Energy loss of a photon moving against gravity

[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?

 

[starthaus]

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?

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.

 

[TriKri]

That was what I started from.

 

[Passionflower]

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.

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.

 

[starthaus]

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.

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.

 

[Naty1]

But for the gravity to act upon it, doesn't it need to have a mass?

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

 

[Austin0]

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.

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??

 

[Naty1]

Posts 4,5,7 are not correct...see here for clarification:
https://www.physicsforums.com/showthread.php?t=144538

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.

 

[Naty1]

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

In a curved spacetime there is simply no way to compare the velocity of two objects unless they are right next to each other. Also, in GR light always follows a null geodesic. Any observer at any event will always measure the relative speed of any local null geodesic to be c.

In GR you can assign any coordinate system you like to any spacetime. In these coordinate systems it is common to have light travel at speeds other than c. But there is no physical significance to the coordinate system.

 

[Naty1]

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

a spaceTIME phenomena..

 

[Jonathan Scott]

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. ...

This is incorrect.

Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by anyone 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.

 

[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?

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

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?

 

[Jonathan Scott]

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?

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).

 

[Passionflower]

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.

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

 

[TriKri]

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

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?

 

[Passionflower]

What does 'flat space' mean?

I meant to say Minkowski spacetime.

 

[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?

 

[Austin0]

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

DaleSPam...In a curved spacetime there is simply no way to compare the velocity of two objects unless they are right next to each other. Also, in GR light always follows a null geodesic. Any observer at any event will always measure the relative speed of any local null geodesic to be c.

In GR you can assign any coordinate system you like to any spacetime. In these coordinate systems it is common to have light travel at speeds other than c. But there is no physical significance to the coordinate system

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

 

[Austin0]

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.

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?

 

[starthaus]

Is this difference in speed measurable ?

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.

Is the difference simply +(down)or - (up)9.7 m/s² either way?

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

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?

The above is incomprehensible.

 

[Austin0]

Is the difference simply +(down)or - (up)9.7 m/s² either way?

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

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.

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?

The above is incomprehensible.

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??

 

[starthaus]

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.

The answer is no.

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??

You are still making no sense.

 

[TriKri]

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?

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?

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.

 

[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

 

[Naty1]

I guess it should be E2 = (mc2)2 + (pc)2, right?

yes..that's the complete expression.

 

[Naty1]

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

yes, locally the speed of light is always "c". "locally" means over a small enough piece of spacetime that it's flat...once curved spacetime has an effect, that is distances become large enough for the curvature of spacetime to have an effect measurements will detect a different speed of light but that an observational phenomena not a physical phenomena as Dalespam posted...

 

[Naty1]

trikri:
Seems like these discussions are far often too confusing...I'd suggest you search some other threads on speed of light or photons in a gravitational field or light in gravitational potential and read those.

Parts of posts #11 and 13 by Scott are in different terms that I have seen here before...I had to reread them two or three times to decide if I understood the intent and if so agreed or not with them... I believe they are accurate but were not obvious to me.

 

[TriKri]

Hehe, discussions like this are always equally prone to lead to misunderstandings :tongue: especially when people are trying to figure everything out themselves while writing their posts. Probably the best thing to do is to just buy a book. ;) But I do appreciate the answers I have got!

 

[Jonathan Scott]

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.

Yes, that's correct.

 

[TrickyDicky]

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.

Hi
Ok, so if we speak about curvature instead of potentials (switch from Newton to GR) and we imagine a spacetime manifold with positive curvature (a hypersphere similar to Einstein model for intance),and events A and B separated by a big enough ds^2, and consider a light signal emmited from A and received by B, would B observe A's photons blueshifted with respect to local photons by the effect of the curvature (different potentials) of the manifold?

 

[Naty1]

so if we speak about curvature instead of potentials (switch from Newton to GR)

inconsistent...no potential effects in Newtonian gravity...

 

[TriKri]

Yes, that's correct.

