[yor_on]

Think Ethos and I share a same opinion there. It would break the geodesic, wouldn't it? If we assume that photons trying to propagate out from a gravity well towards a event horizon then actually 'intrinsically' was losing 'energy'. As the Event horizon could be seen as a limit for that photons 'red shift' it should 'disappear' somewhere around there, if so. The only red shift I know of that's not observer dependent seems to be the redshift in a 'expansion', treating the 'photon' as a wave.
==

Da*n, Alternatively, it shouldn't exist any photons, or geodesics inside a black hole, except from incoming photons, that then from the observation of someone standing at the 'center' should disappear before reaching its 'center'. Or if you define everything inside a event horizon as a 'center', then no geodesics can exist inside that event horizon, as all photons must be 'gone', incoming or not. The redshift as a result of a gravitational potential is strongest at whatever we define as the gravity's 'center', so my first idea turned it about, ah well, probably comes from me reading too much sci fi, you know 'the guy looking out at the stars' sort of (well ok. The 'event horizon' if you now want to be picky:)

Any which way, still think Chiral's idea is interesting, what happens with the energy in a expansion? The thing is that 'photons' do redshift due to an expansion, it's not as some say that they get more 'spaced out' by it. You can measure those photons, and find each one of them of less energy. So even though a wave picture describes it best, 'photons' too loses energy there, independent of any observer dependencies as I gather.

[jeffreyH (OP)]

The reason for posting this was the thought that a gravitational field extends to infinity. If the field removes energy from the photon...

It doesn't. The ascending photon doesn't lose any energy. In similar vein the descending photon doesn't gain any energy. If you send a 511keV photon into a black hole, the black hole mass increases by 511kev/c². Conservation of energy applies. The descending photon appears to be blueshifted because you and your clocks go slower when you're lower.

So if a constant stream of photons of identical wavelength are generated directly away from a black hole with each photon at a regular interval what will be seen? If we then station observation points outward at regular intervals along the photon path to measure the wavelength at each point what do you expect the results to be. All observation points will expect a speed of c which they should record in their local frame. It is the gradual change in wavelength that produce the important data points. The velocity can't change in a vacuum. So why does the wavelength gradually change for different observers at different radial distances?

If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?

[jeffreyH (OP)]

BTW In his derivation de Broglie started with two equations. E = mc^2 and E = hv. He then hypothesized that mc^2 = hv. He then substituted v for c in E = mc^2 so that mv^2 = hv. This was because he wanted a wave that did not only relate to photons but particles of any velocity. The inclusion of such a relationship links energy to the wave equation. So change the wave and change the energy.

[PmbPhy]

He then substituted v for c in E = mc^2 so that mv^2 = hv.

Where did you get that from?

[jeffreyH (OP)]

He then substituted v for c in E = mc^2 so that mv^2 = hv.

Where did you get that from?

It wasn't E= mc^2. It was just mc^2 that was changed otherwise the energy equation would be wrong. mv^2 is of course is related to kinetic energy through (1/2)mv^2.

The page I read through was http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength. The point I was making to John was that energy does change for the photon and what the reasons are.

[yor_on]

That was a very nice link Jeffrey. Succinct and enlightening to how he thought. If you find more links able to compress ideas feel free to share them

[JohnDuffield]

So if a constant stream of photons of identical wavelength are generated directly away from a black hole with each photon at a regular interval what will be seen?

Nothing unusual. What you'd expect.

If we then station observation points outward at regular intervals along the photon path to measure the wavelength at each point what do you expect the results to be.

The observers will give different measurements. Those further out will say the wavelength is longer than those closer in. From that you might conclude that the photons are redshifted. However you could contrive a similar gedankenexperiment with moving observers. The observers moving away from the photon source would report a redshift. But you know that this isn't because the photons are actually getting redshifted, and instead is because the observers are moving.

All observation points will expect a speed of c which they should record in their local frame. It is the gradual change in wavelength that produce the important data points. The velocity can't change in a vacuum. So why does the wavelength gradually change for different observers at different radial distances?

Because the speed of light changes. There's this myth kicking around that it's absolutely constant, but it isn't. Check out the coordinate speed of light, and what Einstein said:

If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?

The energy doesn't change, and the frequency doesn't change. You and your clocks go slower when you're lower, so you measure the frequency to have changed. But it hasn't changed. You and your clocks changed, not the photon. See post #6 in this thread where PmbPhy said the coordinate speed of the photon will change and Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.

[jeffreyH (OP)]

If you consider the de Broglie equations we can relate frequency to energy, frequency to wavelength etc. You say no energy is lost. Why then the relationship between frequency and energy?

The energy doesn't change, and the frequency doesn't change. You and your clocks go slower when you're lower, so you measure the frequency to have changed. But it hasn't changed. You and your clocks changed, not the photon. See post #6 in this thread where PmbPhy said the coordinate speed of the photon will change and Gravitational redshift is only observed when local observers at different positions compare their measurements. The wavelength as measured by  Schwarzschild observers remains unchanged.

