[Eternal Student]

Hi everyone.

   I'm not sure how this thread has progressed, I've been away for a while.
Anyway, it seems that Hamdani's question hasn't been answered.  When I say "answered", I mean pushed around, pulled apart, mulled over and generally considered.  If you (Hamdani) are just after a straight answer, you could probably try a textbook or another forum.

    The moving spaceship thing is quite interesting and I can see how various people can imagine things to be one way or another, coming up with differing explanations as to why redshifts may or may not occur and if energy is actually lost or not.  I've not spent too much time considering the changes in and relationships between the frequency, amplitude, power and time duration of the pulse myself, so I might as well have a look at it now.

    Let's take a particle model for EM radiation to start with.  In this model the laser light is just a discrete collection of particles called photons.

It sends a 1 Watt 600 nm laser pulse, one second long in its frame of reference.

  In the particle model, we can regard the "intensity" of the light as if it is just the intensity of bombardment by photons.  Specifically, we can say that Intensity is proportional to the number of photons received per second.  What is the constant of proportionality?  It's just the energy carried per photon:    Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation.  Without loss of generality, we're going to ignore the area of the receiver in this example:  All emitted photons from the spaceship will eventually be absorbed by the receiver.  Let's say that it's all happening in a narrow beam that is precisely directed from the spaceship to the receiver.   Additionally we aren't going to trouble ourselves too much over the energy or power (measured in Joules or Joules per second).  We are just going to count the number of photons that are received per second.  We know that each photon has an energy E = hf  associated with it,  so if we  know the number of photons per second coming in then we can determine the energy coming in per second if we wanted to just by multiplying by  hf,   where  h is Plancks constant and f is the frequency for the photon(s) being received.
   Just to clarify then, when I say  "Intensity"  I will generally mean the number of photons per second and not the energy per second.  This is what will link to your (Hamdani) mention of the power being 1 Watt,  there is 1 units worth of photons being created per second (or, for simplicity, we could just work with 1 photon per second and scale it up at the end).

  OK.  let's make a start.

1.    The ship has a constant velocity of 0.1 c relative to the receiver.  This produces a gamma factor of
1 / root(0.99)   ≈   1.005.    The gamma factor is the same regardless of whether the ship is moving toward or away from the receiver.

2.  The ship emitted photons for 1 second as far as the crew of the ship were concerned  (in the rest frame of the ship).   Standard time dilation stuff implies the receiver considers the ship to have been emitting photons for  1.005 seconds (in the rest frame of the receiver).   This remains the case for both movement toward or away from the receiver.   This is the first interesting result.  The receiver believes the ship emitted photons for slightly longer (in time) than the crew believe they emitted photons.
   If you (Hamdani) wanted an answer as to how long in time the pulse lasted for the receiver then we can already offer one answer:  The ship was emitting photons for 1.005 seconds as far as the receiver is concerned.  However this is NOT how long (in time) it will take the receiver to collect or absorb all those photons.  If the ship was moving away from the receiver then the last emitted photon travelled from a more distant location than the first emitted photon, so it will take the receiver slightly longer to receive that last photon.   Conversely, if the ship was moving toward the receiver then the last photon has got a bonus of being already brought slightly closer to the receiver at the time of it's emission.  Let's consider this effect next...

3.  The photons eventually come in and bombard the receiver.  We are interested in the time elapsed from the first photon coming in to the last photon coming in.  Without loss of generality set up a standard reference frame for the receiver.  Put the receiver at the origin (at all times, we do want a rest frame for the receiver) and put the ship over on the right hand side which we will consider as having a positive x- co-ordinate.  If the ship is coming towards the receiver then the velocity is negative,  the x- co-ordinate will start large and decrease as time progresses.  If the ship is moving away the opposite applies.  All movement is precisely along the x- axis (if it wasn't then choose a different axis so that it is, that's why we say "without loss of generality").
   Here's a diagram:

Ship-reciever.png (15.42 kB . 1258x648 - viewed 3267 times)

   Now, we know that light always travels at the speed c in any reference frame.
   We also know that the ship has a velocity 0.1c  relative to the receiver's rest frame.

