# Deriving gravitational redshift using the energy change approach

1. Mar 20, 2008

### Crazyevox

An extremely elementary question but....

Lets say we had a spaceship sitting on earth under the influence of gravity.
Light travels from the ship's floor to the ceiling. An observer 'A' stands near the ceiling.

Hence the light loses gravitational potential energy, and therefore decreases in frequency.

m g Δs = h Δf

where m = 'mass' of light, g = gravitational field strength, s = distance between ceiling and floor, h = Planck's constant, f = frequency of light

Replacing 'm' with its equivalent in terms of f, h and c (speed of light), we get,

(h f / c^2) g Δs = h Δf

And we end up with:
Δf / f = g Δs /c^2

--------x------

Now what I dont quite get is the bolded bit.

we use the equations below and substitute for 'm'...
m = E/c^2, E = h f

But why do we use the initial frequency 'f' to substitute for 'm'? Why not Δf (although that would cause the Δf's to cancel out..)? Could someone elaborate?

Or is this an oversimplification?

2. Mar 20, 2008

### 1effect

The short answer is that the derivation is a rather crude oversimplification.
The long answer is that if you used $$\Delta f$$ this would give you $$\Delta E= h \Delta f$$ which, in turn gives you $$\Delta m = \frac {\Delta E}{c^2}$$. But this is not what you wanted , you wanted $$m$$.

3. Mar 20, 2008

### yuiop

There is a point of view that the frequency of light does not alter as it climbs out a gravity well. It works like this. The higher clock is running faster making it look like the frequency of the photon is reduced when in fact it is constant. However, the wavelength of the photon IS getting longer.

4. Mar 21, 2008

### 1effect

Hmm, aren't many "clocks" implemented as just precision oscillators? So, "faster clock" and "higher frequency" are one and the same thing.

The above is contrary to:

$$\lambda f=c$$
c=constant locally in GR
When the frequency goes up $$\lambda$$ must go DOWN (not up)

Last edited: Mar 21, 2008
5. Mar 21, 2008

### yuiop

In my previous post I said the frequency of the photon appears to be reduced which is the same as saying the frequency appears to go DOWN when the photon rises up the gravity well, which corresponds to the wavelength going UP.

I'm a bit rushed at the moment so I will get back to this when I have more time. I was referring to this paper if you want to look it over.

http://arxiv.org/PS_cache/hep-ph/pdf/0010/0010120v2.pdf

6. Mar 21, 2008

### 1effect

But this is not what you said in your original post, you are correcting your error now. Why can't you admit that you made an error? Faster clock rate, higher frequency and lower wavelength are one and the same thing.

Thank you, I've seen it before. Okun says in the conclusion "p increases , $$\lambda$$ DECREASES". Since $$p=hf$$, the above is clearly equivalent with "f increases , $$\lambda$$ DECREASES", or, as you are saying now: "f decreases , $$\lambda$$ increases" . I think we are on the same page now. :-)

Last edited: Mar 21, 2008
7. Mar 21, 2008

### yuiop

This is my original post:

The phrase " the higher clock is running faster" means the clock further away from the massive body has a higher frequency than a clock lower down, from the point of view of an observer at infinity.

So your phrase "faster clock rate, higher frequency and lower wavelength are one and same thing" is not an accurate description of what I said. If I fill in the implied gaps of your phrase it becomes "faster clock rate, higher frequency (of the clock) and lower wavelength (of the photon)" and shows you are getting clocks and photons mixed up.

So to be more explicit in what I said: (Assume stationary clocks at various hights in the gravitational well).

Higher up = faster clock rate (of the clock)
Faster clock rate = higher frequency (of the clock)
Higher frequency of the clock makes the frequency of the photon appear to be lower. (lower frequency of the photon)
Lower frequency of the photon = longer wavelength of the photon. (wavelength of the photon, not the clock)
longer wavelength of the photon = higher wavelength of the photon (not the clock)

As you can see, I said the the wavelength of the photon is getting longer and you claim I said the wavelength is getting lower and demand I admit I made a mistake, when in fact you are misquoting what I said.

