}|$, etc. I would love to be able to do this without needing to revert to any affine coordinates at all. –  kangdon Jan 24 '11 at 2:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.