Let the first photon be emitted by the ship at   x and t co-ordinates   given by the event written as  (x,t)  in the receivers rest frame.   [lower case x and t]
Let the last photon be emitted at the event (X,T) in the same frame   [that is upper case X and T].
   We already know that  (T - t )   =  (gamma factor) . (1 second)  =   1.005  seconds  from  part 1. above (the usual time dilation stuff).
   We also know that  X and x are just the positions of the ship at different times and they are related by the velocity of the ship (in the receivers rest frame) and that elapsed time of 1.005 seconds over which it was emitting laser light (in the receivers rest frame).  Specifically  X = x + 0.1c. (T-t)  for the ship moving away from the receiver.  While X = x - 0.1c . (T-t)  for the ship moving towards the receiver.

   Now the first emitted photon won't just have stayed at position x all the time while the spaceship was moving and completed its emissions.  Instead the first photon will have travelled toward the receiver at speed c over the elapsed time  (T-t).   So at the final time T  in the receiver's rest frame the first photon will actually be located at    x - c(T-t)    in both cases (ship moving toward or away from the receiver).  Meanwhile the final photon emitted is located precisely at the starships position of  x + 0.1c(T-t)   or  x - 0.1c(T-t)  at time T depending on whether it is moving toward or away from the receiver.   All we have to do now is find the difference in x -co-ordinates at time T.... just subtraction....

 x + 0.1c(T-t)  -  [ x -c(T-t) ]   =   1.1 c (T-t)   for the ship moving away.
Similarly we get  0.9 c (T-t) for the ship moving toward the receiver. 
  If you want numbers...  put T-t = 1.005  and we get:     1.1055 light-seconds when the ship is moving away.
    0.9045 light-seconds when the ship is moving toward the receiver.   (For comparison recall the ship's crew, in the ships rest frame, would believe the pulse is physically 1 light-second long since it was travelling at speed c away from their ship and they emitted for 1 second).   That's how long (in distance) each e-m pulse (or the total stream of all the photons) will be at time T, however it stays at that length from then onwards since each individual photon within that stream of photons travels at the same speed and in the same direction.

Anyway, that's how long (in distance) the pulse of e-m radiation will be as it is observed in the receivers rest frame.  If you want to know how long (in time) it takes the receiver to absorb the pulse we can do that easily enough:  e-m radiation always travels at speed c in any reference frame.  So if the pulse is physically  1.1055 light-seconds long then divide that by c and we see it takes 1.1055 seconds to absorb it.     Similarly it takes the receiver 0.9045 seconds to absorb all of the transmission if the ship had been moving toward the receiver.

   This is interesting because it's not what you get if you just applied standard distance contraction stuff to the length of the pulse as observed by the ship's crew.   Recall the ship's crew believe the pulse is physically 1 light-second long.  The standard distance contraction formula would imply this reduces to about 0.9 light second in the receivers rest frame regardless of whether the ship was moving toward or away from the receiver BUT it doesn't.  Why not, or what prevented the formula from working, you may ask?   Well, I think it's best explained by noting that the pulse was NOT moving with the ship,  the pulse had a non-zero velocity relative to the ship.  The standard distance contraction formula only applies to objects that would be moving with the ship or appearing to be stationary to the crew on the ship - but it's interesting and took me a good long moment to think about and check that what we have is actually right.

 4.   What about the energy or intensity of the pulse?

What's the power of the laser that we will receive? what's the wavelength?

