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I admit I am confused and at this point pretty clueless with what's going on so I apologise if the following paragraphs sound messy.

So here's what I know (but could just as well be wrong)

$E = hf$, but relativistic Doppler effect tells us that light will be red/blueshifted due to moving objects/cosmic expansion/gravitational fields etc and as such the frequency will drop. is that an apparent frequency drop or actual frequency drop though? if it's an actual frequency drop wouldn't that mean that conservation of energy is false unless some photon energy "radiate" away but if so how does that work?

like my way of thinking about this right now is that because different photons particles get emitted at different points in space (given motion of the source) the wavelength kinda gets stretched because we perceive light as a wave. hence it seems to have a decrease in energy but it's only apparent not actual. have no clue if this is true or if I missed something though hence my query.

I apologise again if the answer is something obvious but if you know what it is please enlighten me. And I hope I made sense but I'll be more than willing to clarify.

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Commented Feb 19, 2023 at 10:56

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Let's say one matchstick is burned in young universe, and the light is later detected in old universe. The energy of the light appears to be very low. Nothing at all can be said about this case, because energy has no meaning in this situation.

Let's say one matchstick is burned on the surface of a neutron star, and the light is later detected far above that position. The energy of the light appears to be quite low. Which agrees with the conservation of energy, because the energy of the matchstick was quite low compared to a matchstick at much higher position.

Let's say one matchstick is burned, and the light is reflected to one direction where a moving observer detects the light. The energy of the light appears to be quite low to the observer, because the pushing work that the light does on the observer is $F*v*t$, where $F$ is force, $v$ is velocity of observer, $t$ is the pushing time. The energy that the observer detects is the energy of the light - the pushing work. The energy is conserved in this case too.

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But observers in different frames of reference don't agree on the energy that a particle/photon has. Measured energy is not a relativistic invariant.

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