This is by holding the frequency as constant while the coordinate speed changes, yes. But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different. You can't get round it by looking at it from a 'Schwarzschild observer' perspective.

[PmbPhy]

He then substituted v for c in E = mc^2 so that mv^2 = hv.

Where did you get that from?

It wasn't E= mc^2. It was just mc^2 that was changed otherwise the energy equation would be wrong. mv^2 is of course is related to kinetic energy through (1/2)mv^2.

The page I read through was http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength. The point I was making to John was that energy does change for the photon and what the reasons are.

I believe that paper is in error. Let me check on it and get back to you.

[JohnDuffield]

This is by holding the frequency as constant while the coordinate speed changes, yes.

The frequency is constant. So is the energy. When you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Not a gazillion tonnes. Conservation of energy applies.

But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different.

The photon energy doesn't change. The observers change. It takes work to lift a brick. You have to add energy to it. It's the same for an observer. And once you've lifted the observer up, then because you have added energy to that observer, to him, the photon energy appears to have reduced. Even though it hasn't. You could do the same sort of thing by accelerating observers away from a photon source in gravity-free space. 

You can't get round it by looking at it from a 'Schwarzschild observer' perspective.

It isn't a matter of getting around it. It's a matter of getting it right.

[jeffreyH (OP)]

This is by holding the frequency as constant while the coordinate speed changes, yes.

The frequency is constant. So is the energy. When you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Not a gazillion tonnes. Conservation of energy applies.

But we live in a universe where local observations should always concur about measurements. 1 metre will still be 1 metre. The speed of light will still be c. 1 second will still be one second. The energy of the photon for local observers is then different.

The photon energy doesn't change. The observers change. It takes work to lift a brick. You have to add energy to it. It's the same for an observer. And once you've lifted the observer up, then because you have added energy to that observer, to him, the photon energy appears to have reduced. Even though it hasn't. You could do the same sort of thing by accelerating observers away from a photon source in gravity-free space. 

You can't get round it by looking at it from a 'Schwarzschild observer' perspective.

It isn't a matter of getting around it. It's a matter of getting it right.

I think this says it all John.

http://en.wikipedia.org/wiki/Gravitational_redshift
In astrophysics, gravitational redshift or Einstein shift is the process by which electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency, or redshifted, when observed in a region of a weaker gravitational field. This is a direct result of gravitational time dilation - as one moves away from a source of gravitational field, the rate at which time passes is increased relative to the case when one is near the source. As frequency is inverse of time (specifically, time required for completing one wave oscillation), frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential (i.e., equivalently, of lower gravitational field) . There is a corresponding reduction in energy when electromagnetic radiation is red-shifted, as given by Planck's relation, due to the electromagnetic radiation propagating in opposition to the gravitational gradient. There also exists a corresponding blueshift when electromagnetic radiation propagates from an area of a weaker gravitational field to an area of a stronger gravitational field.

[JohnDuffield]

I think this says it all John.

The article is wrong. Gravity is not a force in the Newtonian sense. It doesn't add energy to a falling body, it just converts potential energy into kinetic energy. It doesn't add energy to a descending photon either, or remove energy from an ascending photon. Not only that, the article confuses field and potential.  The opening sentence should say this:

In astrophysics, gravitational redshift or Einstein shift is the process by which electromagnetic radiation originating from a source that is in a gravitational field appears to be reduced in frequency, or redshifted, when observed in a region of a higher gravitational potential.

Maybe I should talk to the guys who've been editing that page about this. The next bit isn't bad, but "is" should be "appears":

This is a direct result of gravitational time dilation - as one moves away from a source of gravitational field, the rate at which time passes is increased relative to the case when one is near the source. As frequency is inverse of time (specifically, time required for completing one wave oscillation), frequency of the electromagnetic radiation is reduced in an area of a higher gravitational potential".

The next bit is just wrong:

(i.e., equivalently, of lower gravitational field) . There is a corresponding reduction in energy when electromagnetic radiation is red-shifted, as given by Planck's relation, due to the electromagnetic radiation propagating in opposition to the gravitational gradient.

If it wasn't wrong, dropping a 511keV photon into a black hole would increase its mass by more than 511keV/c².  The next bit confuses field and potential again:

There also exists a corresponding blueshift when electromagnetic radiation propagates from an area of a weaker gravitational field to an area of a stronger gravitational field.

This has been written by an amateur, Jeffrey.  I recommend you ask around elsewhere about what I've said. Meanwhile ask yourself who you're going to believe, me and Einstein and Pete, or some guy who doesn't know the difference between gravitational potential and gravitational field. 

[PhysBang]

It is amazingly uncontroversial that redshift leads to a loss in energy.

If we just think in terms of time dilation, then we consider the output of energy in the rest frame of the emitter, then consider the time dilated version of that, conservation of energy requires that the energy output over time of the time dilation of the emitter must be lower, since it must be putting out the same energy over a longer period. Whether or not an emitter is time dilated or not is determined entirely by frame of reference, so that can't change the overall output of the emitter between any two given events.

