sheaf said:
My understanding is:
Conformal transformations of the metric preserve the causal structure of spacetime, so the paths of light rays are unchanged. Conformally flat spaces will therefore have the same light cone structure as Minkowski space.
This is correct.
wam_mi said:
Dear all,
In flat conformal space-time,
e.g. \quad g_{\alpha \beta} = e^{4\kappa} \eta_{\alpha \beta}
where \kappa is some function of space-time coordinates.
What sort of paths do photons and massive particles follow?
Photons would simply have the same path as if they were to travel in a Minkowski spacetime. This can be easily proven in the metric you're giving as an example and I suppose you know how to prove it. But in case of massive particles, I have to prove how their paths undergo some change and in fact show how massive particles move along time-like curves. In the following, I take s to be the affine parameter of the time-like curves followed by particles and for simplicity I also take \kappa to be a function of the coordinate x in Cartesian coordinates x^{\alpha}:=(t,x,y,z). The line element is
ds^2=e^{4\kappa(x)}dt^2-e^{4\kappa(x)}(dx^2+dy^2+dz^2).
This can be written as
e^{-4\kappa}=\dot{t}^2-(\dot{x}^2+\dot{y}^2+\dot{z}^2),
where the overdots represent differentiation wrt s. Now working this into the Euler-Lagrangian equations gives the 4 geodesic equations:
\frac{d}{ds}(2\dot{t}e^{4\kappa})=0,
-\frac{d}{ds}(2\dot{x}e^{4\kappa})=4\frac{d\kappa}{ds}e^{4\kappa}[\dot{t}^2-(\dot{x}^2+\dot{y}^2+\dot{z}^2)]=4\frac{d\kappa}{ds},
-\frac{d}{ds}(2\dot{y}e^{4\kappa})=0,
-\frac{d}{ds}(2\dot{z}e^{4\kappa})=0.
Using \dot{\kappa}=\frac{d\kappa}{dt}\dot{t} the first of these 4 equations can be rewritten as
\ddot{t}+4\dot{t}\dot{x}\frac{d\kappa}{dt}=0.
(*)
Now from this and that
\ddot{x}=\dot{\left(\frac{dx}{dt}\dot{t}\right)}=\ddot{t}\frac{dx}{dt}+\dot{\frac{dx}{dt}}\dot{t}=-4\dot{t}\dot{x}\frac{d\kappa}{dt}\frac{dx}{dt}+\dot{\frac{dx}{dt}}\dot{t}.
where we have made use of
(*), the second of the geodesic equations can be cast into the following form:
\frac{d^2x}{dt^2}+2\frac{d\kappa}{dt}e^{-4\kappa}\dot{t}^{-2}=0.
(**)
Integrating the time-related geodesic equation yields
\dot{t}=ae^{-4\kappa}
where a is the integration constant. Introducing this into
(**) gives
\frac{d^2x}{dt^2}=-\frac{2}{a^2}\frac{d\kappa}{dt}e^{4\kappa}
which is the ultimate form of the second geodesic equation. The other two remaining equations are, in a somewhat similar way, found to be
\frac{d^2y}{dt^2}=\frac{d^2z}{dt^2}=0.
You can see that if \kappa=0 then all geodesic equations belong to the Minkowski metric. But since \kappa is a function of the coordinate x, among the spatial geodesic equations only the equation corresponding to the x component of the coordinate 3-acceleration has been modified and the other two are just the same as those of the Minkowski spacetime.
What is the difference between conformally flat space-time and flat space-time physically?
Considering pseudo-Riemannian, these are spacetimes that under a special condition can be mapped into flat spacetimes with metric components being -1 or +1 even though their Riemann tensor before mapping is not necessarily zero. The condition is that the vanishing of the Weyl tensor must be guaranteed for the spacetime to be conformally flat. In such spacetimes, all angles are preserved if the whole of spacetime (if the spacetime is conformally flat not locally conformally flat) gets mapped by a conformal transformation. One of the most significant class of conformally flat spacetimes is the constant-curvature class i.e. spacetimes with a constant curvature.
AB