# Energy loss of a photon moving against gravity

1. Aug 25, 2010

### TriKri

Hi. Since photons lose energy when moving against gravity (because of gravitational redshift), there must be some other energy that is increased, i.e. the energy converts to another form. That other energy must be potential energy, right? But for a photon to obtain potential energy, it must perform a work. And for the photon to perform a work, there must be some force acting on it while it is moving. The only force I can come to think of is the force of gravity. But for the gravity to act upon it, doesn't it need to have a mass?

2. Aug 25, 2010

### starthaus

The photon, while having null rest mass , has energy $$E=hf$$. When a photon "moves against gravity" as you say, it experiences a drop in frequency (see the Pound-Rebka experiment) so, its total energy decreases. So, the gravitational field acts on the photon energy.

Last edited: Aug 25, 2010
3. Aug 25, 2010

### TriKri

That was what I started from.

4. Aug 25, 2010

### Passionflower

When we consider a photon in a gravitational field there are three interesting related factors: speed, wavelength and frequency. If we consider light slows down in a gravitational field then it is very logical to assume the wavelength changes and that the frequency remains the same.

In a gravitational field an observer measures the speed of light to be c at the point where he is located at (e.g. a local measurement). However the speed of light at this point for a different observer, or the (average) speed of light between two points is not c.

5. Aug 25, 2010

### starthaus

True, this falls out of the Schwrazschild metric. The light speed in the Pound-Rebka experiment is not even isotropic, the speed on the way "up" is different from the speed on the way "down". This is again consistent with energy loss when moving "up", vs. energy gain when moving "down". Lev Okum has a good paper on this subject.

6. Aug 25, 2010

### Naty1

no. That's only in classical (Newtonian) gravitational theory. In general relativity, which is more complete and accurate, energy is equivalent to mass (via E = mc2 and consequently has gravitational effects. It turns out pressure also has gravitational effects...

"...the gravitational field acts on the photon energy..." Correct

Last edited: Aug 25, 2010
7. Aug 25, 2010

### Austin0

Could you elaborate on this concept??
Why wouldn't it be logical to think the wavelength stays the same but the frequency changes as frequency is a function of time and speed , no??
Wouldn't wavelength be dependant on orientation??

What would change the wavelength besides a spatial radial contraction??

8. Aug 25, 2010

### Naty1

Posts 4,5,7 are not correct....see here for clarification:

Distant or accelerating observers are coordinate frame based...they can measure anything for the apparent speedof light....physically, photons move locally at constant c = freq x wavelength....distant apparent measured speeds result from spacetime curvature.... space and time vary, not physical photon velocity...all light colors (frequencies) move at fixed speed c....an x ray moves the same speed c as visible light.

Last edited: Aug 25, 2010
9. Aug 25, 2010

### Naty1

I saved this previous post from Dalespam when I was trying to get all this straight in my own head:

10. Aug 25, 2010

### Naty1

a spaceTIME phenomena..

11. Aug 25, 2010

### Jonathan Scott

This is incorrect.

Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by any one observer. In Newtonian terms this is because the combined effects of kinetic energy (from motion) plus the potential energy (from time dilation) result in constant total energy.

The effect of redshift is that an object or photon which starts out at a lower potential has less energy than an identical one which starts at a higher potential. For example, if an atomic transition emits photons at a specific energy, then photons which are emitted from a lower potential will have lower energy and hence appear red-shifted compared with corresponding ones emitted at a higher potential.

When we talk about redshift increasing with potential, what we actually mean is that a series of different local observers at different potentials would observe a photon passing their own location to have different redshift relative to a local reference photon. However, from the point of view of any single observer, the photon has constant frequency and energy when moving freely in a static gravitational field.

In a static gravitational field, using an isotropic coordinate system, the momentum of a free-falling particle increases downwards with time, even if it is a photon, but the energy is constant. This is related to the effect that in such coordinates the speed of light at a location other than the observer's own varies a bit depending on the gravitational potential.

12. Aug 25, 2010

### TriKri

I'm confused ... does the speed of light vary at other locations than the local or not? Jonathan says it does, but Naty1 says it doesn't ... or have I misunderstood anything?

But here m = 0 so E would also = 0. I guess it should be E2 = (mc2)2 + (pc)2, right? Or are you talking about the relativistic mass?

13. Aug 25, 2010

### Jonathan Scott

The speed of light at the observer's own location always has the standard value. However, the speed of light elsewhere is a matter of coordinate systems, because space is curved. As an analogy, consider a flat map of the curved surface of the Earth. You can project the surface of the earth to the map in various different ways, and you can make it match the shape and scale of the earth accurately in any local area, but you cannot make it match the correct shape and same scale everywhere else.

