# Conservation of energy and Doppler effect?

From what I understand, the frequency of light coming from the source moving towards an observer increases. From $E=hv$ , this implies increase in energy of photon.

What really is confusing, is where does that extra energy come from? And also where is the energy is lost during opposite Doppler effect (red shift)? Why doesn't this violate the conservation of energy?

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 the sense that if you measure an amount of energy in one reference frame, and you measure the corresponding amount of energy in a different reference frame, the conservation law tells you nothing about whether those two measured values should be the same or different. If you're going to use conservation of energy, you have to make all your measurements without changing velocity.

In fact, it's kind of misleading to say that energy increases or decreases due to a Doppler shift, because that would imply that there is some physical process changing the energy of the photon. That's really not the case here, it's simply that energy is a quantity for which the value you measure depends on how you measure it.

For more information, have a look at Is kinetic energy a relative quantity? Will it make inconsistent equations when applying it to the conservation of energy equations?.

• Wait, but don't the individual energy changes due to the Doppler Effect cancel out? As in, since light is emitted equally in all directions, you must observe that the light travelling away from you has less energy and the light travelling towards you has more energy, but the overall energy is the same? – Max Jun 5 '17 at 16:11
• Light is not always emitted equally in all directions. I'm not sure offhand, but I suspect the "changes" in different directions would not cancel out. – David Z Jun 5 '17 at 17:02
• Well, yeah, I guess, then. – Max Jun 5 '17 at 17:04
• Consider photons being emitted from a star - they get emitted one at a time from different places on the star. Just consider one photon at a time for simplicity. – Tom Mozdzen Jan 12 '19 at 3:12
• This answer is very incorrect. The conservation of energy applies to all situations, including this one. The initial sentence to the contrary is wrong. Energy conservation applies, it is merely not invariant. I had suggested an edit correcting the error, but you reverted it. – Dale Jan 30 at 1:37

$$f = \sqrt{ \frac{c-v}{c+v} }$$ for (This is the relativistic Doppler shift)

$$E = \gamma mc^2$$ $$\Rightarrow E = \frac{ mc^2 }{ \sqrt{1- \frac{ v^2 }{ c^2 } } }$$

The energy conservation is respected if you use the relativistic Doppler effect.
Basically the energy measured depends on the reference frame where you are.
The same applies to photons emitted by the cosmic background.
Consider that at $$z=1000$$ then the velocity of the emitted light is near the speed of light.
Then a $$3000K$$ black body would appear to have a $$3K$$ wavelength today... simply because the wavelength and hence the energy is measured in our rest frame.

To put it in a simpler form:

$$E= f h$$

$$f_s = \gamma f_o$$

(where $$f_s$$ is the frequency of the source and $$f_o$$ is the frequency of the observer)

$$\Rightarrow E_o = \frac{hf_s}{\gamma}$$

Where $$\gamma = \frac{1}{ \sqrt{ 1 - \frac{v^2}{c^2} } }$$

To put it in wavelength, simply use: $$c = f \lambda$$ (where lambda is the wavelength)

So the energy seems smaller for the observer at rest than for the moving observer.
It is simply a relativistic effect.
Nothing is created or lost.

Assuming a constant power light source that moves:

The moving source pushes the light by force F a distance s. The energy of the source decreases by amount F*s, where s is the distance that the source traveled during the emission of the light.

Force F can be calculated as E/(c*t) , where E is energy of light, and t is the time it took to emit the light.

About reference frames: An observer, like the one that observes the light source does not accelerate, and a reference frame is attached to the observer. Every object in the universe is in this refererence frame, and leaving this reference frame is not possible.

The top answer is correct but incomplete; even within a "bystander" reference frame, it is easy to observe that photons impart more energy onto the recipient when the emission source is moving towards, as opposed to away from, said recipient. Energy is indeed gained or lost by the photons.

This occurs due to radiation pressure. The emission source loses kinetic energy to photons emitted in the direction of motion while gaining kinetic energy from photons emitted in the opposite direction. Similarly, the photons impart kinetic energy onto the recipient, causing loss of kinetic energy if the recipient is moving towards the emission source or gain of energy if the recipient is moving away from the source. As photons always move at the speed of light, the gained / lost energy is observed as a change in wavelength.

I think energy can be created...i agree with david z that energy is not conserved between different reference frames..but i can show you that energy measured in a single frame may wary... Does photon's energy increase when i emit it from a train which is moving at comparable speed and get it reflected back by a mirror on a platform so that i can catch it again in the train? According to doppler effect the frequency of the photon for an observer standing near the mirror is greater to an observer in the train.So to the observer near the mirror, energy of the photon is greater than the energy observed by an observer in train.Similarly when it gets reflected back and reaches the train its frequency is larger than the initial frequency when observed by an observer in train .So obviously energy of photon is increased without doing any work....please correct me if i am wrong somewhere

• @davidz please see my above comment – Iron man Jan 8 '18 at 1:19
• In the frame of the train, the mirror is moving. The mirror does work on the photon. – probably_someone Apr 30 '18 at 17:38
• @probably_someone hi,but I just care about the initial and final energy of the photon. In this case I just get that the final energy of the photon is larger than its initial value. Can you please explain where it came from – Iron man May 19 '18 at 11:55
• @Ironman It comes from the energy of the mirror, which is affected by the momentum transfered from the light. – dmckee --- ex-moderator kitten Sep 30 '19 at 21:23
• @dmckee please see this link where i have shown without the mirror physics.stackexchange.com/questions/505653/… – Iron man Sep 30 '19 at 21:42

Conservation of energy applies to every system. I dont know much about the doppler effect you are talking. I will explain a similar situation. Its called Quantum Jumps. In principle if you tickle any atom slightly by collision with another atom or by shining light on it, the electron may undergo a transition to some other stationary state either by absorbing energy and moving up to a higher energy states or by giving off energy(typically in the form of electromagnetic radiation). In practice such perturbations are always present.

An electromagnetic-wave(light or infrared, ultraviolet etc etc) consists of transverse and mutually perpendicular electric and magnetic fields. An atom in presence of light responds primarily to the electric component. The atom is then exposed to sinusoidally oscillating electric field. In this process atom absorbs energy Eb-Ea = hw0 from the electromagnetic field. We say that it "absorbed photon". Although we treat the field itself classical but the photon really belongs to quantum electrodynamics. So, I suggest you to study Quantum Electrodynamics to understand in more detail.

In electromagnetic fields or generally in fields, the energy is often a misleading term. The energy is present in the fields. Field represents energy. Its difficult to understand but when you study Electrodynamics, you will know it.

• Photon absorption/emission has nothing to do with the effect being asked about, and you certainly don't need to know anything about QED for it. Incidentally, I find it very strange that you are talking about a quantum electrodynamical effect and yet you say you're not familiar with the Doppler shift... – David Z Oct 3 '11 at 8:04