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The energy of electromagnetic radiation is related to frequency; higher the frequency, higher the energy level. If electromagnetic waves have lower frequency when they arrive on Earth than originally emitted due to the Doppler effect, where, according to the conservation of energy, does the excess energy go?

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    $\begingroup$ Recommend also looking at: physics.stackexchange.com/questions/15279/… $\endgroup$ – Bilkokuya May 14 '18 at 12:31
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    $\begingroup$ The answer by Steve Linton is not correct, I'm afraid. The answer by ACascarino is closer to the truth. The correct answer — that energy conservation doesn't apply in an expanding Universe — is explained very well in this article by Tamara Davis. It's behind a paywall, unfortunately, but can be found online if you google for it. $\endgroup$ – pela May 14 '18 at 15:22
  • $\begingroup$ @pela Are you saying that a police Doppler radar gun depends on the expansion of the Universe to operate? $\endgroup$ – user71659 May 15 '18 at 1:37
  • $\begingroup$ @pela so I can't read the article because I don't have access, but that argument somehow makes the most sense per my other comment below. Thank you $\endgroup$ – Dilettanter May 15 '18 at 1:53
  • $\begingroup$ Whether or not energy is conserved in an expanding (or contracting!) universe is somewhat immaterial - as the article you reference even quotes, "[...] the galaxy’s redshift can be interpreted as the result of relative motion, rather than of the expansion of space. Therefore, no energy is lost." when motion is viewed relative to the particle's motion through spacetime; it reduces again down to the fact that conservation of energy is meaningless if you start switching between reference frames without applying the correct transformations. $\endgroup$ – ACascarino May 15 '18 at 8:54
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It's spread out in time. If a source emits a 1W pule of energy lasting 1 second and the receiver is receding so fast that it is Doppler shifted to a frequency which means the power is just 0.5W, then the pulse will take 2seconds to arrive (since the end of it had further to travel).

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    $\begingroup$ I think this is wrong. What happens if the source emits one photon (with a given energy)? If the receiver is retreating so fast that the photon energy is red-shifted to one half, the receiver still only receives one photon. Where has the rest of the energy gone? $\endgroup$ – Martin Bonner May 14 '18 at 11:56
  • $\begingroup$ Yes, this doesn't seem right. A previous post from @bilkokuya was helpful, but to say that "conservation of energy doesn't apply between changing reference frames" feels unsatisfactory when dealing with the 1 photon case. Especially considering if, say, all light in the universe experienced this loss of energy; how could the overall energy in the universe remain constant? $\endgroup$ – Dilettanter May 14 '18 at 12:42
  • $\begingroup$ Even without relativistic corrections, if a machine gun fires some bullets into a receding target, the rate that bullets arrive is reduced by the target motion, and the energy per bullet is reduced even more. One is not caused by the other, they are two separate effects of the motion. Similarly, the change in intensity of a light source is separate from the change in frequency, one does not explain the other. $\endgroup$ – Ken G May 14 '18 at 13:22
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    $\begingroup$ Yes, I think you are all right. My mistake. $\endgroup$ – Steve Linton May 14 '18 at 14:31
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    $\begingroup$ We've all done it most likely, even the people giving answers can learn from this forum! $\endgroup$ – Ken G May 14 '18 at 14:53
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Completely ignoring relativistic effects, it depends on which reference frame you are using; the "missing" energy is seen as kinetic energy in either the emitting or receiving atom as recoil depending on which one you view as moving. Energy is not conserved between reference frames.

If I am travelling at a velocity away from you, and I emit a photon at you which I observe to have a frequency f, then I will assume that photon has an energy E = hf, where h is the Planck constant. I will never observe a different energy for that photon - in my reference frame, energy is conserved. You, however, will observe a different frequency f, and therefore a different energy E. This energy remains constant for you - energy is conserved in your reference frame - but the energy I observe and the energy you observe differ - energy is not conserved between our reference frames; that is to say, the energy is conserved but not invariant

Consider - I drive past stationary you in a car and throw a tennis ball at you. In my perspective, the tennis ball has greater kinetic energy (it is moving at my speed, plus the speed of the ball) than in your perspective. Energy is not invariant in this circumstance either!

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    $\begingroup$ A simpler demonstration that KE isn't invariant is to just look at you in your car and forget the tennis ball altogether. In your frame, you're stationary and you have zero KE; in my frame, you're whizzing by and have a bunch of it. $\endgroup$ – David Richerby May 14 '18 at 15:33
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Conservation of energy doesn't apply to this situation because the energy you measure when at rest with respect to the source and the energy you measure when moving with respect to the source are in different reference frames. Energy is not conserved between different reference frames; in other words, if you're going to use conservation of energy, you have to make all your measurements without changing velocity.

For more information, have a look at https://physics.stackexchange.com/questions/1368/is-kinetic-energy-a-relative-quantity-will-it-make-inconsistent-equations-when-a.

varbatim from David Z's answer for a question on physics SE

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