    The wavelength of the e-m radiation is determined by the usual relativistic Doppler shift formula.
This post is already too long.  So here's the Wiki article about it:   https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
     The photons are red-shifted (have their freqency reduced or the wavelength increased) when the ship is moving away from the receiver.   The converse happens when the ship is moving toward the receiver.
   Now there's an interesting result in special relativity that rarely get's mentioned and just taken for granted.  There is a conservation of events (please note that is not a real name or principle, I made it up).  Specifically the Lorentz transformation  L :  (x,t)  →  ( x', t' )    is injective  or  one-to-one.      If    (x₁, t₁) and (x₂, t₂) are two distinct events  (so that at least one of   x₁ ≠x₂  or  t₁ ≠ t₂  is true)   in reference frame F    then they get mapped to different events  (x₁', t₁') and (x₂', t₂') in the Frame F' .   Why is this useful or important?  It means that events are never lost or squashed down into one event in the other frame, instead they just appear somewhere else and/or at some other time when you change frames of reference but you can always identify an event in one frame uniquely with an event in the other frame.  So we can have a set of events in Frame F   and count them to get their total number,    or   we can consider their corresponding events in F' and count their number in that frame,  we get the same total number of events.
    Anyway, this "conservation of events" idea will hold when the Lorentz transformation is well defined, which just means when the velocity v of the frame F' relative to F is less than c  -   or to paraphrase this further - whenever the situation is actually sensible.   (If you consider frames moving at light speed relative to each other then many events in one frame will be squashed down to the same indistinguishable event in the other frame).
  As I said, this is often taken for granted, events don't just disappear or two events merge into one when you change frames of reference.  Anyway, we're going to use this to count the number of photons in the stream of photons that were emitted by the ship.
   In the rest frame of the ship, there were N total photons emitted.  Each can be identified with an event (x,t) where x = x-co-ordinate of the ship at the time of the photons creation    (so that will be 0 in the rest frame of the ship since the ship can be assumed to always be the origin in that frame).  The time co-ordinate t is the time at which the photon appeared or was created.   These events are mapped by the Lorentz transformation to other events in the rest frame of the receiver but the important thing is that there are still precisely N of these events.  To say that another way there were N photons created in any frame of reference you care to use.   Since we assume all the photons are directed toward the receiver and eventually absorbed the receiver will receive a total of N photons.
   OK, so we know the receiver always absorbs N photons eventually,  the only difference is the amount of time it takes to absorb those photons  (see part 3 discussed above).   We can now determine the bombardment intensity for the case where the ship moves toward or away from the receiver.
   It's   N  /  0.9045      ( I can't recall if I used the right numbers... I got it from part 3 above)   with the ship coming toward the receiver.     While  it's   N / 1.1055  with ship moving away from the receiver.      So it's more intense but lasts less time with the ship coming toward the receiver.

    If you prefer to think about your intensity as a power (in Joules per second) instead of as a number of photons coming in per second then recall that each photon has an energy  hf     with   f  found from the relativistic Doppler shift  (I think I discussed this earlier and I'm just repeating myself, sorry).   Overall, the  intensity (in terms of power) is much greater when the ship is coming toward the receiver but lasts only  0.9045 seconds.      It's a much lower power but lasts longer,  1.1055 seconds when the ship is moving away.

5.   So how does this relate to conservation of energy?
   The total energy received by the receiver    is always  given  by    N . hf    where  N = number of photons received, h is plancks constant and f is the frequency received.
N is the same for the ship moving toward or away (see earlier discussion).  h is obviously always the same.   Meanwhile f changes as discussed above, the radiation received has a higher frequency if the ship is coming toward the receiver.
     So, if the ship is coming toward the receiver then the receiver believes the ship emitted  an amount of energy  Nhf₁    and the receiver will eventually absorb that same amount of energy.
     If the ship was moving away then the receiver believes the ship emitted less energy   Nhf₂  which the receiver will also eventually receive.
     In the ships rest frame, the crew believe that a middling amount of energy was released by them and will be eventually absorbed by the receiver.
     So in all frames of reference, there is a conservation of energy across the whole system when the same reference frame in used throughout the experiment.  However, there are differences in the actual size of the changes in energy that occurred at various places and times between the different frames of reference.
    This is quite conventional in Physics.  Total Energy and the changes in energy that occur are frame dependant, however the conservation of energy holds in all frames.

6.   Caveats and general limitations:
   This thread originally started with some discussion of GENERAL relativity but this post uses only special relativity.  The change in frequency for photons is not given by the conventional relativistic Doppler shift formula when GR is used.  However, locally it all still applies as it is.
   Generalizing the results to GR and a more complicated spatial geometry we should note what was said in item 5. above.    Energy is frame dependant.  Changing frames of reference can adjust the changes in energy that occur in what would physically represent the same place and time within the system.  For a complicated spatial geometry we cannot extend an inertial reference frame to cover all of the universe.  As a result we are forced to use only local frames of reference.  When light has travelled long distances through space, it has been necessary to change local frames many times.   It should, perhaps, come as less of a surprise to find that while a stream of photons required a certain amount of energy E to create in the local frame where they were created, when they are finally absorbed they may only deliver an amount of energy  e < E  to the receiver in the local frame of the receiver.

