# Photon energy changing due to Dopplershift?

1. Jun 26, 2013

### Dreak

How does the energy of a photon change by Doppler shift?

Let's say we have an IR photon, but due to Doppler shift it shifts towards UV wavelengts. Shorter wavelength/higher frequency means more energy, but where does this energy come from?
How do you connect this with Photoelectric effect, which depends on the frequency of the photon?

2. Jun 26, 2013

### Staff: Mentor

It was always there - energy is frame-dependent. To see this more clearly, you might consider two descriptions of the same situation:

1) A 100 gram bullet from an elephant gun strikes a 1000 kg elephant at 500 meters/sec. The total kinetic energy is $\frac{mv_2}{2}$, or 25,000 Joules.
2) But from the point of view of the bullet, it is at rest (albeit experiencing quite a wind blowing from its nose to its tail) when this enormous heavy elephant smashes into it at a speed of 500 meters/sec; total kinetic energy is 125,000,000 Joules.
Its the frequency of the light as observed in a frame in which the metal target is at rest that determines whether the photoelectric effect will kick electrons out of the metal or not. Note that different observers moving at different speeds will see different Dopplers, but they can all do the calculations required to see what the frequency is from the target's point of view, they'll all get the same answer, and they'll all agree about whether an electron is kicked out.

Last edited: Jun 26, 2013
3. Jun 26, 2013

### Staff: Mentor

The same place a baseball or cricket ball gets its extra kinetic energy from, when you run towards the pitcher while he is thowing it.

4. Jun 26, 2013

### Dreak

Very clear answers, thank you very much! :)

5. Jun 26, 2013

### Ookke

How about if a photon starts from a massive object and climbs to higher gravity potential, it would be redshifted?

If escape velocity of the massive object exceeds speed of light, the photon should become more and more redshifted and eventually lose all its energy. Since the photon does not have rest mass, is there anything left of the photon when it has lost all its energy? If not, where the energy has gone?

6. Jun 27, 2013

### jartsa

Let's say we have a large number of identical photons, spread around. The energy of this photon group is E.

Now we collect these photons into a small container, without adding or removing energy, and taking care that the photons stay identical. (by identical I mean same energy)

The container is in empty space and leaks out photons slowly, and the energy E is so large that there's a gravity field caused by all the photons.

After all photons have leaked out, we can measure the energy of the collection of photons, it will be E.

But the photons are now different: the first photon to leak out has the smallest energy, while the last one has the largest energy.

Now it seems quite obvious that the first photon to leave donated some of its energy to the other photons.

(I must add that the energy exchange could not happen by photon losing energy while climbing, so it happened some other way during the process.)

7. Jun 27, 2013

### Ookke

If we have E photons in container each having 1 unit of energy, the total energy is E of these units. When received by an outside collider, the first photon has smallest energy (below than 1) because strongest redshift, the last photon has largest energy which is 1 unit. So the received sum of energy does not match E, but is less.

It could be somehow frame-dependent what the total energy of the photons is, even though the initial container and the collider are at rest. So the total energy of the photons is E, when looked from the initial container, but less than E for the outside collider, because the photons are initially at bottom of a gravity well. That might explain why the outside receives less than amount E of energy: there is no more in the first place, in the colliders frame. Gets complicated, if this is the solution.

8. Jun 27, 2013

### Staff: Mentor

Ok so far.

This is not possible as you state it. There are at least two issues:

(1) If the collected photons in the container have a non-negligible gravity field, then a non-negligible amount of binding energy must have been released in the process of collecting them, since when inside the container the photons are a bound gravitating system, whereas the original photons were not bound.

(2) For the photons to be confined in the container, the container walls must be under stress, and this stress contributes to the externally measured energy of the total system. In other words, the externally measured energy of the system is not entirely "contained" in the photons when they are confined in the container.

No, it won't. First of all, the total energy of the bound system is not E, it's E - B, as above. Second, you are once again neglecting the container; as each photon is emitted, the stress on the container walls decreases slightly, and that needs to be taken into account.

I think this is correct, but it's not as simple as you seem to think, because of the interaction with the container. I would need to do a more complete analysis of the scenario to be sure.

No, it's not at all obvious, because of the issues I raised above.

9. Jun 27, 2013

### Staff: Mentor

If the classical escape velocity of the object exceeds the speed of light, you're dealing with a black hole, so the only observations of the photon are going to be an observer inside the event horizon and heading along with the photon towards the singularity.

But we reframe your question so that the you're just asking where the energy goes as the photon becomes increasingly redshifted....

