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When a photon is emitted from a far away source and then measured by an observer, there is a loss of energy or redshift which takes place. Why does this happen?

I have read this similar post, however the confusion I'm having is due to the invariance of photons, which travel at the speed of light, to non-inertial reference frames. If they are invariant then why would there be a perceived energy loss as it switches from one reference frame to the other?

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There is no loss of energy. If the component of the wave heading towards the Earth is redshifted, then the component of the wave heading away from the Earth is blueshifted. –  Mew Feb 7 '13 at 3:34
    
I understand that there should be no loss of energy by conservation of energy, but why is any shift taking place at all? –  Loourr Feb 7 '13 at 3:36
    
A photon's speed is invariant, yes, but its energy and momentum are not. –  Chris White Feb 7 '13 at 8:21
    
Since red/blue shifting is the Doppler Effect on the wave nature of light, I find it a little strange to hear it described as applying to the particle nature of light. –  kojiro Feb 7 '13 at 12:59
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You need to careful about treating photons as particles like electrons. A photon is best thought of the unit of interaction of the photon field. The light from distant galaxies isn't like a hail of little bullets coming towards us. The photons are delocalised and don't have a particle like position until they interect with your eye/CCD/whatever.

Anyhow, the reason light from distant galaxies is red shifted is that the spacetime in between us and the distant galaxy has expanded since the light was emitted, and that means the energy of the light has been spread out over a larger area.

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Suppose A emits light to Earth at some position X (That is, the rocket is located at position X). If A is stationary with respect to Earth, then if A emits 1 pulse per minute, position X receives 1 pulse per minute (because the rocket is at X), and Earth will receive 1 pulse per minute.

Now A begins to move away from X. At t = 0, A emits a pulse at position X, but at t=1minute, A emits another pulse as before, but this time A's position is X + h (assuming moving at v= h). Now as this pulse travels towards Earth, it doesn't reach position X until time h/c, where c = speed of light.

So while previously the frequency of light at position X was 1 pulse per minute. The gap between the pulses is now (1 + h/c) minutes.

Similarly, as the pulses reach Earth, they will be separated by (1+h/c) minutes. Thus the frequency of the pulses has appeared to decrease (Doppler broadening).

We can see from above, that this broadening is due to light from each consecutive pulse having to travel a greater distance than the previous pulse when the rocket is moving, thus creating a time delay.

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So does redshift not make sense as a concept when only considering individual quanta? Which would mean that the photons only have less energy compared to the ones which came before them? –  Loourr Feb 7 '13 at 3:58
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I used discrete "pulses" to demonstrate the effect. You can think of a "quanta" of light as a tiny packet of a "wave". This "wave" will have peaks and troughs, which can be considered "pulses" in my above analysis. So my analysis certainly does apply to single photons as well. –  Mew Feb 7 '13 at 4:03
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One has to note energy is not "invariant" with lorentz transform. Simply put, if you are at rest, you find yourself with 0 kinetic energy; if an observer B is moving with velocity v against you, he is to find you with kinetic energy $$ 1/2mv^2$$. Energy is naturally different in different frames.

Energy and momentum make up a four vector that transforms with space-time transformation, it is called a energy-momentum vector.

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