7.   Using a wave model of e-m radiation instead of considering it as a stream of individual photons, we need to re-interpret the bombardment intensity of the photons (number of photons incoming per second).  This would now represent an amplitude of the e-m wave received.  So the situation is as follows:
   When the ship moves toward the receiver these things happen:
(i)  The receiver takes only 0.9045 seconds to absorb the entire stream or pulse,  exactly as before.
(ii)  The physical length of the pulse is exactly as stated before, 0.9045 light-seconds.
(iii)  The amplitude of the wave is higher then that measured by the crew of the ship in their own frame.
(iv)  The freqency of the wave is also higher than the crew observe - exactly as stated before  (use relativstic Doppler shift).

... similar stuff for the ship moving away...  the amplitude of the wave is now LOWER than that observed by the crew.
   It's interesting that we usually refer to the Doppler shift as a frequency shift but for light (not usually for sound) there should be an equally interesting shift in the amplitude of the wave that also occurs.   There's some mention of the Doppler effect on intensity here:  https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Doppler_effect_on_intensity

 Anyway, that's it.  It's late on New years eve and I've probably made some mistakes.  Best wishes and happy new year to everyone.

[Eternal Student]

Hi again.

   Request for assistance or verification from someone please.

   There is very little reliable information about the Doppler effect causing a change in the amplitude of an e-m wave only that it causes a shift in frequency.  (Well, not much reliable info that I can find from Google anyway).

    I'd be grateful if someone can check this, provide the benefit of their experience or produce a link to some credible source backing this up.

I wouldn't like to have suggested something that is incorrect.

Thanks,   E.S.

[Halc]

Long time no see ES. Glad to see you back.

I didn't go through the entire post, but it seems wrong to mix a quantum explanation with a classic one.
The power of a light beam from a receding source can be computed by considering a one-second burst, but the answer is, in the end, not a function of the duration of the signal.

    The moving spaceship thing is quite interesting and I can see how various people can imagine things to be one way or another, coming up with differing explanations as to why redshifts may or may not occur and if energy is actually lost or not.

Relative to an inertial frame (which is what we're considering in a scenario with a nearby source of light receding at 0.1c), energy cannot be lost per conservation of energy, but total energy of the one-proper-second signal is frame dependent.  So this is not really discussing the topic of this thread, which is energy actually not being conserved relative to a non-static frame such as the cosmological frame that describes the universe as a whole.

In this model the laser light is just a discrete collection of particles called photons.
...
Specifically, we can say that Intensity is proportional to the number of photons received per second.

This is wrong, since the intensity of light is also a function of the energy of each of those photons, which in turn is frame dependent. You don't seem to take that into account.

The energy of the photon goes down due to redshift, and the power goes down more due to the pulse being spread out over more time than just a second.
So it seems the power is (reduced via redshift energy)/(increased duration).

[Eternal Student]

Hi and thanks Halc.

This is wrong, since the intensity of light is also a function of the energy of each of those photons, which in turn is frame dependent. You don't seem to take that into account.

   Yes, I'm on-board or in agreement with that.

I did mention it here: 

.... We are just going to count the number of photons that are received per second.  We know that each photon has an energy E = hf  associated with it,  so if we  know the number of photons per second coming in then we can determine the energy coming in per second if we wanted to just by multiplying by  hf,   where  h is Plancks constant and f is the frequency for the photon(s) being received....

   ... and later on again, I did specifically mentioned using different frequencies for the different cases...

....(from item 4)...  If you prefer to think about your intensity as a power (in Joules per second) instead of as a number of photons coming in per second then recall that each photon has an energy  hf     with   f  found from the relativistic Doppler shift  (I think I discussed this earlier and I'm just repeating myself, sorry).

However, I fully appreciate that the post was long and no one can read it all.   I deliberately flagged the unusual use of the word "intensity" in the post...

Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation.

   
   
About mixing quantum and classical theories:

I didn't go through the entire post, but it seems wrong to mix a quantum explanation with a classic one.

