<p>One way to answer this is to say that conservation of energy is a law intended to be applied in a single reference frame-- it is not intended to work if you change reference frames.  It is true that general relativity makes this an even more difficult point, because it's not just changing reference frames there, but a simpler question appears even in special relativity.  If a source of light is moving away from you, and that source is turned on for awhile and then turned off, receivers who are moving away from that source will detect less total light energy than those not moving away.  This is true even if the source is a beam that goes entirely into the detector.  Thus you could equally ask, if the detector moves away, and detects less energy, where did the missing energy go?</p>

<p>The answer is, there is not any missing energy.  The detector moving away always reckoned there was less energy in that beam, and that less energy was conserved the whole time.  Conservation of energy is within a given reference frame-- even the energy of a bullet gets much less if you change to the reference frame of the bullet.</p>

CC BY-SA 4.0
2022-08-25T17:01:23.427Z
If I understand correctly, one way to think about the situation is to consider the origin of the photon and the Earth. When photon was emitted, it had given frequency which is defined *relative to time*. As Earth moves at relativistic speeds relative to the original source of the photon, *time* we experience doesn't match the original source. When we measure the frequency of that photon using *our time*, the frequency appears to be red-shifted. If we were able to somehow measure the frequency of the photon relative to the time of the source, it would still have exactly the correct frequency.
Mikko Rantalainen
CC BY-SA 3.0
2016-10-07T14:37:37.687Z
2016-10-07T14:37:37.687Z
Ken G
<p>The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify <em>any</em> inertial frames and certainly not ones that encompass a cosmologically significant volume.</p>

<p>What this means is that if you make a local "box" small enough that it is not affected by the expansion of the universe, then energy conservation will apply. But of course in such a box, a photon would enter and leave with the same energy because the box is unaffected by the expansion of the universe and so there would be no redshifting of the photon.</p>

CC BY-SA 3.0
2016-10-07T10:14:41.820Z
Not just general relativity. Energy isn't even conserved under the Galilean transformation.
John Dvorak
CC BY-SA 3.0
2016-10-06T22:10:28.483Z
2016-10-06T22:10:28.483Z
ProfRob
<p>When talking about the expansion of the universe, it is said that it can be proven by the red-shifting of light.(As we would need higher than lightspeed to get this redshift by the Doppler effect)</p>

<p>I am an amateur, so I am not sure I am correct, but here is what I think.</p>

<p>Redshifting increases the wavelength of the light. higher wavelength = lower frequency = less energy.</p>

<p>So, if my assumptions are correct, where does this energy from the light go?
If not, where did I make an incorrect assumption?</p>

CC BY-SA 3.0
2016-10-07T02:24:49.920Z
This was answered [here](http://physics.stackexchange.com/questions/13577/photons-in-expanding-space-how-is-energy-conserved) great question though.
LaserYeti
<expansion><redshift>
Where does the energy of light go, when it red-shifts?
akaltar
# Where does the energy of light go, when it red-shifts?

(Asked by: username_0 on 2016-10-06)

When talking about the expansion of the universe, it is said that it can be proven by the red-shifting of light.(As we would need higher than lightspeed to get this redshift by the Doppler effect)

I am an amateur, so I am not sure I am correct, but here is what I think.

Redshifting increases the wavelength of the light. higher wavelength = lower frequency = less energy.

So, if my assumptions are correct, where does this energy from the light go?
If not, where did I make an incorrect assumption?

**Tags:** expansion, redshift
> **Comments:**
> **username_1**: This was answered [here](http://physics.stackexchange.com/questions/13577/photons-in-expanding-space-how-is-energy-conserved) great question though.  

---

## 2 answers

### Answer by username_4 on 2016-10-06. Score: 12 | Accepted answer

The problem is that conservation of energy is a slippery concept in General Relativity. There are arguments back and forth but most people accept that conservation of energy is only a local law - it applies only to a local inertial frame and cannot be applied to the universe as a whole. However in an expanding universe it is very difficult to identify *any* inertial frames and certainly not ones that encompass a cosmologically significant volume.

What this means is that if you make a local "box" small enough that it is not affected by the expansion of the universe, then energy conservation will apply. But of course in such a box, a photon would enter and leave with the same energy because the box is unaffected by the expansion of the universe and so there would be no redshifting of the photon.

> **Comments:**
> **username_5**: Not just general relativity. Energy isn't even conserved under the Galilean transformation.  

---

### Answer by username_2 on 2016-10-07. Score: 1

One way to answer this is to say that conservation of energy is a law intended to be applied in a single reference frame-- it is not intended to work if you change reference frames. It is true that general relativity makes this an even more difficult point, because it's not just changing reference frames there, but a simpler question appears even in special relativity. If a source of light is moving away from you, and that source is turned on for awhile and then turned off, receivers who are moving away from that source will detect less total light energy than those not moving away. This is true even if the source is a beam that goes entirely into the detector. Thus you could equally ask, if the detector moves away, and detects less energy, where did the missing energy go?

The answer is, there is not any missing energy. The detector moving away always reckoned there was less energy in that beam, and that less energy was conserved the whole time. Conservation of energy is within a given reference frame-- even the energy of a bullet gets much less if you change to the reference frame of the bullet.

> **Comments:**
> **username_3**: If I understand correctly, one way to think about the situation is to consider the origin of the photon and the Earth. When photon was emitted, it had given frequency which is defined *relative to time*. As Earth moves at relativistic speeds relative to the original source of the photon, *time* we experience doesn't match the original source. When we measure the frequency of that photon using *our time*, the frequency appears to be red-shifted. If we were able to somehow measure the frequency of the photon relative to the time of the source, it would still have exactly the correct frequency.  

---

