Uh-hu! So red-shift doesn't occur at all?
All three photons are the same? Seriously? In any frame of reference?
The three photons cannot be the same in any frame of reference.
Oh! And by the way; red-shift does actually represent a loss of energy.
Capt'n Midnight 00:00
If you match your velocity to that when the photon was emitted all three would be the same. If you don't match your velocity then sure you could accelerate in the opposite direction if you wanted to, or wait until the universe has expanded some more.
But all three photons travel at the speed of light regardless of the velocity of the emitter. There is one blue photon and two red photons; one of the red ones is an old blue one that has been red-shifted and has exactly the same wavelength of the younger red photon that has been travelling for a distance equal to the blue photon.
A detector would not be able to distinguish the age difference between the two red photons in any reference frame since the detector has the photons in its own frame. A comparison of three detectors might yield strange result but any one detector will see two of one and one of the other.
Why am I wrong?
maninasia Registered User
It's all relative my friends.
It's a good point about not being able to tell the photons age.
The photon would see itself as always having the same energy. Red-shifting photons are a feature of co-ordinates.
What would the problem be? The detector will indeed see two of one and one of the other. The photons will all "agree" that the detector sees this, as the detector is not in the "reference frame" of any photon.