Expansion, redshift, and conservation of energy

[TumblingDice]

I've often read posts where the conservation of energy is questioned regarding redshift due to expansion. The question arises because the energy level of photons is directly related to the light's frequency, and this alludes to photons "losing" energy as the frequency lowers. The reasoning I've read most often on PF is that conservation of energy only applies locally, or simply doesn't apply to expansion.

I'd like to approach this from a different perspective. Consider a distant light source that oscillates one second on and one second off. Since it's a thought experiment, why not make it monochromatic - 800THz in low UV.

Now imagine that source was at the appropriate distance at the appropriate time in the past so that today the light is arriving to us with a 100% redshift. The frequency arriving is now 400THz, indicating that when we detect a single photon, it will measure half the energy as we would have detected from a single photon near the 800THz light source.

However, each original one-second pulse is now arriving to us over a period of two seconds - energy is arriving for twice as much time. If we measured the sum total energy of each two-second pulse arriving on an appropriately sized surface area, would it not be equal to the energy arriving in a one-second pulse if expansion didn't occur?

The thought experiment isn't perfect because an "appropriately sized" detector for each setup would be a bit outrageous due to the spherical nature of propagation. I could have suggested an even more outrageous kind of Dyson sphere detector to capture/measure each entire pulse, but in any case I hope the spirit of my question is clear, wanting to learn if energy may actually be conserved when all is considered.

 

[Chronos]

One problem with your scenario is the number of photons remains constant: same number of photons, less energy per photon = lost energy. There may be a way around this conundrum. Light emitted by a compact object [e.g., neutron star] is gravitationally redshifted. It seems pretty obvious where this energy goes - into the gravitational potential of the compact object. That is one of the 'explanations' for where redshift energy goes, it goes into the gravitational potential of the universe at large. Another way to look at it is from the reference frame of an observer traveling at relativistic velocity. From the observers perspective, the universe is blueshifted in the direction of travel, and redshifted in the opposing direction. Where does the extra energy of the blueshifted photons come from?

 

[TumblingDice]

One problem with your scenario is the number of photons remains constant: same number of photons, less energy per photon = lost energy.

You're saying my setup is poor because it's proven or theorized that the "number of photon remains the same". I tried to avoid jumping to that, not mentioning photons except when measuring a quanta of energy as detected upon arrival. You mentioned "same number of photons" quite quickly. Is this a part of a theory, or has been experimentally proven, or both?

There may be a way around this conundrum. Light emitted by a compact object [e.g., neutron star] is gravitationally redshifted. It seems pretty obvious where this energy goes - into the gravitational potential of the compact object. That is one of the 'explanations' for where redshift energy goes, it goes into the gravitational potential of the universe at large. Another way to look at it is from the reference frame of an observer traveling at relativistic velocity. From the observers perspective, the universe is blueshifted in the direction of travel, and redshifted in the opposing direction. Where does the extra energy of the blueshifted photons come from? From the redshifted photons? Surely not.

I tried my best to keep my question clear and focused. I'm not questioning conservation of energy or lack of it, nor looking to reach for other perspectives on resolving any dilemma.

My question is, if we capture/measure all of the EM - all one second at the source and all two seconds at the destination after effects of expansion, would the total energy be the same?

 

[PeterDonis]

The frequency arriving is now 400THz, indicating that when we detect a single photon, it will measure half the energy as we would have detected from a single photon near the 800THz light source.

However, each original one-second pulse is now arriving to us over a period of two seconds - energy is arriving for twice as much time.

You're mixing incompatible models, so of course you're going to get the wrong answer.

If you're using the photon model, then each "pulse" contains some fixed number of photons. It has to, because by your own hypothesis, each pulse contains a fixed amount of energy when it's emitted, and that means a fixed number of photons, since the energy per photon is given by the emitted frequency. And since, by hypothesis, the photons don't interact with anything in flight, the same number of photons must arrive at the detector as were emitted by the source. So the only thing that changes is the frequency. The time it takes to emit or detect the photons is irrelevant to the total energy emitted or received in this model, because each photon is discrete: total energy is just energy per photon times number of photons, with no other variables included.