Energy is observer-relative - refer back to my post #2 to see how this can be. Thus, it's not really true that the "photon is redshifted" - it's redshifted relative to a particular observer, and different observers can observe different redshifts.

(Don't let yourself be fooled into thinking that we're throwing out conservation of energy here - what's conserved is the observer-dependent value, and no observer will ever see conservation of energy violated locally).

10. Jun 27, 2013

### PAllen

For the gravity situation, it is worth mentioning a rocket situation. When a rocket is not accelerating, photons from its back reach its front with the same energy. The greater it thrust (g force), the greater the amount photons from the back on redshifted on reaching the front. No gravity here at all. What gives? The back emits a photon (speed zero compared to an instantly comoving inertial frame). The photon takes some time to reach the front. Due to acceleration, the front is now moving at speed v in this inertial frame away from where the emitter was. In this inertial frame (the instant inertial frame of emission) you have nothing but ordinary Doppler going one.

By the same token, if a planetary gravity situation is analyzed using an 'inertial frame' of the emitter at time of emission, gravitational redshift has become simple Doppler. But, a few caveats:

- There isn't really a true extended inertial frame in GR. However for reasonably small distances and times, in non-extreme gravity, you can construct coordinates that behave essentially like standard coordinates in empty space.

- There are higher order corrections to the redshift you would get using the rocket analogy for a planetary surface. These are due to tidal gravity. However, the major effect (analyzed from a free fall frame comoving with the emitter at time of emission) is the same as the rocket situation.

- The gobal analysis is not equivalent between the rocket and a planet. For the rocket, in every global inertial frame, the photons don't change energy between emission and absorption. For the planet, imagine the whole surface emits light measured locally at some frequency, with a certain power. If you imagine a spherical surface far from the planet such that power received is isotropic, then total power received is less than the power emitted (and the power density is less than predicted by the inverse square law).

Last edited: Jun 27, 2013
11. Jun 28, 2013

### jartsa

A massive lump of matter, with total energy E, turns completely into photons in a microsecond. How much energy is received far away from the lump? Answer: E

A massive lump of matter, with total energy E, turns completely into photons during million years. How much energy is received far away from the lump? Answer: E

Haha the previous paragraph says about the same thing as my complicated thought experiment in my previous post.

12. Jun 28, 2013

### Staff: Mentor

Not quite, because this time you left out two things: the process of assembling the massive object with total energy E out of photons, and the container that would be required to confine the photons when doing so.

(Also, by "total energy E" I assume you mean "total energy at infinity E", since that's the interpretation that makes your statements true. Which means that, if E is large enough that the object has significant gravity, the "locally measured energy" inside the object will *not* be E, but something larger.)

13. Jun 29, 2013

### jartsa

I wonder how does a local inhabitant locally measure the local energy of his planet, or a neutron star?

14. Jun 29, 2013

### jartsa

We shoot the photons into a container, that has open doors all around it, from many direction far away, when the photons are in, we close the container doors.

The stretching container will extract energy equally from all photons, when photons arrive together into the container. When photons leave one by one, the last photon to leave is given the largest energy by the shrinking container.

Well that is a problem.

15. Jun 29, 2013

### Staff: Mentor

Yes, this is more or less what I was thinking.

I think this is ok; but also, the energy that is extracted from the photons must then be radiated away by the container--otherwise the container itself would explode outward, since by hypothesis it has enough energy to escape to infinity (since the photons came in from infinity). The energy radiated away by the container is the binding energy B that I referred to earlier.

I'm not sure that's necessarily the case; that's why I said I wanted to do a more complete analysis of the scenario. But the key point is that there is no guarantee that each photon will be given the *same* energy by the container; so the difference between the photons once they escape can be accounted for by the difference in their interactions with the container.

16. Jul 1, 2013

### jartsa

It was asked where does the energy go when a gravitational redshift happens.

Let's say first one billionth of a massive object is converted into photons, and then the photons are sent to a receiver far away from the massive object. (The photons climb up from the gravity well as a short pulse of light)

The receiver will say: "The amount of matter that was converted into photons in the conversion process that happened on the surface of the massive object, was indeed one billionth of the total mass, but the energy that I received is less than one billionth of the total energy, and also the mass of the massive object has not decreased by one billionth, but less than that, actually the energy of the photons here plus the energy of the massive object there is the same as the original energy of the massive object"

So energy was conserved.

Maybe I should justify why that is correct. Well the conservation of energy is the justification.

Seriously, it's correct. When stuff near a massive gravitating object is removed, the massive gravitating object blueshifts.

Last edited: Jul 1, 2013