   Possibly.   It was done deliberately just to see how the amplitude might be affected.  Evaluating the amplitude of the wave recieved directly using classical theory throughout the derivation was hard while using a particle model for the radiation it just fell off the tree with ease. 
   But if the final result holds, then it should hold regardless of wheteher we look at the e-m radiation as a classical wave or a collection of particles.  If you choose to regard the wave ultimately recieved as a classical e-m wave then you are forced to conclude the amplitude has changed.
   If we look upon this a different way, for a classical (non-quantised) electro-magnetic wave, the energy intesity in the wave is proportional to the square of the amplitude (of either the electric or magnetic field, your choice, or use the product of the two if you prefer) and it turns out to be completely independent of its frequency of oscillation.  I know that sounds weird but we can pull out the references if required.  It seems that to reconcile the notion of energy intesity in a classical e-m wave with the energy content of an individual photon you need to imagine that if you could run a suitable set of detectors over an individual photon then you would observe these two things happening TOGETHER when the energy of the photon is increased:   (i) The frequency of oscillation in the electric and magnetic field you observe will increase with increasing energy of the photon  but also  (ii) The amplitude or peak electiric field strength you would observe will also increase in proportion with the frequency.   To say that another way, more energetic photons not only have a higher frequency they would display (on our imaginary oscilliscope) as if they also had a correspondingly higher amplitude.
     [References available but many of them are books and not online references.   Try  the section around page 347 of Electrodynamics by Giffiths where the Poynting vector is discussed,  if you've got this on your shelf.  The energy flux in the wave is given by  E x B    so has magnitude proportional to   E²    which  varies with time (and position) since E and B can be assumed to be sinusoidal - taking an average intensity over time yields  intensity proportional to Amplitude squared (and the frequency ω matters not the slightest).      For a quick on-line guide (a little less authoritative)  try  physics libretexts  https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/16%3A_Electromagnetic_Waves/16.04%3A_Energy_Carried_by_Electromagnetic_Waves ].

  So, if you do accept that the total energy in the pulse is different in the ships rest frame and the recievers rest frame, it's inescapable that either the classical amplitude is higher in the receivers frame or else the pulse is physically longer and therefore can be absorbed by the reciever for more time.  However, it just isn't longer when the ship is coming toward the receiver, it's shorter (than the 1 light-second in the ships rest frame),  so the classical amplitude is actually forced to be very much higher.

   OK,  that's probably drifted off topic enough, sorry for that.   I've had a bit more time to think about it and I'm a bit happier to leave Hamdani with the impression that the (classical) amplitude of the e-m wave received will change.  I can only apologise if it's incorrect.   (Hamdani, you may want to take some care and double check this yourself).

Best Wishes.

[Halc]

I did mention it here:

OK, I was in a severe rush at the time of that posting and didn't see it. I just saw a statement equating power to photons/sec. Was just leaving on last leg of a big road trip. 4 straight days driving. It's over now 

Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation.

Right. Since the beam was concentrated on a fixed area, that means intensity and power are essentially equivalent. I wasn't considering it being spread out, which would have been even more off-topic.
   

It was done deliberately just to see how the amplitude might be affected.

I've never thought of it in terms of amplitude.

So, if you do accept that the total energy in the pulse is different in the ships rest frame and the recievers rest frame, it's inescapable that either the classical amplitude is higher in the receivers frame or else the pulse is physically longer and therefore can be absorbed by the reciever for more time.  However, it just isn't longer when the ship is coming toward the receiver, it's shorter (than the 1 light-second in the ships rest frame),  so the classical amplitude is actually forced to be very much higher.

Agree with that, but I think the question concerned a receding emitter, not a ship approaching.

The topic was about energy actually disappearing (not being conserved), and that has nothing to do with ships emitting light behind them, but rather has to do with coordinate systems where space is not static over time.

[Eternal Student]

Thanks for your time, Halc.  I very much appreciate it.

    I hope your trip went well.

[Iannguyen]

As the Universe expands, light loses energy, which is then used to expand the Universe itself in the form of labor. You could assume you're making energy when you burn wood to make a fire. But what's really going on is far more subtle:
Breaking and reforming molecular bonds from a less stable configuration (wood and oxygen) to a more stable configuration (ash and water vapor) releases energy.
If you look at the quantity of energy released and apply Einstein's famous conversion, E = mc2, you'll notice that the mass of the product and reactant molecules differs by a very little amount.
In anything like a nuclear event, such as one that occurs in the Sun, the mass difference is considerably more significant. In fact, if you calculated the Sun's mass from birth to now, you'd find that it has lost roughly the mass of Saturn in those 4.5 billion years of energy emission.
It's possible that a stronger definition of energy will emerge in a richer (i.e., quantum) theory of gravity, and we'll be able to tell whether it's conserved or not. However, in the lack of a formal definition, we must make do with what we have, which are the tools and definitions that we currently have. Yes, photons lose energy, but that energy does not vanish forever; the quantity of energy lost (or gained, for that matter) in the expanding (or shrinking) Universe adds up to precisely what it should.

