# A question on the redshift of photons due to cosmic expansion

Given that the universe is expanding over time, in the sense that the (spatial) metric is changing over time, corresponding to the physical distance between objects increasing, naïve intuition leads me to the conclusion that the wavelength of a photon travelling from a distant galaxy (receding us) will be stretched, and consequently its frequency will decrease, leading to the energy of the photon decreasing (with this energy simply being lost, since time translation symmetry is broken due to the big-bang). The problem with this is, relative to another observer, wouldn't the wavelength be stretched by a different amount and hence the redshift the photon will be different, corresponding to a different amount of energy being lost by the photon.

This all leaves me feeling confused on the subject. Does the photon actually get redshifted and lose energy due to cosmic expansion, or is it simply an observer effect (akin to the standard Doppler effect). It does seem a little counter intuitive that a photon would lose energy simply due to its propagation through space?!

Is the whole point of this that it is an observer dependent phenomenon and the energy of an object is an observer dependent quantity (energy is not conserved one moves between two different frames of reference)?!

You need to concentrate on the observer making the measurements. Photons carry energy, but they don't lose energy just because they travel. The "loss" of energy is not the cause of the redshift, only if the photon scatters off something will it lose energy.

However, not all observers will agree that photon has the same amount of energy. Assume you are in a frame in which the photon is observed as green. An observer in a different frame moving relative to yours measures the photon as red.

Because measurements are made in different reference frames, the conservation of energy principle is not violated. Ultimately the energy of a photon is frequency dependent and different observers measure different frequencies.

Analogously, if you toss a coin whilst being driven, the coin has a different velocity to you, as to that measured by a bystander. Keep in mind that energy is conserved within each reference frame. The law of energy states that, in any given reference frame, the amount of energy doesn't change, but it does not apply to the way in which the energy in one frame is related to the energy in another frame.

A good read on this is Preposterous Universe.

• While this is correct, the beginner is still left with the question how nature can accelerate non-co-located observers and their infinitesimally small inertial frames relative to each other without expending energy. Jul 9, 2016 at 20:54
• @count_to_10 So is it essentially like the ordinary Doppler effect, in that the energy of photon hasn't changed from what it was when it was emitted by the source, but it's energy depends on the reference frame in which it is measured in?! Is the point that it is the 4-momentum that is frame independent, and not energy (or momentum) on its own?! Jul 9, 2016 at 20:56
• Answers from people more experienced than me, by a long way. physics.stackexchange.com/questions/214983/… and the questions on the right hand side.
– user108787
Jul 9, 2016 at 21:02
• @count_to_10 Thanks for the links. Mathematically, the frame dependence of energy quantified by $E'=\frac{\partial x'^{0}}{\partial x^{\nu}}p^{\nu}$, right ($p^{\nu}$ is the 4-momentum with respect to the unprimed frame, and $E':=p^{0}$ is the zeroth component of the 4-momentum with respect to the primed frame)? Jul 9, 2016 at 21:31
• @count_to_10 Ah ok, thanks for taking a look. A mixture really. I've been reading Sean Carroll's lecture notes, and Wald's GR book, in particular. Jul 10, 2016 at 10:41