If you're using the "one-second pulse" model, meaning that you're modeling each pulse as taking a certain amount of time to be emitted at the source, then you can't model the pulse as containing photons, because you count photons by number, not by time taken to emit. The "one-second pulse" model is a classical wave model, where you're basically considering the pulse to be a wave packet with a duration of one second when emitted; the total energy carried by the packet is the amplitude squared (modulo some constant factors that don't matter here) times the duration. The redshifting of the packet as it travels keeps the "area under the curve" of the wave packet constant: if the packet duration is doubled, the amplitude is halved. But since energy is amplitude squared times duration, there is an extra factor of two, so the redshifting cuts the total packet energy in half.

 

[Chronos]

I'm saying your logic is flawed.

 

[TumblingDice]

You're mixing incompatible models, so of course you're going to get the wrong answer.

I didn't intend to mix models - I tried to stick with not bringing photons into my question, only recognizing that previous explanations began with the direct relationship between photons and frequency. I thought it might be a "can't recognize the forest by looking at the trees" kind of a thing. Expected I was wrong about this, and wanted to learn.

If you're using the photon model, then each "pulse" contains some fixed number of photons. It has to, because by your own hypothesis, each pulse contains a fixed amount of energy when it's emitted, and that means a fixed number of photons, since the energy per photon is given by the emitted frequency.

I can't see where I made any type of hypothesis. Chronos' and your replies get right into photon counts. I anticipated that my OP would lead me to follow-thru questions in the QM forum, and thought the Cosmo forum would get me started on the right foot. This is why I posed what I thought was a simple question that could be experimentally confirmed for the theory or model being offered.

The "one-second pulse" model is a classical wave model, where you're basically considering the pulse to be a wave packet with a duration of one second when emitted; the total energy carried by the packet is the amplitude squared (modulo some constant factors that don't matter here) times the duration.

This sounds to be the modeling/math I hoped to discuss within.

The redshifting of the packet as it travels keeps the "area under the curve" of the wave packet constant: if the packet duration is doubled, the amplitude is halved. But since energy is amplitude squared times duration, there is an extra factor of two, so the redshifting cuts the total packet energy in half.

And this is the answer I was looking for. Yea for the home team! Thank you, Peter. So the total energy is halved, just as the frequency, and at least in theory, supports the premise of a constant number of photons all the way from emission to detection. I'll go off for a bit and properly digest. I appreciate everyone's patience helping me hammer out the dents and fill the holes in my understanding.

 

[TumblingDice]

I'm saying your logic is flawed.

I do not understand. I posed a setup and asked a question. I didn't suggest an answer, only asked whether the total energy would be the same or not. Can you please explain where you feel my post contained flawed logic?

 

[PeterDonis]

I tried to stick with not bringing photons into my question, only recognizing that previous explanations began with the direct relationship between photons and frequency.

But that relationship requires using the photon model. There's no such relationship if you don't bring photons into the question. Most people explain it using photons because it's easier that way: the direct energy-frequency relationship leads immediately to the energy-redshift relationship.

I can't see where I made any type of hypothesis.

Your OP talks explicitly about photons and the energy-frequency relationship. That makes no sense unless you are using the photon model, as I said above.

This sounds to be the modeling/math I hoped to discuss within.

But this model, as I've already said, has nothing to do with photons or the photon energy-frequency relationship. It can be used to get the right answer, but it does so by a different route.

So the total energy is halved, just as the frequency, and at least in theory, supports the premise of a constant number of photons all the way from emission to detection.

No, it doesn't, because, as I've said, there are no such things as photons in the wave packet model. The fact that this model gives the same answer as the photon model does not mean you can draw any conclusions about photons from the wave packet model. If you want to understand why photon number is constant in this scenario, you have to look at the photon model. (I explained why this model gives constant photon number in my previous post.)

 

[TumblingDice]

But that relationship requires using the photon model. There's no such relationship if you don't bring photons into the question. Most people explain it using photons because it's easier that way: the direct energy-frequency relationship leads immediately to the energy-redshift relationship.