[hamdani yusuf]

In fact, if you calculated the Sun's mass from birth to now, you'd find that it has lost roughly the mass of Saturn in those 4.5 billion years of energy emission.

Have you taken solar wind and solar flares in to account?

[Eternal Student]

Hi @lannguyen.    I'm not sure I've seen your posts before.   Welcome and I hope all is well.

    You've made an interesting post.  Thank you and well done.   There's a lot about your post that is perfectly good and makes solid scientfic sense.

    Usually that doesn't get mentioned, instead someone will just go straight in to challenge one or more points in the post.  So this is where you get a slightly harder time, sorry...

As the Universe expands, light loses energy, which is then used to expand the Universe itself in the form of labor.

   This point for example.   

    Someone could ask for the evidence to show that this is the correct situation.   How were the calculations done?   How do you begin to determine how much energy it takes to expand the universe?   Having got the amount of energy required for that,  how do you determine that the amount of energy lost by all the radiation is sufficient to match that?     Is it actually just a reasonable guess that something like your suggestion is happening (lost energy from radiation is accounted for in the energy required to expand the universe)?
    Finally, does your suggestion mean that the expansion should slow down as the density of radiation falls off and all of it is moved further to the red (low energy) range?  (Because that is precisely the opposite of what does actually seem to be happening).

    Don't worry too much, your post seemed quite sensible and discussion is very much welcomed.  I'm just deliberately focusing in on one detail because it did seem to be presented as if it was a fact rather than just a possibility or a speculation.

Best Wishes.

[alancalverd]

The OP is addressed by the Pound-Rebka experiment.

The gravitational potential of "deep space", infinitely distant from any mass, is zero, and the gravitational potential close to a mass is negative.

A ballistic missile or a photon travelling upwards from the earth's surface gains gravitational potential energy (moving away from the negative potential) and loses an equal amount of kinetic energy. In the case of a photon, this appears as red shift.

If the source and receptor are moving relative to each other, the additional kinetic energy due to that motion is expressed in the Doppler frequency shift. 

[Eternal Student]

Hi.  I hope all is well.   That's a nice post @alancalverd.   

   You could also mention that there was a podcast based on the OP and it is available here:   https://www.thenakedscientists.com/sites/default/files/media/podcasts/episodes/Naked_Scientists_QotW_21.09.13.mp3

The gravitational potential of "deep space", infinitely distant from any mass, is zero, and the gravitational potential close to a mass is negative.

   OK  -  but is there any place in space that is  "infinitely distant from any mass"?
   The usual cosmological principle implies space is quite homogeneous on large scales.  There should be just as much mass and it should be just as close to the photon "out there" in deep space as it was right here.   On average the environment for the photon does not change no matter where it goes or how long it travels.   It's then hard to say that the photon is moving to a region with higher gravitational potential.
   The Pound-Rebka experiment certainly shows that a photon leaving a planet or a star experiences a red-shift for a while - but that effect should stop when it gets into the gravitational well of some other celestial object and it might actually start to experience a blue-shift then.   The red-shift that seems to happen to all light due to the expansion of space isn't just something that happens for a while, it seems to happen eternally.

Best Wishes.

[alancalverd]

   OK  -  but is there any place in space that is  "infinitely distant from any mass"?

Physicists are quite used to dealing with idealised infinities. Indeed I was thinking about this very problem whilst shaving this morning. I can imagine an infinite space with a finite density of particles. On a big scale with a few fluctuations this looks like the universe, and on a small scale with regular spacing it looks like an ideal crystal. We can solve all sorts of equations for an infinite lattice and add boundary conditions for a finite one, so the concept of "deep space" gets us close enough to the observed  energy of photons emitted close to the detector, and a boundary condition of emission from a massive object, or travelling through a finite-density universe, points us in the direction of the Pound-Rebka result. 

P-R does indeed demonstrate blue shift if you invert the apparatus, and there are celestial objects with blue shift too. Problem is that this planet, and anything we make on it, is too small to induce a significant gravitational blue shift of a photon emitted from a distant star, which was necessarily a lot bigger.