Red shifted photons lost energy in which form?

Photons which have experienced a change in frequency (red shift) due to gravity(or other red shifting affects), have necessarily lost energy, total energy is conserved.


Red shifts happen because of various causes.

there also exist blue shifts:

Conversely, a decrease in wavelength is called blueshift and is generally seen when a light-emitting object moves toward an observer or when electromagnetic radiation moves into a gravitational field.

Now on redshifts:

Some redshifts are an example of the Doppler effect, familiar in the change in the apparent pitches of sirens and frequency of the sound waves emitted by speeding vehicles. A redshift occurs whenever a light source moves away from an observer.

Energy is conserved by the motion of the source. Motion =kinetic energy . the red shifted adds to the kinetic energy of the source seen in the rest frame of the obsrver, and the blue adds to the energy of the photon again seen in the rest frame of the observer.

Another kind of redshift is cosmological redshift, which is due to the expansion of the universe, and sufficiently distant light sources (generally more than a few million light years away) show redshift corresponding to the rate of increase in their distance from Earth.

Again the motion takes up the energy balance

Finally, gravitational redshift is a relativistic effect observed in electromagnetic radiation moving out of gravitational fields.

The gravitational field picks up the balance of energy, again in the rest frame of the observer.

  • $\begingroup$ Anna, I was going to ask a similar question, and this one popped up. So considering $E=mc^2$ and $E=h\nu$ fully governs the energy conservation, then for a redshifted gamma photon from the big bang, $\nu$ decreases and so does $m$ since $c$ is constant. Right? Or are there more physics involved? $\endgroup$ – docscience Apr 14 '17 at 16:04
  • $\begingroup$ @docscience E=mc^2 is misleading and we no longer use it in particle physics. This m is just a mathematical description of the extra inertia and confuses things. We only work with the rest mass, ( the "length" of the four vector) and the rest mass of the photon is always zero. So when the energy decreases nu becomes smaller, equivalent to the acoustic doppler shift $\endgroup$ – anna v Apr 14 '17 at 16:21
  • $\begingroup$ Thanks! So then where did the energy go if it didn't change the mass? Considering say one photon by itself as a 'system'. Was it lost to 'space-time'? I'm still researching, but can the redshift be one source of the dark energy? $\endgroup$ – docscience Apr 14 '17 at 16:28
  • $\begingroup$ @docscience For conservation of energy one has to consider the relative velocity of the observer to the photon source,. $\endgroup$ – anna v Apr 14 '17 at 17:00
  • $\begingroup$ But considering the source is a dead star that was formed shortly after the big bang, how could the source matter any more today? Is it because, for the photon, time is 'stopped'? $\endgroup$ – docscience Apr 14 '17 at 17:45

It seems contradictory that red shifted light has lost energy yet total energy is conserved (where did the energy go?). The trick to understanding this is knowing that the energy measured depends on the frame of reference you are measuring from.

Consider a ball flying towards you quite fast and hitting you in the head. From your perspective it has a lot of kinetic energy (and it hurt when it hit).

Now consider yourself flying along at the same speed as the ball. It's stationary compared to you and it has no kinetic energy (compared to you). It can't hit you in the head and it can't hurt you because its not moving towards you any more.

In both cases the ball is doing the same thing (and the energy in the system hasn't changed), its just your frame of reference that has changed and the same is happening with red shifted light. If you move away from light (at a fast enough speed) it will appear red shifted (less energy) and conversely if you move towards it (at a fast enough speed), it will appear blue shifted (appear to have gained energy).

Going back to the ball example, if you drive towards a ball that is flying towards you, it will hurt more (have more energy) if you drive away from it, it will hurt less (have less energy). Nothing has changed in the ball, or it's energy, it's you that has changed (e.g. you've used energy to accelerate towards or away from it).

  • $\begingroup$ Technically, you can't "move away from light." You can move away from a light source, in which case, the spectrum of light that you receive from the source will be shifted toward the red from the spectrum that you expected to see based on your knowledge of the process that produced the light. $\endgroup$ – Solomon Slow Jul 27 '16 at 17:09

Energy is most definitely conserved in the case of gravitation-ally red shifted (GRS) photons. The sun is 4.6 billion years old and has energy output equal to 3.8x10+26 watts. If 1% of the energy output is lost to GRS, an enoumous quantity of energy is missing. If the lost energy were simply hanging out in the surrounding gravitational field, it should be somehow observable by now since the energy has been building up and stored for billions of years. However, that is not the case. Electromagnetic energy can not be "trapped" and somehow stored, but it can be converted to increasingly smaller frequency. All photon energy escapes EVERY gravitational field, but is red shifted. Said differently, a blue photon released from the sun is converted to many red photons as it escapes the gravitational field. Conservation of energy dictates this result. Imagine a transverse wave traveling down a stretched rope. The rope suddenly divides into two ropes. The initial wave is transformed into two smaller waves each of lower energy.


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