Your OP talks explicitly about photons and the energy-frequency relationship. That makes no sense unless you are using the photon model... ...as I've said, there are no such things as photons in the wave packet model. The fact that this model gives the same answer as the photon model does not mean you can draw any conclusions about photons from the wave packet model. If you want to understand why photon number is constant in this scenario, you have to look at the photon model. (I explained why this model gives constant photon number in my previous post.)

Thanks a bunch for the follow-thru to make sure I'm getting it right! I do believe my understandings have not been as far off as you've imagined. I could have, and now see I SHOULD have never mentioned anything about the photon/frequency relationship in the OP. It only served to precipitate replies based on the photon model, which I was purposely trying to avoid. Your classical explanation of energy/amplitude relationship was the simple answer I was looking for. I only mentioned photon counts when thanking you for the classical explanation as an acknowledgment that I was comfortable with the correspondence of the different models.

Terminology and clarity are the largest hurdles I've observed to effective communication on the forums. I'll continue to try to do better.

 

[Drakkith]

Excellent! I had tried to ask this same question a long time ago but couldn't word it right. So whether you view it classically as "wave packets" or quantum mechanically as photons, you still get the same loss of energy, correct? Very interesting...

 

[Lino]

... I anticipated that my OP would lead me to follow-thru questions in the QM forum ...

All, I thought that this question would naturally lead onto the QM forum as well. If I ask the OP question in a different way, are we saying that you can use both / either the photon or classical wave model to explain the apparent increase in energy in a double slit experiment? (I acknowledge that this is Forum is Cosmology and not QM, so I am not asking specifically about the double slit experiment, I just can't think of an equivalent cosmic, but without expansion, example.)

 

[bcrowell]

Energy isn't globally conserved in GR. We have a FAQ on this: https://www.physicsforums.com/showthread.php?t=506985

 

[Lino]

Thanks bcrowell. I am working my way through MTW: Gravitation at the moment, but I'm no where near p.457 yet!

Regards,

Noel.

 

[Chronos]

For further discussion, see Tamara Davis' article "Is the Universe Leaking Energy?" http://www.nature.com/scientificamerican/journal/v303/n1/full/scientificamerican0710-38.html.

 

[Lino]

For further discussion, see Tamara Davis' article "Is the Universe Leaking Energy?" http://www.nature.com/scientificamerican/journal/v303/n1/full/scientificamerican0710-38.html.

Thanks Chronos. I think that I need to do more reading on this!

... Symmetry and conservation ... Sure, the scenery may change depending on where you are standing, when you are standing there and the direction you are looking, but the fundamental underlying laws of physics that dictate how that scenery behaves are independent of your location, orientation and time ... Creative accounting ... This malleability of space implies that the behavior of the universe is not time-symmetric ...

How come the locational variations over time don't destroy physical symmetry in the first example (nature) but do in the second (cosmic)?

Regards,

Noel.

 

[Chronos]

Without descending into the mind numbing depths of killing vectors and such, it's an issue of scale. Locally, conservation of energy works perfectly [at least to our measurement ability]. The global case is where it becomes complicated.

 

[PeterDonis]

Sean Carroll has a good blog post on the issue of energy conservation in GR:

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

Key quote:

ack when you thought energy was conserved, there was a reason why you thought that, namely time-translation invariance. A fancy way of saying “the background on which particles and forces evolve, as well as the dynamical rules governing their motions, are fixed, not changing with time.” But in general relativity that’s simply no longer true. Einstein tells us that space and time are dynamical, and in particular that they can evolve with time. When the space through which particles move is changing, the total energy of those particles is not conserved.

He goes on to explain the difference between local energy conservation, which is still true in GR (the covariant divergence of the stress-energy tensor is zero), and global energy conservation, which is not true in general (though a version of it still holds in spacetimes with a time translation symmetry).

 

[TumblingDice]

Sean Carroll has a good blog post on the issue of energy conservation in GR:
http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

Sean's post covered everything for me, and then some. Simply outstanding. Thank you!

 

[Lino]

Thanks all. Much appreciated.

Regards,

Noel.