Conceptualize radial redshift

[nutgeb]

I find it helpful to conceptualize radial redshift as nothing more than a longitudinal stretching out or elongation of an incoming light beam segment, as measured in the observer's rest frame. For example, a light beam segment which has been redshifted by a factor of 2 has been stretched to twice its length (as measured in the observer's frame) as when it was originally emitted (as measured in the emitter's frame.)

This makes it clear that no energy is "lost" by the light beam segment as a consequence of its redshifting. Instead, the same total energy is spread over twice the length. The energy density striking the observer at each instant is half of the original energy density, but that density decrease is exactly offset by the doubling of the duration over which the total light beam segment is received.

The CMB radiation has been redshift by about 1089 times since it was emitted from the surface of last scattering, and its temperature (kelvin) as measured at Earth has decreased by that factor. The distance between our Earth and that surface of last scattering is constantly increasing approximately at the rate of the cosmic Hubble flow. Each segment (of arbitrarily assigned length) of the CMB radiation beam we observe at Earth has been cumulatively stretched during its transit to what has finally become 1089 times the length of that segment when it was originally emitted. The CMB energy has not been lost as a result of this stretching. Rather, the same total radiation energy merely has been spread out into a configuration of lower energy density.

 

[JesseM]

This page on the difficulties with energy conservation in GR from John Baez's site seems to say something different:

The Cosmic Background Radiation (CBR) has red-shifted over billions of years. Each photon gets redder and redder. What happens to this energy? Cosmologists model the expanding universe with Friedmann-Robertson-Walker (FRW) spacetimes. (The familiar "expanding balloon speckled with galaxies" belongs to this class of models.) The FRW spacetimes are neither static nor asymptotically flat. Those who harbor no qualms about pseudo-tensors will say that radiant energy becomes gravitational energy. Others will say that the energy is simply lost.

Note that in neither case do the physicists say the same amount of energy is still present in the photons themselves.

 

[nutgeb]

Thanks JesseM

The quote from Baez seems to imply the possibility of a different answer, but at best he is waffling in posing alternative interpretations. I think his indecision arises from the fact that the GR theory does not require or perhaps even expect energy conservation in the Friedmann metrics.

But it seems self-evident that the redshifted energy I described has not been lost, merely spread out. Here's another example: Two simultaneous supernova events occur at the distances of z=1 and z=3 from us respectively, involving identical stars. It has been well demonstrated that the supernova event at z=3 will have an observed duration twice as long as as the event at z=1, subject to the normal range of supernova event variability.

Thus although the radiation beam from the z=3 supernova event is twice as redshifted as the z=1 supernova event, the total amount of energy received from each event over its total duration is the same.

Surely Baez wouldn't disagree with that.

 

[JesseM]

It seems self-evident that the energy has not been lost, merely spread out. Here's another example: Two simultaneous supernova events occur at the distances of z=1 and z=3 from us respectively, involving identical stars. It has been well demonstrated that the supernova event at z=3 will have an observed duration twice as long as as the event at z=1, subject to the normal range of supernova event variability.

Thus although the radiation beam from the z=3 supernova event is twice as redshifted as the z=1 supernova event, the total amount of energy received from each event over its total duration is the same.

Surely Baez wouldn't disagree with that.

I'm pretty sure he would, your argument for energy being "spread out" doesn't make sense to me. After all, if the stars are identical, the number of photons we receive from each from the beginning of the supernova to the end should be the same; but the photons from the star at z=3 will have a higher average redshift, and the energy each photon contributes is always given by E=hf where h is Planck's constant and f is the frequency (which gets lower the more the photon is redshifted).

 

[nutgeb]

Yes, your point makes sense, so I stand corrected. The total number of photons in each stream is the same, but the energy of each photon as measured by the observer is halved for the higher-redshift stream. The frequency of each photon in the stream from z=3 is halved and the wavelength is doubled, relative to the stream from z=1. So even though the stream received from z=3 has twice the duration, the total energy of that stream is measured to be half that of the z=1 stream. The instantaneously measured energy density of the z=3 stream at any time must be 1/4 of the energy density of the z=1 stream.

Let's set cosmological redshift aside for a moment and consider classical Doppler redshift using sound waves. The observer is moving away from the emitter of a sound pulse such that the received wavelength measured by the observer is twice the wavelength that was emitted (in the emitter's frame). Again, the observer measures the duration of the sound pulse to be twice what the emitter had measured, and the instantaneously measured energy density to be 1/4 (after factoring out the effect of spatial divergence of the sound waves).

We know that no sound energy has been "lost" in this scenario from the emitter's perspective, even though the observer measures that the sound energy has decreased by 1/2. The measured "loss" of sound energy is simply the result of the observer being in an inertial frame which is moving away from the emitter's inertial frame.

Returning to the cosmological redshift, the fact that the observer measures a "loss" of radiation energy from a redshifted source does not mean that the emitter calculates that this energy has been "lost" as considered from the vantage of his frame. Again the energy loss measured by the observer is a function of the observer and emitter being in different frames. Specifically, the distance between the emitter and observer increases continuously as a cumulative result of the Hubble flow. Of course this is GR so the movement of the two distant local frames relative to each other, and the resulting measured "loss" of energy cannot be calculated using the same classical Doppler mathematics applicable to the soundwave example.

Aside from the complexity of the energy conservation question, I think my other basic point is correct. The physical length (and time duration of reception) of any redshifted lightbeam segment will be measured in the observer's frame to be stretched in the same proportion as the redshift factor (e.g., z+1 for cosmological redshift), as compared to the emitter's frame.

The observations of the stretching of the redshifted lightbeam segments received from supernova events (referred to somewhat confusingly as the "time dilation" of those events) is described in http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.3595v1.pdf" .

 

[Al68]

Why would we expect energy to not depend on relative velocity?

If someone behind me throws a ball at my moving car, neglecting air resistance, it will have less kinetic energy in my frame when it reaches me than it had in the thrower's frame when it left. Not because the ball "lost" energy, but because the energy is frame dependent.

Light (or anything else) received from a star moving away from us would have to have less energy in our frame than the rest frame of the star. Am I missing something?

 

[nutgeb]

Al68, no you aren't missing anything, at least with respect to classical Doppler redshift. That's why I agreed with JesseM that the total energy of a sound pulse measured in the moving observer's frame is indeed lower than the energy emitted in the emitter's frame.

 

[Al68]

Al68, no you aren't missing anything, at least with respect to classical Doppler redshift. That's why I agreed with JesseM that the total energy of a sound pulse measured in the moving observer's frame is indeed lower than the energy emitted in the emitter's frame.

Hi nutgeb,

I apparently wrote that post while you were posting yours. I see you already said the same thing. I'm just surprised it took that long for someone to point out that energy is only conserved in each frame, not equal in every frame.

Al