The most practical coordinate system for a system with a dominant central mass, such as the solar system, is called isotropic coordinates, which are so named because the ratio of ruler distance to coordinate distance is the same in all directions (equivalent to preserving local shape). However, in such coordinates the scale varies with location, which means that the effective coordinate speed of light varies with potential. To be specific, the fractional variation of the speed of light is approximately twice the Newtonian potential in dimensionless units, which means for example that the coordinate speed of light at distance r from the central object of mass m varies approximately proportionally to (1-2Gm/rc2).

14. Aug 25, 2010

### Passionflower

The (average) measured speed of light in an accelerating frame in flat space is also not equal to c.

15. Aug 26, 2010

### TriKri

What does 'flat space' mean?

And this might be a bit of topic, but ... as we were talking about before; the spacetime is curved, which bring up the question: Is there any 'correct' way of representing locations in space? Say for example that we are using Cartesian coordinates, and that we have a representation of Earth and two stars in that system, where the earth has its center at (0, 0, 0), the first star is huge and has its center at (1, 0, 0), and the second star has its center at (2, 0.001, 0). If the first star is massive enough, we will be able to see the second star from earth at two places in the sky. The spacetime is curved, and there is two (actually probably three) different 'straight ways' to the second star.

This got me thinking if it is really possible to represent distant celestial bodies using Cartesian coordinates; I mean, the coordinate representation of the star's location seems to lose its meaning a bit when you realize that there, from earth, exists three different vectors, all of them pointing in different directions, still all of them pointing in direction of the second star. So, is a normal coordinate system the best way to represent locations in space, or is there any better? Is there really any good representation at all of locations in space?

16. Aug 26, 2010

### Passionflower

I meant to say Minkowski spacetime.

17. Aug 26, 2010

### nonequilibrium

Hm I'm a bit confused by all the previous posts taken together... So if we have a photon moving toward a galaxy, for some reason its frequency will drop, okay. So if we put an observer halfway through and it detects the photon, its energy E = hf will have been lower than when it started. Where has this energy gone? (as I believe that was the poster's original question) In Classical Mechanics we have potential U and kinetic K: as here E = hf seems to drop, it seems analogous with U? Is there something analogous to K?

EDIT: Sorry, my last comment was stupid, as when you shine a photon away from a galaxy, its frequency will also drop, although in classical mechanics U rises. I do still wonder where the energy goes though?

Last edited: Aug 26, 2010
18. Aug 26, 2010

### Austin0

Does this imply there is a physical significance to local measurements of c ?????

19. Aug 26, 2010

### Austin0

Is this difference in speed measurable ???
Is the difference simply +(down)or - (up)9.7 m/s2 either way???
Clock dilation at top and bottom???
It seems like both effects would be infinitesimal but also that they would be complementary, with the dilated rate at the bottom increasing the coordinate speed for a "down" measurement and vice versa. Perhaps???

Last edited: Aug 26, 2010
20. Aug 26, 2010

### starthaus

No, the speed of light in vacuum is measured "locally" to always be "c".
This is just a difference in coordinate speed predicted by theory.

We are talking speed, so the difference cannot be measured in m/s^2.

The above is incomprehensible.

21. Aug 26, 2010

### Austin0

This was meant as "down".. s= c + ( ( 9.7 m/s)* dt) ,,,"up".. s = c - ( ( 9.7 m/s)* dt)
or something on this order.

If the clock at the bottom of the tower is running slower this would mean less elapsed time for any speed trial , yes??
If the clock at the top of the tower is running faster it would mean greater elapsed time for any speed trial , Yes??
If you make the tower high enough to have significant effects then the dilation and the g acceleration would compound each other , amplifying the difference between the up and down measured coordinate speeds , no??
Is this comprehensible??

22. Aug 26, 2010

### starthaus

You are still making no sense.

23. Aug 26, 2010

### TriKri

I think, from what I can conclude from the previous posts, that the energy of the photon does not change within one reference system. However, if the photon starts at A and travels to B, an observer at A will measure the photon to have another frequency, thus also another energy, than an observer at B. However, these observers are located in different reference systems, where times passes by differently quickly in relation to each other. Is that so? I'm not really sure if I have got this correctly.

24. Aug 26, 2010

### nonequilibrium

Oh I see, much like someone moving along with a moving ball sees zero kinetic energy. So the key principle here is that you can't detect light unless it falls into your eyes? Meaning you can't detect it more than once in one reference frame, and thus its energy content can't change? Somehow of course this seems fishy :p

25. Aug 26, 2010

### Naty1

yes..that's the complete expression.