Ok. Another thing that bugs me, and which really was the original reason for me to start this thread: It is said that a body with sufficient mass can prevent light from escaping from it. But according to my physics book, the gravitational redshift for a photon moving against gravity equals to gh/c² (where g is the gravitational acceleration, h is the distance the light is moving against gravity, and c is the speed of light), thought only for gh/c² << 1. This means that for a distance h that the light has traveled, you will get a factor (1 - gh/c²) which the frequency is multiplied with, and this factor can never < 0, since it's only valid for gh/c² << 1. So how is it possible that gravitational redshift can take the light out completely?

This is how I think it should be (you can tell me where I am going wrong): For large gh/c², you need to divide the distance the photon is traveling into n smaller parts, with distances h_i, 1 < i < n, so that g_ih_i/c² << 1 for each i. Here g_i is the average gravitational acceleration within the i:th distance. When the photon is sent out from the source, an observer O₀ measures the frequency of the photon to be f₀ > 0. It then travels the first distance, and an observer O₁ at the end of the first distance measures the frequency to be

f₁ = (1 - g₁h₁/c²)·f₀​

Then the photon travels the second distance and observer O₂ at the end of that distance measures the frequency to be

f₂ = (1 - g₂h₂/c²)·f₁​

Finally, the photon reaches observer O_n, at the end of the last distance, who measures the frequency to be

f_n = (1 - g_nh_n/c²)·f_n-1 = (1 - g₁h₁/c²)·(1 - g₂h₂/c²)·...·(1 - g_nh_n/c²)·f₀
​

Logarithm both sides:

ln(f_n) = ln(1 - g₁h₁/c²) + ln(1 - g₂h₂/c²) + ... + ln(1 - g_nh_n/c²) + ln(f₀)
​

Since g_ih_i/c² is so small for each i, ln(1 - g_ih_i/c²) is approximately = -g_ih_i/c², so

ln(f_n) = -(g₁h₁/c² + g₂h₂/c² + ... + g_nh_n/c²) + ln(f₀) = ΔV/c² + ln(f₀)
​

where ΔV is the gravitational potential that differs between the source and the destination. Reverse the logarithm again to get

f_n = e^ΔV/c2·f₀ > 0​

This means that no matter the distance or how big the gravitational field is, an observer far away will always be able to see the photon and register a positive frequency for it. This is in contradiction to very many predictions that great masses can prevent light from escaping. So, where do I go wrong?

 

[TrickyDicky]

inconsistent...no potential effects in Newtonian gravity...

This answer is a bit... ,hmmm, inconsistent?

 

[Jonathan Scott]

Hi
Ok, so if we speak about curvature instead of potentials (switch from Newton to GR) and we imagine a spacetime manifold with positive curvature (a hypersphere similar to Einstein model for intance),and events A and B separated by a big enough ds^2, and consider a light signal emmited from A and received by B, would B observe A's photons blueshifted with respect to local photons by the effect of the curvature (different potentials) of the manifold?

Curvature and potential are different things.

Firstly, curvature has multiple meanings anyway.

One type of curvature is analogous to that of a conical surface, which is locally flat but adds up to less than flat around a complete circuit. This is loosely like the way a gravitational field curves space-time in empty space.

Another type is analogous to that of part of a sphere or the top of a hill. This is loosely like the curvature produced by mass and energy which gives rise to fields.

The potential in a static field in Newtonian gravity (when expressed in dimensionless units, that is potential energy per rest energy, rather than potential energy per rest mass) is exactly equivalent to the time-dilation effect in General Relativity, which can effectively be described in term of a position-dependent scale factor between local time and the observer's coordinate time.

If you start considering points which are not a fixed distance apart in a static gravitational field, then comparisons of frequency and energy also have to take into account the effect of relative motion, so you get Doppler effects and time dilation due to relative speed.

 

[Jonathan Scott]

inconsistent...no potential effects in Newtonian gravity...

I think you've got something muddled up there.