redshift and conservation of energy

fffranklin · Dec 1, 2007

If the expansion of space redshifts light, where does the lost energy go? In other words, if light of a lower frequency has less energy, then how is energy conserved when the light is redshifted to a lower frequency? thanks for any help with this!

adrenalynn · Dec 1, 2007

If there were apparent losses, why wouldn't it be radiated? But from a more practical guess, I'd suggest thinking about it relativistically. It's redshifted to us, the observer. If an observer were traveling faster than the expansion, it would be blue-shifted, right? <div class="Discussion_UserSignature"> .bipartisan (bī-pär'tĭ-zən, -sən) [Adj.] Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.IMPEACH OBAMA! </div>

SpeedFreek · Dec 2, 2007

I think the standard view is that the energy isn't actually lost. With cosmological redshift, the wavelength has been stretched by the expansion of space as the light travelled through it. With a redshift of z=1 we find that light has been stretched to double its original "length" (the universe is now double the size it was when objects at z=1 emitted the light we are now seeing), so light that was emitted over 1 second will take 2 seconds for us to receive (we have observed this "time-dilation" of light in the durations of type1a supernovae). So the energy in that light during 1 second of emission is spread across 2 seconds when we receive it. The energy wasn't lost, it just took longer to arrive! <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

fffranklin · Dec 5, 2007

Adrenalynn, I'm not sure what you mean by "radiated." If you mean radiated as a electromagnetic radiation, I don't think photons can radiate more photons (they are bosons with spin 1 so don't interact). I'm not sure about the relativistic thing you mentioned, except that maybe in GR energy is not conserved.

fffranklin · Dec 5, 2007

Speedfreak, imagine a single photon being redshifted by the expansion of space. It is a quanta of energy, so I don't think the time it would take to recieve it would make any difference in regards to its energy. I see how what your saying might make sense in a classical sense with particles that have mass: how energy is force applied over some distance, so more time applied means more distance, hence you get the same energy as a greater force over a shorter distance. But in a quantum mechanical sense, with individual photons, I don't see how that would work.

SpeedFreek · Dec 5, 2007

Sorry to disappoint you, but we cannot consider redshift in the context of a single photon. Why not? Well, how can a single photon have a redshifted wavelength? A wavelength is found when measuring light as a wave, over a given period of time, and that is why my explanation is correct. <img src="/images/icons/smile.gif" /> edit: or not - see further down the thread - I must remember to check my answers before I post! I will leave my original erroneous explanation intact though ! <img src="/images/icons/wink.gif" /> To use your example, we would have to use a "stream" of individual photons, which would each have the same energy as it started with, but the photons in that stream are further apart than when they started their journey. The stream of photons released during 1 second at the emitter would be stretched by the expansion of space and thus received over more than 1 second at the destination. <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

Mee_n_Mac · Dec 5, 2007

The stream of photons released during 1 second at the emitter would be stretched by the expansion of space and thus received over more than 1 second at the destination. So putting it another way the power received (photons/sec) is less than the power transmitted and the energy measured by integrating over any "short" time period is also less but the total energy, assuming the stream has a finite length, remains the same. <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

SpeedFreek · Dec 5, 2007

Exactly! <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

larper · Dec 5, 2007

<blockquote>In reply to:<hr />Why not? Well, how can a single photon have a redshifted wavelength? A wavelength is found when measuring light as a wave, over a given period of time, and that is why my explanation is correct. <hr /></blockquote> Wrong. Single photons are redshifted. Wavelength has nothing to do with a stream of photons. Each photon has a frequence and a wavelength. Light travels at the speed of light. It isn't being slowed down by redshifting. <blockquote>In reply to:<hr />The stream of photons released during 1 second at the emitter would be stretched by the expansion of space and thus received over more than 1 second at the destination. <hr /></blockquote> Now, THAT is correct, which is basically the definition of time dilation. Time dilation has nothing to do with redshifting. Each photon in that stream is redshifted. <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Dec 5, 2007

Oops! I messed up a little there, didn't I? <img src="/images/icons/laugh.gif" /> At least I got part of it right! So how do we explain the loss of energy (lower frequency) of each photon then? <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

larper · Dec 5, 2007

The loss of energy goes into the expansion of the universe. <img src="/images/icons/smile.gif" /> Now, is that really true? I don't know. The universe does not have to be considered a closed system, and therefore energy does not need to be conserved. <div class="Discussion_UserSignature"> Vote Libertarian </div>

SpeedFreek · Dec 5, 2007

Well, I learn something new every day, and hope to continue doing so! From http://www.faqs.org/faqs/astronomy/faq/part9/section-19.html I.17 Since energy is conserved, where does the energy of redshifted photons go? The energy of a photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is its wavelength. The cosmological redshift indicates that the wavelength of a photon increases as it travels over cosmological distances in the Universe. Thus, its energy decreases. One of the basic conservation laws is that energy is conserved. The decrease in the energy of redshifted photons seems to violate that law. However, this argument is flawed. Specifically, there is a flaw in assuming Newtonian conservation laws in general relativistic situations. To quote Peebles (_Principles of Physical Cosmology_, 1995, p. 139): Where does the lost energy go? ... The resolution of this apparent paradox is that while energy conservation is a good local concept ... and can be defined more generally in the special case of an isolated system in asymptotically flat space, there is not a general global energy conservation law in general relativity theory. In other words, on small scales, say the size of a cluster of galaxies, the notion of energy conservation is a good one. However, on the size scales of the Universe, one can no longer define a quantity E_total, much less a quantity that is conserved. <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

MeteorWayne · Dec 5, 2007

Wow, thanks for the explanation. This whole conversation has been making my brain hurt all day. I feel better now <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

