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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?

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3 Answers 3

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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.

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?.

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can you explain me one thing? how does a photon have zero rest mass? -- just any logic ... the above question comes directly as a friend of mine tried to explain why .. –  Santosh Linkha Oct 3 '11 at 16:25
    

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.

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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.

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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

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