Quote:
Originally Posted by Nesti
Surely that can only be so from the reference frame of the photon?! Because from any other frame it must traverse all topology along its geodesic pathway, ie light getting bent as it moves along its geodesic within a gravitational field, true? (I'm assuming we don't have two convenient sets of rules out there for how a photon behaves...like in QM)
So, if the photon must traverse its geodesic path, then surely gravity waves must get affected also (assuming that there are many local gravitational fields), otherwise we could easily get a situation where a gravity wave arrives ahead of the photon...which just isn't possible.
|
Whether a photon has a "frame of reference" has always been a debateable point. Since the world line is a null geodesic (ds^2=0), in the photon's frame it can simultaneously occupy every point along the worldline in the space-time of the wave, or it can be interpreted as a Lorentz contraction to zero length of the photon path.
From an observer's frame of reference the null geodesic is straightforward, it means the photon travels at c irrespective of the topology of space-time or the photon's trajectory.
Quote:
To an observer's frame of reference
Again, are you talking purely from the photon's frame of reference?
|
No. The field equations for gravitational waves reduce to Helmholtz equations. The solutions are sinusoidal functions of constant amplitude and frequency and travelling at c. The shape of the wave doesn't change, hence the metric for the space-time of the wave doesn't alter.
Regards
Steven