8. Mar 22, 2008

### 1effect

But this is not what you said. Here is exactly what you said:

"There is a point of view that the frequency of light does not alter as it climbs out a gravity well. It works like this. The higher clock is running faster making it look like the frequency of the photon is reduced when in fact it is constant. However, the wavelength of the photon IS getting longer."

You can't have the "photon frequency constant and the wavelength getting longer" because, as you well know, $$\lambda f=c=constant$$.
Do you understand your mistake ?

9. Mar 22, 2008

### yuiop

Ok, I see where we are at cross purposes. Below I have restated my original statement with my unspoken assumptions added and highlighted in brackets.

"There is a point of view that the frequency of light does not alter as it climbs out a gravity well. (That point of view is the point of view of an observer at infinity using coordinate measurements) It works like this. The higher clock is running faster (from the POV of the observer at infinity. -To the local observer the local clock appears neither fast or slow) making it look like the frequency of the photon is reduced (to the local observer) when in fact it is constant (to the observer at infinity.) However, the wavelength of the photon IS getting longer (from the point of view of the local observer AND the observer at infinity)."

I grant that I should have spelled things out more clearly and was being a bit lazy :P

Now for the local observer c is constant and the expression $$\lambda f=c=constant$$ is true as far as a local observer is concerned.

For the observer at infinity the vertical coordinate speed of light is not constant and varies as $$c ' = c_o/ \gamma^2$$ as can be comfirmed by analysing the Shwarzchild metric. From the POV of the infinity observer the frequency of the photon is constant (an so is the energy because E =hf) while the coordinate wavelength of the photon is varying by $$\gamma^2$$. This is consistent with $$(\lambda / \gamma^2) f= (c_o / \gamma^2 )$$ for the infinity observer. Here $\gamma$ is the gravitational gamma factor $1 / \sqrt{1-GM/(Rc^2)}$.

The infinity observer concludes that the local observer low down in the gravity well sees the frequency of the falling photon as increasing because the local observer is using a clock that running slower than that of observer at infinity. The infinty observer also concludes that the local observer sees the wavelength of the falling photon as getting smaller only by a factor gamma rather than the factor of gamma squared that the infinty observer sees, because the local observer is using a ruler that is length contracted by gamma relative to the ruler of the observer at infinity. Overall, the infinity observer concludes that the local observer sees the speed of light in a gravity well as locally constant due to the combined effects of using slower clocks and shorter rulers (as far as the infinity observer is concerned).

Now, I am not saying that the infinty observer or the local observer has a preferred point of view. They are simply different points of view and as in special relativity each observer is is entitled to their own point of view and entitled to consider the measurements of their own clocks and rulers as the "proper" point of view.

It should be noted that to the infinity observer, the amount of energy that a photon adds to a black hole is the same as the energy the photon had at infinty. To successive local observers the energy of the photon is continuously increasing as it falls and by extrapolation each falling photon adds an infinite amount of energy to the black hole.

10. Mar 22, 2008

### 1effect

Yes, you were very sloppy :-)
There is no observer at infinity in the experiment discussed by the OP. In effect, the OP is discussing an (over)simplified explanation of the Pound-Rebka experiment.

11. Mar 22, 2008

### yuiop

.. and where is the OP in all this or is he just another of those people that pulls the pin on a grenade and rolls it along the aisle while slipping away into the night?

:p

12. Mar 22, 2008

### 1effect

I guess he/she got his/her answer to his/her question and didn't stick around through the unnecessary diversions about observers at infinity, black holes and infalling photons. All he/she was interested was the derivation of the frequency variation, maybe he/she didn't care for the proof as why frequency is constant for the observer at infinity.

13. Mar 22, 2008

### yuiop

I guess he/she didn't feel the need to thank anyone for help answering his/her question.