Say you have 2 identical LED monochromatic flashlights. The battery lasts exactly 1 hour. We are able to calculate the energy emitted as photons from the LED by Planck's Law: $E = h\nu$.

Now if we were to move one of our flashlights though a vacuum at half the speed light, we'd be able to calculate the change in in frequency $\Delta \nu$ using the Doppler Effect for an observed frequency: $\nu = (1 + \frac{\frac{1}{2}c}{c})\nu_0 = 1.5\nu_0$, meaning that each photon from our flying flashlight is blue shifted so that it carries 1.5 times the energy of the stationary flashlight.

Accounting for time dilation, a stationary observer would only observe the flying battery/LED emitting light for $1 hour * \sqrt{1-\frac{0.5c^2}{1c^2}} ≈ 0.866 hour$, but I feel this is a moot point because photons should be emitted from the flying flashlight at the same rate as the stationary flashlight, from an observer moving/standing with either flashlight. That is to say both flashlights will emit the same number of photons before the battery dies, it's just that the stationary flashlight will emit them over 1 hour and the flying flashlight will emit them over a period of 0.866 hours to a stationary observer.

Accounting for the energy for moving the flashlight, Newton's first law of motion states that an object in motion stays in motion and I feel that the energy required to accelerate the flashlight to $\frac{1}{2}c$, does not need to be accounted for here, as that energy is conserved until the flashlight needs to be slowed back down to stationary.

This says to me that the total energy emitted from the flying flashlight would increase just because it's moving? If the battery and LED in the flashlight were 100% efficient (in theory) while stationary, how could it emitting 1.5x as much energy while moving from the perspective of a stationary observer?

  • $\begingroup$ Is that energy needed to accelerate the flashlight conserved, though? Where might it be going? $\endgroup$
    – Obie 2.0
    May 25, 2018 at 16:09
  • 4
    $\begingroup$ Energy is not conserved between different reference frames, so while we would observe light with different energies from light sources moving at different speeds, this does not mean that they are actually releasing different amounts of energy. $\endgroup$ May 25, 2018 at 16:10
  • $\begingroup$ Energy is not a scalar, the momentum will change too as a result of the process. $\endgroup$
    – user196418
    May 25, 2018 at 16:32
  • $\begingroup$ energy is conserved only in the same inertial frame en.wikipedia.org/wiki/Inertial_frame_of_reference . $\endgroup$
    – anna v
    May 25, 2018 at 18:32

2 Answers 2


The correct expression for the relativistic Doppler effect can be found here.

An object has a different energy when observed from a different reference frame, relativistic or Galilean. Photons are qualitatively speaking no different than tennis balls in this respect.


I feel that the energy required to accelerate the flashlight to $\frac 12 c$, does not need to be counted for here, as that energy is conserved until the flashlight needs to be slowed back down to stationary.

Actually, it does need to be counted. The energy(mass) of the flashlight is reduced after it emits the photons. So the energy you can recover from slowing down the flashlight to stationary is less that was put in to accelerate it. I think you'll find that it equals the gain in the energy to the beam of light.


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