Similarly, a blueshift leads to an increase in energy. But this is not really amazing or unexpected, since when one speaks of an object falling to the ground, one speaks of the potential energy that the object converts into other forms of energy. One cannot simply speak of a photon falling in to a black hole without giving relevant details and expect a coherent physical picture.

If one asks of a photon from a distant galaxy if there is a difference from its frequency when it was emitted versus when it was received, then the answer, "yes," makes a great deal of sense.

[PmbPhy]

Gravity is not a force in the Newtonian sense.

Wrong.

It doesn't add energy to a descending photon either, or remove energy from an ascending photon.

Wrong yet again! It's very easy to prove too:

Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

[yor_on]

Maybe one can think of in form of a 'field' with photons as 'excitations' in it. Then the 'field' is what determines the energy you will find measuring somewhere in a photon 'propagation'. In other words it should be about the 'relations' defining it. If I was to argue that a black hole is time dilated intrinsically I also will be able to do that for any gravitational potential. That should then give us different speeds measured depending on the inertial platforms density (invariant mass). That would put into question any definition of a constant being true, not compensated for the gravitational potential that exist where you measure it. In effect, we should actually get different answers depending on mass. Do we?
=

the way around it would then be to assume a length 'elongation' balancing it up Meaning that the reason it stays a constant locally measuring (presuming some 'slower time/clock') is that it locally measured now finds it a longer 'way' to propagate, balancing the slower clock, giving us a constant speed of light in a two mirror experiment, 'inertially' and uniformly measuring. You can't apply a Lorentz contraction locally on that as that would give a opposite effect.

But it depends, you could argue that if 'c' and the arrow (local clock) is equivalent, it won't be noticeable, as the speed of light then 'slows down' equivalently, locally measured. What that gives us is a constant that theoretically may differ but locally and measurably is invariant. In the end it comes down to what is most simple. Constants, or no constants? And naturally, how you want to define physics? As being the same everywhere? And 'repeatable experiments' too.

[jeffreyH (OP)]

Gravity is not a force in the Newtonian sense.

Wrong.

It doesn't add energy to a descending photon either, or remove energy from an ascending photon.

Wrong yet again! It's very easy to prove too:

Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

Thanks for that Pete I will read it later. I am done with John.

[jeffreyH (OP)]

That was a very nice link Jeffrey. Succinct and enlightening to how he thought. If you find more links able to compress ideas feel free to share them

Pete is looking into it. It may be wrong so beware.

[jeffreyH (OP)]

Gravity is not a force in the Newtonian sense.

Wrong.

It doesn't add energy to a descending photon either, or remove energy from an ascending photon.

Wrong yet again! It's very easy to prove too:

Jeff: Please see: http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

I have seen that page before and it was very informative. Now that I am getting into Lagrangians the Schwarzschild metric equation makes much more sense.

[evan_au]

arrive at a destination that was 100 ly away when they were emitted

A position 100 light-years away is in our galaxy, in fact in the same spiral arm of the Milky Way. This is too close to experience cosmic redshift.

Even the Andromeda Galaxy, at 2.5 million LY distance is part of our local galaxy cluster, and does not show cosmic redshift (in fact, it is moving towards a collision with our galaxy, and so exhibits Doppler blue-shift).

To see cosmic redshift, you need to go outside our local cluster of galaxies.

However, gravitational redshift has actually been demonstrated here on Earth, with photons "climbing" out of Earth's gravitational well, but the effect is incredibly small. It is also very slight and hard to measure on the Sun, since the high temperature of the Sun's surface means that the thermal motion of highly ionised atoms in the Sun's atmosphere is large compared to the gravitational redshift. Neutron stars would exhibit higher gravitational redshift, but their incredibly high surface temperature would completely ionise the atoms, and would impart a huge thermal spread, so it would be hard to collect a spectrum.

[Pedantic, I know; I also didn't answer the question about redshift of photons & electrons that was asked by chiralSPO [V] ]

[chiralSPO]

I think some pedantry is acceptable here, but allow me to ask the same question, adding a few orders of magnitude to the distance and speed of the electron: if the distance traveled were 10000000 ly, and the electron were moving at 0.9999 c (now arriving 1000 years later than the photon, if I counted the digits correctly), would there be a noticeable difference in the red-shift of each of the waves?

Is it expected just to be an issue of Doppler shift based on the velocity of the source when the wave was emitted and the velocity of the detector when the wave is measured? Or does the red-shift manifest as a result of the expansion of space through which the wave is propagating?

If the former is true, then a photon that has traveled infinitely far could have finite non-zero energy (or even infinite energy) as measured by an observer traveling at infinite speed towards the source of the light (don't tell me this is impossible, we are already talking about a photon that has traveled an infinite distance--if we replace "infinite distance" for "arbitrarily long distance" and "infinite speed" with "arbitrarily close to c" the results will be the same until the distance is great enough that the light will never be able to reach the detector because spacial expansion will have overtaken it, in which case... what photon?)