SpeedFreek · Dec 5, 2007

I should have done my research at the beginning, rather than at the end of the thread! I hadn't really looked into this question before, and had a somewhat skewed understanding of how the properties of light are affected by the metric expansion of space... <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

MeteorWayne · Dec 5, 2007

Hey sometimes we all learn as we go along. Questions that people ask inspire us to do more research. That's what makes SDC such fun and so educational!! <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

fffranklin · Dec 5, 2007

Of course you can consider redshift in the context of a single photon, if you couldn't there'd be no such thing as quantum mechanics! <img src="/images/icons/smile.gif" /> I looked around online and found this same question asked and answered in some physics forums, and the clearest answer seems to be that conservation of energy is frame dependent (doh!). Just like if you drop a ball from a train, from the intertial frame of the train, it has no kinetic energy, but to an observer at a train station it is moving as fast as the train and therefore has kinetic energy. Now think of a person with a flashlight on the train. They emit one photon. From the trains frame, the photons energy (or wavelength) doesn't change. From the frame of the train station, the photon was emitted with a higher energy (lower wavelength). It would be a little different with space expansion, because instead of just a doppler shift kind of change, you also have space itself stretching, not just a receding emitter. The point though that is the energy measured depends on your reference frame. The energy isn't "lost." Adrenalynn had the right idea.

SpeedFreek · Dec 5, 2007

Of course you can consider redshift in the context of a single photon, if you couldn't there'd be no such thing as quantum mechanics! Yup, I goofed! I looked around online and found this same question asked and answered in some physics forums So did I after larper put me right, but there were people like me posting, which meant there were many different interpretations! I decided to look to science journals instead and found the peebles reference, which is explained more clearly at faqs.org than anywhere else I could find. My description of the time dilation of the light is correct, we receive the light over a longer period than it was emitted. But my assumption about the energy was wrong. and the clearest answer seems to be that conservation of energy is frame dependent Yes, but although reference frames are a factor in the redshift measurement, they only represent relativistic redshift, the component of the overall redshift that is caused by the relative inertial motions of the emitter and the observer. Observers in different inertial frames of reference will measure different relativistic redshifts of a third object. Whilst it might be true in a theoretical sense that "If an observer were traveling faster than the expansion, it would be blue-shifted, right?" I felt this was a little bit of a red herring in the context of this question. Adrenalynn was right of course, but the answer needed looking into a little deeper, I thought! The thing is, relativistic redshift can be considered an apparent change in wavelength. But cosmological redshift is more like an absolute change, to which the relativistic and gravitational redshift is factored i <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

adrenalynn · Dec 5, 2007

Thanks for the nod, but reading the other responses, and thinking it through, I'm not certain I was "on the right track" (to leverage your example). It still feels right to me that it's shifted relative to the observer. It makes logical sense, but I'm not certain. (So much of cosmology seems to fly in the face of, and defeat, basic logic...) The other references posted almost demand further thought. Regardless - welcome to SDC and any and all of the thought-provoking and difficult questions like this! <div class="Discussion_UserSignature"> .bipartisan (bī-pär'tĭ-zən, -sən) [Adj.] Maintaining the ability to blame republications when your stimulus plan proves to be a devastating failure.IMPEACH OBAMA! </div>

fffranklin · Dec 6, 2007

Interesting stuff. On my last post I hit reply but didn't actually enter a reply for a while as I was doing other things, so I never got a chance to see all the other replies to your post speedfreek. I see what your getting at (I think) when you are differentiating between doppler redshift and cosmological redshift. I'm not sure about the terms "absolute" change, etc. I don't think there is an absolute, correct, reference frame, it's all relative. I think I see what you mean though, in the sense that if you could travel along with the photon at c, you would watch its wavelength change from expansion. I chose to just take reference frames as an explanation because I think if we want to take it any farther than that it may require complex math and not the words we are used to using in classical/relativistic physics. For example this explanation: http://www.math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html Using only hand waving arguments, I'm not sure there is one correct way to phrase an explanation to the redshift from the metric expansion of space using the physical terms we are used to using. Then again I could be wrong. <img src="/images/icons/smile.gif" /> I just have never worked equations in reference frames that are themselves actually expanding, so it isn't clear to me what you do in those cases, although it looks pretty complex.

SpeedFreek · Dec 6, 2007

Yes, I was reluctant to use the term "absolute" myself. But I'm thinking that if the absorption and emission lines in the spectrum of a distant object are shifted by the expansion of space as the light passes through that space, that change would be "absolute". The light would be actually changed by a definite amount (dependent on the rate of expansion whilst the light travelled through that space). There may not be an absolute, correct, reference frame, as you say, and so measurements of that light would be contaminated by the relative inertial movements and gravities of the emitter and the observer. But the amount that the expansion of space had changed that light by would be a definite figure. The difference from that definite figure in an observers measurement would be due purely to the gravitational and relativistic doppler component of the redshift. In modern cosmology, once any "normal" object (i.e. not a black hole or something) is redshifted over z=0.1 we consider that nearly all that redshift is cosmological in nature - the relativistic doppler and gravitational components would be only a tiny fraction of the overall shift, and the rest is due to the expansion of space. Once we look past z=1.5 we have an apparent recession speed that is superluminal, so it wouldn't matter what the observers relative inertial velocity was - whatever frame of reference they are in they would measure the object to be receding faster than light. (Even if you travel at 0.9999c you see light move away from you at c, receding from you at 300,000 km/s). The concept of the metric expansion of space is that distances only apparently increase outside of gravity-bound systems. Where gravity binds objects, the expansion causes no increase in the relative distances of those objects. In effect this means that space only measurably expands in the voids in between the galactic clusters. All the other clusters outside of our <div class="Discussion_UserSignature"> _______________________________________________ SpeedFreek </div>

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