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Help me out, physicists: redshift violates conservation of energy?

I'm missing something. Here's my reasoning so far, which has left me in a hole:
The doppler effect is observed with light waves, changing the observed frequency. The energy in a single particle of light equals h times frequency. So, if the observer is moving away from the light source, then the energy they receive from a single particle of light will be less than was sent. Or more if they're moving toward it.

This, of course, doesn't make sense. Please explain what and where I've screwed up.

Tags: conservation of energy

Views: 19

Replies to This Discussion

Its about entropy I believe... the energy is the same its just spread out over greater distance.

What your seeing in the shift is the apparent spread of the light... the further away the bigger the spread... thats how I understand it to be. I could be wrong, I am a interested amateur when it comes to physics.
You haven't screwed up. Cosmologists and theoretical physicists are trying to determine where this energy "goes."

Some say that energy isn't conserved locally, but universally (generally for expansionary redshift). Some say that the energy is converted to gravitational energy by employing pseudo-tensors.

It seems to me that no energy is lost, though I haven't any degree in the field. Since light is NOT purely a particle, the "loss" of energy is because of the expansion or motion. This means the energy doesn't lessen, it's just spread out over a greater area. In that sense, we're measuring the energy density, not the absolute energy of Newton's corpuscles.

Which makes more and more sense as I consider it. That same energy would "reappear" as soon as you started moving toward the emitter (or expansion reversed). So clearly it hasn't left at all.
I used something like the Doppler effect when I think about how this one works its a analogy and probably operates differently but it helps me visualize the idea in my mind. I don't really understand in any mathematical way.
Energy is the zeroth component of the momentum four-vector. If you move into a different inertial frame, the four momenta change. In special relativity - which is a viable in low gravity - only the so called "invariant length" is invariant under frame change (which is called boost). That is defined as the following for four momenta: m^2 = E^2 - px^2 - py^2 - pz^2 . This invariant quantity is the mass of the photon, which is zero for free/real/propagating photons. So the energy changes after a boost/inertial frame change, and the three momenta as well. But there is no problem, the mass remains the same. Changing the frame is not a physical/kinematical thing, it is conceptual.
I have not yet researched this, but here is a stab off the top of my head. Consider a pitcher standing on the ground throwing a baseball to a catcher squatting on a flat car. The energy that the catcher must absorb when catching the ball is KE = 0.5mv^2, where m is the mass of the baseball and v is the velocity of the baseball. If the train is moving away from the pitcher, the v in the equation is the reduced; it is now the velocity of the baseball relative to the flatcar, v = v(as pitched) - v(train). The energy the catcher absorbs is less than before. If the train is moving toward the pitcher, the KE is higher as the relative velocity is higher than v(as pitched). Lower KE corresponds to a red shift as lower frequency means lower energy. This analogy may be quite wrong, but that is what these groups are for, right?


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